[EM] methods based on cycle proof conditions
I. BDR or Bucklin Done Right: Use 4 levels, say, zero through three. First eliminate all candidates defeated pairwise with a defeat ratio of 3 to 1. Then collapse the top two levels, and eliminate all candidates that suffer a defeat ratio of 2 to 1. If any candidates are left, among these elect the one with the greatest number of positive ratings. snip This seems to be even more Approvalish than normal Bucklin. 65: A3, B2 35: B3, A0 (I assume that zero indicates least preferred) Forest's BDR method elects A, failing Majority Favourite. In response to the above, Abd Lomax-Smith wrote (3 June 2010): snip Now, who would use BDR with only two candidates? It's like using Range with only two candidates. Why would you care about majority favorite if you decide to use raw range. I wonder why the A faction even bothered to vote with that pattern of utilities (ratings). That's what is completely unrealistic about this kind of analysis. snip I was content to simply prove that the method simply fails Majority Favourite, but to appease Abd here is a similar example with three candidates: 60: A3, B1, C0 35: B3, A0, C1 05: C3, A2, C0 A is the big majority favourite and the big voted raw range winner, and yet B wins. Chris Benham Election-Methods mailing list - see http://electorama.com/em for list info
[EM] methods based on cycle proof conditions
Forest Simmons wrote (1 June 2010): snip I. BDR or Bucklin Done Right: Use 4 levels, say, zero through three. First eliminate all candidates defeated pairwise with a defeat ratio of 3 to 1. Then collapse the top two levels, and eliminate all candidates that suffer a defeat ratio of 2 to 1. If any candidates are left, among these elect the one with the greatest number of positive ratings. snip This seems to be even more Approvalish than normal Bucklin. 65: A3, B2 35: B3, A0 (I assume that zero indicates least preferred) Forest's BDR method elects A, failing Majority Favourite. Chris Benham Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] methods based on cycle proof conditions
At 02:54 PM 6/3/2010, Chris Benham wrote: Forest Simmons wrote (1 June 2010): snip I. BDR or Bucklin Done Right: Use 4 levels, say, zero through three. First eliminate all candidates defeated pairwise with a defeat ratio of 3 to 1. Then collapse the top two levels, and eliminate all candidates that suffer a defeat ratio of 2 to 1. If any candidates are left, among these elect the one with the greatest number of positive ratings. snip This seems to be even more Approvalish than normal Bucklin. 65: A3, B2 35: B3, A0 (I assume that zero indicates least preferred) Forest's BDR method elects A, failing Majority Favourite. As will any method which optimizes expressed utility, assuming that these numbers are a rough expression of utility. Because they can be some kind of artifact, I would suggest that a vote like this calls for a runoff. Note, in Bucklin, those approvals would be spread, and, of course, A wins in the first round. In real elections, though, with numbers like this, and no interfering candidate, if the 65 actually do rate B with a 2, they are expressing that they don't much care, so the election can legitimately be decided by the voters who do care. This is normal democratic practice! If a runoff where held with these primary numbers, I would expect low turnout and B wins by a landslide. But if the A faction was somehow duped into voting like that, the reverse will happen. Now, who would use BDR with only two candidates? It's like using Range with only two candidates. Why would you care about majority favorite if you decide to use raw range. I wonder why the A faction even bothered to vote with that pattern of utilities (ratings). That's what is completely unrealistic about this kind of analysis. For many, for years, to note that a method failed the Majority criterion (Majority favorite) was the same as saying that it was totally stupid. But real, direct, human decision-making doesn't satisfy the criterion unless people just want a fast decision and don't care much. If they don't care very much, and somebody does care and makes a big fuss, what happens? From the answer you can then tell what kind of society it is, its sense of coherence and unity, its ability to negotiate consensus and thus natural operational efficiency, etc. If I let you have your way when you care and I don't, then you let me have my way when I care and you don't. Therefore utility maximization systems will generally improve outcomes for *everyone*. The overall game is not a zero-sum game. Single-winner elections appear to be only when they are divorced from the context. There is something related in comparing Approval and Range. I found this odd effect, studying absolute expected utility in a zero-knowledge Range 2 election voted sincerely in the presence of a middle utility candidate. The individual expected utility for the voter was the same for the approval-style vote vs the Range vote. (And, of course, it was the same if you voted for the middle candidate or not, the situation is symmetrical). But if the *method* was approval, the expected utility was lower than if the method was Range. Compared to Range, Approval was lowering everyone's expected utility! So everyone votes Approval style, the expected utility of the outcome must be lower than if at least one person votes Range style. (Otherwise Range is the same as Approval, if nobody actually uses the intermediate rating.) Nobody has bothered to confirm or disconfirm this result. Warren validated some of it, but not that part, and to really nail it down required more math than I could easily do. In order to insist on the Majority criterion, we lower the expected utility for nearly everyone, certainly for most people, when we consider many elections. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] methods based on cycle proof conditions
A couple of other possibilities for methods based on cycle proof conditions: I. BDR or Bucklin Done Right: Use 4 levels, say, zero through three. First eliminate all candidates defeated pairwise with a defeat ratio of 3 to 1. Then collapse the top two levels, and eliminate all candidates that suffer a defeat ratio of 2 to 1. If any candidates are left, among these elect the one with the greatest number of positive ratings. II. SSCPE or Six Slot Cycle Proof Elimination Use six levels, zero through five. First eliminate all candidates with a pairwise defeat ratio of five to one. Then allowing only those ballot comparisons with strength 2 or greater (i.e. the preferred candidate is rated at least two levels above the other), eliminate all candidates with a defeat ratio of two to one. Then allowing only comparisons with strength three or greater, eliminate all candidates beaten with a defeat ratio of one to one, i.e. all defeated candidates. If there are two or more undefeated candidates, elect the one with the greatest number of positive ratings. III. SPE or Strong Preference Elimination: Use 2n levels. First eliminate all of the candidates that are defeated when the only ballot preferences counted are of strength n or greater, i.e. the rating of the preferred one is at least n levels greater than the rating of the other. If there are two or more unbeaten candidates, collapse the bottom three levels to zero, decrement the other levels by two, and decrement 2n to 2(n-1) and repeat the process recursively. The idea of SPE is that the most important eliminations are done by strong preferences, and weaker preferences are invoked only to break ties. More than one tie breaker step is needed only to ensure the technical compliance with Pareto. Random ballot could be used as a tie breaker just as well where ever voters are not allergic to it. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] methods based on cycle proof conditions
How about Condorcet-compliant Range-2 with Runoffs: use 3 levels, 0 through 2 (these numbers matter). Find the largest number of approvals for any candidate. Any candidate with a range score higher than that number makes it into the runoff. This cannot eliminate the CW (or indeed, any member of the Smith set. To extend this property to more than two levels, you have to implicityly collapse to 2 levels first.) Then use a condorcet-compliant method for the runoff. JQ 2010/6/1 fsimm...@pcc.edu A couple of other possibilities for methods based on cycle proof conditions: I. BDR or Bucklin Done Right: Use 4 levels, say, zero through three. First eliminate all candidates defeated pairwise with a defeat ratio of 3 to 1. Then collapse the top two levels, and eliminate all candidates that suffer a defeat ratio of 2 to 1. If any candidates are left, among these elect the one with the greatest number of positive ratings. II. SSCPE or Six Slot Cycle Proof Elimination Use six levels, zero through five. First eliminate all candidates with a pairwise defeat ratio of five to one. Then allowing only those ballot comparisons with strength 2 or greater (i.e. the preferred candidate is rated at least two levels above the other), eliminate all candidates with a defeat ratio of two to one. Then allowing only comparisons with strength three or greater, eliminate all candidates beaten with a defeat ratio of one to one, i.e. all defeated candidates. If there are two or more undefeated candidates, elect the one with the greatest number of positive ratings. III. SPE or Strong Preference Elimination: Use 2n levels. First eliminate all of the candidates that are defeated when the only ballot preferences counted are of strength n or greater, i.e. the rating of the preferred one is at least n levels greater than the rating of the other. If there are two or more unbeaten candidates, collapse the bottom three levels to zero, decrement the other levels by two, and decrement 2n to 2(n-1) and repeat the process recursively. The idea of SPE is that the most important eliminations are done by strong preferences, and weaker preferences are invoked only to break ties. More than one tie breaker step is needed only to ensure the technical compliance with Pareto. Random ballot could be used as a tie breaker just as well where ever voters are not allergic to 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
