> > Here’s a method that seems to have the important properties that we have > been worrying about lately: > > (1) For each ballot beta, construct two matrices M1 and M2: > In row X and column Y of matrix M1, enter a one if ballot beta rates X > above Y or if beta gives a top > rating to X. Otherwise enter a zero. > IN row X and column y of matrix M2, enter a 1 if y is rated strictly above > x on beta. Otherwise enter a > zero. > > (2) Sum the matrices M1 and M2 over all ballots beta. > > (3) Let M be the difference of these respective sums > . > (4) Elect the candidate who has the (algebraically) greatest minimum > row value in matrix M. > > Consider the scenario > 49 C > 27 A>B > 24 B>A > Since there are no equal top ratings, the method elects the same candidate > A as minmax margins > would. > > In the case > 49 C > 27 A>B > 24 B > There are no equal top ratings, so the method gives the same result as > minmax margins, namely C wins > (by the tie breaking rule based on second lowest row value between B and > C). > > Now for > 49 C > 27 A=B > 24 B > In this case B wins, so the A supporters have a way of stopping C from > being elected when they know > that the B voters really are indifferent between A and C. > > The equal top rule for matrix M1 essentially transforms minmax into a > method satisfying the FBC. > > Thoughts? >
To me, it doesn't seem like this fully solves our Approval Bad Example. There still seems to be a chicken dilemma. Couldn't you also say that the B voters should equal-top-rank A to stop C from being elected: 49 C 27 A 24 B=A Then A wins, right? But now the A and B groups have a chicken dilemma. They should equal-top-rank each other to prevent C from winning, but if one group defects and doesn't equal-top-rank the other, then they get the outright win. Am I wrong? ~ Andy
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