It looks very nearly a tie, which complicates things. :) In the weighted proportional method I described in another email, the results would be (for a slate of 3 candidates, and ten votes for each pair mentioned in the problem).
ABC = 60 ABD = 60 ABE = 60 ACD = 50 ACE = 50 ADE = 50 BCD = 50 BCE = 50 BDE = 50 CDE = 60 The possible winning slates are ABC, ABD, ABE, and CDE. So it picks the same groups as mentioned before. Not much help there. If we look at each winning group, A and B appear in three possibilities, while C, D, and E appear in 2 apiece. So it seems reasonable to make sure that the winning group has both A and B in it. If we are limited to 3 people, that leaves one spot. So flip a fair 3-sided coin, and pick the third winner. :) Mike PS with a weighted system, A would get 25 votes (out of 60), B would get 25, and (C,D, or E) would get 10. Of course, in a 3 person legislature, this is equivalent to giving each the same power, since any two could outvote the third. > Forest and I were discussing PR last week and the following situation > came > up. Suppose there are five candidates, A, B, C, D, E. A and B evenly > divide the electorate and, in a completely orthogonal way, C, D, and E > evenly divide the electorate. That is: > > One-sixth of the electorate approves A and C. > One-sixth of the electorate approves A and D. > One-sixth of the electorate approves A and E. > One-sixth of the electorate approves B and C. > One-sixth of the electorate approves B and D. > One-sixth of the electorate approves B and E. > > It is obvious that the best two-winner representative body is A and B. > What > is the best three-winner representative body? > > CDE seems to be the fairest. As Forest said, it is "envy-free". > > Some methods would choose ABC, ABD, or ABE, which seem to give more total > satisfaction. > > Is one unequivocally better than the other? > > I tend to feel that each representative should represent one-third of the > voters, so CDE is a much better outcome. Certain methods, like STV, > Monroe, > and AT-TV (I think) can even output a list of which voters are represented > by each candidate, which I really like. > > I also note that if there was another candidate, F, approved by everybody, > it is probably true that ABF would be an even better committee than CDE. > Is > there a method that can choose CDE in the first case and ABF in the second > case? > > Andy > ---- > Election-Methods mailing list - see http://electorama.com/em for list info > ---- Election-Methods mailing list - see http://electorama.com/em for list info
