I'm still writing the proof. I'll have it done in the next week. However, I realized a consequence of the result that exactly one method satisfies strong FBC for 3 candidates:
Recall the idea of a voting machine that takes each voter's preference order and assigns him an optimal strategy given everybody else's assigned strategies and some election method. Suppose that the machine assigns everybody his optimal strategy in approval voting. Since there's never a disincentive to approve your favorite, the only question that the machine needs to answer for each person is "Should this voter also approve his second favorite?" One would think that such a machine would never flunk strong FBC, because a voter always wants #1 to get a vote. There may be an incentive to give the machine an insincere preference order (as Gibbard and Satterthwaite proved) but it would seem that the incentive can only be to swap #2 and #3, to prevent a strong #2 from beating #1 (in cases where #3 is weak). However, any method satisfying strong FBC for 3 candidates must be equivalent to "Top 2 Voting" (give one vote each to your favorite and runner-up, zero to your least favorite). Such a machine will not, in general, give the same result as top 2 voting. The reason is the ambiguity in deciding what strategies to assign to the voters. It's easy to say what my optimal strategy is if I know everybody else's. But what if the machine must also specify everybody else's strategies? It has a conflict of interest. One other interesting observation: The method of "top 2 voting" is equivalent to approval if we allow truncation. Interesting that a method that (sort of) satisfies strong FBC would be equivalent to Approval. Alex ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
