Re: [EM] Good references on voting theory
Dear Remi, welcome on this list! When you dig a little into the archives of it, you will find that it provides loads of thoughts on and examples of probabilistic voting systems and strategic voting, certainly more than you will find in any book. Some of those discussions eventually lead to this Social Choice and Welfare article: http://www.springerlink.com/content/5xr0220678805288/ Although I'm not able to recommend a good recent book, someone else on the list might and hopefully will! Joyeuses Fêtes! Jobst Am 22.12.2011 22:44, schrieb Rémi: /Disclaimer:/ I am quite sorry that my e-mail does not really fit the aims of this mailing list, but I found no other solution to contact specialists of voting theory... Hi everyone, I am a French researcher in mathematics (rather a specialist of probability actually); I have decided to write a popularization paper on the mathematical aspects of voting theory, as a contribution to a French website called /Images des Mathématiques/ (http://images.math.cnrs.fr/), which website is devoted to popularization of (contemporary) mathematical research. My aim is to explain the main problematics linked to voting theory, with a stress on the mathematical (or game-theoretical) aspects of this theory. After an introduction dealing with the general aims of voting and the ambiguousness of the concept of collective choice, my article would explain the main criteria desirable for devising a voting system, state some great impossibility results, and finally compare some particular voting systems. Among other things, I would like to handle the (linked) questions of strategic voting and probabilistic voting systems. My article would remain focused on the single-winner problem and would contain no discussion at all on the practical aspects of designing a voting protocol. Of course, I have found much valuable information on these topics over the Internet (just the Wikipedia voting system page is quite remarkable), and certainly I already have the material to write some good and rather complete article. However, I would also like to get a good reference on the topic, typically a (not-too-old?) book written by a researcher, with the triple goal of (i) getting more complete, better-organized information, (ii) giving greater authority to my article by quoting a book rather than just Wikipedia or other webpages, and (iii) suggesting a good reading for those readers of my future article who would like to go further. The problem is, I have found several titles of such books, but I have no idea of which ones are better-written, more complete, better fitted to my goals, more classical or easier to find... This is why I am asking you which references you would advise to me as the main source for my future article (obviously I will not be able to or even read all of them, so I have to make a choice...). Hoping for your kind answers! Sincerely yours, Rémi Peyre Assistant Professor, École des Mines de Nancy (France) remi.pe...@iecn.u-nancy.fr http://www.normalesup.org/~rpeyre/pro/index-en.html Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Better Than Expectation Approval Voting (2nd try readable format)
Forest wrote: Now for the interesting part: if you use this strategy on your approval ballot, the expected number of candidates that you would approve is simply the sum of the probabilities of your approving the individual candiates, i.e. the total score of all the candidates on your score ballot divided by the maximum possible score (100 in the example). Suppose that there are n candidates, and that the expected number that you will approve is k. Then instead of going through the random number rigamarole, just approve your top k candidates. So we could justifiably call this strategy the honest approval strategy, since if preferences are sufficiently mixed and all voters use this strategy, the outcome is the same as the one with sincere Range ballots, i.e., the option with the highest total rating! Happy Holidays from Jobst to all of you! Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Arrow's Theorem
Dear Stephen, you wrote: democratic implies majority rule This is by no means clear. You might want to consult this list's archives to find some discussion of this. My opinion is rather that democracy and majority rule are incompatible since the latter assigns all power to 50%+epsilon in the worst case, allowing a mere majority to oppress all other voters. Democratic group decisions rather require some amount of chance to give all voters a fair share of the power. There is an article on this by Forest and me in Soc Choice Welf (http://www.springerlink.com/content/5xr0220678805288/, http://edoc.gfz-potsdam.de/pik/get/4618/0/.../4618oa.pdf) Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A Comparison of the Two Known Monotone, Clone Free Methods for Electing Uncovered Alternatives
Hi again,
I wrote:
Unfortunately, I fear Short Ranked Pairs might not be monotonic. One
would habe to check. And I'm not sure your description of an algorithm
for Short Ranked Pairs is valid -- after all, I only defined it
abstractly by saying that one has to find the lexicographically maximal
short acyclic set without giving an algorithm to find it.
Maybe my gut-feeling was incorrect. Short Ranked Pairs is probably
monotonic. As it is also clone-proof and uncovered, we really have a
third type of method with those three properties, just as Kristofer
remarked!
Now why should it be monotonic? Assume that...
s(x,y) is the strength of the defeat xy,
s1s2s3... is the descending sequence of all defeat strengths,
x1y1,x2y2,x3y3,... are the corresponding defeats,
D is the lexicographically maximal short acyclic set of defeats,
w is the top option of D
and I is the set of ranks i of defeats in the above numbering that
belong to D, so that
D = { xiyi : i in I }.
Then no defeat xw is in D, hence all defeats wy are in D since
otherwise adding them won't construct a cycle.
Now assume some defeat wa is reinforced, moving it up in the ordering
x1y1,x2y2,x3y3,... by one place. Let j be the index of wa in this
list before the reinforcing, so that w=xj, a=yj. Also introduce a new
numbering of the defeats corresponding to the new ranking of defeats:
x'1y'1,x'2y'2,x'3y'3,...
Then x'i=xi and y'i=yi for all i different from j and j-1,
and x'j=x(j-1), y'j=y(j-1), x'(j-1)=xj=w, y'(j-1)=yj=a.
D consists of those defeats x'iy'i with ranks i in the new set
J = I with j and j-1 exchanged.
We will prove that after the reinforcing, D is still lexicographically
maximal. Assume that it is not. Then there is another short acyclic set
of defeats D' that is lexicographically larger than D in the new ranking
of defeats. Hence there is some new rank k such that the defeat x'ky'k
is in D' but not in D, and such that for all ik, the defeat x'iy'i is
either in both D' and D or in neither of the two.
If k were different from j-1, then D' would be already lexicographically
larger than D in the original ranking of defeats, which was not the case
as D was maximal there.
But k can't be j-1 since the defeat x'(j-1)y'(j-1) is the defeat wa
and thus belongs to D. Hence there is no such D' as assumed, D is still
lexicographically maximal, and w is still the winner after the
reinforcement of the defeat wa. That is, Short Ranked Pairs is monotonic.
I propose to proceed as follows: Check how that lexicographically
maximal short acyclic set can be found in the simpler case in which we
define defeat strength as approval of defeating option.
This is still open and I don't see a simple algorithm coming to my mind...
This will also
allow us to compare the method to DMC since DMC is the result of
applying ordinary Ranked Pairs with this definition of defeat strength,
so applying Short Ranked Pairs to them should not be too much different.
The resulting short acyclic set will contain all defeats from the
approval winner to other options, but I don't see immediately whether
one can somehow continue to lock in defeats similar to ordinary Ranked
Pairs, skipping certain defeats that would destroy the defining
properties of a short acyclic set. I somehow doubt that since that
defining property is not that some configurations must not exist but
than some configurations must exist. Anyway, in the case where defeat
strength is approval of defeating option, all might be somewhat simpler.
Jobst
I also guess you could make methods with properties like the above by
constraining monotone cloneproof methods to the Landau set (whether by
making something like Landau,Schulze or Landau/Schulze). I'm not sure of
that, however, particularly not in the X/Y case since the elimination
could lead to unwanted effects.
Is one of the two methods you mention UncAAO generalized?
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Re: [EM] A Comparison of the Two Known Monotone, Clone Free Methods for Electing Uncovered Alternatives
Hi Kristofer, you wrote: I think Short Ranked Pairs also passes all these. To my knowledge, Short Ranked Pairs is like Ranked Pairs, except that you can only admit XY if that will retain the property that every pair of affirmed candidates have a beatpath of at most two steps between them. See http://lists.electorama.com/pipermail/election-methods-electorama.com/2004-November/014255.html . The definition of a short acyclic set of defeats was later changed, and the new definition is at http://lists.electorama.com/pipermail/election-methods-electorama.com/2004-November/014258.html Thanks for unearthing this old idea of mine -- I had forgotten them completely already. Unfortunately, I fear Short Ranked Pairs might not be monotonic. One would habe to check. And I'm not sure your description of an algorithm for Short Ranked Pairs is valid -- after all, I only defined it abstractly by saying that one has to find the lexicographically maximal short acyclic set without giving an algorithm to find it. I propose to proceed as follows: Check how that lexicographically maximal short acyclic set can be found in the simpler case in which we define defeat strength as approval of defeating option. This will also allow us to compare the method to DMC since DMC is the result of applying ordinary Ranked Pairs with this definition of defeat strength, so applying Short Ranked Pairs to them should not be too much different. The resulting short acyclic set will contain all defeats from the approval winner to other options, but I don't see immediately whether one can somehow continue to lock in defeats similar to ordinary Ranked Pairs, skipping certain defeats that would destroy the defining properties of a short acyclic set. I somehow doubt that since that defining property is not that some configurations must not exist but than some configurations must exist. Anyway, in the case where defeat strength is approval of defeating option, all might be somewhat simpler. Jobst I also guess you could make methods with properties like the above by constraining monotone cloneproof methods to the Landau set (whether by making something like Landau,Schulze or Landau/Schulze). I'm not sure of that, however, particularly not in the X/Y case since the elimination could lead to unwanted effects. Is one of the two methods you mention UncAAO generalized? Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Fair and Democratic versus Majority Rules
Dear Rob, you wrote: here's a fundamental philosophical question: why is it better, even in a two-candidate race, to elect the majority winner? I think the question is ill-posed in at least two ways: First, you say better but not better than what. Second, after you settled for an alternative, you should not ask why is it better? but is is better?. My answer: The alternatives here are not only elect the majority candidate or elect the minority candidate but also elect both with just probabilities reflecting their share of the vote. And the latter is obviously the only democratic of the three since otherwise one part of the electorate can easily oppress the rest. Athenians knew why they filled offices by lot in their democracy... Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Fair and Democratic versus Majority Rules
Dear Abd ul-Rahman, some short remarks to your claims: Random ballot does nothing to encourage compromise! Perfectly true. That's why nobody suggests it to be used for decisions. We suggest to use it instead as a benchmark any reasonable method must improve upon. More precisely, we want that the actually used decision process is preferred to the Random Ballot lottery by all voters. Majoritarian methods do not produce outcomes that everyone prefers to the Random Ballot lottery. Our methods do. The goal of government is not to be fair, per se, but to maximize social welfare. Nice try. So we're back to the problem of deciding what we mean by social welfare. Most definitions would agree that in this very simple and common situation, C maximizes social welfare: 55 voters assign utilities A 100, C 80, B 0, and 45 voters have B 100, C 80, A 0. That was my challenge back in 2007. Obviously, majoritarian methods spectacularly fail to maximize social welfare. Our consensus methods choose C in this example. Democracy begins with majority rule. Democracy begins with Athenian Democracy in which offices were filled by lot, which is essentially equivalent to Random Ballot. Majority rule has nothing to do with democracy, its pleocracy. It does not end there, but random ballot discards majority rule, which leads to, inevitably, minority rule, at least part of the time. Nope. What you don't seem to understand is that neither the majority nor the minority need to rule. Ruling means you can be sure your choice will be realized. In our consensus method, no group that is smaller than 100% can guarantee that their favourite wins, so there is no ruling but rather the cooperative choice of a consensus. And the damage from that can be enormous. Suppose 10% of the population believes that using nuclear weapons to get rid of enemies is a fine idea. Would we, to provide this faction with a fair opportunity to exercise decision-making power, give them the button 10% of the time? No we wouldn't. We are well-advised to prevent such options from being considered at all in the first place since they might easily be approved by large majorities as well (Examples: death penalty, Rwanda, Hitler, near-extinction of native Americans, etc.). For this reason it makes no sense to repeat bringing up ridiculous examples with extreme options. Our goal is to find good methods to decide between feasible not infeasible options. The latter can only be safely excluded legally in a constitution. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Uncovered set methods (Re: How close can we get to the IIAC)
Hi again Markus, it's different. The goldfish winner can be really strange. If the defeat strength are A B C D A - 2 3 0 B 0 - 1 4 C 0 0 - 5 D 6 0 0 - then Beatpath and Tideman give B, River gives C, and Goldfish gives D since the table evolves like this: B C D B - 1 0 C 0 - 0 D 2 3 - B D B - 0 D 2 - D D - Yours, Jobst Am 31.10.2010 18:35, schrieb Jobst Heitzig: Hi Markus, on 29.04.2010 20:33 you asked: is Jobst Heitzig's river method identical to Blake Cretney's goldfish method? I'm sorry that I have not read any list posts for months, so this caught my attention just now. I will check the differences! You probably refer to the method from Blake's Aug 12, 1998 post I cite below? Yours, Jobst On Aug 12, 1998, Blake Cretney wrote: Here's my entry for single-winner system of the week. It was motivated by my desire to make a method that would be easy to program. To this end, it does not require cycles or the Smith set to be found. I'll call it Goldfish until someone shows me a previous mention under a different name. The idea of goldfish is that the candidates seem to eat each other, becoming bigger and bigger, until only one is left swimming in the electoral fish bowl. Goldfish definition: Successively find the worst defeat and eliminate the pair-wise loser. Any win achieved by the pair-wise loser is now scored as if it was achieved by the pair-wise winner, provided it is larger than the one already scored by him, or he is currently scored a loss. Start by making a victory table. For each row, enter the votes against each column's candidate, if the row's candidate wins pair-wise. Otherwise enter a 0. The best way to resolve ties is for a chairman, president, or random voter to enter a special ballot. This ballot must not be truncated. Repeat until only one candidate is left: FIND: Find the highest value in the table. Call this cell i,j. If more than one row share this value, choose the row that is higher in the special ballot. MERGE: Here's where the big fish eats the little one. For each cell in the i row, if there is a higher value for that column in the j row, copy it over. For each cell in the i column, if there is a zero for that row in the j column, copy it over. Do not change the empty cells on the diagonal. ELIMINATE: Remove the j candidate and its row and column from consideration. I'm going to use the word beats to mean defeats pair-wise and eats to mean is chosen to defeat and merge with. MIIAC -- Candidates outside the Smith set are always beaten by members of the Smith set. When they eat them, the rows and columns are merged, but this provides nothing of use for beating other Smith members, because candidates outside the Smith set only have losing scores against those inside, and the merge rule does not copy losing scores. This is because only 0 values are copied from column to column. GITC -- If someone outside a clone set eats a clone, all the clones will be eaten on successive rounds, just as if there was only one. If a clone eats someone outside, the merge occurs. Because the outsider loses to the clone, it can provide no help in defeating other clones. It does not matter which clone eats an outsider, because eventually all clones will be eliminated, or one will eat all the others, and merge with them. GMC -- Because candidates are removed in order of votes against, and because removal does not eliminate a majority vote against a candidate, but merely copies it, candidates with a majority against will be removed first. Elimination methods frequently have the problem that it is possible to help elect a candidate by ranking it lower. This happens when you can reduce the amount by which a victory is obtained, so that a candidate is not eliminated, and can carry on to defeat your enemies. The merge step in Goldfish makes this strategy unnecessary. The winner ends up beating the same candidates as the loser, and by as much. Lower losing values are not copied, but having another candidate in the race with lower losing values is not helpful. This seems like a pretty good system and is fairly easy to program. With a couple of tweaks, it can be converted to Tideman. Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Uncovered set methods (Re: How close can we get to the IIAC)
Hi Markus, on 29.04.2010 20:33 you asked: is Jobst Heitzig's river method identical to Blake Cretney's goldfish method? I'm sorry that I have not read any list posts for months, so this caught my attention just now. I will check the differences! You probably refer to the method from Blake's Aug 12, 1998 post I cite below? Yours, Jobst On Aug 12, 1998, Blake Cretney wrote: Here's my entry for single-winner system of the week. It was motivated by my desire to make a method that would be easy to program. To this end, it does not require cycles or the Smith set to be found. I'll call it Goldfish until someone shows me a previous mention under a different name. The idea of goldfish is that the candidates seem to eat each other, becoming bigger and bigger, until only one is left swimming in the electoral fish bowl. Goldfish definition: Successively find the worst defeat and eliminate the pair-wise loser. Any win achieved by the pair-wise loser is now scored as if it was achieved by the pair-wise winner, provided it is larger than the one already scored by him, or he is currently scored a loss. Start by making a victory table. For each row, enter the votes against each column's candidate, if the row's candidate wins pair-wise. Otherwise enter a 0. The best way to resolve ties is for a chairman, president, or random voter to enter a special ballot. This ballot must not be truncated. Repeat until only one candidate is left: FIND: Find the highest value in the table. Call this cell i,j. If more than one row share this value, choose the row that is higher in the special ballot. MERGE: Here's where the big fish eats the little one. For each cell in the i row, if there is a higher value for that column in the j row, copy it over. For each cell in the i column, if there is a zero for that row in the j column, copy it over. Do not change the empty cells on the diagonal. ELIMINATE: Remove the j candidate and its row and column from consideration. I'm going to use the word beats to mean defeats pair-wise and eats to mean is chosen to defeat and merge with. MIIAC -- Candidates outside the Smith set are always beaten by members of the Smith set. When they eat them, the rows and columns are merged, but this provides nothing of use for beating other Smith members, because candidates outside the Smith set only have losing scores against those inside, and the merge rule does not copy losing scores. This is because only 0 values are copied from column to column. GITC -- If someone outside a clone set eats a clone, all the clones will be eaten on successive rounds, just as if there was only one. If a clone eats someone outside, the merge occurs. Because the outsider loses to the clone, it can provide no help in defeating other clones. It does not matter which clone eats an outsider, because eventually all clones will be eliminated, or one will eat all the others, and merge with them. GMC -- Because candidates are removed in order of votes against, and because removal does not eliminate a majority vote against a candidate, but merely copies it, candidates with a majority against will be removed first. Elimination methods frequently have the problem that it is possible to help elect a candidate by ranking it lower. This happens when you can reduce the amount by which a victory is obtained, so that a candidate is not eliminated, and can carry on to defeat your enemies. The merge step in Goldfish makes this strategy unnecessary. The winner ends up beating the same candidates as the loser, and by as much. Lower losing values are not copied, but having another candidate in the race with lower losing values is not helpful. This seems like a pretty good system and is fairly easy to program. With a couple of tweaks, it can be converted to Tideman. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Geometric Condorcet cycle example, improved
Hi folks, I just recalled that four years ago I constructed a sophisticated example which is somewhat similar: http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/2005-May/015982.html Happy New Year! Jobst Warren Smith schrieb: This point-set also works: A=(1,0) B=(0,4) C=(3,5) D=(9,2) Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A Proportionally Fair Consensus Lottery for which Sincere Range Ballots are Optimal
Folks, you probably overlooked that I have already described a variant which works *completely* without Random Ballot and will definitely elect one of the top-3 range options (as determined from the 'strategic' ballot): Method Range top-3 runoff (RT3R) === 1. Each voter separately supplies a nomination range ballot and a runoff range ballot. 2. From all nomination ballots, determine the options A,B,C with the top-3 total scores abc. 3. Let L be the lottery in which B wins with probability p = max(0,(2b-a-c)/(b-c)) and C wins with probability 1-p. 4. Let q be the proportion of nomination ballots on which the lottery L has an expected rating below the rating of A on that ballot. 5. Option A wins if, on at least the same proportion q of all runoff (!) ballots, the lottery L has an expected rating below the rating of A on that ballot. Otherwise B wins with probability p and C wins with probability 1-p. So if you want it to be as deterministic as possible, you can do it like this or similarly. If you modify it further and set q=1/2, you even get a majoritarian version if you want that. Yours, Jobst Warren Smith schrieb: --yep. Only reason I did what I did was simplicity (kind of a pain if voters have to submit both a range-type and a condorcet or approval-type ballot). But your way is better in that it tends to yield a better winner than my way. Also, note -- which is even more obnoxious -- we could have each voter submit TWO ratings-style ballots, the honest range ballot and the dishonest range ballot; then the HRB is used to decide between DHR and random ballot... On 11/20/09, Raph Frank raph...@gmail.com wrote: This is effectively performing random ballot and then giving the voters the option to roll the dice a second time. Any single seat method could be used to select the first candidate. If you used a good single seat method to pick the compromise winner, then the random ballot would rarely if ever be activated. For example. 1) Voters submit ratings ballot and also a ranked or an approval ballot 2) Determine the winner using condorcet or approval (or other method) 3) Determine the random ballot odds for each candidate 4) If a majority prefer the winner in 2 to the expectation in 3), then the winner from 2 wins. 5) Otherwise, use the random method Ofc, using a majority instead of a unanimous decision breaks some of the properties of the pure consensus method. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A Proportionally Fair Consensus Lottery for which Sincere Range Ballots are Optimal
Even simpler is this: Method Top-3 approval sincere runoff (T3ASR) == 1. Each voter separately supplies a nomination approval ballot and a runoff range ballot. 2. From all nomination ballots, determine the options A,B,C with the top-3 approval scores abc. 3. Let p be the proportion of nomination ballots which approve of C but not of B. 4. If on at least half of all runoff ballots we have a rating r(A) p*r(C) + (1-p)*r(B), then option A wins. 5. Otherwise draw a nomination ballot. If it approves of C but not of B, C wins, otherwise B wins. Most of the time, this will elect one of the top-2 approval options, and only rarely the 3rd placed. One can then also compute and publish some kind of index of sincerity by comparing the submitted approval and range ballots. The method is majoritarian, since any majority can rule by bullet voting on both ballots. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A Proportionally Fair Consensus Lottery for which Sincere Range Ballots are Optimal
Hello Forest, Most of the credit should be yours; in fact, the proof and all of the ingredients are yours. I hurried to post the message this morning, because I was sure that you were going to beat me to it! I would certainly believe you if you said that you had already thought of the same thing but didn't have time to post the message before I did. Well, I don't. I was never before thinking of asking separately some additional information from the voters which is already contained in the ratings part of their ballot (namely what the favourite is), in order to transfer the strategic incentives away from the ratings to this separate information. And that is exactly the genial part! Now that the general technique is clear, we can easily derive a lot of similar methods, also some that only incorporate a very small amount of chance, for those who don't like chance processes in voting methods. The general technique is this: Ask for ratings and some additional information. Use the additional information and to determine -- independently from all ratings -- two possible winners or winning lotteries, at least one of which must be a lottery of at least two options in which the probabilities can vary. Then use the ratings to decide between these two possibilities in some monotonic way (e.g., using unanimity as in your proposal, or some qualified majority, or even Random Ballot, or whatever). Then strategy-freeness in the ratings part follows from the fact that they are only used in a monotonic binary choice between lotteries which are not known before. For example: Method Range top-3 runoff (RT3R) === 1. Each voter separately supplies a nomination range ballot and a runoff range ballot. 2. From all nomination ballots, determine the options A,B,C with the top-3 total scores abc. 3. Let L be the lottery in which B wins with probability p = max(0,(2b-a-c)/(b-c)) and C wins with probability 1-p. 4. Let q be the proportion of nomination ballots on which the lottery L has an expected rating below the rating of A on that ballot. 5. Option A wins if, on at least the same proportion q of all runoff (!) ballots, the lottery L has an expected rating below the rating of A on that ballot. Otherwise B wins with probability p and C wins with probability 1-p. Notes: p is so designed that it can take all values between 0 and 1 but will be the larger the lower c is, in order to get a large expected rating of the final winner. Of course, the formula for p could be modified in all kinds of ways. The expected total nomination rating of the final winner is at least a - 2(a-b), so if the race between A and B is close (i.e., a-b is small), we have quite an efficient outcome. On the other hand, if A clearly beats B, it will win with a high probability since the true proportion of voters who prefer A to B and C is probably larger than can be seen from the strategically used nomination ballots. What do you think? Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A Proportionally Fair Consensus Lottery for which Sincere Range Ballots are Optimal
Dear folks, although Forest's posting comes along so matter-of-factly, let's make it absolutely clear that it is an ENORMOUS MILESTONE! Why so? He describes a very SIMPLE, EFFICIENT, and FAIR method which REVEALS THE TRUE UTILITY VALUES of all voters who are rational in the sense of von Neumann and Morgenstern. The only other known methods which have this revelation property are not only more artificial and complex but are much less efficient or require monetary taxes to be paid and destroyed (like the Clarke tax). Very simple proof that sincere ratings are optimal: My ratings are only relevant in a specific situation. In this situation a fall-back lottery has already been determined from all the labels (thus not dependent on my ratings), and a possible consensus option has been nominated from the circle on a drawn ballot (thus also independently from my ratings). If my ratings are relevant, they will decide between this given fall-back lottery and this given nominated consensus option, but I will not know beforehand which lottery and which nominated option they will be (except if I knew all other ballots, which is impossible in a secret poll). So the only way to make sure that my ratings will lead to the fall-back lottery when I prefer it over the nominated consensus option, and that they will lead to the nominated consensus option when I prefer it to the fall-back lottery, is to give ratings that reflect my true preferences, in other words, to specify a set of sincere utility values. Note that this is not only true in some equilibrium situation but NO MATTER HOW THE OTHERS VOTE! In other words, it is always a dominant strategy. Now, that does not mean, however, that the whole method is strategy-free, since the other part of the ballot, namely the circle and the label, are strategic. I may, for example, have incentives to label a more extreme option as favourite than my true favourite, in order to lower the expected rating of the fall-back lottery and make a consensus more probable. However, every such strategic behaviour would be visible from the ballot since the labelled favourite would not have the highest rating. That is a very interesting property which I have never seen before in any method: you have the incentive to vote strategically, but you cannot hide if you do so! My guess is that we will soon find a similar method in which a single voter cannot prevent the consensus completely but only lower its probability... Forest: EXCEPTIONALLY WELL-DONE! Jobst fsimm...@pcc.edu schrieb: A proportionally fair lottery is a lottery method in a which any faction of the voters can unilaterally guarantee that their common favorite will be elected with a probability proportional to the size of their faction. A consensus candidate is any candidate that would be liked at least as much as the random favorite by 100 percent of the voters (assuming all voters to be rational). A consensus lottery is a method that elects consensus candidates with certainty (again, assuming rational voters). I won't attempt to define sincere range ballot here, but the meaning will be apparent from this method: Ballots are range style (i.e. cardinal ratings). Each voter rates the candidates, circles one of the names as a proposed consensus candidate, and labels another (or perhaps the same) name as favorite or favourite. Have I overlooked anything? The ballots are collected and the probabilities in the random favorite lottery are determined. These probabilities are used to determine and mark a random favorite rating expectation on each range ballot. A ballot is then drawn at random. If the circled name on the randomly drawn ballot has a rating above the random favorite rating expectation, on any ballot (including the one in play), then another ballot is drawn, and the indicated favorite of the second ballot is elected. Otherwise, the proposed consensus candidate whose name was circled on the first drawn ballot is elected. That's it. Note that any voter has the power to turn the election into random favorite by giving only one candidate (favorite=consensus) a positive non-zero rating. But whenever that is optimal rational strategy, sincere range yields the same expectation, and is therefore optimal, too. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] A Proportionally Fair Consensus Lottery for which Sincere Range Ballots are Optimal
Dear Raph, you wrote: Likewise, you might as well pick your favourite as favourite. This is, unfortunately, not true: The labelled favourite influences the expected ratings against which possible consensus options are compared on each ballot, so you can have the incentive to exaggerate by labelling a more extreme candidate than your true favourite in order to lower those ratings and make your preferred consensus more likely! But this, I guess, will not decrease but rather increase the method's efficiency in realistic examples. The consensus candidate is different. It is inherently strategic. There is the possibility for group chicken effects. For example, a party could say that all of their supporters are going to rate candidate X at minimum, so there is no point in nominating that candidate. This could cause the other partys' supporters to disregard that candidate as a potential consensus candidate. Also, I wonder if it might be worth having a rule that allows additional consensus attempts. For example, if 10% refuse, then the other 90% would be given the option of choosing the consensus candidate. The 2 choices in that case would be Option 1) Full random ballot Option 2) 90% chance of consensus candidate 10% chance of random ballot (only the ballots outside the 90% are considered) This would probably break the strategic purity of the single stage method. I guess so, too, but I think we can overcome the unanimity requirement in a different way. Let me think about it... Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Anyone got a good analysis on limitations of approval andrange voting?
Dear folks, there is another assumption in Arrow's theorem which people almost always forget: Determinism. Methods which use some amount of chance can easily meet all his other criteria, the most trivial example of this being again Random Ballot (i.e. pick a ballot uniformly at random and copy its ranking as the group's ranking). Some people think this violates the no-dictator requirement, but it doesn't since a dictator would be a person determined *beforehand*. Yours, Jobst Raph Frank schrieb: The theorem states (from wiki) that there is no method which has the following properties: * If every voter prefers X over Y, then the group prefers X over Y. * If every voter prefers X over Y, then adding Z to the slate won't change the group's preference of X over Y. * There is no dictator. All 3 of those conditions are met for range. The only problem is that adding Z could cause renormalisation changes in how people vote. A voter who votes A: 100 B: 0 might change vote to: A: 100 B: 50 Z: 0 after Z is added. Thus changing the difference between A and B for that ballot. Ranked systems allow full ranking. Adding another candidate just requires that you insert the candidate into the rank order. With range this might not be possible. If the candidate has a rating outside the max and min, a voter may have to rescale their prior preferences. If the assumption is that voters are just allowed add a rating for Z and not change any of their other ratings, then it meets the 3 conditions and thus is a counter example to Arrow's theorem. Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Anyone got a good analysis on limitations of approval andrange voting? (long)
Dear Matthew, you wrote: Suite of complicated systems that strive to reach Condorcet ideals. 1. No regular bloke would ever trust 'em because you can't explain how they work in one or two sentences. Well, here's a very simple Condorcet system which can easily be explained in two sentences: 1. Let voters vote on candidate pairs, successively replacing the loser with a different candidate, in a random order of candidates. 2. As soon as all candidates have been included, the winner of the last pair is declared the overall winner. This system is arguably the earliest example of a Condorcet system. It was devised by Ramon Llull in the 13th century and was successfully used for elections in monastaries. It is easily understood by the common man since it resembles a procedure frequently used in child play. Yours, Jobst Matthew Welland schrieb: Thanks all for the discussion and pointers. I still can't concretely conclude anything yet but here are some rambling and random thoughts based on what was said and my prior experiences. Plurality 1. Leads to two lowest common denominator parties which are not accountable to the voters. This conclusion supported by real world observation. 2. Feels right to the non-critical mind, one man, one vote 3. Very fast at the polls Approval 1. Encourages participation of minor parties and thus should keep the big guys paying attention to a wider base. 2. Almost zero marginal implementation cost. Hanging chads count just fine :) 3. Understandable by anyone but feels wrong at first not fair, you get more than one vote. 4. Apparently has a terrible flaw but no one seems to be able to articulate it in layman terms. No real world experience available to illustrate the problem. Here is where I need to learn more. Data provided to date is unconvincing to me. 5. Does not meet the desire of some to be able to differentiate between I like, I like a lot etc. (note: this seems like perfectionism to me. Large numbers of voters and opinions all over the bell curve should make individual expression at the greater level of granularity irrelevant.) 6. Very fast at the polls. Pick yer favorites and head home for beer and telly. Range 1. Can break the vicious cycle of plurality 2. Not voting for someone at all can have a strong influence on election outcome. This is very non-intuitive and would take some getting used to. 3. Allows for nuanced voting. 4. Pain in the ass at the polls (relatively speaking). You can't safely disregard the candidates you don't care about so you *have* to assign everyone a ranking, possibly addressable by defaulting to zero for all candidates? This is considered a feature and I agree it has merit. But in reality it is a deal breaker for joe six pack and co. (and for lazy sobs like me). IRV 1. Demonstrably broken. 'nuff said. Suite of complicated systems that strive to reach Condorcet ideals. 1. No regular bloke would ever trust 'em because you can't explain how they work in one or two sentences. 2. Technically superior to other systems. 3. Not clear what problem with approval they would solve. Unless you are a perfectionist and insist that individuals express nuances of opinion... Some time ago I put together a site (primitive and unfinished[i]) to promote approval voting and in the process I spent a lot of time trying different systems on the web and repeatedly testing my own site. I noticed some interesting things from all that playing around. 1. It was very uncomfortable to go back to plurality after trying other systems. It feels unfair and broken. 2. It was very tedious voting in any of the ranking systems. 3. Approval felt boring but good. I have checked in on this list now and then and I admit I don't have the time or skills to follow all the arguments but it strikes me that approval voting is good enough to break the deadlock, at least in US politics and that it doesn't have any major flaws. The very understandable desire to be able to articulate in a finer grained way in your vote is perfectionism. With millions of voters, for every person on the fence about a particular candidate there will be some to either side who will essentially make or break the vote. If you are on the fence, approve or disapprove, it won't matter. So, to re-frame my question. What is the fatal flaw with approval? I'm not interested in subtle flaws that result in imperfect results. I'm interested in flaws that result in big problems such as those we see with plurality and IRV. [i] www.approvalvote.org Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Anyone got a good analysis on limitations of approval andrange voting? (long)
Dear Robert, you wrote: Round Robin tournament, Ranked Ballot: The contestant who wins in a single match is the candidate who is preferred over the other in more ballots. The candidate who is elected to office is the contestant who loses to no one in the round robin tournament. that's two sentences and two labels. But it's incomplete as well... Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Anyone got a good analysis on limitations of approval and range voting?
Hello Kristofer, you wrote: However, my point was that Range goes further: a minority that acts in a certain way can get what it wants, too; all that's required is that the majority does not vote Approval style (either max or min) and that the minority does, and that the minority is not too small. It is in that respect I mean that Range is more radical, because it permits a minority to overrule a majority that otherwise agrees about which candidates it prefers. For those who mean that elections have to be, at least, majoritarian, Range may contain a surprise. That's true. Methods in which a group can suppress the rest are certainly bad, even more so when the group can be small... You could probably devise a whole class of SEC-type methods. They would go: if there is a consensus (defined in some fashion), then it wins - otherwise, a nondeterministic strategy-free method is used to pick the winner. The advantage of yours is that it uses only Plurality ballots. The hard point is, I think, to define what actually a potential consensus option is. And here the idea was to say everything unanimously preferred to some benchmark outcome qualifies as potential consensus. The benchmark then cannot be any feasible option but must be a lottery of some options, otherwise the supporters of the single option would block the consensus. But which lottery you take as a benchmark could be discussed. I chose the Random Ballot lottery since it seems the most fair one and has all nice properties (strategy-freeness, proportional allocation of power). I suppose the nondeterministic method would have to be bad enough to provide incentive to pick the right consensus, yet it shouldn't be so bad as to undermine the process itself if the voters really can't reach a consensus. Although I can hardly imagine real-world situations in which no consensus option can be found (maybe be combining different decisions into one, or using some kind of compensation scheme if necessary). Assume (for the sake of simplicity) that we can get ranked information from the voters. What difference would a SEC with Random Pair make, with respect to Random Ballot? This sounds interesting, but what exactly do you mean by Random Pair? Pick a randomly chosen pair of candidates and elect the pairwise winner of them? I will think about this... It would lead to a better outcome if the consensus fails, but so also make it more likely that the consensus does fail. Or would it? The reasoning from a given participant's point of view is rather: do I get something *I* would like by refusing to take part in consensus -- not, does *society* get something acceptable. I'm not sure I know what you mean here. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Anyone got a good analysis on limitations of approval and range voting?
Dear Kristofer, both Approval Voting and Range Voting *are* majoritarian: A majority can always get their will and suppress the minority by simply bullet-voting. So, a more interesting version of your question could be: Which *democratic* method (that does not allow any sub-group to suppress the rest) has (usually or on average or in the worst case) the least Bayesian Regret. I conjecture that at least when the nomination of additional options is allowed, the method SEC described recently is a hot candidate for this award, since it seems that SEC will lead to the election of the option at the *mean* (instead of the median) voter position, and I guess that in most spacial utility models the mean position is in many senses better and will in particular have less Bayesian Regret than the median position. (Recall that in a one-dimensional spacial model where additional options can be nominated, all majoritarian methods likely lead to median positions being realized and are thus basically all equivalent.) Yours, Jobst -Ursprüngliche Nachricht- Von: Kristofer Munsterhjelm km-el...@broadpark.no Gesendet: 08.11.09 10:23:11 An: Warren Smith warren@gmail.com CC: election-methods election-meth...@electorama.com Betreff: Re: [EM] Anyone got a good analysis on limitations of approval and range voting? Warren Smith wrote: It seems to me that approval and range voting eliminate most of the strategic opportunity in single winner elections and the marginal improvement of other methods is fairly small. Can anyone point me to analysis, preferably at a layman level, that contradicts or supports this assertion? Or, in succinct terms, what are the strategic flaws in approval or range voting? Thanks, Matthew Welland --well... there is the whole rangevoting.org website... my more-recent papers at math.temple.edu/~wds/homepage/works.html discuss range voting including some ways it is provably better than every rank-order voting system for either honest or strategic voters... --but those are not exactly succinct... OK Let me try: 1. Range for 100% honest voters behaves better than IRV, Borda, Condorcet and it is pretty intuitively clear why -- strength of preference info used, not discarded. There is, of course, the flipside of that property. If one wants a voting method where the majority wins, then Range won't work, simply because a minority of strong opinions can outweigh a majority of weak ones. You might argue that that is no bug at all (strong opinions *should* outweigh weak ones), but for those for which Majority compliance is a must-have, it should be mentioned - particularly since that is supposed to be one aspect of the fairness of traditional democracy. In that sense, moving to Range (and perhaps Approval - depends on how you interpret it) is a more radical proposal than, for instance, moving to Condorcet (which passes Majority). (And now I wonder which election method that passes Majority has the least Bayesian regret.) Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] SEC quickly maximizes total utility in spatial model
Dear Peter, I claimed that SEC... make sure option C is elected in the following situation: a% having true utilities A(100) C(alpha) B(0), b% having true utilities B(100) C(beta) A(0). with a+b=100 and a*alpha + b*beta max(a,b)*100. (The latter condition means C has the largest total utility.) ...to which you correctly replied: Still, I have the very strong feeling that that claim is not part of your above mentioned paper and also it is not true. Obviously, I made a typical copy-and-paste error from an earlier post here. The correct condition under which SEC makes sure that C is elected in the above situation is instead the following: alpha a and beta b This means that all voters prefer C to the Random Ballot lottery. All these don't make the proposals necessarily look bad in my eyes. It looks promising wherever high-value compromises exist, and it looks logical they often do. I think they do exist usually. In the described spatial model they do. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
[EM] SEC quickly maximizes total utility in spatial model
Dear folks, earlier this year Forest and I submitted an article to Social Choice and Welfare (http://www.fair-chair.de/some_chance_for_consensus.pdf) describing a very simple democratic method to achieve consensus: Simple Efficient Consensus (SEC): = 1. Each voter casts two plurality-style ballots: A consensus ballot which she puts into the consensus urn, and a favourite ballot put into the favourites urn. 2. If all ballots in the consensus urn have the same option ticked, that option wins. 3. Otherwise, a ballot drawn at random from the favourites urn decides. This method (called the basic method in our paper) solves the problem of how to... make sure option C is elected in the following situation: a% having true utilities A(100) C(alpha) B(0), b% having true utilities B(100) C(beta) A(0). with a+b=100 and a*alpha + b*beta max(a,b)*100. (The latter condition means C has the largest total utility.) Since then I looked somewhat into spatial models of preferences and found that also in traditional spatial models, our method has the nice property of leading to a very quick maximization of total utility (the most popular utilitarian measure of social welfare): Assume the following very common spatial model of preferences: Each voter and each option has a certain position in an n-dimensional issue space, and the utility a voter assigns to an option is the negative squared distance between their respective positions. Also assume that voters can nominate additional options for any in-between position (to be mathematically precise, any position in the convex hull of the positions of the original options). Traditional theory shows that, given a set of voters and options with their positions, total utility is maximized by the option closest to the mean voter position, but many traditional voting methods fail or struggle to make sure this option is picked. With our method SEC, however, total utility will be maximized very quickly: If the optimal option X located at the mean voter position is already nominated, every voter will have an incentive to tick X on her consensus ballot since she will prefer X to the otherwise realized fall-back lottery that picks the favourite of a randomly drawn voter. If X is not already nominated, every voter will have an incentive to nominate X for the same reason. This makes sure X is elected and thus total utility is maximized. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] SEC quickly maximizes total utility in spatial model
Dear Abd ul-Rahman, you wrote: Well, I find it hard to believe how wrong-headed this is. Well, thank you very much. In a real society that is large enough, the consensus urn will never choose a winner unless there is a true consensus process already in operation, people will not naturally agree on a large scale, and, while in small organization, 100% consensus is attainable, attaining it in very large ones is next to impossible. With 100,000 voters, at least one of them, even if they all agreed, would accidentally mark the wrong choice. Of course. The method is not suggested for large groups. The cited paper includes suitable variations for that case (using thresholds and the like). It is traditional in democracies that no collective action can be taken without the consent of a majority. And that precisely makes those democracies undemocratic since it gives majorities the power to ignore minorities. While random choice has an appeal, where deliberation is impossible and where results over many elections will average out, what if 1% of the electorate wants to elect a crazy who will start a nuclear war? Could we afford to take a 1% chance of that? Of course not. But such an option must never appear on a ballot in ANY voting method, since such options could easily reach majority support as well, as history has proven over and over again. Exclusion of such options is a different topic which in my view cannot be addressed by voting methods but must be addressed with legal measures. The rest of your post does not seem to be related to mine, and I wonder how you were able to write this much in such short time. Sorry if I don't have the time to read it. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] (Possibly) new method/request for voting paradoxes. :)
Dear Michael, very interesting, I don't think I saw anything like this before. When trying do evaluate a new method, I always try to check very simple criteria first, like neutrality and anonymity (obviously fulfilled here), Pareto efficiency, monotonicity, etc. Concerning the latter two, I was not able to verify them for your method yet. I think you should focus on those before checking more complex things like Condorcet efficiency and so on! Also, immediately a ratings-based generalization came to mind (using ratings differences instead of rank differences). Finally, when only seeking a single winner, you could alternatively build a score for each option X by adding all entries of the final matrix in rows labeled XY and columns labeled with positive differences. Yours, Jobst Michael Rouse schrieb: As usual with such posts, there is a good chance someone has come up with the same (or very similar) method, but I thought it had interesting properties, and was wondering what glaring voting paradoxes it had. In addition, the number of possible orders is overwhelming if there are a large number of candidates, and I'm not sure that can be simplified. Finally, Thunderbird sometimes seems to have weird formatting issues in email, which may screw up the following into unreadability. With that in mind, here it is. Step 1: For each ranked ballot, create a matrix for each pairwise vote, based on the distance and direction between each candidate. For example, on the ballot ABC, you would get: -2-1012 AB0 0010 BA0 1000 AC0 0001 CA1 0000 BC0 0010 CB0 1000 Taking the rows in order, this shows that A is one position higher than B on this ballot, which conversely makes B one position lower than A on the same ballot. Also, A is two positions above C (C is two positions below A), and B is one position above C (or C is one position below B). Such detail may be unnecessary -- simply looking at position 1 and above is sufficient, if you don't allow ties -- but I wanted to show the symmetry. Step 2. Add all matrices together. As a simple example, let's consider the following 12 votes in a circular tie (to make it interesting): 5: ABC 3: BCA 4: CAB Taking the first line, ABC = -2-1012 AB0 0010 BA0 1000 AC0 0001 CA1 0000 BC0 0010 CB0 1000 Multiplied by 5 gives you: -2-1012 AB0 0050 BA0 5000 AC0 0005 CA5 0000 BC0 0050 CB0 5000 Taking the second line, 3: BCA = -2-1012 AB3 0000 BA0 0003 AC0 3000 CA0 0030 BC0 0030 CB0 3000 And finally, taking the third line, 4: CAB -2-1012 AB0 0040 BA0 4000 AC0 4000 CA0 0040 BC4 0000 CB0 0004 Adding these together gives you (5: ABC, 3: BCA, 4: CAB) = -2-1012 AB3 0090 BA0 9003 AC0 7005 CA5 0070 BC4 0080 CB0 8004 Step 3: This is where a bit of a curve is thrown in. Looking at each position on the matrix, determine which is less -- the sum of the numbers above the current position, or the sum of the numbers below the current position -- and add that number to the value of the current position. In essence, this is adding the value of a position to the value of all other positions away from the median. Output that number to a new matrix in the appropriate spot. Using the numbers above, you end up with: -2-1012 AB3 3390 BA0 9333 AC0 7555 CA5 5570 BC4 4480 CB0 8444 Step 4: Looking at each possible preference order, add the values for the appropriate positions on the matrix, and choose the preference order with the highest score. Using the above values and excluding ties: ABC = (AB) [one position] + (BC) [one position] + (AC) [two positions] = 9+8+5 = 22 BCA = 8+7+3 = 18 CAB = 7+9+4 = 20 ACB = 5+4+0 = 9 CBA = 4+3+0 = 7 BAC = 3+5+0 = 8 So the ranking of possible orders is ABC = 22 CAB = 20 BCA =18 ACB = 9 BAC = 8 CBA = 7 ABC is the most preferred order. CBA is the least preferred order. As another example (ignoring a few pages of work), here is
Re: [EM] Schulze definition (was: information con tent, game theory, cooperation)
Dear Raph, Schulze and ranked pairs are the only methods that meet clone independence and the condorcet rule. Nope. River, too, of course, meets all three criteria... Does ranked pairs fail the Smith criterion? I would change B to If there is a group of candidates all preferred over all candidates outside the group, then only those candidates may win and the candidates outside the group may have no effect on the result. If you don't restrict the winner to the Smith set (which your rules don't necessarily), then you could end up with a non-condorcet method. Also, just because the popular/proposed condorcet methods are excluded by your definition doesn't mean that some other weird method can't be found that also meets the rule. It might be better to just include the reasons that you like Sculze and use those rules rather than trying to select Sculze by a process of elimination. BTW it would be nice if the wikipedia page would actually contain something describing Schulze method, not just the heuristics. The best I have found so far is: http://rangevoting.org/SchulzeExplan.html Therefore, my aim was to find a method that satisfies Condorcet, monotonicity, clone-immunity, majority for solid coalitions, and reversal symmetry, and that tends to produce winners with weak worst pairwise defeats (compared to the worst pairwise defeat of the winner of Tideman's Ranked Pairs method). Yeah. Though, ofc, Schulze isn't allow to edit the article. Could someone on this list give a brief outline or the formal rule ( actually his statutory rules are probably it)? Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] [RangeVoting] A Consensus Seeking Method Based on Range Ballots
Hi friends, Raph wrote: On Sat, Feb 14, 2009 at 7:29 PM, fsimm...@pcc.edu wrote: Note well that in the case of a tie among lotteries the tie is broken by a voter, not the voter's ballot, and the tie breaking voter only decides among the tied lotteries, not directly picking the winner. It might be worth collecting 2 ballots from each voter then. A voter could give a strategic ballot and an honest ballot. To avoid this, the tie-breaking rule could also be this: Amoung those lotteries in the tie (i.e., those maximizing the product of all ballot expectations), use the one with the smallest entropy or smallest sum of ballot variances or something like that. This would mean you don't have to find the voter later to ask them to pick the winning lottery. Under this ULVM, rational voters in the above scenario will vote 60: [100, 0, 60] 40: [0, 100. 40], which turns out to be the Nash equilibrium with the highest voter expectations. Is this true in the general (3 candidate) case, i.e. is the Nash equilibrium always for the highest utility candidate to win? What do you mean by highest utility? Also, sometimes with Nash equilibria, it can be possible to obtain a better result by committing to a stance. The result of the other 'players' adjusting to your solid position can yield higher results for you then that Nash equilibrium. But that would require the other voters to be quite ignorant about your true utilities, otherwise they would not adjust to your solid position. This is because then they would know that if they did adjust to your solid position, they would produce a situation which is not a Nash equilibrium and in which you would have an incentive to deviate from your claimed solid position. I wonder if the whole process could be implemented as a DSV method. You give your honest utilities and then method casts your vote for you based on those utilities. It then adjusts your vote based on how others vote. The rule would be to find the best Nash equilibrium. Best Nash equilibrium in what sense? Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Fwd: Some chance for consensus revisited: Most simple solution
Hi Raph, this was suggested before, but I can't remember by whom. The problem is, as you have noted, that the method gives everyone an incentive to reduce her approvals for non-favourites as long as this doesn't change the approval order. For example, consider our example in which the true utilities are 55: A(100) C(70) B(0) 45: B(100) C(70) A(0). With the suggested method, the resulting approval score of C will not be 100 (which would be desirable and would be the result under D2MAC, e.g.) but would rather be only 56. This is because if it would be larger than 56, any voter who approves of C will have an incentive to not approve of C as this would transfer some probability from C to her favourite. Your 50%-modification will not change the above flaw, I'm afraid, since C is nobody's favourite anyway. Yours, Jobst Raph Frank schrieb: I have including the EM alias On Mon, Feb 9, 2009 at 11:55 PM, fsimm...@pcc.edu wrote: I like this concept, but I'm not sure exactly how to combine it with the above tyranny free method. I'm sure it is worth pursuing. I was thinking: - Each voter marks a favourite and approved compromises 1) Find the approval winner 2) Place all ballots which approve that winner in a pile and assign it to the approval winner 3) Repeat the process excluding any ballots which have been assigned to a pile 4) When all the ballots are assigned to a pile, pick a ballot at random 5) The candidate associated with the pile the ballot was picked from is declared the winner Notes: Step 4 could just be pick a pile with each pile having a weighting equal to the number of ballots in the pile Piles smallers than a threshold could be excluded. In that case, the voters corresponding to those piles should be allowed to reassign their probability. For example, they could rank the candidates and their ballot would be assigned to the pile of the highest ranked candidate. A bullet voter would automatically create/be assigned to a pile for his candidate, since he hasn't approved any other candidate, so the ballot cannot be assigned to a different pile. The favourite marking doesn't actually have any effect in this method. However, it doesn't handle the generic case well though. 55: A(100) B(70) C(0) 45: A(0) B(70) C(100) If the 45 faction votes 5: B+C 40: C and the 55 faction compromises 55: A+B The results are Round 1 A: 55 B: 60 C: 40 B wins, B pile becomes 5: B+C 55: A+B Remaining 40: C Round 2 C wins, C pile becomes 40: C The result is B: 60% C: 40% By compromising, the 55 faction effectively transferred its weight from A to B, without the 45 faction having to respond in kind. A rule that would help would be that a ballot can only be placed in a pile for a candidate who isn't the favourite on the ballot, if ballots with that candidate as favourite represent less than half of the ballots in that pile. This would change the outcome: 5: B+C* 40: C* 55: A*+B The results are Round 1 A: 55 B: 60 C: 40 B wins, B pile becomes 5: B+C* 55: A*+B A* ballots are 91% of the ballots The A favourite ballots are more than 50% of the pile, so 50 of them are removed to balance the pile. B pile becomes 5: B+C* 5: A*+B A* ballots are 50% of the ballots Remaining 50: A*+B 40: C Round 2 50: A*+B 40: C A wins (B pile has already been created) A pile is 50: A* + B Round 3 40: C C wins C pile is 40: C Result A: 50% B: 10% C: 40% The 50% rule means that the 55 faction has effectively said that they will only move one ballot from A to B for every ballot the 45 faction moves from C to B. If both factions were to compromise, the result would be B wins round 1 and has pile 55: A*+B 45: B+C* 10 A*+B ballots are removed to make it 50% B pile 45: A*+B 45: B+C* Remainder 10: A*+B Remainder goes to A's pile. Result: 90%: B 10%: A This is reasonably fair, as it recognises the fact that faction A is larger. Also, it is an improvement over the previous non-compromise result from the perspective of both factions. Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Some chance for consensus revisited: Most simple solution
Hi Raph, The odds of it actually working are pretty low. For it to work, all voters must be aware that C is a valid compromise. Sure, that's the flipside of it being so ultimately simple. The easiest way to safeguard against a small number of non-cooperative voters would be to require only, say, 90% of the consensus ballots to have the same option ticked in order for that option to be elected. I guess that's what you mean by threshold: In practice, there needs to be a reasonable threshold. There is always going to be a need to balance tyranny of the (N%) majority against the hold-out problem. Even with a 90% threshold, a tyranny of a 90% majority can be avoided, but this requires another slight modification: Instead of on two separate ballots, every voter marks her favourite and consensus options on one ballot using markers 1 and 2. Then a ballot is drawn at random. If at least 90% of all ballots mark the same option 2 as this drawn ballot does, then that option wins. Otherwise the option marked 1 on the drawn ballot wins. In this way, a bullet-voting faction of, say, 5%, allocates at least 5% winning probability to their favourite (as required by my interpretation of democratic method). Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Some chance for consensus revisited: Most simple solution
Dear folks, I want to describe the most simple solution to the problem of how to make sure option C is elected in the following situation: a% having true utilities A(100) C(alpha) B(0), b% having true utilities B(100) C(beta) A(0). with a+b=100 and a*alpha + b*beta max(a,b)*100. (The latter condition means C has the largest total utility.) The ultimately most simple solution to this problem seems to be this method: Simple Efficient Consensus (SEC): = 1. Each voter casts two plurality-style ballots: A consensus ballot which she puts into the consensus urn, and a favourite ballot put into the favourites urn. 2. If all ballots in the consensus urn have the same option ticked, that option wins. 3. Otherwise, a ballot drawn at random from the favourites urn decides. Please share your thoughts on this! Yours, Jobst Jobst Heitzig schrieb: Hello folks, I know I have to write another concise exposition to the recent non-deterministic methods I promote, in particular FAWRB and D2MAC. Let me do this from another angle than before: from the angly of reaching consensus. We will see how chance processes can help overcome the flaws of consensus decision making. I will sketch a number of methods, give some pros and cons, starting with consensus decision making. Contents: 1. Consensus decision making 2. Consensus or Random Ballot 3. Approved-by-all or Random Ballot 4. Favourite or Approval Winner Random Ballot: 2-ballot-FAWRB 5. Calibrated FAWRB 6. 4-slot-FAWRB 7. 5-slot-FAWRB 1. Consensus decision making The group gathers together and tries to find an option which everyone can agree with. If they fail (within some given timeframe, say), the status quo option prevails. Pros: Ideally, this method takes everybody's preferences into account, whether the person is in a majority or a minority. Cons: (a) In practice, those who favour the status quo have 100% power since they can simply block any consensus. (b) Also, there are problems with different degrees of eloquence and with all kinds of group-think. (c) Finally, the method is time-consuming, and hardly applicable in large groups or when secrecy is desired. Let us address problem (a) first by replacing the status quo with a Random Ballot lottery: 2. Consensus or Random Ballot - Everybody writes her favourite option on a ballot and gives it into an urn. The ballots are counted and put back into the urn. The number of ballots for each option is written onto a board. The group then tries to find an option which everyone can agree with. If they fail within some given timeframe, one ballot is drawn at random from the urn and the option on that ballot wins. Pros: Since the status quo has no longer a special meanining in the process, its supporters cannot get it by simply blocking any consensus - they would only get the Random Ballot result then. If there is exactly one compromise which everybody likes better than the Random Ballot lottery, they will all agree to that option and thus reach a good consensus. Cons: Problems (b) and (c) from above remain. (d) Moreover, it is not clear whether the group will reach a consensus when there are more than one compromise options which everybody likes better than the Random Ballot lottery. (e) A single voter can still block the consensus, so the method is not very stable yet. Next, we will address issues (b), (c) and (d) by introducing an approval component: 3. Approved-by-all or Random Ballot --- Each voter marks one option as favourite and any number of options as also approved on her ballot. If some option is marked either favourite or also approved on all ballots, that option is considered the consensus and wins. Otherwise, one ballot is drawn at random and the option marked favourite on that ballot wins. Pros: This is quick, secret, scales well, and reduces problems related to group-think. A voter has still full control over an equal share of the winning probability by bullet-voting (=not mark any options as also approved). Cons: (b') Because of group-think, some voters might abstain from using their bullet-vote power and also approve of options they consider well-supported even when they personally don't like them better than the Random Ballot lottery. Also, (e) from above remains a problem, in particular it is not very likely in larger groups that some options is really approved by everyone. Now comes the hardest part: Solving problems (b') and (e) by no longer requiring full approval in order to make it possible to reach almost unanimous consensus when full consensus is not possible. In doing so, we must make sure not to give a subgroup of the electorate full power, so that they can simply overrule the rest. Instead, we must make the modification so
Re: [EM] Some chance for consensus revisited: Most simple solution
You're absolutely right, Juho -- I modified the condition a number of times and didn't realize the last version did not imply both factions prefer C to Random Ballot. The correct set of situations for which SEC is a solution is characterized by both factions prefering C to Random Ballot. The latter is in particular true when alpha=beta and C has the largest total utility. Sorry for the mistake, Jobst Juho Laatu schrieb: Makes sense but doesn't this allow also 50: A(100) C(40) B(0) 50: B(100) C(70) A(0) where 50*40 + 50*70 max(50,50)*100 but the A supporters may prefer random ballot from the favourites urn to the possible consensus result (C) and therefore vote (e.g.) for A in their consensus ballot. Juho --- On Sun, 1/2/09, Jobst Heitzig heitzi...@web.de wrote: From: Jobst Heitzig heitzi...@web.de Subject: [EM] Some chance for consensus revisited: Most simple solution To: election-methods@lists.electorama.com Date: Sunday, 1 February, 2009, 11:02 PM Dear folks, I want to describe the most simple solution to the problem of how to make sure option C is elected in the following situation: a% having true utilities A(100) C(alpha) B(0), b% having true utilities B(100) C(beta) A(0). with a+b=100 and a*alpha + b*beta max(a,b)*100. (The latter condition means C has the largest total utility.) The ultimately most simple solution to this problem seems to be this method: Simple Efficient Consensus (SEC): = 1. Each voter casts two plurality-style ballots: A consensus ballot which she puts into the consensus urn, and a favourite ballot put into the favourites urn. 2. If all ballots in the consensus urn have the same option ticked, that option wins. 3. Otherwise, a ballot drawn at random from the favourites urn decides. Please share your thoughts on this! Yours, Jobst Jobst Heitzig schrieb: Hello folks, I know I have to write another concise exposition to the recent non-deterministic methods I promote, in particular FAWRB and D2MAC. Let me do this from another angle than before: from the angly of reaching consensus. We will see how chance processes can help overcome the flaws of consensus decision making. I will sketch a number of methods, give some pros and cons, starting with consensus decision making. Contents: 1. Consensus decision making 2. Consensus or Random Ballot 3. Approved-by-all or Random Ballot 4. Favourite or Approval Winner Random Ballot: 2-ballot-FAWRB 5. Calibrated FAWRB 6. 4-slot-FAWRB 7. 5-slot-FAWRB 1. Consensus decision making The group gathers together and tries to find an option which everyone can agree with. If they fail (within some given timeframe, say), the status quo option prevails. Pros: Ideally, this method takes everybody's preferences into account, whether the person is in a majority or a minority. Cons: (a) In practice, those who favour the status quo have 100% power since they can simply block any consensus. (b) Also, there are problems with different degrees of eloquence and with all kinds of group-think. (c) Finally, the method is time-consuming, and hardly applicable in large groups or when secrecy is desired. Let us address problem (a) first by replacing the status quo with a Random Ballot lottery: 2. Consensus or Random Ballot - Everybody writes her favourite option on a ballot and gives it into an urn. The ballots are counted and put back into the urn. The number of ballots for each option is written onto a board. The group then tries to find an option which everyone can agree with. If they fail within some given timeframe, one ballot is drawn at random from the urn and the option on that ballot wins. Pros: Since the status quo has no longer a special meanining in the process, its supporters cannot get it by simply blocking any consensus - they would only get the Random Ballot result then. If there is exactly one compromise which everybody likes better than the Random Ballot lottery, they will all agree to that option and thus reach a good consensus. Cons: Problems (b) and (c) from above remain. (d) Moreover, it is not clear whether the group will reach a consensus when there are more than one compromise options which everybody likes better than the Random Ballot lottery. (e) A single voter can still block the consensus, so the method is not very stable yet. Next, we will address issues (b), (c) and (d) by introducing an approval component: 3. Approved-by-all or Random Ballot --- Each voter marks one option as favourite and any number of options as also approved on her ballot. If some option is marked either favourite or also approved on all ballots, that option is considered the consensus and wins. Otherwise, one ballot is drawn at random
Re: [EM] A new simulation
Hi Adb ul-Rahman, still, Asset Voting is majoritarian and therefore not democratic. The reason why we have been studying methods with chance components (that is, non-deterministic methods) is that we wanted to find a democratic method, i.e. one that does not give any subset of the voters (no matter how many they are) total decision power. Of course, Asset Voting can easily be modified to achieve this: After the inter-candidate negotiations, not Plurality but Random Ballot is used for the final decision. Yours, Jobst Abd ul-Rahman Lomax schrieb: At 05:15 PM 12/4/2008, Kristofer Munsterhjelm wrote: If you think the risk is too great even so, have a preface adjustment where all candidates that fall below a threshold of first votes are eliminated. The threshold should be very low, say 10. This will introduce some compromising incentive, but again, that incentive should be very slight. Sure. However, if we are going to discuss pie-in-the-sky alternatives, why not start with one that is spectacular in what it could accomplish, that finesses the whole issue. Asset Voting. No need for any vote threshold. Total representation in further process by choice. For starters. Election of an Assembly that is fully representative, where every seat was chosen directly or indirectly (through chosen electors) by N voters, through voluntary cooperation rather than competition. And as to the dregs, represented by those electors who can't get it together to cooperate sufficiently, they are *still* represented in votes, if the electors -- who are identified public voters -- can still vote directly. Existence of a large fully representative body (the electoral college), which can handle special elections, for example, efficiently and cheaply. Possibility of recall of seats who no longer represent those who chose them. Possibility of deciding all single-winner elections deliberatively, which is probably, properly done, better than any known single-ballot method. And on and on. And not a new idea. Asset Voting was described by Lewis Carroll (Charles Dodgeson) in the early 1880s as a tweak on Single Transferable Vote, to deal with exhausted ballots, allowing people without sufficient information to rank more than one to still participate even if they don't vote for one of the potential winners. Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Wiki on Electorama no longer maintained?
Does anybody know why the wiki on www.electorama.com is still not working properly? Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] name of multi-winner method
Of course! Thank you very much. I was not aware of the poorest first variant and thought that asset voting always involves a negotiation after the voting. So, it could be called automatic bottom-up asset voting or so... Diego Santos schrieb: I think this method is Warren Smith's multiwinner poorest firstasset voting with predefined lists. 2008/11/16 Jobst Heitzig [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] Hi Kristofer, That's just the multiwinner adaptation of IRV. I don't think so! The point is that the *candidates* provide the ranking from which the vote transfers are determined. The idea is to keep it maximally simple for the voters: they still vote for only one candidate. I now realize that it is also different from STV in an important respect: there is no transfer of excess votes! What I want with this method is a maximally simple multi-winner method that does not rely on lists but is focussed on individual candidates and that makes sure that all large-enough minorities are represented. It is not important that it results in proportionality, hence it needs no transfer of excess votes. Perhaps this is actually a new method? If so, what would we call it? Jobst I don't think it has a formal name, but here's how I've defined (naive) multiwinner adaptations in my simulations: Take single-winner method X. Produce a social ordering, then if there are k winners to be elected, pick the k highest ranked on the social ordering, and elect them. The social ordering of IRV is opposite of its elimination order: the one who's eliminated in the first round is ranked last, and so on. -- That is, unless the next candidate on her list is the next candidate on the *candidate*'s list, in which case it would be a sort of STV-Asset/Party-list hybrid. Election-Methods mailing list - see http://electorama.com/em for list info -- Diego Renato dos Santos Mestrando em Ciência da Computação COPIN - UFCG Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Some chance for consensus (was: Buying Votes)
Dear Forest, you wrote: This reminds me of your two urn method based on approval ballots: Initialize with all ballots in the first urn. While any ballots are left in the first urn ... find the approval winner X of these remaining ballots circle candidate X on all of the ballots in the first urn that approve candidate X, and then transfer them to the second urn. End While Elect the circled candidate on a randomly drawn ballot from the second urn. Yes, it's inspired by that method which was due not to me but to someone else, I think. The important difference is, though, that in the above method there will never be full cooperation since as long as the two largest approval scores are more than 1 point apart, every voter approving but not favouring the approval winner has an incentive to remove her approval for the approval winner. In particular, the equilibria in our 55/45 situation would look like the following, with x+y=56: 55-x: A x: A+C y: B+C 45-x: B So with the above method, C would win only with probability 56%. In the new suggestion, in contrast, the voters can make sure that the contract to elect C only becomes effective when all voters cooperate: Knowing that everybody prefers C to the Random Ballot lottery, they can all rate C at 1. It looks like your newest method is a variation where Approval is interpreted as positive rating, partial approval I would call it. The rating can be interpreted not as a utility value but as a limit on non-cooperation. Or, the other way around, the value 100 minus the rating can be interpreted as a cooperation threshold. I got the idea for this when reading the Wikipedia article on the National Popular Vote Interstate Compact which has a very similar provision making sure that signing the contract is safe even when it is not known in advance who exactly the other participants will be! and X is circled only on those ballots that rate X sufficiently high* relative to the number of ( remaining) ballots that do not approve X. ...That do not approve X *at least as strongly*. This is important! Otherwise there would alway be an incentive to use only the values 100 (for the favourite only), 1, and 0: Rating an option 1 would then lead to other voters who have a higher rating transfer their probability, without me transferring it, too. Therefore the requirement is that a voter with rating R only transfers her probability if more than 100-R percent of all voters do so, too! This gives me a possibility to specify a safe rating without exactly knowing who will be the other cooperating voters. If 99% of the (remaining) ballots do not approve X, then X is circled only on those ballots that rate X above 99%. If less than 1% of the ( remaining) ballots do not approve X, then even a ballot that rates X at a mere 1% would get a circle around X. Right! The exact relation between the required rating relative to the lack of approval (on the remaining ballots) can be played with to get variations of this method. True, but I guess as long as you use a monotonic transformation for this, the result will be equivalent. Except that the order in which the options are processed would vary. It is charming to be able to simply explain it this way: If you rate C at R, your vote will not be transferred to C whenever R or more voters rate C less than R. In this method there is no need to rate any candidate that the voter cannot conceive of as a compromise. Therefore it seems quite natural to consider positive rating as some level of approval. Right. With this interpretation, options are considered in order of decreasing partial approval score. It is important not to use the rating sum instead since then there would be a conflict between the necessity of using a small score to be safe and using a larger score to make sure the compromise is considered before the polar favourite options! *Some provision must be made for ties and for the case where no ballot rates the current X high enough to get transfered into the second urn. When no ballot rates X high enough, then R must be considered to be infinity, hence no ballots get transferred and the next option is considered. As for ties, I usually think of them late. However: Ties are only relevant for the order in which the options get processes here. The natural tiebreaker would be this: If two options have the minimum number of zero ratings, consider the number of 1-ratings next, then (if still equal) the number of 2-ratings and so on. What do you think of the following name for this method: EC6 (Equal Chances Choice with Controlled Cooperation for Consensus or Compromise) Is this silly or smart? Yours, Jobst Does that capture the idea? Forest - Original Message - From: Jobst Heitzig Date: Thursday, November 6, 2008 3:37 pm Subject: Re: [EM] Some chance for consensus (was: Buying Votes) To: [EMAIL PROTECTED
Re: [EM] Some chance for consensus (was: Buying Votes)
the compromise's winning probability... What do you make of this? Yours, Jobst (The former method was this:) The method is surprisingly simple: 1. Each voters assigns a rating between 0 and 1 to each option. 2. For each option, the mean rating is determined. 3. A ballot is drawn at random. 4. For each option, the score of the option is the option's rating on the drawn ballot times its mean rating determined in step 2. 5. The winner is the option with the highest score. In case of ties, the mean rating is used to decide between the tied options. First of all, the method is obviously monotonic by definition since both the mean and product operations are monotonic. Also, if a faction of p% bullet-votes for some option X (giving it 100 and all others 0), that option gets at least p% winning probability since whenever one of those ballots is drawn, X gets a score 0 and all others get a score of 0. Now let us analyse the equilibria in the above situations. Situation 1: 55% A(100)C(70)B(0) 45% B(100)C(70)A(0) Put a = .55 b = .45 x = a / sqrt(a²+b²) y = b / sqrt(a²+b²) The claimed equilibrium is this: The first 55% of the voters rate A(1)C(x)B(0), the other 45% of the voters rate B(1)C(y)A(0). The mean ratings are A: a*1+b*0 = a B: a*0+b*1 = b C: a*x+b*y = sqrt(a²+b²) If one of the first 55% ballots is drawn, the scores are A: 1*a = a B: 0*b = 0 C: x*sqrt(a²+b²) = a so C wins since it has a larger mean rating than A. Likewise, if one of the other 45% ballots is drawn, the scores are A: 0*a = 0 B: 1*b = b C: y*sqrt(a²+b²) = b so C wins since it has also a larger mean rating than B. No voter has an incentive to change her rating of C: increasing it doesn't change a thing; decreasing it would make C's score smaller than the favourite's score no matter what ballot is drawn, so the resulting winning probabilities become A(.55), B(.45), C(0) which is not preferred to A(0), B(0), C(1) by anyone. Situation 2: 30% A(100)C(70)B,D(0) 25% B(100)C(70)A,D(0) 45% D(100)A,B,C(0) Put a = .30 b = .25 x = a / sqrt(a²+b²) y = b / sqrt(a²+b²) The claimed equilibrium is this: 30% rate A(1)C(x)B,D(0) 25% rate B(1)C(y)A,D(0) 45% rate D(1)A,B,C(0) The mean ratings are A: a*1 = a B: b*1 = b C: a*x+b*y = sqrt(a²+b²) D: .45 If one of the first 30% ballots is drawn, the scores are A: 1*a = a B: 0*b = 0 C: x*sqrt(a²+b²) = a D: 0*.45 = 0 so again C wins since it has a larger mean rating than A. It's similar for the 25%. When one of the 45% is drawn, D is elected. Again, no voter can gain anything by increasing or decreasing her C-rating. Situation 3: 32% A(100)C(40)B,D(0) 33% B(100)C(40)A,D(0) 35% D(100)C(40)A,B(0) Put a = .32 b = .33 d = .35 x = a / sqrt(a²+b²+d²) y = b / sqrt(a²+b²+d²) z = d / sqrt(a²+b²+d²) The claimed equilibrium is this: 32% rate A(1)C(x)B,D(0) 33% rate B(1)C(y)A,D(0) 35% rate D(1)C(z)A,B(0) The mean ratings are A: a*1 = a B: b*1 = b C: a*x+b*y+d*z = sqrt(a²+b²+d²) D: d*1 = d If one of the first 32% ballots is drawn, the scores are A: 1*a = a B: 0*b = 0 C: x*sqrt(a²+b²+d²) = a D: 0*d = 0 so still C wins since it has a larger mean rating than A. It's again similar for the other voters, and still no voter can gain anything by increasing or decreasing her C-rating. Problems: (a) The stated equilibria are not exactly stable since already a deviation on the part of one faction gives a different faction the means to manipulate the scores so that C wins when a ballot from the first faction is drawn but not when a ballot from their own faction is drawn. This problem might be bigger or smaller when faction don't know their respective sizes. (b) The meaning of the asked-for ratings is not clear. Maybe their meaning can only be defined operational by pointing out how they are used in the method. It seems they cannot naively be interpreted as utilities. All this makes it difficult to tell what a sincere rating would be. (c) It would be nice if the score formula could somehow be changed so that the equilibrium ratings would not include the normalization factor 1/sqrt(...). But I fear that this is not possible. I tried to use the minimum of the individual and the mean score instead of their product, but that did not result in any equilibria at all. Using a sum or maximum instead of the product would destroy the bullet-voting property. Also, using (individual rating)^(some exponent) * (mean rating) does not help. More ideas I did not have yet. Perhaps the mean rating must be replaced by some other location statistic such as the median rating. Or perhaps we somehow include the Q-quantile of the ratings, where Q is the rating on the drawn ballot... Any thoughts? Jobst Jobst Heitzig schrieb
Re: [EM] Will to Compromise
Dear Greg, you wrote: Nondeterminism is a delightful way of skirting the Gibbard-Satterthwaite theorem. All parties can be coaxed into exposing their true opinions by resorting or the threat of resorting to chance. Actually, if I remember correctly, that theorem just said that Random Ballot was the only completely strategy-free method (given some minor axioms such as neutrality and anonymity), so it's not really skirting it but just taking it seriously. However, it seems some minor possibilities for strategizing are acceptable when they allow us to make the method more efficient. FAWRB tries to be a compromise in this respect. I don't dispute that. The nondeterminsitc methods I have seen appear to be designed to tease out a compromise because a majority cannot throw its weight around. Right, that's the main point. The abilities of nondeterministic methods to generate compromises is formidable, but since we speak of utility, I would like to point something out. 1) Using Bayesian utility, randomness is worse than FPTP. Two answers: i) Please cite evidence for this claim, ii) Bayesian utility is not a good measure for social utility in my opinion. We had lengthy discussions on this already a number of times on this list, so I won't repeat them. Instead, I will produce evidence from simulations this weekend which shows that no matter what measure of social utility is used, Random Ballot does not perform much worse than optimal. 2) False compromises are damaging What do you mean by false? If a proposed compromise fails to be desirable by most voters over the Random Ballot lottery, it will not get much winning probability. If it is, on the other hand, it is not a false but a good compromise. The simulations I will report about this weekend show that usually we can good compromises to exist which have quite large social utility. The reduced power of a majority means that at any choice with a greater-than-random-ballot average utility is a good compromise Notice how lousy the Bayesian utility of random ballot is and you begin to see my point. See above. In simulations with well-known preference models, Random Ballot results are not lousy at all. Also note that the method for determining the compromise is majoritarian (to the extent that approval is) so the intermediate compromise procedure is a red herring that produces some nasty side-effects. The compromise is determined to be the most-supported at-least-above-average candidate. How does this avoid the original criticism of majoritarian methods? You are right in that the majority still has some special influence on the *nomination* of the compromise. But the important difference to majoritarian methods is that they can't make any option get more winning probability than their share without the minority cooperating in this. So, yes, they can present the minority with a compromise they value only slightly better than Random Ballot. This is not perfect yet, but it guarantees the minority to get a better-than-average result where a majoritarian method doesn't guarantee a minority anything! Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Some chance for consensus (was: Buying Votes)
Dear Raph, you wrote: I was thinking of a 'stable marriage problem' like solution. Good idea! If it works, the main difficulty will be to make the whole process monotonic, I guess... Yours, Jobst Each voter rates all the candidates. Each voter will assign his winning probability to his highest choice (probably split equally if he ties 2 candidates for first). If 2 voters 'marry', then the candidate with the highest score sum is the compromise candidate. Solve the stable marriage problem. It might be necessary to randomly split the ballots into 2 'genders' to guarantee that a stable solution exists. Using the above example: G1: A1(100) A(70) A2(0) G2: A1(0) A(70) A2(100) G3: B(100) G4: C(100) (unnamed options are rated zero) If a member of G1 'marries', then the compromises are G1: A1 (+0) G2: A (+40), i.e. 100-70 (-30) and 0-70 (+70) G3: A1 and B tie (+0) .. effectively not a 'marriage' G4: A1 and C tie (+0) .. effectively not a 'marriage' Thus rankings are G1: G2G1=G3=G4 Similarly G2: G1G2=G3=G4 G3: all equal G4: all equal Thus the 25 G1s will 'marry' the 25 G2s and compromise on A. The result being A: 50% B: 25% C: 25% Also, what about an iterative method. If the candidate with the lowest probability has less than 1/3 probability, eliminate him and re-run the calculations (and probably rescale the ratings). This is kind of similar to the requirement that a candidate has 1/3 approval before being considered. As an added complication, in the above, it might be worth doing a second pass. Once all the marriages are stable, you could have 'suitors' propose to 'engaged' voters and make an offer with a different compromise candidate. For example, if two voters has ratings, A1(100) A2(90) A3(75) A4(55) A5(0) A1(0) A2(55) A3(75) A4(90) A5(100) The possible compromises are A2, A3 and A4. However, A2 favours the first voter and A4 favours the 2nd voter. It might be the case that after being refused, a 'suitor' could sweeten the deal by offering a better option. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Will to Compromise
Dear Kristofer, you wrote: With more candidates, a minority might find that it needs to approve of a compromise with just slightly better expected value than random ballot, if the majority says that it's not going to pick a compromise closer to the minority than that just-slightly-better candidate. That is, it would give an incentive to compromise early, under the threat that to do otherwise might make the method fall back to random ballot, and the compromise is better than random ballot even if it's not all that much better. True. But for the minority, Random Ballot is usually already much better than the majority preference, so that would be OK, right? Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Some chance for consensus (was: Buying Votes)
Dear Forest, good to hear from you again! You said: Not quite as important, but still valuable, is achieving partial cooperation when that is the best that can be done: 25 A1AA2 25 A2AA1 25 B 25 C Here there isn't much hope for consensus, but it would be nice if the first two factions could still cooperate on gettiing A elected, say 25% of the time. (50% seems too much to hope for) That's absolutely true! We both tried to achieve this during the last year. But it is very difficult to make this happen with strategic voters. Perhaps I find the time this weekend to write a summary of what we tried in this respect, so that perhaps someone can build on that an come up with a new idea. It seems to me that if we require our method to accomplish the potential cooperation in this scenario while achieving consensus where possible, the ballots would have to have more levels, and there would have to be an intermediate fall back between the consensus test and the random ballot default. That could work, but I wouldn't bet on it yet. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Some chance for consensus (was: Buying Votes)
Dear Raph and Forest,
I have a new idea which might be monotonic, generalizing the 2-voter-marriage
idea to larger groups of voters.
I will define it as an optimization problem: basically, the idea is to find the
socially best lottery which can
be produced by starting from the Random Ballot lottery and allowing for one set
of voters to reach a contract in which they transfer their share of the
winning probability from their favourite options to other options. More
precisely, the suggested method is this:
1. Each voter submits a cardinal rating for each option.
2. Amoung all possible lotteries that assign winning probabilities to the
options, we determine the feasible ones. In order to determine whether a
given
lottery L is feasible, we do the following:
a) Compare L with the Random Ballot lottery, RB, and find the set S of options
which have a lower winning probability under L than under RB. Mathematically:
S = { options X with L(X) RB(X) },
where L(X) = probability of option X in lottery L.
b) For each option X in S, determine the number N1(X) of voters who favour X
and like L at least as much as RB, judging from their submitted ratings.
Mathematically:
N1(X) = no. of voters V with V(L) = V(RB),
where V(L) = sum of V(X)*L(X) over all options X
and V(X) = rating voter V assigned to option X.
c) Also, determine the number N2 of those voters who favour X which must agree
to transfer their share of the winning probability from X to other options in
order to produce L. Mathematically:
N2(X) = (RB(X)-L(X)) * N,
where N is the no. of all voters.
d) Then check whether N2(X)=N1(X) for all X in S. If this is fulfilled, then
this means that a group of voters exists who have both the means and the
incentices to change RB into L by transferring winning probability from their
respective favourite options to other options. So, if the condition is
fulfilled, L is considered feasible.
3. Finally, find amoung the feasible lotteries the one that maximizes a given
measure of social utility, e.g. total utility or Gini welfare function or
median voter utility or whatever. Apply this socially optimal feasible
lottery to determine the winner.
With sincere voters, the method achieves what we desire:
1. With 55 having A(100)C(70)B(0) and 45 having B(100)C(70)A(0), the
optimal lottery L would be L(A/B/C)=0/0/1. This is feasible since it has
S={A,B}, N1(A)=N2(A)=55, and N2(B)=N2(B)=45.
2. With 25 having A1(100)A(90)A2(70)B(0), 25 having
A2(100)A(90)A1(70)B(0), and 50 having B(100)A,A1,A2(0), the optimal lottery
L would be L(A/A1/A2/B)=.5/0/0/.5 with S={A1,A2},
N1(A1)=N2(A1)=N1(A2)=N2(A2)=25.
I did not yet analyse the strategic implications, though. So we need to check
that and the hoped-for monotonicity. The crucial point for the latter will be
what happens when some voter changes her favourite, I guess.
Some final notes:
- There are always feasible lotteries since the Random Ballot lottery itself is
feasible by definition (with the set S being empty).
- For the same reason, the method gives no lower social utility than Random
Ballot.
- Geometrically, the set of feasible lotteries is a closed, star-shaped
polyeder, but it is usually not convex. (It would be convex if more than one
contracting group of voters were allowed.)
What do you think?
Jobst
-Ursprüngliche Nachricht-
Von: Raph Frank [EMAIL PROTECTED]
Gesendet: 31.10.08 15:35:30
An: Jobst Heitzig [EMAIL PROTECTED]
CC: [EMAIL PROTECTED], election-methods@lists.electorama.com, [EMAIL
PROTECTED], [EMAIL PROTECTED]
Betreff: Re: Some chance for consensus (was: [EM] Buying Votes)
On Fri, Oct 31, 2008 at 11:17 AM, Jobst Heitzig [EMAIL PROTECTED] wrote:
Dear Raph,
you wrote:
I was thinking of a 'stable marriage problem' like solution.
Good idea! If it works, the main difficulty will be to make the whole
process monotonic, I guess...
Yours, Jobst
I think the method which eliminates the lowest probability candidate
will be non-monotonic.
In the single run case, the fundamental problem is that bilateral
monopolies can exist. You can gain by not offering compromises.
However, assuming competition, you might be 'outbid' by another
voter/party if you do that.
-Ursprüngliche Nachricht-
Von: Raph Frank [EMAIL PROTECTED]
Gesendet: 31.10.08 15:35:30
An: Jobst Heitzig [EMAIL PROTECTED]
CC: [EMAIL PROTECTED], election-methods@lists.electorama.com, [EMAIL
PROTECTED], [EMAIL PROTECTED]
Betreff: Re: Some chance for consensus (was: [EM] Buying Votes)
On Fri, Oct 31, 2008 at 11:17 AM, Jobst Heitzig [EMAIL PROTECTED] wrote:
Dear Raph,
you wrote:
I was thinking of a 'stable marriage problem' like solution.
Good idea! If it works, the main difficulty will be to make the whole
process monotonic, I guess...
Yours, Jobst
I think the method which eliminates the lowest probability candidate
will be non-monotonic
Re: [EM] Buying Votes
Dear Greg, you wrote: I wish I knew how FAWRB worked... I will give a new step-by-step exposition of FAWRB in a new thread during this day. Most people say the majority criterion is a good one. I, for one, doubt its importance. I was merely saying that a counter-criterion inconsistent with the majority criterion hasn't really been offered. That counter-criterion has been offered: Every voter should have full control over the same amount of winning probability. Methods complying with this then should try to increase the efficiency by providing some means for anonymous cooperation. umm... in any electoral method some subset will have absolute power... What makes you think so? The simplest counter-example to your claim is Random Ballot. Also, in ANY electoral method you can get useless votes. Again: The simplest counter-example to this claim is Random Ballot. For example, if I have 999 people supporting me under some non-majoritarian method, 1 person voting against it won't make a hell of a lot of difference. The 999 have complete power! The question is not whether it makes a hell of a lot of difference but whether it makes a difference at all. In FAWRB, you would have incentive to provide that 1 person with some small amount of compensation in order to get her vote, too. This is a gripe with democracy not with majoritarianism. Nope. See above. It's all about *establishing* democracy! Thinking outside the box, they could secede. Yes, and that's a threat to society which could ultimately lead to anarchy. Athenian democracy doesn't work. Sorry. Using a non-determinstic method incurs massive flaws in real life. Give examples, please. Athenian democracy worked well for the Athenians for two centuries. Applying the same logic to majoritarian methods, every opinion has a given probability of being a majority. That probability being zero when the opinion is the minority opinion. Yay we agree OMOV is trivial. Why do you keep bringing up violations of OMOV then? I did never talk about OMOV, it was you who brought it up. Probably you misunderstood me. Denying voters influence is an OMOV flaw. No it's not. Most majoritarian methods comply with OMOV and still deny the minority any influence. I'm asking myself how often I will have to repeat this obvious fact. If there are a group of int(.5*voters)+1 people who vote X non-X and X is one and only one candidate, how does the minority express its opinion w/o violating OMOV. Er? You ask how they express their opinion? Well, on the ballot, of course? They ARE better, just not intrinsically so. Specific violation of a given property does not a perfect voting system make. Nobody is claiming that since it would be ridiculous. Is there ever an incentive not to secret contract? If so, is it significant? To avoid this is exactly the task of the method designer. The method should be designed so that there are positive incentives to cooperate when the compromise is good enough. I see... is this FAWRB? Candidates have three statuses on a ballot: Favorite, Approved, Disapproved I pick two ballots at random. If there is a candidate that is favorite on both ballots, pick it. If there is a candidate that is favorite on the first and approved on the second, pick it. If there is a candidate that is approved on the first and favorite on the second, pick it. If there is a candidate that is approved on both the first and second, pick it. Otherwise pick the favorite on the first. That's more or less what we call D2MAC, which is similar in spirit to FAWRB and achieves the task of electing a good compromise, too. The main difference is that with D2MAC compromise options which are only good but not *very* good have fewer chances than with FAWRB. Ok now the actual criticism. I know that FAWRB is nondeterministic. Here is why that is bad. Factions (both unwilling to compromise): A 55% B 45% you view A as gaining a 55% chance of victory. This reasoning is flawed. Instead of viewing A as getting .55 victory units, think of it as a random choice between two possible worlds: A-world and B-world A-world is 10% more likely to occur, however they share remarkable similarities. In both worlds =45% of the people had no say whatsoever. They had a say in that they had a fair chance. In reality, chance occurs everywhere anyway. Picking an option is always a risky thing. When economists talk of utility and analyse preferences they almost always mean *expected* utility because the actual utility is the result of a random process. Anyway, the point is that in your example people (if they use the right decision method) will have an incentive to find a compromise C which everyone likes a lot better that the 55%/45% lottery. So they will usually reduce the randomness by cooperation. Now, you're reasoning apppears to stem from a simple observation... If A achieves one more vote, its chance of victory
Re: [EM] Buying Votes
Dear Greg, you wrote: I'm not speaking about majoritarianism in this case, although you are correct that it alleviates many of the problems. What I meant was there is the potential for vote buying under any voting method where voting is verifiable and non-unanimity can pass a policy. OK, I agree. You continued, answering me: In every reasonable voting method (remember democracy is distinct from consensus), ... I can only remember what I believe to be true. This claim is not! democracy is distinct from consensus? Of course it is! I can win under any reasonable voting method by pleasing less-than-everyone. I think I misunderstood you there. What I meant was: Sophisticted democratic decision methods can lead to almost complete consensus. FAWRB promises to do so by giving everybody an incentive to search for a good compromise and making sure it wins with very large probability. This is indeed very near to reaching consensus! Only majoritarian (and thus undemocratic) methods lead to results often far from consensus. For this reason, I tend to find majoritarian methods relatively unreasonable :-) I attempted to explain in You Can't Have it Both Ways that a voting system cannot and should not be designed to protect rights... but I digress. A voting system should not be designed to protect rights? Of course it should! It should be designed to protect the right of everybody to have an equal amount of power in decisions! I have never seen any method lauded so much for disobeying a criterion. ; ) What's a criterion? Usually it is a sentence which can either be true or false about any decision method. Whether passing or failing a criterion is the good thing depends on values. My most important values are first equality and individual rights, then efficiency. Hence failing the majority criterion is mandatory for any reasonable decision method since majoritarian ones disrespect my most important values. You're totally right. This is the best motivation for giving each voter the same voting power instead of giving some majority all of the power. Then the majority has something to trade. In order to get my proposed option elected, I need their cooperation which I must buy by taking their preferences into account in my proposal. I don't follow. If I reward a majority, then that does nothing to prevent future majorities from forming. I don't want to prevent any majority from forming. That would be ridiculous as people have every right to have the same opinion as others. The point is not whether there are majorities or not. The point is that majorities must not be given 100% of the decision power in any single decision. Majoritarianism isn't some complete shift of power to whoever can muster 51%... Er? That's *exactly* what majoritarianism is! What else than a shift of power would you call it when 51% of the people need not care what the other 49% want in some individual decision because they can safely establish what they want? Obviously they have the complete power in that decision when a majoritarian method is used. Every voter has the same capacity to influence the election. If you believe this is the case with your favourite majoritarian method, then please show me how the latter 45% in the following quite common situation have the same capacity to influence the election: 55% wanting A, 45% wanting B. Just tell me what the 45% can do to avoid getting A for certain. I can tell you what they can do when FAWRB is used: They can just vote for B and thus give B a 45% winning chance, compared to a 55% winning chance for A. But even better: They can also propose a good compromise option C which everybody prefers much to the 55%/45% lottery. If they suggest such a C, everybody will have an incentive to mark C as approved under FAWRB, so that C will be elected with certainty. OMOV and majority are not in conflict. No rules says that a majority method is automatically non-OMOV. OMOV is a purely formal requirement which is so trivial that I cannot remember a single decision method having been discussed here that not fulfilled it when interpreted correctly. OMOV Interpreted correctly means the only information about the voter used in the decision process must be his or her preferences as revealed by him or her on the ballot. What OMOV does *not* guarantee is that everybody has an influence on the decision. Obviously, majoritarian methods are OMOV but make it impossible for as much as 49% of the voters to influence the decision. I don't think that non-majoritarian methods are intrinsically better. If you don't think democracy is important... Right... voting is non-contractual. THAT IS THE POINT! If it were contractual (read verifiable), No, I don't read contractual as verifiable. FAWRB makes it essentially contractual in providing safe ways to cooperate anonymously without having to reveal my preferences to anybody. Voters do not
Re: [EM] Who comes second in Ranked Pairs?
Dear Steve, in my old post http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/2004-April/012735.html, some simulations are reporterd. There the following sentence is found: Judging from who beats whom, max. length, mean length, or sum of defeats, we get MAM River+ Beatpath. By this I meant that the MAM winner beat the River winner more often than vice versa. That simulation was apparently with 4 or 20 options and a heterogeneous electorate where there was no linearity in the sense that preferences of different voters were not correlated. Yours, Jobst Steve Eppley schrieb: Hi, For a lot more information about the second place finisher, follow the link to the Immunity from Majority Complaints criterion (and related criteria) at the following website: http://alumni.caltech.edu/~seppley (It's best viewed with the Internet Explorer browser, I think, since it uses a Microsoft character set where Greek symbols are used. For instance, my Firefox browser renders the epsilon symbol, which means is an element of, as an I with a caret on top. Sorry.) That webpage includes a proof that the alternative that would be elected by MAM (called Ranked Pairs by some people) if the winner were deleted from all ballots is the alternative that finishes second in the ordering that can be constructed from the acyclic set of majorities MAM identified to select the winner. (Which also happens to be the same as the ordering that minimizes the largest thwarted majority in the minileximax sense, which is proved elsewhere in the website.) I use the name Maximize Affirmed Majorities (MAM) rather than Ranked Pairs, because Ranked Pairs is the name coined by Nicolaus Tideman for a different voting method he proposed in 1987/1989. The most important difference is that Tideman's Ranked Pairs measures the size of a pairwise majority for X over Y to be the number of voters who ranked X over Y minus the number of voters who ranked Y over X. Due to that difference, Tideman's Ranked Pairs fails several criteria met by MAM. I think the potential for confusion if the name Ranked Pairs is used for a methods other than Tideman's justifies using a different name, hence my preference for the name MAM. Jobst, is there any justification for the criterion you mentioned below that's satisfied by Beatpath but not by MAM or River? For each beatpath against the winner, there is a stronger (or equally strong) beatpath back. In scenarios where a losing alternative has a stronger beatpath to the MAM winner and/or the River winner, what is the harm? If there is no harm, why should anyone care about this criterion? I never ran a simulation comparing MAM vs River to see which method's winner is ranked over the other's winner more often over the long run. Has anyone run that simulation? The MAM vs Beatpath simulation showed that MAM winners are ranked over Beatpath winners over the long run. For more details, follow the link at the website above to MAM vs PathWinner. Regards, Steve -- Jobst Heitzig wrote: Hi folks, you're right, the option ending up second in the ranking constructed in RP is also the one that wins when you exclude the first winner. Even more so, the whole new ranking is just the old one with the top removed. There is, by the way, a related major difference between RP and the otherwise very similar methods Beatpath (aka Schulze) and River: In those two the 2nd place winner (= winner when 1st place winner is removed) cannot nearly as easily be determined from the original result. And in those two methods it can even happen that the 2nd place winner is an option which defeats the 1st place winner pairwise. To see this, consider a situation with four options A,B,C,D and the following pairwise defeats in order of descending defeat strength: BD, AC, CD, DA, AB, BC. In this situation Beatpath will elect B 1st and will elect A when B is removed. Also River will elect B 1st, locking in the defeats BDAC only, and will elect A when B is removed. So under both methods the 2nd place winner A defeats the 1st place B. But RP will lock in all defeats but DA, making A the 1st place winner and B the 2nd place. This property is one of the very few in which RP, Beatpath, and River differ. The other two such properties are (ii) that only River fulfills IPDA and ISDA, and (iii) only Beatpath fulfills (by definition) the requirement that for each beatpath against the winner there is a stronger (or equally strong) beatpath back. So each of the three similar methods has at least one desirable property the other two don't share. Yours, Jobst Kristofer Munsterhjelm schrieb: Scott Ritchie wrote: I'm writing a ranked pairs counter as practice for learning python, and I realized I don't know the answer to this question. Suppose I want to know who comes in second in a ranked pairs
Re: [EM] Who comes second in Ranked Pairs?
Dear Steve, you wrote: Jobst, is there any justification for the criterion you mentioned below that's satisfied by Beatpath but not by MAM or River? I personally don't think so. In the beginning, I liked the criterion however for its mathematical beauty, and since it's a natural generalization of the requirement of immunity, but that's all. I don't see any practical relevance for it. I never ran a simulation comparing MAM vs River to see which method's winner is ranked over the other's winner more often over the long run. Has anyone run that simulation? I think I have but I will have to look it up. I will answer tomorrow. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range Voting vs Condorcet (Greg Nisbet)
Dear Diego, you wrote: The risk of minority will remains. How does FAWRB perform in binary issues? What you mean by risk of minority? That a minority favourite may win? Well, that is just the *feature* of FAWRB: It gives each part of the electorate full control over an equal share of the winning probability. This is the requirement of democracy. So, when 55% prefer A and 45% prefer B and both groups do not care to look for a good compromise C or do not cooperate in electing such a good compromise by using FAWRBs cooperation mechanism, then indeed A will win with 55% probability and B will win with 45% probability - which is just fair and what a democratically thinking person would expect. This also answers your question about the binary case. However, let me point out that in most real-world issues, there is a possibility to come up with a good compromise option. Sometimes, for example, this can be achieved by side payments, that is, C is A plus some payments (or other forms of compensation) from the A supporting group to the B supporting group. Once a good compromise is found, using FAWRB makes it probable that this compromise is also elected. Majoritarian methods fail here since with them, the majority has no incentive at all not to bullet vote for A and thus overrule the rest. If a consensus exists between the factions, then this danger would be too rare. There`s no gain for any faction to leave the issue undecided. I don't think so. In my experience of politics it is often the case that one faction strongly wants to stick with the status quo, so they would have a strong incentive to refuse cooperation under your scheme. Not always we can find an unanimity... Yes, that's exactly the reason why sometimes we need to resort to a chance process in order to give every voter their fair right to influence the decision. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range Voting vs Condorcet (Greg Nisbet)
Dear Greg,
you wrote:
Group membership is difficult to define. With ranked ballots it's
simple, but in the majority criterion debate, I argue that a score of
60% represents 60% of a first preference, not the preference between 59%
and 61%.
Sorry, I don't get your meaning here.
However, it seems to me that there is a confusion about the usage of the
term majority in the context of our debate. In the majority criterion
and related criteria, we usually don't speak of *the* majority,
referring to one specific subset of the electorate, but we refer to *a*
majority, by which we mean *any* subgroup consisting of more than half
of the voters.
For example, consider the classical cycle of true preferences, where
voter X ranks ABC
voter Y ranks BCA
voter Z ranks CAB.
In this situation, there are three different majorities: {X,Y}, {Y,Z},
and {Z,X}. Of course these groups are not disjoint and it makes no sense
to speak of the majority. Rather, the majority criterion only requires
that each of these groups, should they decide to do so, can overrule the
third voter. That is, X and Y can cooperate in overruling Z and making
sure B wins. Likewise (but not at the same time of course), Y and Z
could agree to elect C. So, it usually makes no sense to speak of the
majority since most often there are lots of majorities - it all depends
on which of these groups happens to make the deal to overrule the rest.
... we have not settled the issue of simultaneous
majorities.
See above for clarification. There is no issue of simultaneous
majorities, the criterion simply requires that each subgroup of more
than half of the voters has a way of overruling the rest. It does not
require that two such subgroups can do so at the same time, which is
obviously impossible.
I continued:
While of course civil rights are very important to make sure that
no-one's basic *rights* are violated, they cannot make sure that
everybody's *preferences* are have a fair chance of influencing
decisions that are made *within* the limits the civil rights pose.
to which you replied:
Let me explain my point. I set the bar fairly high for tyranny of
majority i.e. it must constitute actually oppressing me and not merely
annoying or inconveniencing me to be labelled tyranny.
I don't care for the label tyranny. My point is that when a majority
is able to overrule the rest with certainty, then that's not democratic.
You talk about the destruction of
democracy.
Did I? I don't think so. I don't think there has been any large-scale
truly democratic system yet. Only some families and small groups often
decide in an approximately democratic way when they make sure that each
member of the group makes a decision at some point in time, for example
by letting the members decide in turn.
That democracy is an all-or-nothing type thing. I am arguing
that a good constitution will prevent a majority from acting in such a
way that democracy itself is subverted.
Not when the constitution allows the majority to decide all issues
without having to be concerned about other peoples wishes.
If you argue instead that
suboptimal results come about, yes I agree with you.
My point is not the optimality of results, whatever that may mean. To
define and ensure optimality is a large but different task than to
ensure the democratic right to influence the decision.
For example, some philosophers argued that it would be optimal if some
highly intelligent, well-informed and impartial person (the
philosopher-king) decided all issues. Though I tend to agree that this
might give optimal results, such a system would obviously be not a bit
democratic.
On the other hand, simply drawing a random ballot to decide is perfectly
democratic since it gives each voter exactly the same power regardless
of factions. However, that method would not give optimal results at
all since compromise options would get no chance at all. What is missing
here is an incentive to cooperate.
So, whether a method is democratic and whether it leads to optimal
results are just two questions which are in large part (but not totally)
independent. This is why we developed FAWRB, a method which gives each
voters the same power but gives them also strong incentives to cooperate
in finding and electing good compromise options.
Again, you speak about actively preventing the majority from doing
something that violates the rights of minority. Such cannot be prevented
by any voting method!
Excuse me! Of course it can. I have demonstrated this over and over.
With FAWRB, the worst a majority of, say, 55% of the electorate can do
to the minority is to bullet-vote for the option considered worst to the
other 45%, thus assigning 55% of the winning probability to that option.
But this is not violating the minorities rights since at the same time
those 45% of the voters can assign the remaining 45% of the winning
Re: [EM] Range Voting vs Condorcet (Greg Nisbet)
Dear Raph, you answered to me: a) FAWRB is not a random but a very specific and quite sophisticated method. It only uses a certain amount of chance, just as many things in our life do. Chance should not be mixed up with arbitrariness. Used in a rational way, FAWRB will usually elect good compromise options with near certainty, not leading to significant amounts of randomness. I know, but it does have randomness. I includes a chance process just as many sophisticated things in our life do. It does not include arbitrariness. It will most often lead to a certain winner (one option getting 100% winning probability). Here's some evidence that the perceptions that chance processes are evil and that deterministic processes cannot lead to random results is wrong: 1. Some time ago I challenged you all by asking for a method which elects C with certainty in the 55/45-example. The only methods which achieved this seeminly simple goal included a chance process. 2. Every majoritarian method leads to a severe kind of randomness when there's no Condorcet Winner! This is because in all these situations there is no group strategy equilibrium, that is, whatever the winner is, there will be some majority having both the incentive and the means to change the winner to an option they like better. So, where the strategic process will end is mostly random since it cannot settle on an equilibrium. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range Voting vs Condorcet (Greg Nisbet)
Dear Greg, I will focus on the question of majoritarianism in this message. First my working definition of majoritarian method: A method is majoritarian if for every option X and every group G consisting of more than half of the voters, there is a way of voting for G which makes sure X wins regardless of how the voters outside G vote. In other words: Any majority can overrule the rest if that majority votes in a certain way. Now for the discussion. I said: That leads me to the main problem with Range (as with any other majoritarian method): It is simply not democratic. It cannot be because every majoritarian method gives 100% of the power to less than 100% of the people (the demos in greek). Often, about 60% of the people can consistently impose their will on the other 40% without the latter being given any means at all by the majoritarian method to influence the decision. Of course, this is a problem of most popular election methods, but that does not mean the problem cannot be solved. Democratic decisions are possible but not with majoritarian methods. To which you replied: Interesting point. I would argue that a compromise candidate is better than a polarizing but barely passing candidate (like FPTP with primaries tends to produce). I'd say this isn't a voting-issues question, but a civil rights question. A nice constitution will help protect you from tyranny of the majority. While of course civil rights are very important to make sure that no-one's basic *rights* are violated, they cannot make sure that everybody's *preferences* are have a fair chance of influencing decisions that are made *within* the limits the civil rights pose. Advocates of majoritarianism argue that majority decision making is intrinsically democratic and that any restriction on majority decision making is intrinsically undemocratic. I wonder how they do so. It's as simple as that: When any group of people, be it a single person (dictatorship) or a small group (oligarchy) or a large group (majoritarianism) can overrule the rest, that's not democratic since democracy in its main sense requires that *all* people must have a means to influence decisions. If democracy is restricted by a constitution which cannot be changed by a simple majority decision then yesterday's majority is being given more weight than today's; We may later discuss shifting majorities, but please let us first continue discussing a single decision since that is complicated enough. You continue to ask: ... if not the majority, then who decides? Simple answer, contained in the definition of democracy: It's not a subgroup of the voters which decides but its *all* voters who decide. I guess your real question is not who decides but how they do it. If you delegate the responsibility to some group (even yourself) to judge what is best for society, then you are imposing your will on people. Right. That would be much worse. But essentially majoritarianism *does* delegate the decision to some group (the majority that finally overrules the rest). The only difference is that it does not prescribe who belongs to this group. Rather, any willing majority can establish itself as this deciding group. But this is not much better because some group overrules the rest anyway. The whole point of democracy is that *no* group can overrule the rest, neither a predefined group nor a group that establishes itself as a majority. Arguments both for and against majoritarianism both tend to boil down to rights. Do you have the right to non-interference from the majority? Does the majority have the right to non-interference from you? Please don't shift the focus. The question is not whether some group can intefere but whether some group can overrule. So, the right everyone should have is the right not to be overruled by a majority without my preferences having any chance to influence the result. Probably you still think, how on earth could this be achieved? But it is very easy to see that real democratic decisions are possible. Just imagine everyone marks their favourite option and then a ballot is drawn at random to decide the winner. Of course I don't suggest to use this method called Random Ballot. It is only to illustrate that the requirement of democracy can be met. The real task now is to find methods which are not only democratic but also satisfy other criteria (like anonymity, neutrality, monotonicity, clone-proofness etc.) and are efficient in electing good compromise options. This is achieved by the methods D2MAC and FAWRB for example - you make look them up in the archives. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range Voting vs Condorcet (Greg Nisbet)
Dear Raph, you replied to me: That leads me to the main problem with Range (as with any other majoritarian method): It is simply not democratic. It cannot be because every majoritarian method gives 100% of the power to less than 100% of the people (the demos in greek). They do have an equal vote. The move the median in their direction. First, what does an equal vote help when the other group (the majority) can elect whomever they want regardless of what you do? Nothing. And, the median claim is plain wrong: When you're already on one side of a median, moving further away from it does never change that median. Basic statistics. However, you do get degenerate societies where there is a majority that is a bloc. In Northern Ireland, for example, the Unionists have a majority. This led to discrimination of the Nationalist minority. That's exactly my point. There are lots of such examples which all show clearly that majoritarianism is not democratic. The problem with this majority is that it is solid and unchanging. Ideally, majority should just mean the group of more than 50% on a particular issue. Every person should sometimes be part of the majority and sometimes part of the minority. That doesn't help because then the majority on issue A will still overrule the rest in every single decision on that issue. So a compromise option for that issue will have no chance. If a certain group of people are always part of the minority, then this leads to a poorly functioning society. A split society will only function poorly when a majoritarian method is used. When they use a method like FAWRB instead, they will function well because then they will care what the other faction wants, will try to devise good compromise options, and will vote in a way which makes sure the good compromises are elected instead of the random ballot result. This is possible *precisely* because with a non-majoritarian method the majority cannot simply ignore the minority but has to figure out how to get them to approve a compromise that is sufficiently near to their favourite. Non-majoritarian methods encourage discourse and cooperation. Germany has 'eternal' provisions. Some amendment proposals can be blocked by their Constitutional Court. This I think is undemocratic. The eternal provisions relate to fundamental rights, which is their reasoning. The reason why we have such should be obvious from history. It saves us for example from such restrictions of civil rights our American fellows experience since 9/11. Someone wrote: Then let me challenge you right away: I don't understand at all what those numbers a range-ballot asks me for are supposed to mean. They are not explained but instead it is simply assumed naively that each voter will be able to assign meaningful numbers to options. That someone was me. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range Voting vs Condorcet (Greg Nisbet)
Hi Terry, although FAWRB can be found in the lists archives, I use the opportunity to give the current definition of ... My favourite version of FAWRB (Favourite or Approval Winner Random Ballot) -- 1. Each voter rates each option as either harmful, not agreeable, agreeable, good compromise or favourite, the default being agreeable. Only one option may be marked favourite. 2. Those options which are rated harmful by more than, say, 90% of voters get excluded. (This security provision is only necessary when there is danger of really harmful options which are not already excluded by other mechanisms) 3. That options which is rated agreeable or better on the largest number of ballots is the nominated option. 4. A die is tossed. If it shows a six, 15 ballots are drawn at random, otherwise only 3 ballots. 5. If the nominated option is rated good compromise or better on all those ballots, it wins. Otherwise wins the option rated favourite on the first of the drawn ballots. (Some unimportant details for tie breaking need to be added) Although this seems pretty much randomness, my claim is that in practise, it will actually not be very random since opposing factions will cooperate in electing good compromise options with very high probability. In my 55/45-example of 55% of voters having A 100 C 80 B 0 and 45% of voters having B 100 C 80 A 0, the strategic equilibrium under FAWRB is when the first 55% vote A favourite, C good compromise, B bad and the other 45% vote B favourite, C good compromise, A bad in which case C is the sure winner without any randomness involved. This is because no voter gains anything in rating C lower. If you want to try FAWRB, you can use this demo which even adds a delegable proxy component to it: http://62.75.149.22/groucho_fawrb_dp.php Yours, Jobst Terry Bouricius schrieb: What does FAWRB stand for? Terry Bouricius - Original Message - From: Jobst Heitzig [EMAIL PROTECTED] To: Greg Nisbet [EMAIL PROTECTED]; Raph Frank [EMAIL PROTECTED] Cc: election-methods@lists.electorama.com Sent: Thursday, October 16, 2008 10:38 AM Subject: Re: [EM] Range Voting vs Condorcet (Greg Nisbet) Dear Raph, you replied to me: That leads me to the main problem with Range (as with any other majoritarian method): It is simply not democratic. It cannot be because every majoritarian method gives 100% of the power to less than 100% of the people (the demos in greek). They do have an equal vote. The move the median in their direction. First, what does an equal vote help when the other group (the majority) can elect whomever they want regardless of what you do? Nothing. And, the median claim is plain wrong: When you're already on one side of a median, moving further away from it does never change that median. Basic statistics. However, you do get degenerate societies where there is a majority that is a bloc. In Northern Ireland, for example, the Unionists have a majority. This led to discrimination of the Nationalist minority. That's exactly my point. There are lots of such examples which all show clearly that majoritarianism is not democratic. The problem with this majority is that it is solid and unchanging. Ideally, majority should just mean the group of more than 50% on a particular issue. Every person should sometimes be part of the majority and sometimes part of the minority. That doesn't help because then the majority on issue A will still overrule the rest in every single decision on that issue. So a compromise option for that issue will have no chance. If a certain group of people are always part of the minority, then this leads to a poorly functioning society. A split society will only function poorly when a majoritarian method is used. When they use a method like FAWRB instead, they will function well because then they will care what the other faction wants, will try to devise good compromise options, and will vote in a way which makes sure the good compromises are elected instead of the random ballot result. This is possible *precisely* because with a non-majoritarian method the majority cannot simply ignore the minority but has to figure out how to get them to approve a compromise that is sufficiently near to their favourite. Non-majoritarian methods encourage discourse and cooperation. Germany has 'eternal' provisions. Some amendment proposals can be blocked by their Constitutional Court. This I think is undemocratic. The eternal provisions relate to fundamental rights, which is their reasoning. The reason why we have such should be obvious from history. It saves us for example from such restrictions of civil rights our American fellows experience since 9/11. Someone wrote: Then let me challenge you right away: I don't understand at all what those numbers a range-ballot asks me for are supposed to mean. They are not explained
Re: [EM] Range Voting vs Condorcet (Greg Nisbet)
Dear Raph, you wrote: The thing is that in such a case, it isn't really a single 'demos'. It is two groups voting as one. Do you mean to say democracy is only for societies which are sufficiently homogeneous? That doesn't help because then the majority on issue A will still overrule the rest in every single decision on that issue. So a compromise option for that issue will have no chance. You can still have compromises. Only if the majority for some reason prefers to elect the compromise than their favourite. But in that it seems the favourite was just not the true favourite of the majority but the compromise was. So, still the minority has no influence on the decision but can only hope that the majority is nice enough to decide for the compromise. In fact, it can be helpful if multiple issues are voted as a single unit. This allows negotiation between factions in order to make up the majority. This common behavious is a pretty artificial construct to overcome the discussed drawbacks of majoritarian rules. A faction can make compromises on issues that it doesn't care about in order to get things that it does. This requires there is no solid bloc though. And when both factions care about both issues? A split society will only function poorly when a majoritarian method is used. When they use a method like FAWRB instead, they will function well because then they will care what the other faction wants, will try to devise good compromise options, and will vote in a way which makes sure the good compromises are elected instead of the random ballot result. This is possible *precisely* because with a non-majoritarian method the majority cannot simply ignore the minority but has to figure out how to get them to approve a compromise that is sufficiently near to their favourite. Non-majoritarian methods encourage discourse and cooperation. Sounds reasonable, the problem is that a) people don't like random methods b) it will result in certain outlier elements in society getting some power. a) FAWRB is not a random but a very specific and quite sophisticated method. It only uses a certain amount of chance, just as many things in our life do. Chance should not be mixed up with arbitrariness. Used in a rational way, FAWRB will usually elect good compromise options with near certainty, not leading to significant amounts of randomness. Perhaps a threshold could be set before a candidate can participate. Yes, I agree. The version I just proposed to Terry incorporates such a threshold. Also, the citizens of the US didn't get to vote on the restrictions of civil rights directly. It was handled by Congress. Using majority rule? That someone was me. Sorry, Greg didn't include your name in his post (or I couldn't find it). No need to be sorry. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range Voting vs Condorcet (Greg Nisbet)
Dear Diego, But randomness of FAWRB can cause institutional conflicts, especially if the minority faction leader was the winner. My focus has always been to decide issues, not to elect people. My suggestion if your scenario exists is: 1. Perform simultaneously an approval election and a PR election for an electoral college 2. If the approval winner has approval higher than a threshold (e.g. 2/3), s(he) is elected. 3. Otherwise the electoral college performs a multi-round approval election until some candidate has a score higher than the threshold. OK, we need a game-theoretic analysis of this. My guess is that because of the multi-round provision there is the danger of not getting a decision in any predetermined fixed time. Also, there are probably a number of strategic equilibria and it so the impact of my vote will be difficult to foresee. And, what is most important: It does not solve the problem at all, it only shifts the threshold for overruling the minority from 1/2 to 2/3. That's still not nearly democratic. You may suggest a much higher threshold, but then I guess no decision will be made at all... Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] IRV vs Condorcet vs Range/Score
Dear Kristofer, you wrote: Since Condorcet is a majoritarian method (as it needs to be in order to be a good single-winner method) ... A good single-winner method *must not* be majoritarian but must elect C in the situation of 55% voters having A 100 C 80 B 0 and 45% voters having B 100 C 80 A 0 :-) Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range vs Condorcet Overview
Dear Kevin, your wrote: The problem is that if you do not guarantee the majority that they will get their favorite if they vote sincerely, then they will stop telling you who their compromise choices are. No. In D2MAC there is no such guarantee (since it is not majoritarian) and this fact is the *very* reason that under D2MAC majorities *will* tell you what their compromise is (if it's a good compromise) since that is the only way to get the compromise elected instead of ending up with a random ballot lottery! No majoritarian method will elect the compromise in the simple 55/45-example I posted several times, only non-majoritarian methods succeed here. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Fixing Range Voting
Hi, you wrote: encourages people to vote honestly What makes you believe this? Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Who comes second in Ranked Pairs?
Hi folks, you're right, the option ending up second in the ranking constructed in RP is also the one that wins when you exclude the first winner. Even more so, the whole new ranking is just the old one with the top removed. There is, by the way, a related major difference between RP and the otherwise very similar methods Beatpath (aka Schulze) and River: In those two the 2nd place winner (= winner when 1st place winner is removed) cannot nearly as easily be determined from the original result. And in those two methods it can even happen that the 2nd place winner is an option which defeats the 1st place winner pairwise. To see this, consider a situation with four options A,B,C,D and the following pairwise defeats in order of descending defeat strength: BD, AC, CD, DA, AB, BC. In this situation Beatpath will elect B 1st and will elect A when B is removed. Also River will elect B 1st, locking in the defeats BDAC only, and will elect A when B is removed. So under both methods the 2nd place winner A defeats the 1st place B. But RP will lock in all defeats but DA, making A the 1st place winner and B the 2nd place. This property is one of the very few in which RP, Beatpath, and River differ. The other two such properties are (ii) that only River fulfills IPDA and ISDA, and (iii) only Beatpath fulfills (by definition) the requirement that for each beatpath against the winner there is a stronger (or equally strong) beatpath back. So each of the three similar methods has at least one desirable property the other two don't share. Yours, Jobst Kristofer Munsterhjelm schrieb: Scott Ritchie wrote: I'm writing a ranked pairs counter as practice for learning python, and I realized I don't know the answer to this question. Suppose I want to know who comes in second in a ranked pairs election. Is it: 1) Run ranked pairs algorithm on the ballots, find that candidate A wins, then purge A from all the ballots and rerun the algorithm to find a new winner and call him the second place candidate OR 2) Run ranked pairs algorithm on the ballots, lock in all pairs in their order that don't create cycles, then look at whom is second in the graph (ie, whoever beats all but A) Or will these two always be the same? It'd be nice if I could see an example where that's not the case. I think they're the same. I don't have proof of this, but I think it was given in an earlier post on this list. In any event, your #2 answer is the right one. When you lock in victories, you'll either go through all 0.5*n*(n-1) possible victories, or enough that you have a complete ordering. In either case, you use that ordering as the final result. For instance, if you have a situation with candidates A, B, and C, and 1. A B 2. B A 3. A C 4. B C 5. C B in that order, you lock A B, A C, B C, which gives A B C. Thus B comes in second. Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range vs Condorcet Overview
Dear Greg, I appreciate your trying to sum up the recent discussion. Some remarks: You wrote: Range Voting: There are two types of arguments against this system: There is another one - in my opinion the most serious problem of Range Voting: It fails to achieve exactly that which it claims to be designed to achieve, namely the maximization of social utility (or, if you like, the minimization of Bayesian regret). Just look at the following very simple and common situation of two factions and a good compromise option: 55% of voters have these utilities: A 100, C 80, B 0 45% of voters have these utilities: B 100, C 80, A 0 Obviously, utilitarians would want C to win, but A will be elected most certainly under Range (as under any other majoritarian method including Condorcet methods, but Condorcet methods don't claim to maximize social utility). 1) Ratings themselves are useless/unreasonable/illogical/not indicative of reality My point was not that they are useless/unreasonable/illogical/not indicative of reality but that we may not assume that a typical voter can easily come up with a meaningful set of ratings! Even if there were a working definition of the meaning of a rating in terms of happiness levels measured biochemically or in whatever way, it would still be practically impossible to come up with more than a very vague assignment of ratings. I wonder if any supporter of ratings has really ever honestly asked himself whether he can justify why he has assigned to some option a rating of, say, 61 instead of 62 on a scale from 0 to 100. I doubt it! a. The concept of comparing candidates along a single dimension is more intuitive and hence more meaningful to voters than making O(n^2) binary decisions Is there any scientific evidence to support this claim? For me it is surely a simpler task to answer questions of the sort do you prefer A or B? than to answer questions of the sort where would you place A, B, and C along an interval from 0 to 100?, even more so when no indication is given as to what these numbers are supposed to mean. a. Does zero-info in this case mean a) lack of info about of the behavior of other voters or b) (a) and lack of info about other candidates as well? Either way, if the problem can be ameliorated by adding info, then add info. Nice idea. Perhaps you could tell us how this should be done when at the same time votes are secret? Or do you suggest vote secrecy is not necessary? b. Is this behavior even a good thing? If the majority isn't exercising its influence and a compromise candidate is elected instead, do you really want a polarizing candidate or a compromise one? At least to utilitarians that should be obvious: If the compromise is a good one (leading to more social utility than the polarizing ones), he should clearly be the winner. There are methods which make this highly probable since they are not majoritarian methods. RP: 1) This is a system I initially cited as an example of a reasonable Condorcet method, it hasn't really been argued about. At least not recently. One problem with RP (as with Beatpath aka Schulze) is that they fail Independence of Pareto Dominated Alternatives: adding a clearly bad candidate - in this case one to which some other candidate is preferred by each and every voter - should have no impact on the result. But with RP and Beatpath such a candidacy can affect the result while with River it cannot. Framing the debate: Debating the specific merits of Range Voting or Condorcet Method X is meaningless unless we can agree on some kind of metric. No. Assuming there would be a single metric that summarizes all the conflicting goals and criteria one can rightly ask for in an election method is just as wrong as assuming voters can always place options on a single dimension. There is just no evidence for either claim but plenty of evidence against them. Debates about which properties are important don't really lead anywhere. There are a few we can probably agree upon. Let's see how often it satisfies those properties. I advocate moving away from a binary framework and focusing on how often certain properties are satisfied. I like the Bayesian Regret metric because it's nice and quantifiable. Loving quantifiable things is among the worst diseases in science. As a mathematician I know what I'm talking about here. As long as there is no evidence that some notion is quantifiable, the safe assumption is always to assume that it is *not* quantifiable, or at least not quantifiable in a *single* dimension. So, let me again suggest that we try to work based on evidence and not on wishful thinking. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Range Condorcet (No idea who started t his argument, sorry; I am Gregory Nisbet)
Dear Greg, Reasons why Range is better and always will be. I would like to end the truce. That won't work I guess. Using the term better alone is a major flaw of many discussions here. Obviously, it all depends on what goals a method is expected to achieve. I'll be generous to the Condorcet camp and assume they suggest something reasonable like RP, Schulze or River. As you might guess, I appreciate this, of course :-) Property Related: favorite betrayal, participation and consistency. Implications: 1) It is always good to vote and it is always good to rate your favorite candidate 100. The only Condorcet method to satisfy favorite betrayal is an obscure variant of Minmax which I'll ignore because of its glaring flaws (clone dependence *cough*) 2) How does it make sense to be able to divide a region into two constituencies each electing A if B is the actual winner? Condorcet methods are not additive, this calls into question the actual meaning of being elected by a Condorcet method. No, it calls into question the actual meaning of being elected in a region. The misunderstanding arises only when you interpret the election of A in a region as meaning that A is best in some sense. But Condorcet methods are based on a different logic than measuring goodness of candidates. They have more to do with stability, for example: Electing a candidate other than the Condorcet Winner always faces immediate opposition by some majority who prefers the Condorcet Winner. So, if you consider majorities significant (which you seem to judging from your reasoning further down), you should consequently not accept different winner when a Condorcet Winner is available. answers to potentital majority rule counterarguments: 1) Range voting isn't a majority method. answer: any majority can impose their will if they choose to exercise it. I greatly appreciate your making this clear! Warren has often argued that Range is *not* really majoritarian when I pointed this out. The main point is of course the question of whether one assumes intelligent voters who vote strategically or dumb voters who vote honestly against their own interest (whatever honestly means with a ratings-based method - more on this below). When we assume intelligent voters, Range is clearly a majoritarian method. That leads me to the main problem with Range (as with any other majoritarian method): It is simply not democratic. It cannot be because every majoritarian method gives 100% of the power to less than 100% of the people (the demos in greek). Often, about 60% of the people can consistently impose their will on the other 40% without the latter being given any means at all by the majoritarian method to influence the decision. Of course, this is a problem of most popular election methods, but that does not mean the problem cannot be solved. Democratic decisions are possible but not with majoritarian methods. Voter Experience: Range Voting (based on the existence of Amazon product ratings, youtube video ratings, hotornot.com, the number of movies rated out of stars.) I cannot find a single instance of Condorcet methods besides elections in various open source communities. It doesn't qualify as mainstream. That may be right but is irrelevant for the question of what is a good method and what is not. Understandability: Range Voting (I dare anyone to challenge me on this) Then let me challenge you right away: I don't understand at all what those numbers a range-ballot asks me for are supposed to mean. They are not explained but instead it is simply assumed naively that each voter will be able to assign meaningful numbers to options. Some even suggest that voters should apply their gut feelings to derive the numbers - as if voting were about diffuse emotions and not hard facts. In real world situations it is difficult enough to decide whether I *prefer* A to B or B to A or neither to the other. Often enough it turns out that A is preferable in some aspects and B is preferable in other aspects. Suggesting to weigh the aspects first is of no help since it lifts the problem of coming up with meaningful numbers only to a higher, more abstract level. Also, Range advocates tell us that the numbers are not simply supposed to be monetary values but rather such things as degrees of utility or perhaps happiness. Simple question: Can you enumerate y our happiness with, say, having saved a species from extinction? And even if some people can, it is obviously not justified to simply assume that every voter should be able to do such magic without at least providing some serious scientific evidence for such a bold claim. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
[EM] Implementation of FAWRB combined with Delegable Proxy (was: Test implementation of FAWRB)
Hello all, thanks for your comments so far. I have not yet improved the interface much yet but have instead implemented a second web service. While the first one was for immediate use by groups who are already convened in one place and provided no secrecy, this second implementation is for distributed groups who want secrecy and the ability to delegate parts of their ratings to other voters. This demonstrates how FAWRB can easily be combined with Delegable Proxy in a secure way. Please be so kind to test the FAWRB/DelegableProxy service at http://62.75.149.22/groucho_fawrb_dp.php It should work with Firefox and Opera browsers and hopefully also with Internet Exploder (if not, please report problems!) Before setting up your own poll, you might just want to have a quick look at how a typical ballot would look like: http://62.75.149.22/polls/what_movie_ballot.html Any comments are welcome! Yours, Jobst Am Mittwoch, 27. August 2008 06:01 schrieb Brian Olson: I'd like to see some things that would make the process more understandable. First, expand FAWRB somewhere on the site and explain it. Part of the explanation could be showing all of the stages of the form at once instead of making it a surprise as each stage shows up. Alternately, or additionally, there could be a side-bar flow chart that lists all the stages, probably vertically. Bonus points if the stages light up as the process moves through them. On Aug 25, 2008, at 4:10 PM, Jobst Heitzig wrote: Dear folks, feel free to test my simple implementation of FAWRB at http://www.impro-irgendwo.de/groucho/fawrb.html It's really easy and I hope it is some fun, too. And please send comments or corrections if some of my English is too bad :-) Yours, Jobst pgppaAUIsylMo.pgp Description: PGP signature Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] [english 94%] PR favoring racial minorities
Hello all, although I did not follow all of the discussion so far, the following question strikes me: Why the hell do you care about proportional representation of minorities when the representative body itself does not decide with a method that ensures a proportional distribution of power? It is of no help for a minority to be represented proportionally when still a mere 51% majority can make all decisions! So, if you really care about the rights of minorities, the consequence would be to also promote some non-majoritarian, truly democratic decision method for the representative body itself. Examples of such methods have been discussed here. Yours, Jobst Warren Smith schrieb: Instead it is based on the mathematical theorem that if people vote in a racial manner, then if X percent of people vote black, and enough black candidates are available, then we get X percent black winners. Dopp: And this theorem while it may sound believable at first glance is instead of course obvious mathematical nonsense and easily rebutted with any one of a countable infinite number of counterexamples. --I reply... How fascinating that Kathy Dopp has refuted over a century of research into PR voting systems. But rather than keeping her infinite number of obvious counterexamples secret, how about exhibiting one? I mean, I'll settle for just one. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] [english 91%] Re: Representative Range Voting with Compensation - a new attempt
Dear Forest, This is indeed astonishing from a slightly different point of view: if I rate the winner well below the benchmark lottery, my action tends to cause other voters' accounts to lower, but does not cause mine to increase, yet the total is constant. Magic! Not so astonishing after all, since I am now sure that my claim concerning the expected value was plain false. The expected value will be zero. One can see this when writing down the precise sums. I now believe also that this part cannot be fixed, so what remains is the strategy-freeness and the conservation of voting money, while the third goal (compensating those who liked a Random Ballot lottery better) probably cannot be achieved. So, I will post a simpler version tomorrow night, and that we can analyse further. - As to your median idea: I am doubtful whether that could work. Actually, I think the *only* aggregation function for individual ratings for which a similar taxing mechanism can make the method strategy-free, is the sum of ratings. Assume the winner is the option X for which f(R(X,1),...,R(X,N)) is maximal, where f is some (symmetric) aggregation function. When we try to adjust voter 1's account by some amount determined only from the other voters' ratings, and hope that the sum of true utility and adjustment is larger with honest voting, then it seems we end up with a condition like R(X,1) + f(R(X,2),...,R(X,N)) R(Y,1) + f(R(Y,2),...,R(Y,N)) whenever f(R(X,1),...,R(X,N)) f(R(Y,1),...,R(Y,N)). And this most surely implies that f is just the sum plus perhaps some constant value. Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
[Election-Methods] Representative Range Voting with Compensation - a new attempt
Dear folks,
I must admit the last versions of RRVC (Representative Range Voting with
Compensation) all had a flaw which I saw only yesterday night. Although
they did achieve efficiency and strategy-freeness, they did not achieve
my other goal: that voters who like the winner more than the random
ballot lottery compensate voters who liked the random ballot lottery
more than the winner. In short, the flaw was to use the three randomly
drawn voter groups for only one task each, either for the benchmark, or
the compensation, or the decision.
I spare you the details and just give a new version which I think may
finally achieve all three goals: efficiency, strategy-freeness, and
voter compensation.
The basic idea is still the same: Partition the voters randomly into
three groups, let one group decide via Range Voting, and use each group
to benchmark another group and to compensate still another group.
To make an analysis more easy, I write it down more formally this time
and assume the number of voters is a multiple of 3.
DEFINITION OF METHOD RRVC (Version 3)
=
Notation:
-
X,Y,Z are variables for options
i,j,k are variables for voters
f,g,h are variables for groups of voters
Input:
--
All voters give ratings and mark a favourit. Put...
R(X,i) := the rating voter i gave option X
F(i) := the option marked favourite on ballot of voter i
A(i) := balance on voter i's voting account before the decision
Tally:
--
Randomly partition the N voters into three groups of equal size.
The winner is the range voting winner of group 1.
The voting accounts are adjusted as follows. Put...
S := N/3
Q := (S-1)/S
G(i) := group in which voter i landed
T(X,f) := total rating group f gave option X
= sum { R(X,i) : i in group f }
W(g) := range voting winner of group g
= that W with T(W,g)T(X,g) for all X other than W
P(X,h) := proportion of group h favouring X
= probability of X in group h's random ballot lottery
= # { i in group h : F(i)=X } / S
D(f,g,i) := rating difference on voter i's ballot
between the range voting winner of group f
and the random ballot lottery of group g
= R(W(f),i) - sum { P(X,g)*R(X,i) : X }
E(f,g,h) := total rating difference in group h
between the range voting winner of group f
and the random ballot lottery of group g
= sum { D(f,g,i) : i in group g }
For each voter i, add the following amount to her voting account C(i):
If i is in group 1:
deltaC(i) := E(1,2,1)-D(1,2,i) - E(2,2,2) - Q*E(3,3,2) + E(3,3,3)
If i is in group 2:
deltaC(i) := E(3,3,2)-D(3,3,i) - E(3,3,3) - Q*E(1,1,3) + E(1,1,1)
If i is in group 3:
deltaC(i) := E(1,1,3)-D(1,1,i) - E(1,1,1) - Q*E(1,2,1) + E(2,2,2)
(Remark: E(1,2,1) and D(1,2,1) are not typos!)
(END OF METHOD RRVC)
Analysis:
-
1.
The sum of all C(i) remains constant, so voting money retains its
value. To see this, note that
sum { E(1,2,1)-D(1,2,i) - E(2,2,2) : i in group 1 }
= S*E(1,2,1) - E(1,2,1) - S*E(2,2,2) )
= S*( Q*E(1,2,1) - E(2,2,2) )
= sum { Q*E(1,2,1) - E(2,2,2) : i in group 3 }
and analogous for the other terms in the above sums.
2.
Note that the terms E(1,2,1)-D(1,2,i), E(3,3,2)-D(3,3,i), and
E(1,1,3)-D(1,1,i) in the above sums do not depend on voter i's ratings!
Hence the only way in which the ballot of voter i can affect her own
voting account is trough the dependency of W(1) on her ratings, and
this is only the case for voters in group 1, the deciding group.
So, as only voters in group 1 can influence their outcome, an analysis
of individual voting strategy is only required these voters. For such a
voter i the net outcome, up to some constant which is independent of
i's behaviour, is this:
O(i) := sum { R(W(1),j) : j other than i } + U(W(1),i)
where
U(X,i) := true value of X for i.
If voter i is honest and puts R(X,i)=U(X,i), this simply adds up to
O(i) = T(W(1),1) (if i is honest).
Now assume this honest voter i thinks about changing the winner from
W(1) to some other option Y by voting dishonestly. The net outcome for
i after this manipulation would be
O'(i) = sum { R(Y,j) : j other than i } + U(Y,i)
= T(Y,1)-R(Y,i) + U(Y,i)
= T(Y,1)
T(W(1),1) = O(i).
So after all, i has no incentive to manipulate the outcome because she
would have to pay more than she gains from this.
3.
Now consider a large electorate of honest voters, and think about what a
voter can expect, before the random process of drawing the three groups
is applied, of how much her voting account will be adjusted. If I got
it right this time, this expected value of deltaC(i) should be, up to
some constant term which is equal for all voters, just
the rating difference on voter i's ballot
between the random ballot lottery
and the winner of the decision,
Re: [Election-Methods] Representative Range Voting with Compensation -a new attempt
A first typo: It must read C(i) instead of A(i) under Input...
--
Dear folks,
I must admit the last versions of RRVC (Representative Range Voting with
Compensation) all had a flaw which I saw only yesterday night. Although
they did achieve efficiency and strategy-freeness, they did not achieve
my other goal: that voters who like the winner more than the random
ballot lottery compensate voters who liked the random ballot lottery
more than the winner. In short, the flaw was to use the three randomly
drawn voter groups for only one task each, either for the benchmark, or
the compensation, or the decision.
I spare you the details and just give a new version which I think may
finally achieve all three goals: efficiency, strategy-freeness, and
voter compensation.
The basic idea is still the same: Partition the voters randomly into
three groups, let one group decide via Range Voting, and use each group
to benchmark another group and to compensate still another group.
To make an analysis more easy, I write it down more formally this time
and assume the number of voters is a multiple of 3.
DEFINITION OF METHOD RRVC (Version 3)
=
Notation:
-
X,Y,Z are variables for options
i,j,k are variables for voters
f,g,h are variables for groups of voters
Input:
--
All voters give ratings and mark a favourit. Put...
R(X,i) := the rating voter i gave option X
F(i) := the option marked favourite on ballot of voter i
A(i) := balance on voter i's voting account before the decision
Tally:
--
Randomly partition the N voters into three groups of equal size.
The winner is the range voting winner of group 1.
The voting accounts are adjusted as follows. Put...
S := N/3
Q := (S-1)/S
G(i) := group in which voter i landed
T(X,f) := total rating group f gave option X
= sum { R(X,i) : i in group f }
W(g) := range voting winner of group g
= that W with T(W,g)T(X,g) for all X other than W
P(X,h) := proportion of group h favouring X
= probability of X in group h's random ballot lottery
= # { i in group h : F(i)=X } / S
D(f,g,i) := rating difference on voter i's ballot
between the range voting winner of group f
and the random ballot lottery of group g
= R(W(f),i) - sum { P(X,g)*R(X,i) : X }
E(f,g,h) := total rating difference in group h
between the range voting winner of group f
and the random ballot lottery of group g
= sum { D(f,g,i) : i in group g }
For each voter i, add the following amount to her voting account C(i):
If i is in group 1:
deltaC(i) := E(1,2,1)-D(1,2,i) - E(2,2,2) - Q*E(3,3,2) + E(3,3,3)
If i is in group 2:
deltaC(i) := E(3,3,2)-D(3,3,i) - E(3,3,3) - Q*E(1,1,3) + E(1,1,1)
If i is in group 3:
deltaC(i) := E(1,1,3)-D(1,1,i) - E(1,1,1) - Q*E(1,2,1) + E(2,2,2)
(Remark: E(1,2,1) and D(1,2,1) are not typos!)
(END OF METHOD RRVC)
Analysis:
-
1.
The sum of all C(i) remains constant, so voting money retains its
value. To see this, note that
sum { E(1,2,1)-D(1,2,i) - E(2,2,2) : i in group 1 }
= S*E(1,2,1) - E(1,2,1) - S*E(2,2,2) )
= S*( Q*E(1,2,1) - E(2,2,2) )
= sum { Q*E(1,2,1) - E(2,2,2) : i in group 3 }
and analogous for the other terms in the above sums.
2.
Note that the terms E(1,2,1)-D(1,2,i), E(3,3,2)-D(3,3,i), and
E(1,1,3)-D(1,1,i) in the above sums do not depend on voter i's ratings!
Hence the only way in which the ballot of voter i can affect her own
voting account is trough the dependency of W(1) on her ratings, and
this is only the case for voters in group 1, the deciding group.
So, as only voters in group 1 can influence their outcome, an analysis
of individual voting strategy is only required these voters. For such a
voter i the net outcome, up to some constant which is independent of
i's behaviour, is this:
O(i) := sum { R(W(1),j) : j other than i } + U(W(1),i)
where
U(X,i) := true value of X for i.
If voter i is honest and puts R(X,i)=U(X,i), this simply adds up to
O(i) = T(W(1),1) (if i is honest).
Now assume this honest voter i thinks about changing the winner from
W(1) to some other option Y by voting dishonestly. The net outcome for
i after this manipulation would be
O'(i) = sum { R(Y,j) : j other than i } + U(Y,i)
= T(Y,1)-R(Y,i) + U(Y,i)
= T(Y,1)
T(W(1),1) = O(i).
So after all, i has no incentive to manipulate the outcome because she
would have to pay more than she gains from this.
3.
Now consider a large electorate of honest voters, and think about what a
voter can expect, before the random process of drawing the three groups
is applied, of how much her voting account will be adjusted. If I got
it right this time, this expected value of deltaC(i) should be, up to
some constant term which is equal for all voters, just
the rating difference on
Re: [Election-Methods] a strategy-free range voting variant?
, do the following:
a) Determine the total rating T(X,S) the ballots in SD gave to X.
b) Determine the total rating B(X,S) the ballots in SB gave to X.
c) Determine the probability P(X,S) that X is the favourite on a
ballot drawn randomly from SB.
4. For each such partition S, let...
a) W(S) be that W with T(W,S)T(X,S) for all X other than W.
b) Z(S) be that Z with B(Z,S)B(X,S) for all X other than Z.
5. For each such partition S and each voter i, put...
a) alphaC(i,S) := 0 if i is not in SD, otherwise
alphaC(i,S) :=
sum { P(X,S)*( B(X,S)-T(X,i,S) ) : X } + T(W,i,S)-B(Z,S),
where T(X,i,S) = T(X,S)-R(X,i).
b) gammaC(i,S) := 0 if i is not in SC, otherwise
gammaC(i,S) := - sum { alphaC(i,S) : i } / (N-2D),
where the sum is over all voters i in SD.
6. Finally, for each voter i, add the following amount to her voting
account C(i):
deltaC(i) := sum { alphaC(i,S)+gammaC(i,S) : S } / sum { 1 : S }
where the sums are over all partitions S.
I have not yet calculated by what factor the variance of deltaC(i)
will shrink because of the averaging. That might be a bit difficult
since the values of alphaC(i,S) are not independent for different S.
My guess, however, is that the averaging will lead to the standard
deviation having an order of at most O(1) instead of O(sqrt(D)) or
even O(D).
Perhaps someone can analyse this averaging in more detail and come up
with am estimation of that standard deviation?
Yours, Jobst
Jobst Heitzig schrieb:
Another small remark:
With N voters total and B benchmark voters, the size D of the deciding
group should probably be O(sqrt(N-B)).
This is because the amount transferred to an individual deciding voter's
account is roughly proportional to D times a typical individual rating
difference, hence the total amount transferred to the deciding group is
proportional to D² times a typical individual rating difference. The
same total amount is payed by the group of at most N-B-D compensating
voters. Each of them should not be required to pay more than a constant
multiple of a typical individual rating difference, hence D²/(N-B-D)
should be O(1).
Jobst
Election-Methods mailing list - see http://electorama.com/em for list info
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] a strategy-free range voting variant?
I performed a quick little simulation for version 2:
With K options and N voters, I drew the all K*N ratings independently
from a standard normal distribution and then applied the method with
D=sqrt(N)/2.
However, instead of using all partitions as suggested, I only used N/2D
partitions. More precisely, I ordered the ballots in a random way in
groups of size D, and then first used groups 1 and 2 as the benchmark
and deciding group, afterwards used groups 3 and 4 for this, then used
groups 5 and 6, and so on. In other words, the account adjustments were
averaged not over all possible partitions but only over these sqrt(N)
many groups.
I did this 100 times for each of a number of different pairs (K,N) and
evaluated the standard deviation of the individual account adjustments.
It turned out that for K=2 this standard deviation was approximately
0.2 / sqrt(sqrt(N))
and only slightly larger for K=16 or K=128.
Since this is quite small when compared to the standard deviation of the
original ratings, which is 1 of course, this averaging in version 2
indeed looks promising! (Without it, the standard deviation of the
individual account adjustments would grow not shrink with growing N.)
Jobst
Jobst Heitzig schrieb:
Dear folks,
this night I had two additional ideas for RRVC, so here's two new
versions of it.
In the first version, the fee F is determined from the benchmark ballots
so that the expected price a deciding voter has to pay from her
voting account is just that voter's rating difference between the winner
and the random ballot lottery:
RRVC - New Version 1
0. Each voter i is assumed to have a voting account whose balance is
denoted C(i).
1. All N voters fill in a range ballot and additionally mark their
favourite in case of a top-rating tie. Voter i can use ratings
0...C(i) only. If C(i) is negative, she can use the rating 0 only
(but still mark her favourite). Let R(X,i) be the rating voter i
gave to option X.
2. Put D = sqrt(N) (rounded up), and draw D deciding ballots. For
each option X, determine the total rating T(X) these deciding
ballots gave to X. The winner W of the decision is that option whose
total rating is maximal, i.e. that option W for which T(W)T(X) for
all X other than W.
3. From the remaining ballots, draw D benchmark ballots. For each
option X, determine the total rating B(X) these benchmark ballots
gave to X, and determine the probability P(X) that X is the
favourite on a ballot drawn randomly from these benchmark ballots.
(I.e., P(X) is the fraction of benchmark ballots favouring X).
Let Z be that option whose total rating is maximal in this group, i.e.
that option Z for which B(Z)B(X) for all X other than Z.
4. For each voter i whose ballot is amoung the deciding ballots, add
the following amount to her voting account C(i):
deltaC(i) := sum { P(X)*( B(X)-T(X,i) ) : X } + T(W,i)-B(Z),
where the sum is over all options X, and where
T(X,i) = T(X)-R(X,i)
is the total rating of X amoung all deciding ballots of voters other
than i.
5. The remaining N-2D voters are the compensating voters. For each
compensating voter j, add the following to her voting accout C(j):
deltaC(j) := - sum { deltaC(i) : i } / (N-2D),
where the sum is over all deciding voters i.
Remarks for version 1:
Since the deciding and benchmark groups are of equal size, the expected
values of T(X) and R(X) are the same, and it is also likely that
Z=W. This implies that the expected value of deltaC(i) given that i
is a deciding voter and all voters report sincere ratings, is just
sum { P(X)*R(X,i) : X } - R(W,i).
In other words, when ratings are sincere a deciding voter can expect to
pay exactly her rating difference between the winner and the Random
Ballot lottery. (This is a major difference to the Clarke tax where this
take Random Ballot as a benchmark philosophy is not incorporated).
Also note that the standard deviation of deltaC(i) under these
assumptions is of an order somewhere between O(sqrt(D)) and O(D),
depending on how correlated the individual voters' ratings are.
Still, the actual price payed by voter i is independent of her ratings
as long as she does not manage to change the winner. Hence there is
still no incentive to bargain for a lower price by misrepresenting my
ratings.
Assuming the true value of W for voter i is U(A,i)=R(W,i), the net
outcome for i is
U(W,i) + deltaC(i)
= sum { P(X)*( B(X)-T(X,i) ) : X } + T(W)-B(Z).
Now assume voter i thinks about changing the winner to A, originally
having a total of T(A)T(W). Since this manipulation does not change
the values T(X,i), the net outcome for i after this manipulation
would be
U(A,i) + sum { P(X)*( B(X)-T(X,i) ) : X } + T(A,i)-B(Z)
= sum { P(X)*( B(X)-T(X,i) ) : X } + T(A)-B(Z).
Since this differs from
Re: [Election-Methods] [english 97%] Re: [english 80%] CTT voting
Dear Warren and list members, you wrote: The main differences, however, are these: - Voting money is transferred on a regular basis, not only in the very rare case of swing voters, thus making the strategic incentive much stronger. --this is good. However, in one of your revised schemes you made the deciders become a very small fraction of the population instead of (as originally) 1/3. That means very few voters are connected to reality via (your equivalent of) Clarke taxes. I think you have a DILEMMA: A* If you make there be many deciders, then the compensators tend to pay too much money. B* If you make there be too few deciders, then a voter is unlikely to be a decider, and hence her economic incentives to vote honestly, are diminished. B was also a problem with CTT voting - the Clarke payments were rare, divorcing voters from reality. Need to seek the best tradeoff between A B. That could be another reason for using the averaging process of version 2, since it treats all voters alike as potential deciders, so it distributes the incentive evenly. - The transfers are so designed that they compensate those deciding voters who get a worse result than with the random ballot lottery, thus keeping with the philosophy that that lottery is to be considered the benchmark for every more efficient choice. --the random ballot lottery (unfortunately?) is a pretty low benchmark. Low only in terms of efficiency. But the focus is not on efficiency here (that is solved by electing the option maximizing the average rating) but on equality and fairness. In my opinion, every voters has a priori a right to distribute a share of 1/N of the winning probability. So if a method takes away this control, it has to compensate the voter. (You might remember that we had this benchmark discussion already about a year ago... :-) You could also consider the random candidate lottery (elect each with probability 1/K, where K candidates in all). That is an even-worse benchmark than random ballot, but it is simpler and does not depend on the votes at all. Because it does not depend on votes, you then would not need your benchmarker class of voters, only deciders and compensators. See above for the motivation why it is exactly the Random Ballot process which has to serve as the benchmark in my view. Also, a random candidate lottery is not clone-proof. - Voting money is not destroyed and then gradually refilled but is always preserved, thus keeping its value constant. --Is there a worry that over a sequence of elections, we get unfairness? That might occurr if the variance of the adjustments is too large, so that it could happen that only by sheer bad luck a voter's account becomes negative and the voters cannot influence the next decision until her account is positive again. Also for this reason the variance of the adjustments due to the random assignment of groups should be as small as possible. In election #1, suppose you are correct and so everybody wants to vote honestly. Excellent. BUT now in election #2, some voters have more money than others, hence have more power. COULD IT HAPPEN THAT this inequality could be CORRELATED with the politics? If so, then in election #2, the Democrats would win, unfairly, purely because Democrats have more power. If this could somehow feed back (so that in election #3, democrats also tended to have more money, and so on forever) this could be extremely bad. Hmm... I hope the feedback will always be negative, partly because the benchmark against which compensations are determined is not the majority choice but the average choice (Random Ballot). Also, your benchmarkers could have motivation to vote DIShonestly, thus defeating the purpose. CONCRETE SCENARIO TO WORRY ABOUT: The benchmarker votes affect the payouts. If benchmarkers dishonestly rank the Democrat top, that tends to cause those who rate Democrats low, to get lower payoffs. That causes Democrats to have more money in election #2. That causes this strategy to work even better in election #2. Feedback. Disaster. (I'm not sure whether this scenario is a real problem. But it might be.) A REPAIR: Note that in the random-candidate-lottery there are no benchmarkers, hence the worry I just described perhaps no longer exists. -wds Let us think about this some further. My hope is that because (i) a voter does not know whether she is a benchmark voter, and (ii) the benchmark group will usually be a representative sample of the electorate, and (iii) the averaging process of version 2 treats all voters alike when it comes to adjustments, the incentive described by you will be averaged out if it exists in the first place. But surely this deserves more detailed research... Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] a strategy-free range voting variant?
Dear Warren,
you wrote:
But I do not fully understand it yet and I think you need to
develop+clarify+optimize it further... plus I'd like you to unconfuse me!
I'll try...
Of course, this is far from being a new idea so far, and it is not yet
the whole idea since it has an obvious problem: although it obviously
manages to elect the better option (the one with the larger total
monetary value), it encourages both the seller and the buyer to
misrepresent their ratings so that the gap between R2(B)-R2(A) and
R1(A)-R1(B) becomes as small as possible and hence their respective
profit as large as possible. In other words, this method is not at all
strategy-free.
--QUESTION:
if they make the gap small, then the buyer pays little to the seller.
Yes, that is better for the buyer.
But doesn't the seller have the opposite incentive?
It is not clear to me the incentive you say exists here, really does exist.
If it doesn't, then you do not need to fix this problem
because there is no problem. It'd help to clarify this point.
Isn't that the usual situation when bargaining? Given that the buyer
would be willing to pay more than the seller would minimally accept as a
price, the seller tries to maximize the price as long as he thinks the
buyer is willing to pay it, and the buyer tries to minimize her offer as
long as she thinks the seller is willing to accept it. So, both work to
minimize the gap between the demanded and the offered payment.
5. Finally, the voting accounts are adjusted like this:
a) Each deciding voter's account is increased by an amount equal to the
total rating difference between the winner and the benchmark lottery
among the *other* deciding voters, minus some fixed fee F, say
10*N^(1/2). (Note that the resulting adjustment may be positive or
negative.)
QUESTION:
I'm confused about this whole benchmarking thing.
You said the benchmark voters were being benchmarked, but now you
say the deciding
voters are being benchmarked. ???
That might be a language problem for my part. What I mean is this: In my
thinking, democracy demands equal decision power for every voter. Random
Ballot accomplishes this in a way, but is not efficient. But the Random
Ballot lottery can still serve as a benchmark for other, more
efficient choices. In my suggested method, the benchmark voters are
needed only to estimate what the Random Ballot lottery amoung all voters
would be. The individual ratings for the actual winner of the election,
who is only determined by the deciding voters, is then compared to the
individual ratings for this benchmark (i.e. of the estimated Random
Ballot lottery) in order to the individual transfers of voting money.
The higher a deciding voter rated the benchmark and the lower she rated
the winner, the more voting money is transferred to her account (or,
rarely, the less is transferred *from* her account).
In mathematical terms: Let p(X) be the probability of X being the
highest rated option when we draw one of the benchmark voter's ballots
uniformly at random. (So the p's define our benchmark lottery)
Let r(i,X) be the rating deciding voter i specified for X. Put
r0(i) := sum { p(X)*r(X,i) : X }
(over all options X), i.e., the expected rating deciding voter i
specified for the lottery outcome. Then put
t(X) := sum { r(X,i) : i }
(over all deciding voters i) and
t0 := sum { p(X)*t(X) : X }.
Assume W is the range voting winner of the deciding ballots, i.e.,
t(W) t(X) for all X other than W
Now the voting account of deciding voter i is changed by this amount:
sum { r(W,j)-r0(j) : j different from i }
(over all deciding voters j different from i),
which is equal to
(t(W)-t0) - (r(X,i)-r0(i))
The higher you rated the winner (i.e., the higher your r(X,i)) and the
lower you rated the average favourite of the benchmark voters (i.e., the
lower your r0(i)), the less voting money you get.
What does total rating difference between the winner and the
benchmark lottery among the *other* deciding voters MEAN precisely???
This is not clear english... the winner's rating is a number but
the benchmark lottery is not a number. You need two numbers.
It means
sum { r(W,j)-r0(j) : j different from i }
(see above).
The compensating voter's accounts are decreased by the same total
amount as the deciding voter's accounts are increased, but in equal
parts. (This may also be positive or negative)
--this seems to hurt poor voters. I.e. if there are rich voters who
vote +-100
and poor voters who vote +-1 then the poor voters will need to pay the same fee
in 5b as the rich voters. They may therefore have incentive to avoid
being in the electorate at all, in which case the electorate will
become biased (rich-dominated).
Yes, that might be a problem. So, being in the electorate (meaning
amoung the whole number of N voters) should not be something one can
choose. In other words, we put N to be the number of all eligible
voters, no matter whether they choose
Re: [Election-Methods] a strategy-free range voting variant?
Another small remark: With N voters total and B benchmark voters, the size D of the deciding group should probably be O(sqrt(N-B)). This is because the amount transferred to an individual deciding voter's account is roughly proportional to D times a typical individual rating difference, hence the total amount transferred to the deciding group is proportional to D² times a typical individual rating difference. The same total amount is payed by the group of at most N-B-D compensating voters. Each of them should not be required to pay more than a constant multiple of a typical individual rating difference, hence D²/(N-B-D) should be O(1). Jobst Election-Methods mailing list - see http://electorama.com/em for list info
[Election-Methods] a strategy-free range voting variant?
Dear folks, some time ago we discussed shortly whether it was possible to design a strategy-free ratings-based method, that is, a method where voters give ratings and never have any incentive to misrepresent their true ratings. If I remember right, the methods that were discussed then were only of academic use since they were far from being efficient and often elected bad options unwanted by most of the voters. Several days ago, I had a new idea how range voting could be modified to get a method both strategy-free and efficient. A bit of research revealed that much of it resembles the ideas in the paper http://mpra.ub.uni-muenchen.de/627/, but not all of it. I will first describe the basic idea and then the method. Disclaimer: All of what follows is suitable only for the case where one can assume that voters can sincerely attribute some numerical utility to all options, which is an assumption I personally don't believe to hold generally :-) Anyway, here's the... Basic Idea --- In order to understand the basic idea, consider a decision problem with two options, A and B, and two voters, V1 and V2, who are able to attribute some monetary values U1(A)U1(B), U2(B)U2(A) to these options. (We will not need to assume monetary values later on, but the idea is easier to grasp this way) Now consider the following method: Both voters fill in a ratings ballot for A and B, giving ratings R1(A)R1(B), R2(B)R2(A). Then a coin is tossed to decide which of the two voters is the seller and which is the buyer. Let's assume throughout the following that V1 turns out to be the seller. Now the winner is determined like this: If R2(B)-R2(A) = R1(A)-R1(B) then A wins. Otherwise, that is, if R2(B)-R2(A) R1(A)-R1(B), then V2 buys the decision from V1: B wins but V2 pays an amount of ( R2(B)-R2(A) + R1(A)-R1(B) ) / 2 to V1. If this deal happens, V2 profits from it if and only if this price for getting B instead of A, ( R2(B)-R2(A) + R1(A)-R1(B) ) / 2, is at most U2(B)-U2(A). Fortunately, she can ensure that the deal happens exactly when this is fulfilled: she only needs to specify her sincere ratings by putting R2(A)=U2(A) and R2(B)=U2(B). If she does so, the deal happens if and only if U2(B)-U2(A)R1(A)-R1(B), which is equivalent to ( U2(B)-U2(A) + R1(A)-R1(B) ) / 2 U2(B)-U2(A), so the deal happens if and only if it is profitable for V2. Moreover, V2 can ensure this independently of V1's behaviour! Analogously, V1 profits from the deal if the price is at least U1(A)-U1(B), and she can also ensure that the deal happens exactly when it is profitable for her: she specifies her sincere ratings by putting R1(A)=U1(A) and R1(B)=U1(B), no matter what V2 does. Of course, this is far from being a new idea so far, and it is not yet the whole idea since it has an obvious problem: although it obviously manages to elect the better option (the one with the larger total monetary value), it encourages both the seller and the buyer to misrepresent their ratings so that the gap between R2(B)-R2(A) and R1(A)-R1(B) becomes as small as possible and hence their respective profit as large as possible. In other words, this method is not at all strategy-free. However, there is a simple modification which makes it strategy-free! The reason for the strategic incentives is that the ratings V1 (and analogously V2) gives not only influence whether the deal happens but also how much V1 profits from the deal when it happens. This is no longer the case when we change the method so that V1's profit depends on V2's ratings only and vice versa: If the deal happens, that is, when R2(B)-R2(A) R1(A)-R1(B), then B wins instead of A, V1 gets an amount of R2(B)-R2(A) but V2 only pays an amount of R1(A)-R1(B). As before, both voters can ensure that the deal happens exactly when they profit from it by voting sincerely. The difference is that now they no longer have any incentive to narrow the gap between R2(B)-R2(A) and R1(A)-R1(B) since a voter's profit is independent of her ratings! There is just a minor problem with this: The balance of the money transfers is positive, so where is this extra money supposed to come from? Obviously, we cannot let V1 and V2 each pay half of the required extra money since that would make the method identical to the original method. Solving the extra money problem A solution to this extra money problem becomes clear when we now increase the number of voters and assume 3 instead of 2 voters. Consider this method next: Each voter fills in a ratings ballot for the options A,B. We draw at random one default option, say A, and one compensating voter, say V3. The other voters (here V1,V2) are the deciding voters. That option whose total ratings from the deciding voters is maximal wins. If this is not the default option (so if it's B), the following money transfers happen: - Each deciding voter gets an amount equal to the total rating difference between the
Re: [Election-Methods] Challenge Problem
Hi again. There is still another slight improvement which might be useful in practice: Instead of using the function 1/(5-4x), use the function (1 + 3x + 3x^7 + x^8) / 8. This is only slightly smaller than 1/(5-4x) and has the same value of 1 and slope of 4 for x=1. Therefore, it still encourages unanimous cooperation in our benchmark situation 50: A(1) C(gamma) B(0) 50: B(1) C(gamma) A(0) whenever gamma (1+1/(1+(slope at x=1)))/2 = 0.6, just as the other methods did. The advantage of using (1 + 3x + 3x^7 + x^8) / 8 is that then there is a procedure in which you don't need any calculator or random number generator, only three coins: ** ** Method 3-coin-FAWRB ** - ** Ballots: Approval with one option marked favourite. ** Tally: ** 1. Determine the approval winner, X. ** 2. Draw a ballot at random; ** if it does not approve of X, its favourite, Y, wins. ** 3. Otherwise, toss three coins; ** if they show no heads, X wins. ** 4. If it's one head, draw one further ballot; ** if it's two heads, draw another seven ballots; ** if all three show heads, draw another eight ballots. ** 5. If all drawn ballots approve of X, she wins; ** otherwise, Y wins. ** Isn't this guaranteed fun? Jobst I wrote: Dear Forest, well - thanks. Anyway, there is still room for improvement. Our last version was this: Let x be the highest approval rate (=approval score divided by total number of voters). Draw a ballot at random. With probability 1/(5-4x), the option with the highest approval score amoung those approved on the drawn ballot wins. Otherwise the favourite of that ballot wins. We saw that this method performs well in a large number of situations. But it seems to me that, with more than three options, it can be hard to find the optimal strategies because approving a non-approval-winner can be bad. For example, consider this case: 33: A B C=D=E 34: C B=D A=E 33: E D A=B=C Here the C faction can either cooperate with the A faction to give B a high probability of winning, or with the E faction to give D a high probability of winning. But when the A faction approves of B but the C and E factions approve D, it would have been better for the A faction to have bullet-voted. The following even simpler method, however, makes it safe to approve of an option which does not turn out the approval winner: ** ** Method FAWRB (Favourite-or-Approval-Winner Random Ballot): ** - ** Everyone marks a favourite and may mark any number of also approved ** options. The approval winner X and her approval rate x are ** determined. A ballot is drawn at random. If the ballot approves of X, ** X wins with probability 1/(5-4x). Otherwise, or if the ballot does ** not approve of X, its favourite option wins. ** FAWRB is again monotonic and solves the original challenge problem in the same way as the other methods we discussed recently. But in the above situation it makes it safe for the A and C factions to approve of B and D since only one of the two factions will actually partially transfer their winning probability from their favourite to the compromise option. I guess it should be possible to analyse FAWRBs strategic implications in detail since the method is so extremely simple! I'm pretty sure already that with FAWRB you will never have an incentive to misrepresent your favourite, and seldom or never to approve of one option while not approving of all more preferred options as well. With the other methods these variations of order reversal would occur more often I think. Yours, Jobst PS: I have not yet thought much about your most recent proposals. Only it seems that they won't elect any compromise option that's not the favourite of anyone, right? [EMAIL PROTECTED] schrieb: Jobst wrote ... What do you think about this? I think you have the golden touch! Forest Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Challenge Problem
Dear Forest, well - thanks. Anyway, there is still room for improvement. Our last version was this: Let x be the highest approval rate (=approval score divided by total number of voters). Draw a ballot at random. With probability 1/(5-4x), the option with the highest approval score amoung those approved on the drawn ballot wins. Otherwise the favourite of that ballot wins. We saw that this method performs well in a large number of situations. But it seems to me that, with more than three options, it can be hard to find the optimal strategies because approving a non-approval-winner can be bad. For example, consider this case: 33: A B C=D=E 34: C B=D A=E 33: E D A=B=C Here the C faction can either cooperate with the A faction to give B a high probability of winning, or with the E faction to give D a high probability of winning. But when the A faction approves of B but the C and E factions approve D, it would have been better for the A faction to have bullet-voted. The following even simpler method, however, makes it safe to approve of an option which does not turn out the approval winner: ** ** Method FAWRB (Favourite-or-Approval-Winner Random Ballot): ** - ** Everyone marks a favourite and may mark any number of also approved ** options. The approval winner X and her approval rate x are ** determined. A ballot is drawn at random. If the ballot approves of X, ** X wins with probability 1/(5-4x). Otherwise, or if the ballot does ** not approve of X, its favourite option wins. ** FAWRB is again monotonic and solves the original challenge problem in the same way as the other methods we discussed recently. But in the above situation it makes it safe for the A and C factions to approve of B and D since only one of the two factions will actually partially transfer their winning probability from their favourite to the compromise option. I guess it should be possible to analyse FAWRBs strategic implications in detail since the method is so extremely simple! I'm pretty sure already that with FAWRB you will never have an incentive to misrepresent your favourite, and seldom or never to approve of one option while not approving of all more preferred options as well. With the other methods these variations of order reversal would occur more often I think. Yours, Jobst PS: I have not yet thought much about your most recent proposals. Only it seems that they won't elect any compromise option that's not the favourite of anyone, right? [EMAIL PROTECTED] schrieb: Jobst wrote ... What do you think about this? I think you have the golden touch! Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Challenge Problem
as favourite, then this doesn't affect x, so monotonicity follows from the monotonicity of Random Ballot in this case. (ii) If noone changes the favourite but someone adds an approval of some option A which has not the maximal approval score, again x is unaffected and monotonicity follows from the monotonicity of Random Approval Ballot this time. (iii) If, finally, noone changes the favourite but someone adds an approval of some option A which already has the maximal approval score, then all of the three values x, f(x), and the winning probability of A under Random Approval Ballot are increased. Thus, some probability is shifted from Random Ballot towards Random Approval Ballot. But since A is an Approval winner, her winning probability under Random Approval Ballot exceeds the one under Random Ballot, so her overall winning probability is increasing as required by monotonicity. QED. Yours, Jobst [EMAIL PROTECTED] schrieb: Dear Jobst, Yes, I meant for more RABMAC probability as the doc increases. If doc were just the max approval (as a percentage of the voters) would the monotonicity problem go away? But then that does nothing to encourage the lesser factions to cooperate. I think you got the essence of the idea here: Looks like a good idea to make the probability of using Random Ballot depend on some degree of cooperation. Perhaps we can salvage it. Forest - Original Message - From: Jobst Heitzig Date: Sunday, June 1, 2008 2:57 pm Subject: Re: [english 90%] Re: Challenge Problem To: [EMAIL PROTECTED] Cc: election-methods@lists.electorama.com Dear Forest, glad you find time to delve somewhat deeper into these questions. Looks like a good idea to make the probability of using Random Ballot depend on some degree of cooperation. Two notes as of now: I guess you meant RABMAC*doc^M + RB*(1-doc^M) instead of RB*doc^M + RABMAC*(1-doc^M), right? And I'm not so optimistic about monotonicity: Consider n voters with these ballots: n-3: A favourite B also approved 2: B favourite 1: C favourite With large n, RB elects A almost surely while RABMAC elects B almost surely. When the last voters switches to approving A also, this is still the case, but the degree of cooperation you defined will increase by almost 1/n. So your method will move probability from the RB- winner A to the RABMAC-winner B while monotonicity demands an increase in A's probability. Details: Before: doc = ((n-3)²+2(n-1)+1)/n² = 1 - 4/n + 8/n² P(A) = 0*doc^M + (n-3)/n*(1-doc^M) After: doc' = ((n-2)²+2(n-1))/n² = 1 - 2/n + 2/n² P(A)' = 1/n*doc'^M + (n-3)/n*(1-doc^M) For M=1: P(A)'-P(A) = 1/n - 2/n² + 2/n³ + (1-3/n)*(-2/n+6/n²) = -1/n + 10/n² - 16/n³ which is negative for n10, right? Maybe this can be fixed somehow be altering the definition of degree of cooperation? Yours, Jobst [EMAIL PROTECTED] schrieb: Dear Jobst, I have a another solution to the challenge problem: p: ACB q: BCA Here's the method: 0. The ballots are approval with favorite indicated. 1. First the total approvals are counted and the candidates listed in order of approval. 2. For each ballot B, let f(B) be the approval score (as a percentage of the number of ballots cast) of the highest candidate on the list approved by B. 3. Let doc (for degree of cooperation) be the average value of f(B) over all ballots. 4. Determine the winner with the lottery RB*doc^M + RABMAC*(1-doc^M), i.e. decide whether to use Random Ballot or Random Approval Ballot Most Approved Candidate on the basis of the lottery (doc^M, 1-doc^M), where M will be specified below. In the challenge problem, let's suppose that C is rated at R and S in the respective factions, with R p and S q, and that the approvals are x: AC x': A only y: BC y': B only with x + x' = p, and y + y' = q. Then as long as x + y max(p, q) we have doc = (x')^2 + (x+y)^2 + (y')^2, and Prob(A wins) = doc^M*x'+ (1-doc^M)*p, Prob(B wins) = doc^M*y' + (1-doc^M)*q, Prob(C wins) = doc^M*(x+y). For later use, note that doc=1 @ (x, y) = (p, q), and that the partial derivative of doc w.r.t. x or y @ (p, q) is 2, so that the partial of doc^M w.r.t. x or y @ (p, q) is simply 2*M. The utility expectation for the first faction is E1 = P(A wins) + R*P(C wins) If we evaluate the partial derivative w.r.t. x of E1 at the full cooperation point (x, y) = (p, q), we get -1 - 2*M*p + R*(1 + 2*M) = 2*M*(R - p) - (1 - R) which is greater than zero when M is sufficiently large, since R p. Similarly the partial derivative of E2 w.r.t. y at the same point is 2*M*(S - q) - (1 - S), which is greater than zero when M is sufficiently large. Therefore, local unilateral defection from full cooperation
Re: [Election-Methods] Challenge Problem
Dear Forest, I think the following modification of your method is both monotonic and performs better in the 33/33/33-situation: 1. We use approval ballots with favourite option indicated. We determine all approval scores. Assume the highest approval score divided by the number of voters is x. 2. One ballot is drawn. 3. With a probability of P=f(x), that option which has the highest approval score amoung those approved on this ballot wins, otherwise the favourite of that ballot wins. In other words: Perform Random Approval Ballot with a probability of P=f(highest approval rate), else perform Random Ballot. The function f(x) is chosen so that in important reference situations full cooperation is encouraged. Let us assume we want to encourage full cooperation in the following situation: 50: A (1) C (gamma) B (0) 50: B (1) C (gamma) A (0) Here full cooperation is an equilibrium when the expected utility of the A-voters given that all B voters cooperate, is maximal for x=1, and vice versa. As this expected utility is f(x)*(x*gamma + (1-x)*1) + (1-f(x))*1/2, which equals gamma for x=1, the condition on f is this: f(x) = (2gamma-1) / (1 - (2-2gamma)*x) E.g., for gamma=0.6 we can choose f(x) = 1/(5-4x). Let's compare this choice f1(x)=1/(5-4x) with the function f2(x)=x^4 which you used in your method and which will also encourage full cooperation in the above situation with gamma=0.6. More precisely, let's look how they perform in this different situation: 33: A1 A A2 B 33: A2 A A1 B 33: B A1=A2=A If all A-voters cooperate (i.e. approve of A), A will win with a probability of 2/3 * f(2/3) which is 2/7 if f=f1 but only 32/243 if f=f2. It seems f1 will usually give the compromise a larger probability than f2. Finally, why is the above method monotonic? (i) If noone changes approvals but someone marks a different option as favourite, then this doesn't affect x, so monotonicity follows from the monotonicity of Random Ballot in this case. (ii) If noone changes the favourite but someone adds an approval of some option A which has not the maximal approval score, again x is unaffected and monotonicity follows from the monotonicity of Random Approval Ballot this time. (iii) If, finally, noone changes the favourite but someone adds an approval of some option A which already has the maximal approval score, then all of the three values x, f(x), and the winning probability of A under Random Approval Ballot are increased. Thus, some probability is shifted from Random Ballot towards Random Approval Ballot. But since A is an Approval winner, her winning probability under Random Approval Ballot exceeds the one under Random Ballot, so her overall winning probability is increasing as required by monotonicity. QED. Yours, Jobst [EMAIL PROTECTED] schrieb: Dear Jobst, Yes, I meant for more RABMAC probability as the doc increases. If doc were just the max approval (as a percentage of the voters) would the monotonicity problem go away? But then that does nothing to encourage the lesser factions to cooperate. I think you got the essence of the idea here: Looks like a good idea to make the probability of using Random Ballot depend on some degree of cooperation. Perhaps we can salvage it. Forest - Original Message - From: Jobst Heitzig Date: Sunday, June 1, 2008 2:57 pm Subject: Re: [english 90%] Re: Challenge Problem To: [EMAIL PROTECTED] Cc: election-methods@lists.electorama.com Dear Forest, glad you find time to delve somewhat deeper into these questions. Looks like a good idea to make the probability of using Random Ballot depend on some degree of cooperation. Two notes as of now: I guess you meant RABMAC*doc^M + RB*(1-doc^M) instead of RB*doc^M + RABMAC*(1-doc^M), right? And I'm not so optimistic about monotonicity: Consider n voters with these ballots: n-3: A favourite B also approved 2: B favourite 1: C favourite With large n, RB elects A almost surely while RABMAC elects B almost surely. When the last voters switches to approving A also, this is still the case, but the degree of cooperation you defined will increase by almost 1/n. So your method will move probability from the RB- winner A to the RABMAC-winner B while monotonicity demands an increase in A's probability. Details: Before: doc = ((n-3)²+2(n-1)+1)/n² = 1 - 4/n + 8/n² P(A) = 0*doc^M + (n-3)/n*(1-doc^M) After: doc' = ((n-2)²+2(n-1))/n² = 1 - 2/n + 2/n² P(A)' = 1/n*doc'^M + (n-3)/n*(1-doc^M) For M=1: P(A)'-P(A) = 1/n - 2/n² + 2/n³ + (1-3/n)*(-2/n+6/n²) = -1/n + 10/n² - 16/n³ which is negative for n10, right? Maybe this can be fixed somehow be altering the definition of degree of cooperation? Yours, Jobst [EMAIL PROTECTED] schrieb: Dear Jobst, I have a another solution to the challenge problem: p: ACB q: BCA
Re: [Election-Methods] [english 90%] Re: Challenge Problem
Dear Forest, glad you find time to delve somewhat deeper into these questions. Looks like a good idea to make the probability of using Random Ballot depend on some degree of cooperation. Two notes as of now: I guess you meant RABMAC*doc^M + RB*(1-doc^M) instead of RB*doc^M + RABMAC*(1-doc^M), right? And I'm not so optimistic about monotonicity: Consider n voters with these ballots: n-3: A favourite B also approved 2: B favourite 1: C favourite With large n, RB elects A almost surely while RABMAC elects B almost surely. When the last voters switches to approving A also, this is still the case, but the degree of cooperation you defined will increase by almost 1/n. So your method will move probability from the RB-winner A to the RABMAC-winner B while monotonicity demands an increase in A's probability. Details: Before: doc = ((n-3)²+2(n-1)+1)/n² = 1 - 4/n + 8/n² P(A) = 0*doc^M + (n-3)/n*(1-doc^M) After: doc' = ((n-2)²+2(n-1))/n² = 1 - 2/n + 2/n² P(A)' = 1/n*doc'^M + (n-3)/n*(1-doc^M) For M=1: P(A)'-P(A) = 1/n - 2/n² + 2/n³ + (1-3/n)*(-2/n+6/n²) = -1/n + 10/n² - 16/n³ which is negative for n10, right? Maybe this can be fixed somehow be altering the definition of degree of cooperation? Yours, Jobst [EMAIL PROTECTED] schrieb: Dear Jobst, I have a another solution to the challenge problem: p: ACB q: BCA Here's the method: 0. The ballots are approval with favorite indicated. 1. First the total approvals are counted and the candidates listed in order of approval. 2. For each ballot B, let f(B) be the approval score (as a percentage of the number of ballots cast) of the highest candidate on the list approved by B. 3. Let doc (for degree of cooperation) be the average value of f(B) over all ballots. 4. Determine the winner with the lottery RB*doc^M + RABMAC*(1-doc^M), i.e. decide whether to use Random Ballot or Random Approval Ballot Most Approved Candidate on the basis of the lottery (doc^M, 1-doc^M), where M will be specified below. In the challenge problem, let's suppose that C is rated at R and S in the respective factions, with R p and S q, and that the approvals are x: AC x': A only y: BC y': B only with x + x' = p, and y + y' = q. Then as long as x + y max(p, q) we have doc = (x')^2 + (x+y)^2 + (y')^2, and Prob(A wins) = doc^M*x'+ (1-doc^M)*p, Prob(B wins) = doc^M*y' + (1-doc^M)*q, Prob(C wins) = doc^M*(x+y). For later use, note that doc=1 @ (x, y) = (p, q), and that the partial derivative of doc w.r.t. x or y @ (p, q) is 2, so that the partial of doc^M w.r.t. x or y @ (p, q) is simply 2*M. The utility expectation for the first faction is E1 = P(A wins) + R*P(C wins) If we evaluate the partial derivative w.r.t. x of E1 at the full cooperation point (x, y) = (p, q), we get -1 - 2*M*p + R*(1 + 2*M) = 2*M*(R - p) - (1 - R) which is greater than zero when M is sufficiently large, since R p. Similarly the partial derivative of E2 w.r.t. y at the same point is 2*M*(S - q) - (1 - S), which is greater than zero when M is sufficiently large. Therefore, local unilateral defection from full cooperation won't pay if M is sufficiently large. If I am not mistaken, the method is monotone and satisfies your property about proportional probability for those factions that steadfastly approve only their favorite. On a technical note, if we replace the lottery (doc^M, 1-doc^M) for deciding which kind of random ballot to use with the lottery (g(doc), 1 - g(doc)), where g(t)=1 - (1 - t)^(1/2), then we don't have to worry about M. This works because (no matter how large M) the slope of g(t) eventually dominates the slope of t^M as t approaches 1. Nevertheless, for the sake of simplicity I suggest using t^5 instead of g(t). My Best, Forest Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] [english 94%] Re: D(n)MAC/RB
Dear Forest, a quick calculation for your suggestion (please check!) gives: Winning probability for A under full cooperation of the A1 and A2 voters: (16+4*8)/81 + 8/27*1/2*2/3 = 56/81 = approx. 70% (OK) Gain in expected utility for the A1 voters when reducing their cooperation by an infinitesimal epsilon: epsilon*( 4*3*(2/3)²*1/3*( 1/2*1/3*(alpha1-alpha) + 1/2*1/3*(alpha2-alpha) + 2/3*(beta-alpha) ) + 8/27*1/2*( (alpha1-alpha) + 1/3*1/(2/3)*(alpha1-alpha) ) ) = epsilon/27*(14alpha1+8alpha2+16beta-38alpha) In this, alpha[1|2] and beta are the utilities of A[1|2] and B for the A1 voters. We may assume that alpha1=1 and beta=0. The A1 voters have no incentive to reduce their cooperation as long as the above gain is 0, i.e. when alpha(7+4alpha2)/19. The latter is always true when alpha58%. Similarly, the A2 voters will fully cooperate when they rate A at least at 58% the way from B to their favourite A2. This is good, however with pure Random Approval as fallback it is even better, it seems: The gain then changes to epsilon*( 4*3*(2/3)²*1/3*( 2/3*(beta-alpha) ) + 8/27*( (alpha1-alpha) + 1/3*1/(2/3)*(alpha1-alpha) ) ) = epsilon/27*(12alpha1+16beta-28alpha) which is negative even when alpha3/7 only! (Please double-check these calculations!) Yours, Jobst [EMAIL PROTECTED] schrieb: Jobst, After thinking about your recent example: 33: A1AA2 B 33: A2AA1 B 33: B A1,A2,A and the 66 A-voters try to cooperate to elect A by unanimously approving of her, then they still get A only with a low probability of 16/81 (approx. 20%) while A1 and A2 keep a probability of 64/243 (approx. 25%) each. A I have two ideas for incremental improvement: 1. For the fall back method, flip a coin to decide between Random Ballot and Random Approval Ballot. Note that if Random Approval Ballot were used exclusively, then there could be insufficient incentive for the first two factions to give unanimous support to A. 2. Reduce the approval requirement from 4 of 4 to 3 out of 4 matches. The the fall back method would be used less frequently, since the 3 of 4 requirement is more feasible for a candidate approved on two thirds of the ballots. Of course, this doesn't solve the general problem, and I'm afraid that any attempt to automate these kinds of adjustments might be vulnerable to manipulation by insincere ballots. - Original Message - From: Jobst Heitzig Date: Saturday, May 24, 2008 10:04 am Subject: Re: [english 94%] Re: D(n)MAC To: [EMAIL PROTECTED] Cc: election-methods@lists.electorama.com Dear Forest, your analysis was right from the beginning while mine in the last message was wrong unfortunately: I claimed that already your D(2)MAC/RB would elect a 52%-compromise, but I got the numbers wrong! So, we really need n2, as you said, and I think that perhaps n=4 could be a good choice. However, another similar but slightly different method really needs only n=2, but that method is again non-monotonic like AMP, and therefore sometimes gives incentive to order-reverse. Anyway, here's that variant: Each voter marks one favourite and at most one compromise. Two ballots are drawn. If they have the same option marked as compromise (not favourite!), that option is the winner. Otherwise the favourite of a third drawn ballot is the winner. Under that method, full cooperation is indeed an equilibrium in our example situation P: ACB Q: BCA as long as everyone prefers C to the Random Ballot lottery. This is because when everyone else marks C as compromise, my not marking her as compromise changes the outcome exactly in those situations in which mine is one of the first two ballots, in which case it takes the win from C and gives it to the Random Ballot lottery. QED. One point still troubles me with n2: In situations where not the whole electorate but a subgroup seeks to cooperate, such a method performs badly. For example, when n=4, the preferences are 33: A1AA2 B 33: A2AA1 B 33: B A1,A2,A and the 66 A-voters try to cooperate to elect A by unanimously approving of her, then they still get A only with a low probability of 16/81 (approx. 20%) while A1 and A2 keep a probability of 64/243 (approx. 25%) each. AMP performs better here in giving A the complete 66% probability, but AMP is considerably more complex and non-monotonic... Yours, Jobst [EMAIL PROTECTED] schrieb: Dear Jobst, Thanks for your encouragement. And D(n)MAC/RB it shall be! I also wanted to speak my admiration of your outline of a computationally effective way of carrying out your trading method: quite a tour d'force with many beautiful mathematical ideas elegantly applied. Here's a further partial result, that might interest you. Suppose that we have 2QPQ0, P+Q=100, and factions P: ACB Q: BCA with C
Re: [Election-Methods] [english 94%] Re: D(n)MAC
Dear Forest, your analysis was right from the beginning while mine in the last message was wrong unfortunately: I claimed that already your D(2)MAC/RB would elect a 52%-compromise, but I got the numbers wrong! So, we really need n2, as you said, and I think that perhaps n=4 could be a good choice. However, another similar but slightly different method really needs only n=2, but that method is again non-monotonic like AMP, and therefore sometimes gives incentive to order-reverse. Anyway, here's that variant: Each voter marks one favourite and at most one compromise. Two ballots are drawn. If they have the same option marked as compromise (not favourite!), that option is the winner. Otherwise the favourite of a third drawn ballot is the winner. Under that method, full cooperation is indeed an equilibrium in our example situation P: ACB Q: BCA as long as everyone prefers C to the Random Ballot lottery. This is because when everyone else marks C as compromise, my not marking her as compromise changes the outcome exactly in those situations in which mine is one of the first two ballots, in which case it takes the win from C and gives it to the Random Ballot lottery. QED. One point still troubles me with n2: In situations where not the whole electorate but a subgroup seeks to cooperate, such a method performs badly. For example, when n=4, the preferences are 33: A1AA2 B 33: A2AA1 B 33: B A1,A2,A and the 66 A-voters try to cooperate to elect A by unanimously approving of her, then they still get A only with a low probability of 16/81 (approx. 20%) while A1 and A2 keep a probability of 64/243 (approx. 25%) each. AMP performs better here in giving A the complete 66% probability, but AMP is considerably more complex and non-monotonic... Yours, Jobst [EMAIL PROTECTED] schrieb: Dear Jobst, Thanks for your encouragement. And D(n)MAC/RB it shall be! I also wanted to speak my admiration of your outline of a computationally effective way of carrying out your trading method: quite a tour d'force with many beautiful mathematical ideas elegantly applied. Here's a further partial result, that might interest you. Suppose that we have 2QPQ0, P+Q=100, and factions P: ACB Q: BCA with C rated at R% by all voters. If R = Q/2^(n-1) + P, then (under D(n)MAC/RB) the common strategy of each faction approving C on exactly Q ballots is a global equilibrium , so that the winning probabilities for A, B, C become 1-2Q%, 0, and 2Q%, respectively. This can be thought of as implicit trading, since the second faction moves C up to equal-first on all Q of its ballots, while the first faction moves C up to equal-first on Q of its ballots, as well. The expected utilities for the two factions are EA = (100-2Q)+R*(2Q)%, and EB = R*(2Q)%. For example, if P=60, Q=40, n=3, and R=70, the equation R=Q/2^(n-1)+P is satisfied, so the global equilibrium strategy is for both to approve C on 40 ballots, yielding the winning probabilities 20%, 0, and 80%, respectively, so that the expectations are EA= 76, and EB=56, compared with the benchmarks of 60 and 40, respectively. With these values of P, Q, and R, the number n would have to be 4 (or more) in order to get unanimous support for C. In that case we would have EA=EB=70. Here's the key to my calculations: Let X and Y be the number of ballots on which C is approved in the respective factions. Then the probability that no candidate is approved on all n of the drawn ballots is given by the formula g = 1 - q^n - p^n - (x+y)^n + x^n + y^n, where p=P/(P+Q), q=Q/(P+Q), x=X/(P+Q), y=Y/(P+Q). So g is the probability that an additional random ballot will have to be drawn to decide the winner. Then the respective probabilities for wins (under D(n)MAC/RB) by A, B, and C are ... pA = p^n+g*p - when(X+YP, x^n, else 0) pB= q^n+g*q - when(X+YQ, 0, else y^n) pC=(x+y)^n - when(X+YP, 0, else x^n) - when(X+YQ, y^n, else 0). Miraculously, these probabilities add up to 1 ! The two faction expectations are EA = pA + R*pC, and EB = pB + R*pC From there, it's all downhill. My Best, Forest - Original Message - From: Jobst Heitzig Date: Thursday, May 22, 2008 3:45 pm Subject: Re: ID(n)MAC To: [EMAIL PROTECTED] Cc: election-methods@lists.electorama.com Dear Forest, your's is the honour of having solved the method design challenge in the most convincing way! To see this, one can also look at it a little differently and perhaps even simpler than in your reasoning: First of all, let's keep in mind that your class of methods is not really a direct generalization of D2MAC since in D2MAC, when the two drawn ballots have no compromise, the deciding ballot is not drawn freshly but is simply the first of the two already drawn. For original D2MAC, this had two effects: First, a faction of p% size has complete control over p% of the winning probability (which is not true with your class of methods
Re: [Election-Methods] method design challenge +new method AMP
Dear Juho, you wrote: Yes, but as I see it the reasons are different. In a typical non- deterministic method like random ballot I think it is the intention to give all candidates with some support also some probability of becoming elected. Not at all! At least in those non-deterministic methods which I design the goal is to make it probable that the voters implement a strategic equilibrium in which a compromise option (instead of the favourite of a mere majority) will be elected with (near) certainty. But for such an equilibrium to exist in the first place, the method cannot be majoritarian, since then the majority would have no incentive at all to cooperate. Instead, all voters must have some power, not only those belonging to the majority, and therefore each voter is given control over an equal amount of winning probability. Still, the goal is not that they assign this amount to their favourite option but that they trade it in some controlled way, in order to elect a compromise which makes all the cooperating voters better off than without the trading! Since at the same time, voting shall be secret, the trading cannot be expected to be performed by open negotiations between the voters, but it must be facilitated by some mechanism which trades winning probabilities automatically depending on the preference information on the voters' ballots. If then in certain situations it happens that not much trading actually takes place, so that the winning probabilities remain with the voters' favourites, then this is only an indication that no sufficiently attractive compromise options existed in that situation. But whenever such an option does exist, the goal of non-deterministic methods like DFC, D2MAC, and AMP is that voters recognize that they are better off with the compromise than with the benchmark Random Ballot solution, and that they can bring about the election of the compromise by safely indicating their willingness to trade their share of the winning probability, without running the risk of being cheated by the other faction(s). D2MAC is quite good at this if only the compromise option is sufficiently attractive, but not in a situation which is as narrow as the one I gave at the beginning of this thread. AMP is better there, but it is not monotonic unfortunately. Yours, Jobst In the deterministic methods electing some non- popular extremist is typically an unwanted feature and a result of the method somehow failing to elect the best winner. *No* election or decision method should be applied without first checking the feasibility of options with respect to certain basic requirements. This sorting out the constitutional options cannot be subject to a group decision process itself since often the unconstitutional options have broad support (Hitler is only the most extreme example for this). In other words, without such a feasibility check *before* deciding, also majoritarian methods can produce a very bad outcome (think of Rwanda...). Ok, this looks like an intermediate method where one first has one method (phase 1) that selects a set of acceptable candidates and then uses some other method (phase 2) (maybe non-deterministic) to elect the winner from that set. There is need for pure non-deterministic methods like random ballot, and pure deterministic methods, and also combinations of different methods may be useful. Also in the case where the no-good candidates are first eliminated I see the same two different philosophies on how the remaining candidates are handled. Either all remaining candidates (with some support) are given some probability or alternatively one always tries to elect the best winner. The intention was thus not to say non- deterministic methods would not work properly but that there are two philosophies that are quite different and that may be used in different elections depending on the nature of the election. Due to this difference I'm interested in finding both deterministic and non-deterministic solutions for the challenge. Juho Yours, Jobst ___ ___ _ EINE FÜR ALLE: die kostenlose WEB.DE-Plattform für Freunde und Deine Homepage mit eigenem Namen. Jetzt starten! http://unddu.de/? [EMAIL PROTECTED] ___ Inbox full of spam? Get leading spam protection and 1GB storage with All New Yahoo! Mail. http://uk.docs.yahoo.com/nowyoucan.html Election-Methods mailing list - see http://electorama.com/em for list info pgpUbFPDuaKZJ.pgp Description: PGP signature Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] method design challenge + new method AMP
Dear Juho, you wrote: One observation on clone independence and electing a centrist candidate using rankings only and when one of the extremists has majority. ... It is thus impossible for the algorithm in this case and with this information (rankings only) to satisfy both requirements and to be fully clone independent. D'accord. This is a good reason to consider rankings insufficient, since from rankings only one cannot determine whether to apparent clones are truly clones in the sense that they are (nearly) equivalent in all relevant aspects. From ratings information, however, one can see this. Therefore I would not at all consider A1,A2 clones in your ratings example: A=100 C=55 B=0 = A1=100 A2=56 C=54 B=0 B=100 C=55 A=0 = B=100 C=56 A1=54 A2=0 For A1,A2 to be considered clones, the ratings would have to be something like 51: A1 100 A2 99 C 55 B 0 49: B 100 C 55 A1 1 A2 0 You also seem to think so, since you wrote: One approach to try to avoid this problem would be to use a more limited clone concept: candidates that are ranked/rated equal with each others. But that would never really occur in practice. I think one should define the notion clone like this: A1,A2 are clones if and only if on each ballot, the difference in ratings between any pair of options is smallest for the pair A1,A2. (Analogously, a set S of options should be called a clone set if and only if on each ballot, all rating differences between two options in S are smaller than all rating differences between other pairs of options. Even more generally, a system Y of disjoint sets S1,...,Sk of options could be called a clone partition if and only if on each ballot, all rating differences between two options which are contained in the same member of Y are smaller than all rating differences between other pairs of options.) With this definition, the problem you described cannot really occur: Assume the rankings are 51: X1X2X3X4 49: X4X3X2X1 If X1,X2 are clones, X2 cannot be considered a good compromise since 49 voters don't like her. Similarly, if X3,X4 are clones, X3 cannot be considered a good compromise since 51 voters don't like her. Yours, Jobst pgp9sEhTbdSjT.pgp Description: PGP signature Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] method design challenge + new method AMP
Dear Raphfrk, you wrote There needs to be some system for providing an incentive for people to give their honest ratings.? A random system with trading seems like a reasonable solution. I am glad that I am no longer alone with this opinion... If a majority has a 100% chance of getting their candidate elected, then there is no incentive for them to trade.? If the voters are 100% strategic, they will know this. Yes, although some Range Voting supporters try hard to convince us of the opposite, it seems. OTOH, a support of a majority should be better than support of a minority. Absolutely! Usually I consider Random Ballot a benchmark method for this very reason: the default winning probability of a candidate should equal the proportion of the voter who favour her. Any deviances from this default distribution should be justified somehow, for example by an increase in some measure of social utility. (The underlying rationale for methods like D2MAC or AMP is even stronger: every voter should have full control over her share of the winning probability, so that in particular when she bullet votes, this share must goes to her favourite. Only such methods are truly democratic.) Optimal utility via trade requires that voters have something to trade, and fractions of a win probability seems to be quite a reasonable solution. I cannot really imagine any other thing unless we consider money transfers... Yours, Jobst pgpQdRGiMeeWy.pgp Description: PGP signature Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] [english 95%] Re: [english 94%] Re: method design challenge+new method AMP
Dear Raphfrk, it did not think through all you wrote yet, but one point troubles me: Also, it is majority compliant. If a majority support a candidate first choice (i.e. first choice and nominate him), then he cannot lose. If that is true, your method cannot be a solution to the given problem, since any majoritarian method will elect A in the situation I described -- remember that voters are strategic! Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] [english 89%] Re: [english 95%] Re: [english 95%] Re: [english 94%]Re: method designchallenge+new method AMP
Dear Raphfrk, I also see no obvious way how the Anti-STV approach might become clone-proof when voters (or factions) can add options. So, the method AMP (and variants thereof) still seems to be the only solution yet... I wonder if anyone comes up with a different approach. In particular, every utilitarian should be interested strongly in solving this problem, I guess :-) Yours, Jobst [EMAIL PROTECTED] schrieb: Jobst wrote: Do you think one could modify the Anti-STV approach in a different way to overcome the cloning problem without making the method majoritarian? It is hard to see how to force a majority to give information about lower preferences without having some form of candidate control. However, if you give the candidate control to the voters, then the majority can put up a majority of the candidates. Raphfrk Interesting site what if anyone could modify the laws www.wikocracy.com AOL's new homepage has launched. Take a tour http://info.aol.co.uk/homepage/ now. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] [english 95%] Re: [english 94%] Re: method design challenge +new method AMP
Dear Juho,
this sounds nice -- the crucial point is that we'll have to analyse what
strategic voters will vote under that method! Obviously, it makes no
sense to the A voters to reverse their CB preference since that would
eliminate C instead of B and will result in B winning instead of C...
Did you look deeper into the strategic implications yet?
Yours, Jobst
P.S. It is quite easy to use also other methods than STV since the
combinatorics are not a problem. There are only n different possible
outcomes of the proportional method (if there are n candidates). In
this example it is enough to check which one of the sets {A,B}, {A,C}
and {B,C} gives best proportionality (when looking at the worst
candidates to be eliminated from the race).
Juho
On May 2, 2008, at 23:59 , Juho wrote:
Here's an example on how the proposed method might work.
I'll use your set of votes but only the rankings.
51: ACB
49: BCA
Let's then reverse the votes to see who the voters don't like.
51: BCA
49: ACB
Then we'll use STV (or some other proportional method) to select 2
(=3-1) candidates. STV would elect B and A. B and A are thus the
worst candidates (proportionally determined) that will be eliminated.
Only C remains and is the winner.
- I used only rankings = also worse than 52 point compromise
candidates would be elected
- I didn't use any lotteries = C will be elected with certainty
Juho
On May 2, 2008, at 22:29 , Jobst Heitzig wrote:
Dear Juho,
I'm not sure what you mean by
How about using STV or some other proportional method to select
the n-1 worst candidates and then elect the remaining one?
Could you give an example or show how this would work out in the
situation under consideration?
Yours, Jobst
Juho
On Apr 28, 2008, at 20:58 , Jobst Heitzig wrote:
Hello folks,
over the last months I have again and again tried to find a
solution to
a seemingly simple problem:
The Goal
-
Find a group decision method which will elect C with near
certainty in
the following situation:
- There are three options A,B,C
- There are 51 voters who prefer A to B, and 49 who prefer B to A.
- All voters prefer C to a lottery in which their favourite has 51%
probability and the other faction's favourite has 49% probability.
- Both factions are strategic and may coordinate their voting
behaviour.
Those of you who like cardinal utilities may assume the following:
51: A 100 C 52 B 0
49: B 100 C 52 A 0
Note that Range Voting would meet the goal if the voters would be
assumed to vote honestly instead of strategically. With strategic
voters, however, Range Voting will elect A.
As of now, I know of only one method that will solve the problem
(and
unfortunately that method is not monotonic): it is called AMP
and is
defined below.
*** So, I ask everyone to design some ***
*** method that meets the above goal! ***
Have fun,
Jobst
Method AMP (approval-seeded maximal pairings)
-
Ballot:
a) Each voter marks one option as her favourite option and may
name
any number of offers. An offer is an (ordered) pair of options
(y,z). by offering (y,z) the voter expresses that she is
willing to
transfer her share of the winning probability from her
favourite x to
the compromise z if a second voter transfers his share of the
winning
probability from his favourite y to this compromise z.
(Usually, a voter would agree to this if she prefers z to
tossing a
coin between her favourite and y).
b) Alternatively, a voter may specify cardinal ratings for all
options.
Then the highest-rated option x is considered the voter's
favourite,
and each option-pair (y,z) for with z is higher rated that the mean
rating of x and y is considered an offer by this voter.
c) As another, simpler alternative, a voter may name only a
favourite
option x and any number of also approved options. Then each
option-pair (y,z) for which z but not y is also approved is
considered
an offer by this voter.
Tally:
1. For each option z, the approval score of z is the number of
voters
who offered (y,z) with any y.
2. Start with an empty urn and by considering all voters free for
cooperation.
3. For each option z, in order of descending approval score, do the
following:
3.1. Find the largest set of voters that can be divvied up into
disjoint
voter-pairs {v,w} such that v and w are still free for
cooperation, v
offered (y,z), and w offered (x,z), where x is v's favourite and
y is
w's favourite.
3.2. For each voter v in this largest set, put a ball labelled
with the
compromise option z in the urn and consider v no longer free for
cooperation.
4. For each voter who still remains free for cooperation after
this was
done for all options, put a ball labelled with the favourite
option of
that voter in the urn.
5. Finally, the winning option is determined by drawing a ball
from the
urn.
(In rare cases, some tiebreaker may be needed in step 3 or 3.1.)
Why this meets the goal: In the described
Re: [Election-Methods] [english 94%] Re: method design challenge + new method AMP
Dear Juho,
I'm not sure what you mean by
How about using STV or some other proportional method to select the
n-1 worst candidates and then elect the remaining one?
Could you give an example or show how this would work out in the
situation under consideration?
Yours, Jobst
Juho
On Apr 28, 2008, at 20:58 , Jobst Heitzig wrote:
Hello folks,
over the last months I have again and again tried to find a
solution to
a seemingly simple problem:
The Goal
-
Find a group decision method which will elect C with near certainty in
the following situation:
- There are three options A,B,C
- There are 51 voters who prefer A to B, and 49 who prefer B to A.
- All voters prefer C to a lottery in which their favourite has 51%
probability and the other faction's favourite has 49% probability.
- Both factions are strategic and may coordinate their voting
behaviour.
Those of you who like cardinal utilities may assume the following:
51: A 100 C 52 B 0
49: B 100 C 52 A 0
Note that Range Voting would meet the goal if the voters would be
assumed to vote honestly instead of strategically. With strategic
voters, however, Range Voting will elect A.
As of now, I know of only one method that will solve the problem (and
unfortunately that method is not monotonic): it is called AMP and is
defined below.
*** So, I ask everyone to design some ***
*** method that meets the above goal! ***
Have fun,
Jobst
Method AMP (approval-seeded maximal pairings)
-
Ballot:
a) Each voter marks one option as her favourite option and may name
any number of offers. An offer is an (ordered) pair of options
(y,z). by offering (y,z) the voter expresses that she is willing to
transfer her share of the winning probability from her favourite
x to
the compromise z if a second voter transfers his share of the winning
probability from his favourite y to this compromise z.
(Usually, a voter would agree to this if she prefers z to
tossing a
coin between her favourite and y).
b) Alternatively, a voter may specify cardinal ratings for all
options.
Then the highest-rated option x is considered the voter's favourite,
and each option-pair (y,z) for with z is higher rated that the mean
rating of x and y is considered an offer by this voter.
c) As another, simpler alternative, a voter may name only a
favourite
option x and any number of also approved options. Then each
option-pair (y,z) for which z but not y is also approved is
considered
an offer by this voter.
Tally:
1. For each option z, the approval score of z is the number of
voters
who offered (y,z) with any y.
2. Start with an empty urn and by considering all voters free for
cooperation.
3. For each option z, in order of descending approval score, do the
following:
3.1. Find the largest set of voters that can be divvied up into
disjoint
voter-pairs {v,w} such that v and w are still free for cooperation, v
offered (y,z), and w offered (x,z), where x is v's favourite and y is
w's favourite.
3.2. For each voter v in this largest set, put a ball labelled with
the
compromise option z in the urn and consider v no longer free for
cooperation.
4. For each voter who still remains free for cooperation after this
was
done for all options, put a ball labelled with the favourite option of
that voter in the urn.
5. Finally, the winning option is determined by drawing a ball from
the
urn.
(In rare cases, some tiebreaker may be needed in step 3 or 3.1.)
Why this meets the goal: In the described situation, the only
strategic
equilibrium is when all B-voters offer (A,C) and at least 49 of the
A-voters offer (B,C). As a result, AMP will elect C with 98%
probability, and A with 2% probability.
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[Election-Methods] method design challenge + new method AMP
Hello folks,
over the last months I have again and again tried to find a solution to
a seemingly simple problem:
The Goal
-
Find a group decision method which will elect C with near certainty in
the following situation:
- There are three options A,B,C
- There are 51 voters who prefer A to B, and 49 who prefer B to A.
- All voters prefer C to a lottery in which their favourite has 51%
probability and the other faction's favourite has 49% probability.
- Both factions are strategic and may coordinate their voting behaviour.
Those of you who like cardinal utilities may assume the following:
51: A 100 C 52 B 0
49: B 100 C 52 A 0
Note that Range Voting would meet the goal if the voters would be
assumed to vote honestly instead of strategically. With strategic
voters, however, Range Voting will elect A.
As of now, I know of only one method that will solve the problem (and
unfortunately that method is not monotonic): it is called AMP and is
defined below.
*** So, I ask everyone to design some ***
*** method that meets the above goal! ***
Have fun,
Jobst
Method AMP (approval-seeded maximal pairings)
-
Ballot:
a) Each voter marks one option as her favourite option and may name
any number of offers. An offer is an (ordered) pair of options
(y,z). by offering (y,z) the voter expresses that she is willing to
transfer her share of the winning probability from her favourite x to
the compromise z if a second voter transfers his share of the winning
probability from his favourite y to this compromise z.
(Usually, a voter would agree to this if she prefers z to tossing a
coin between her favourite and y).
b) Alternatively, a voter may specify cardinal ratings for all options.
Then the highest-rated option x is considered the voter's favourite,
and each option-pair (y,z) for with z is higher rated that the mean
rating of x and y is considered an offer by this voter.
c) As another, simpler alternative, a voter may name only a favourite
option x and any number of also approved options. Then each
option-pair (y,z) for which z but not y is also approved is considered
an offer by this voter.
Tally:
1. For each option z, the approval score of z is the number of voters
who offered (y,z) with any y.
2. Start with an empty urn and by considering all voters free for
cooperation.
3. For each option z, in order of descending approval score, do the
following:
3.1. Find the largest set of voters that can be divvied up into disjoint
voter-pairs {v,w} such that v and w are still free for cooperation, v
offered (y,z), and w offered (x,z), where x is v's favourite and y is
w's favourite.
3.2. For each voter v in this largest set, put a ball labelled with the
compromise option z in the urn and consider v no longer free for
cooperation.
4. For each voter who still remains free for cooperation after this was
done for all options, put a ball labelled with the favourite option of
that voter in the urn.
5. Finally, the winning option is determined by drawing a ball from the
urn.
(In rare cases, some tiebreaker may be needed in step 3 or 3.1.)
Why this meets the goal: In the described situation, the only strategic
equilibrium is when all B-voters offer (A,C) and at least 49 of the
A-voters offer (B,C). As a result, AMP will elect C with 98%
probability, and A with 2% probability.
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Re: [Election-Methods] utility theory lesson for a ve ry confused rob brown
Dear Clay, you wrote: the point is that we know it exists. a very simple economic concept called revealed preference demonstrates this. it works like this. say you prefer apples to oranges to bananas. i give you a guarantee of having to eat an orange, or a 50/50 chance of having to eat an apple or a banana. if utility_apple-utility_orange is less than utility_orange-utility_banana, then you'll choose the orange - because you like it more than the average of the other two fruits. but say we change those odds to 60/40. well, then you have to ask whether orange- banana is at least 60% as much as apple-banana. if so, stick with orange, otherwise take the gamble. by offering you enough different options, we can force you to reveal your true magnitude of preference (unless you get more utility out of lying to us about your preference than you do out of the fruit outcome, but that is obviously besides the point). so it's clear that utility exists, and has levels of intensity that are empirical reality. Are you aware that this is no existency proof at all? Assuming one can urge individuals to answer to certain questions about preferences between a sure outcome and some lottery does not prove anything about what those answers will look like and on what reasoning they will be based. I urge you to read this posting which makes absolutely clear under what conditions one could infer some additive utilities from those answers about preferences: http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/2007-February/019584.html So, unless you can somehow show that the answers the individuals will give to your questions will always meet the conditions (Tot), (Trans), (Comp), (Decomp) and (Archi) of the posting, you will not have proved anything. Even if those conditions are met, one can only speak of *individual* utilities. It is even more problematic to argue that they be comparable amoung different individuals. We had this discussion before on this list, of course... Yours, Jobst _ In 5 Schritten zur eigenen Homepage. Jetzt Domain sichern und gestalten! Nur 3,99 EUR/Monat! http://www.maildomain.web.de/?mc=021114 Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Simple two candidate election
Dear Juho! You wrote: I could imagine a voting system that might address this issue for larger groups, but it isn't Range. One could have elections that take into account e.g. proportionality in time (that could be called one kind of reciprocity) (favour a republican after a democrat, favour other pizzas after pepperoni). Try D2MAC! It does precisely that in the long run :-) Yours, Jobst _ Der WEB.DE SmartSurfer hilft bis zu 70% Ihrer Onlinekosten zu sparen! http://smartsurfer.web.de/?mc=100071distributionid=0066 Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] elect the compromise
Dear Forest! Binding agreements will not solve the problem completely I think. Assume the situation is this, with 4 candidates A,B,C,D of which 3 (A,B,D) have each received 1/3 of the vote, and with the following preferences over lotteries: A: A 100, C 80, BD 0 B: B 100, C 80, AB 0 D: D 100, C 80, AB 0 (The number 80 in the first row means A prefers getting C for certain to getting A with 80% and B or D with 20% probability.) Now if one of the three bluffs by claiming not to consider C a good compromise, the other two still gain by signing an agreement to transfer their probability share to C. Hence each of the three has an incentive to bluff if she can hope the other two will probably sign the agreement anyway. The only solution seems to be that at least one of them announces that she won't sign an agreement with only one other but only with both others. But such an announcement would only be credible if that candidate would at the same time represent her rating for the compromise as something between 34 and 49 instead of 80. In any case, it seems that also with binding agreements it depends on what information the candidates have about the other's preferences... Yours, Jobst Forest W Simmons schrieb: Jobst, I'm not sure how to define rational in this context, either. As for the prisoner's dilemma problem, I wonder if the possibility of defection could be eliminated by having trading parties sit down and sign binding agreements during formal trading. My Best, Forest Jobst Heitzig wrote: Dear Forest, Perhaps candidates should be required to publish their range ballots before the election, and their trading of assets should be required to be rational relative to these announced ratings? I had this idea, too. But upon closer inspection, it is not quite easy to define what in this case rational means, since in this form of trading there easily arise situations similar to the prisoner's dilemma and situations in which bluffing could work... Yours, Jobst Or perhaps, a randomly chose jury of candidate X supporters should have some say in the candidate X proxy decisions? Forest Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] elect the compromise
Dear Forest, Perhaps candidates should be required to publish their range ballots before the election, and their trading of assets should be required to be rational relative to these announced ratings? I had this idea, too. But upon closer inspection, it is not quite easy to define what in this case rational means, since in this form of trading there easily arise situations similar to the prisoner's dilemma and situations in which bluffing could work... Yours, Jobst Or perhaps, a randomly chose jury of candidate X supporters should have some say in the candidate X proxy decisions? Forest Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Challenge: Elect the compromise when there'reonly 2 factions
Dear Abd ul-Rahman, I dislike, by the way, describing voters as selfish if they vote in their own interest. That's the default, they *should* vote in their own interest. That is probably a language problem again. I thought selfish was a synonym for acting in my own interest only, is it not? However, the latter was what I meant to say. Yes, it is a synonym for that. However, the implication here is that not only is one acting in one's own self-interest, it is a narrow self interest that does not care if nearly half the electorate ends up with a maximally unsatisfactory outcome, as long as they personally gain a dime. This is actually sociopathy, someone who truly thinks like this and who is not afraid of consequences would slit your throat for pocket change. And yet you think they *should* vote in their own interest? In a Range poll, social utility is maximized if everyone votes *absolute* utilities, accurately. Only if social utility is defined so that your statement becomes true by definition (and becomes a triviality thus). Welfare economics, however, does not define social utility as the sum of individual utility, it rather defines social welfare in some more sophisticated ways which we already discussed earlier several times. What I ended up suggesting was that the problem is resolved if the voters negotiate. It's possible to set up transfers of value (money?) such that the utilities are equalized, and that the benefit of selecting C is thus distributed such that the A voters do *not* lose by voting for C. If they vote for A, they get A but no compensation. If they vote for C, they get C plus compensation. If the utilities were accurate -- Juho claimed that they were *not* utilities, but that then makes the problem incomprehensible in real terms -- then overall satisfication is probably optimized by the choice of C with compensation to the A voters, coming from the C voters. Certainly the reverse is possible, that is, the A voters could pay the C voters compensation to elect A, but it would have to be much higher compensation! I understood this. But I consider it quite absurd that the A voters should be compensated for anything. This is because you refuse to look at the underlying utilities. Because you don't believe in utility, in particular in *commensurable* utilities, you have only preference left, and from the raw preferences it appears that C is the best compromise. I love to look at utilities. I did just that to infer that C is a good compromise in the example I gave. By the same reasoning (which I will not repeat again here) it also follows that C would be *no* good compromise had the ratings been 55 voters: A 100, C 20, B 0 45 voters: B 100, C 20, A 0 Do you still think only the rankings matter? I don't and never did. Indeed, if that is all the information we have, C is the best compromise. But what has been overlooked, which is precisely what makes the arguments about compensation mysterious to Jobst, is that compromise means that all parties lose something, compared to the ideal for them. Yes, *all* parties, that's exactly the point! So no one of them has to compensate the other, since neither can hope to get their will for certain. They have to compromise. After all, that's what societies are about. By the way, compensation is no mystery at all for me, it is simply not justified in the situation at hand. Suppose it is realized before the election that B is not a viable candidate, and we do not consider B at all. What we have left is 55: AC 45: CA What is the optimal outcome? For ranked methods, it is obvious. You think so? May I assume then that your obvious best outcome is the same as mine, namely electing A with 55% probability and C with 45%? Because this would make it quite attractive to all of them to search for a compromise that all would like better than this lottery. (This is how far I got into your post.) Yours, Jobst For Range and selfish voters, it is also obvious. Only the introduction of the irrelevent candidate makes it appear not obvious. But we do have more information than the ranks. *If* we assume commensurable utilities in the original votes, then we can say much more. There is a relative preference strength, commensurable, of 100:80 for the A voters and 80:0 for the original B voters. The majority has a weak preference and the minority a strong one. There is a complication, if this is a real election. The majority will have reduced motivation to turn out, so if we actually get a 55:45 preference in the final poll, the *real* preference would be greater than that, generally. Forcing all voters to turn out warps elections unnaturally, causing true weak preference to become equal to strong preference. The common argument that strong preference is somehow selfish is seriously flawed, because true knowledge will cause strong
Re: [Election-Methods] Challenge: Elect the comprom ise when there'reonly 2 factions
Dear Adb ul-Rahman, I dislike, by the way, describing voters as selfish if they vote in their own interest. That's the default, they *should* vote in their own interest. That is probably a language problem again. I thought selfish was a synonym for acting in my own interest only, is it not? However, the latter was what I meant to say. What I ended up suggesting was that the problem is resolved if the voters negotiate. It's possible to set up transfers of value (money?) such that the utilities are equalized, and that the benefit of selecting C is thus distributed such that the A voters do *not* lose by voting for C. If they vote for A, they get A but no compensation. If they vote for C, they get C plus compensation. If the utilities were accurate -- Juho claimed that they were *not* utilities, but that then makes the problem incomprehensible in real terms -- then overall satisfication is probably optimized by the choice of C with compensation to the A voters, coming from the C voters. Certainly the reverse is possible, that is, the A voters could pay the C voters compensation to elect A, but it would have to be much higher compensation! I understood this. But I consider it quite absurd that the A voters should be compensated for anything. This would be only justified if something was taken from them which in a sense belonged to them rightfully. What my arguing is all about is that I don't think the A voters have such a right to the certain election of A, at most one could perhaps say the have a right to A getting at least 55% winning probability. So, if they would prefer to have A with 55% and B with 45% over having C with 100%, only then one could perhaps argue that they should be compensated if C was to be elected with certainty. Yours, Jobst ___ Jetzt neu! Schützen Sie Ihren PC mit McAfee und WEB.DE. 3 Monate kostenlos testen. http://www.pc-sicherheit.web.de/startseite/?mc=00 Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Challenge: Elect the compromise when there're only 2 factions
Dear Abd ul-Rahman, I am most concerned about majority *consent.* Jobst is ignoring the fact that I'm suggesting majority *consent* for decisions; What exactly is majority consent? In my understanding consent means *all* voters share some opinion... what do you call it when a minority imposes its will on a majority? It is not democratic whenever some group can impose its will on the others (in the sense of making their preferred outcome certain). No matter whether that group is a majority or a minority. From this it follows that a method which is always deterministic cannot possibly be democratic. Question: if the majority explicitly consents to this for a specific election, does the election method satisfy the Majority Criterion? If the system would have allowed the majority to decide otherwise, the *system* is majoritarian. I'm not sure at all what a just share of power is. Me neither. But no power at all is definitely not a just share of power. By posting on this topic I hope a discussion on this will eventually begin. What I pointed out here was that the ratings given did not contain sufficient information to determine justice. Yes it does. I gave a reasoning why I consider C the more just solution because everyone prefers it to the democratic benchmark. Again, without defining justice, but relying upon common understanding of it, we can easily construct scenarios that fully explain the ratings as sincere, but which have quite different implications regarding justice. In the challenge election, to repeat, we have 55: A 100, B 0, C 80 45: A 0, B 100, C 80 It was assumed that the ratings were sincere, though that was not defined. I gave at least two interpretations of this, so it was defined. I prefer the preferences over lotteries interpretation. Now, it's obvious that C is what we would ordinarily understand as the best winner. But a majority will disagree, and thus the challenge. I don't recall the exact wording, but is there a method which, if adopted, would cause C to win, even if the A and B voters are selfish, and we might assume, the A voters know that they are in the majority? The answer given was Borda with equal ranking prohibited. Now, when I first read this, I did not properly understand it. I should repeat what I did before, only correctly. Let me be explicit about how this could elect C. I will modify the way Borda count from how it is usually stated to make it equivalent to a Range 2 election (CR 3). Sincere votes. 55: ACB 45: BCA Counts: A, B, C 55: 2 0 1 45: 0 2 1 totals: A 110, B 90, C 100. This does not elect C. However the B voters, if they understand the situation, can vote 45: CBA or counts A, B, C: 45: 0 1 2 totals: A 110, B 45, C 145. C wins, so it appears a quite desirable strategy for the B voters, as we would understand the sincere ratings. Is there a counter-strategy? What if the A voters reverse their second and third preferences? 55: 2 1 0 With the strategic votes from the other side the totals are A 110, B 100, C 90; they defeat the compromise attempted by the B voters. However, the gain is relatively small, it would seem (but there is an assumption that a gain of 20 in rating is small. Not necessarily.) and with the original sincere Borda votes from the B voters, this counterstrategy would give us totals A 110, B 135, C 90. So, somewhat off the topic, but interesting nevertheless, the B voters, being not only selfish, but clever, mount a secret campaign to get all the B voters to vote the strategy. However, they also arrange to leak this information to the A voters, and, *supersecretly*, they are not going to do that, they are going to vote sincerely. If the A voters fall for it and vote strategically, to defeat the nefarious stratagem of the B voters, and the B voters then simply vote sincerely, B prevails, which is a disaster for the A voters and a total victory for the B voters. The A voters are *probably* better off simply voting sincerely. And that was Jobst's point. I don't think that was my point. In order to get a stable situation, i.e. a group strategy equilibrium, all voters should order reverse to make sure the other faction cannot reverse the outcome to their advantage. So the A voters are better off voting CAB. For this reason, I consider Borda a possible but not a good solution to the problem. Juho's suggestion to use weights like 1.4, 1, and 0 improves this since with them C is already elected with sincere ballots. Explicitly, Jobst stated that the ratings given were not utilities, and that he doesn't believe in utilities as having any meaning. Again, this is not true. I only stated that I don't believe in *measurable* utilities or, most importantly, even in *commensurable * ones. That does not mean I regard the term utility as meaningless. When someone prefers some A to some B, I think we can
Re: [Election-Methods] Elect the Compromise
Dear Abd ul-Rahman, the ratings that Jobst fed us as a distraction. You're doing it again -- please stop it. That was not an insult. It made the challenge more interesting. I'm sorry that you thought it critical. I don't think it was insulting. You just repeatedly attribute opinions or intentions to me which are not mine. I explained several times how the ratings are meant to have more meaning than just rankings. I also don't think you were critical here, but I certainly invite you to be it! The only thing I wish is that you try to be post shorter messages, since I really have trouble to read that much! Yours, Jobst Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Challenge: Elect the comprom ise when there're only 2 factions
Dear Abd ul-Rahman, No. It's an understanding of what utilities mean. If you think so... If A does not win, the supporters of A lose something. They are in the majority. If each of them grabs a B supporter and wrestles with him, or her, I suppose, the excess A supporters can then arrange things the way they like. A drastic picture, but actually part of the theory behind majority rule. That's more or less the point I try to make over and over again: A democratic decision system should not reproduce what would happen in an anarchic world such as you describe but should instead protect the weaker parts of society against the majority by giving them their just share of power instead of letting the majority always overrule them. If C wins, the B supporters gain 60% utility, that's large. If they pay the A voters the equivalent of the A loss, 20%, they are still way ahead. You still assume that their is a loss to the A voters. But that is just wrong: the A voters have no right to the election of A, it is not their property which they can loose. It is a very good deal for the B voters No, they would have to pay for a solution which I think they have a right to! Jobst regards it as unjust that the majority should be paid by the minority to get an outcome he regards as more just. However, he isn't looking at the utilities No. Why must I repeat over and over again that I don't believe in measurable utility. I interpret the numbers I gave in the example in the way I describes several times: as representing preferences over lotteries! The actual consequences of the election are irrelevant to him. What do you think you do here? Where did I say such a thing? The actual consequences should of course be that the obvious compromise solution C should be elected without anyone having to pay for it! But this is a democracy. What is the this you are referring to? Sure, one can imagine systems where majority rule is not sufficient for making decisions, I cannot imagine a system where majority rule *is* sufficient for making really *democratic* decisions. Contrary to what Jobst might assume, I have a lot of experience with consensus communities, both positive and negative. I don't assume anything about your experience and have never said so. But please keep in mind that consensus is a much different thing from majority rule. I should think my example makes this very clear: No consensus about A nor about B, only consensus about B being nearly as good! However, when you get down to the nuts and bolts of a system, *including how the system is implemented,* majority rule has proven itself to be practical *and* sustainable. Could you give any evidence for this fact? Point is, when you don't have majority rule, you have decisions being made by something *other* than the majority, even if it is only the default decision to change nothing. And a determined minority can then hold its right to withhold consent over the rest of the community, in order to get what it wants. Again, it would never, in that context, blatantly do this, but it happens, social dynamics do not disappear in consensus communities. Therefore I don't consider consensus as a parcticable idea in all situations. There is nothing magic about 50%, it is simply the point where there are more people on one side than another, there are more saying Yes to a motion than No. Or the reverse. In real communities, other than seriously unhealthy ones, the majority is restrained. It does not make decisions based on mere majority, ordinarily, it seeks broader consent, and deliberative process makes this happen. You repeat this, but could you give evidence for this claim? The original conditions assume commensurability of utilities, No, definitely not! I would never propose such a thing! I only said that those who believe in such measures may interpret the given numbers in that way... If the utilities are not commensurable, then there is no way to know who is the best winner. If Jobst does not understand that, if he does not understand how normalization -- and these are clearly normalized utilities, can distort the results, we could explain it for him. I gave a reasoning why C is the better solution than A. Commensurable utilities are nonsense in my opinion. Nice for use in models but no evidence for them. Essentially, the C-election 20% preference loss of the A voters could have an absolute value greater than the 60% gain by the C voters. A negotiation would expose that, because a negotiation, You give us this in exchange for that causes the utilities to be translated to commensurable units, the units of the negotiation. As I mentioned, it does not have to be money. So what unit will it be then if not money? Please be more precise, The assumption that Jobst easily makes, that the C option is more just, is based on an assumption of
Re: [Election-Methods] Challenge: Elect the compromise
Dear Steve,
Although Jobst may not have intended this assumption, I will continue to
make the assumption that the B minority's preference intensity for the
compromise C over A is much greater than the A majority's preference
intensity for A over C.
Sorry, I had just not read carefully the first time. Of course that
interpretation is consistent with what I had in mind, although I do not believe
that preference intensities belonging to different persons can be compared.
(I am NOT saying there is a way to measure or
compare sincere preference intensities or utilities suitable for input
into a good vote tallying algorithm.) Without an assumption like this,
we would have no reason to believe C is better than A for the society.
I think we have! The reasoning is this: 55% like A best, 45% like B best.
Therefore the democratic benchmark solution with which we should compare
prospective solutions is the lottery that elects A with 55% probability and B
with 45% probability. Now, all voters prefer C to this benchmark, but only 55%
prefer A to this benchmark and only 45% prefer B to the benchmark. From this
point of view C is a better solution than A is.
But I hope that also without this kind of reasoning it should be obvious that a
compromise which everybody likes almost as most as her favourite is a better
election outcome than one of the polar favourites...
In other words, I believe
we should confine ourselves to solving the Tyranny of the Nearly
Indifferent Majority but not try to solve the Tyranny of the
Passionate Majority.
You suggest not to solve the problem of the Tyranny of the Passionate
Majority? Why? Shouldn't problems be solved?
In the real world, it is much easier to elect a compromise than Mr.
Lomax seems to be saying below, because in the real world the set of
alternatives is not fixed to {A,B,C} by nature (nor by Jobst). Most
procedures allow a very small minority to add an alternative to the set
being voted on. (Under Robert's Rules of Order, for instance, only two
people are required: one to propose alternative D and the other to
second the proposal.)
It seems you and Adb ul-Rahman try to convince us that the problem I posed does
not exist in the real world. Well, if you really think so, I can't help it.
Anyway, it would be nice if you could still give a hint what kind of method you
would suggest to solve the stated problem *assuming* that the problem exists :-)
Yours, Jobst
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Re: [Election-Methods] Challenge: Elect the comprom ise when there're only 2 factions
Dear Abd ul-Rahman, Okay, here is my solution. The B voters gain some very substantial advantage for the election of C over the favorite of the A voters, who have only a substantially smaller preference for A over C. So the B voters offer something of value to the A voters to compensate them for their loss. That is certainly an interesting proposal. It seems to be based on the assumption that the just solution is to elect A and that in order to get the compromise, the minority should pay for it. Although that would probably solve the problem, this is not how I think society should work: I don't think nearly half of the electorate should pay the other half for getting what is the more just solution in my eyes. Perhaps that is a difference in culture? The original conditions assume commensurability of utilities, No, definitely not! I would never propose such a thing! I only said that those who believe in such measures may interpret the given numbers in that way... Yours, Jobst __ Erweitern Sie FreeMail zu einem noch leistungsstärkeren E-Mail-Postfach! Mehr Infos unter http://produkte.web.de/club/?mc=021131 Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Elect the Compromise
Dear Forest, The main thing I overlooked was vote trading. So there are two main devices for solving the challenge: vote trading and randomness. There is a third one! One of the oldest voting methods that have been studied can also solve it at least in part. I wonder who will first see what I mean :-) Jobst suggested a way of combining them: asset voting with random ballot as the base method, so that probabilities are traded. Right. That could solve the problem: With Random Ballot, A and B will win with 55% and 45% probability, respectively. If Candidates A and B agree to trade their power by transferring their complete share of the probability to C, both factions will gain. There is only one problem left: If candidates are allowed to trade also parts of their power, C will not be elected with certainty since then A and B will only offer to transfer a part of their probability large enough so that the other faction will still gain somewhat (details to come). We could also combine them into a DYN version of D2MAC. The basic ballots are DYN ballots. Voters decide Yes/No for each candidate that they feel sure about, and then Delegate the remaining Y/N votes to one of the candidates, presumably their favorite. After all of the Y/N votes have been completed by the proxies, two ballots are drawn at random. If there is a candidate that was (either directly or by proxy) voted Yes on both ballots, then the common Yes candidate with the greatest number of Y's (from the other voters or their proxies) is elected. Otherwise the favorite (i.e. proxy candidate) of the first drawn ballot chooses the winner. That's just an idea meant to stimulate exploration of further possibilities. A very nice idea in my view! One could even let the candidates know what the direct votes are and communicate with each other and let them sign contracts what candidates they will approve of. This would give them also some means of asset trading... Yours, Jobst _ Der WEB.DE SmartSurfer hilft bis zu 70% Ihrer Onlinekosten zu sparen! http://smartsurfer.web.de/?mc=100071distributionid=0066 Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Elect the Compromise
Dear Kevin, Hi, It seems to me there might be a use for something like the method that was proposed awhile ago that had to do with offering voters incentives to give sincere ratings. For example, the majority would give the sincere score to their compromise in exchange for their vote having greater effect in reducing the win odds of their least favorite candidate. If they rate their compromise too high, the loss of win odds from favorite to compromise outweighs the value of the corresponding lessened odds of the worst candidate. And vice versa for rating the compromise too low. Hmm, I don't think I understand how this works, either. It would have to be a non-majoritarian method in order to solve the problem, of course. Yours, Jobst ___ Jetzt neu! Schützen Sie Ihren PC mit McAfee und WEB.DE. 3 Monate kostenlos testen. http://www.pc-sicherheit.web.de/startseite/?mc=00 Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] Challenge: Elect the compromise
Dear Steve! However, assuming the intensity difference between the A faction's 100 and 80 is much less than the intensity difference between the B faction's 80 and 0, That was not the assumption I wanted anyone to make. Those of you who believe in measurable utility: please assume that the ratings reflect utilities in the *same* units. All others: please interpret the ratings A 100, C 80, B 0 as saying that the person would prefer C over each lottery that elects A with a probability of p less than 80%, and B with a probability of 1-p, and that the person would prefer over C each lottery that elects A with a probability of p above 80%, and B with a probability of 1-p. Yours, Jobst Forest S replied: Under strategic voting with good information, any decent deterministic method (including Approval) would elect the Condorcet Winner A . Uncertainty as to the faction sizes could get C elected, but not necessarily. So some randomness is essential for the solution of this problem. The indeterminism has to be built into the method in order to make sure that it is there in all cases. Jobst's D2MAC would work here because the compromises' 80 percent rating is above the threshold for sure election when the two faction sizes differ by ten percent or more, if I remember correctly. If the compromise had only a 60 percent rating, for example, optimal strategy might give A a positive chance of winning. It is paradoxical that randomness, usually associated with uncertainty, is the key to making C the certain winner. Look up D2MAC in the archives for a more quantitative analysis. I hope that this doesn't prematurely take the wind out of the challenge. Forest From: Jobst Heitzig [EMAIL PROTECTED] Subject: [Election-Methods] Challenge: Elect the compromise when there'reonly 2 factions To: election-methods@lists.electorama.com Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=iso-8859-15 A common situation: 2 factions 1 good compromise. The goal: Make sure the compromise wins. The problem: One of the 2 factions has a majority. A concrete example: true ratings are 55 voters: A 100, C 80, B 0 45 voters: B 100, C 80, A 0 THE CHALLENGE: FIND A METHOD THAT WILL ELECT THE COMPROMISE (C)! The fine-print: voters are selfish and will vote strategically... Good luck have fun :-) Election-Methods mailing list - see http://electorama.com/em for list info Election-Methods mailing list - see http://electorama.com/em for list info ___ Jetzt neu! Schützen Sie Ihren PC mit McAfee und WEB.DE. 3 Monate kostenlos testen. http://www.pc-sicherheit.web.de/startseite/?mc=00 Election-Methods mailing list - see http://electorama.com/em for list info
[Election-Methods] Challenge: Elect the compromise when there're only 2 factions
A common situation: 2 factions 1 good compromise. The goal: Make sure the compromise wins. The problem: One of the 2 factions has a majority. A concrete example: true ratings are 55 voters: A 100, C 80, B 0 45 voters: B 100, C 80, A 0 THE CHALLENGE: FIND A METHOD THAT WILL ELECT THE COMPROMISE (C)! The fine-print: voters are selfish and will vote strategically... Good luck have fun :-) Jobst _ In 5 Schritten zur eigenen Homepage. Jetzt Domain sichern und gestalten! Nur 3,99 EUR/Monat! http://www.maildomain.web.de/?mc=021114 Election-Methods mailing list - see http://electorama.com/em for list info
