Kevin, good work!
On Thu, Oct 10, 2013 at 9:23 AM, Kevin Venzke <step...@yahoo.fr> wrote: > Hi Forest, > ... > > Unfortunately, I realized that an SFC problem is possibly egregious: > > 51 A>B > 49 C>B > > B would win easily, contrary to SFC (which disallows both B and C). But > more alarmingly it's a majority favorite problem. > So it is non-majoritarian in the same sense that Approval is. In this case the count is too close for approval voters to drop their second preferences, so B will be the Approval winner. Of course with perfect information, they would bullet, and A would win. Philosophically, in this situation I sympathize with electing the candidate broader support (the "consensus candidate") over the mere majority favorite, which is why Approval's failure of (one version of) the Majority Criterion has never bothered me. Jobst and I have gone to a lot of trouble to contrive methods that make B the game theoretic winner in the face of such preferences. I'm sure you remember his challenge to find a method that makes B the perfect information game theoretic winner when utilities are given by (say) 60 A(100), B(70) 40 C(100), B(50) It seems that only lottery methods can solve this challenge in a satisfactory way. We co-authored a paper with the double entendre title of "Some Chances for Consensus" on this topic for the benefit of people who take the "tyranny of the majority" problem seriously. In sum, my non-majoritarian leanings allow me to bid adieu to SFC without too much regret. Now here is a proof of the fact that you observed ... that not all candidates can have greater MPO than IA: Suppose to the contrary that (for some ballot set) every candidate has a greater MPO than IA. Let C(1), C(2), C(3), ... be a sequence of candidates such that C(n+1) gives max opposition to C(n). Since the MPO for C(n) is greater than the IA for C(n), and the IA for C(n+1) is at least as great as the opposition of C(n+1) to C(n), we can conclude that the IA for C(n+1) is greater than the IA for C(n). Therefore,IA increases along the sequence C(1), C(2), C(3), ... so the sequence cannot cycle. therefore there must be an infinite number of candidates. This impossibility shows that the existence of the posited ballot set.was a false assumption. In light of this fact I propose the following variation on our method: 1. Eliminate all candidates that have higher MPO than IA. 2. Elect the remaining candidate with the greatest difference between its IA and its MPO. I like differences better than ratios in this context, but I used ratios in IA/MPO because I worried about people who couldn't easily agree that (25 - 30) > (72 - 90) , for example. But now that we know eliminating all of the negative differences is possible without eliminating all of the candidates, let's switch to differences. Forest
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