Here's why I believe that no voting method based on ranked ballots can satisfy both the Favorite Betrayal Criterion and the Majority Criterion:
Suppose that sincere preferences are given by x:A>B>>C y:B>C>>A z:C>A>>B and that none of the three factions has a majority. Suppose (by way of contradiction) that method M satisfies both the Majority Criterion and the FBC. If all of the voters have confidence that M satisfies the FBC, then the voted ballots (in an election governed by method M) will be x1:AB x2:AC x3:A y1:BC y2:BA y3:B z1:CA z2:CB z3:C where x1+x2+x3=x, y1+y2+y3=y, and z1+z2+z3=z. Now suppose (w/o loss in generality) that M gives the win to A. Then, since the members of the B faction consider A to be much worse than C, the B faction members would have been better off giving a majority to their second preference C. Next suppose that in a similar election (with the same sincere preferences as before) the members of the A and C factions still have confidence in the FBC, and intend to vote accordingly, and that the front runner is still A. It seems to me that rather than trusting in the FBC through fine tuning the values y1, y2, and y3, the B faction should abandon its favorite B, and just vote CBA. If they do so, then the A faction will end up regretting its trust in the FBC, since no amount of twiddling of x1, x2, and x3 can save it from the majority attack of the B and C factions. In summary, the B faction's favorite betrayal might have been unnecessary, but once it decided to go in that direction, then favorite betrayals (by members of both the A and C factions) become the A faction's only hope. So method M cannot actually satisfy the FBC after all. I realize that this argument is not air tight, but I believe that it shows the essential difficulty in attempting to satisfy both the FBC and the Majority Criterion. Forest ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
