Re: [Election-Methods] [english 95%] Re: [english 94%] Re: method design challenge +new method AMP
On May 9, 2008, at 13:39 , Jobst Heitzig wrote: Dear Juho, you wrote: (Roughly the question is if one wants to give Stalin and other unwanted fellows a small probability to become elected or a zero probability.) I don't think this is the point. To the contrary, bringing up such examples is quite misleading, I think, because extreme options are not at all a problem of non-deterministic methods only. 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. 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
Re: [Election-Methods] [english 95%] Re: [english 94%] Re: method design challenge +new method AMP
One observation on clone independence and electing a centrist candidate using rankings only and when one of the extremists has majority. Votes: 51: ACB 49: BCA C is the winner. A will be cloned. The votes could be: 51: A1A2CB 49: BCA2A1 C should still be the winner. B will be cloned. The votes could be: 51: ACB1B2 49: B2B1CA C should still be the winner. The problem is that these two sets of votes are identical: 51: X1X2X3X4 49: X4X3X2X1 In the first set of votes the intended winner C is X3 and in the latter X2. 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. Similar conclusions could be drawn at least for normalized ratings. 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 or A=100 C=55 B=0 = A=100 C=56 B1=54 B2=0 B=100 C=55 A=0 = B2=100 B1=56 C=54 A=0 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. Juho 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 95%] Re: [english 94%] Re: method design challenge+new method AMP
Jobst wrote: 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 Right, under the assumption of perfect strategy.? However, only a small number of voters need to nominate C as compromise for C to win. Are you assuming that the 51% block of voters knows that they have the majority ? In the case you give 51: ACB 49: BCA It seems to me that the voters, will not be sure which of the main candidates have a chance. Assuming the odds are 50/50 and considering a voter in the 51 block and how they cast their nominate vote: Nominate C This will either have no effect or cause C to win. A-C shift: -48 B-C shift: +52 expected utility: +2 Nominate A or B This will either have no effect or shift away from C C-A shift: +48 C-B shift: -52 expected utility: -2 Thus, it is worth it for A and B supporters to nominate C instead of A. Only, 3% of them need to actually do it to elect C. Btw, I think your original proposal is pretty cool too.? I wonder what the effects of putting a threshold would be on the strategic effects. For example, if a candidate represents more than 90% of the balls in the urn, they are declared the winner without drawing any. Raphfrk Interesting site what if anyone could modify the laws www.wikocracy.com AOL's new homepage has launched. Take a tour at 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
