This new method (Top Tier Pairwise Rule modified MinMax) gives a satisfactory resolution to "Kevin's Bad MMPO Example:"
49 A 01 A=C 01 B=C 49 B It yields a tie between A and B. Shall we call the method MinMax(TTPR)? ----- Original Message ----- From: Date: Friday, December 2, 2011 4:12 pm Subject: This might be the method we've been looking for: To: [email protected], > 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? > ---- Election-Methods mailing list - see http://electorama.com/em for list info
