Suppose we have an election method M that elects the CW when one exists and uses some unspecified auxillary procedure when no CW exists.
I will argue that this method cannot satisfy the requirement that no voter ever has an incentive to rank another candidate above or equal to his favorite. By assuming strict preference orders with no indifference between candidates we eliminate weak FBC from consideration (no incentive to rank another candidate above your favorite). I know that earlier I argued that no monotonic method which always elect the favorite of a majority can satisfy strong FBC, but not all Condorcet methods are montonic (e.g. IRV-completed Condorcet). Begin with 2m+1 voters: 1 A>B>C m B>C>A m C>A>B Here we have a cycle. If M elects C then the A>B>C voter has an incentive to switch his vote to B>C>A, in which case B is the CW. The only way to satisfy strong FBC while leaving that incentive intact is to say that the A>B>C voter could cause A or B to win by specifying the preference A>C>B. In that case the voter has no reason to betray his favorite because another strategy can give him an equally good or better outcome. However, if the first voter in the example reports A>C>B then C is the Condorcet winner. Hence, the method M cannot elect C in this case. Now suppose that we have the situation 2 A>B>C m-1 B>C>A m C>A>B Once again, M can't elect C. One of the voters with the preference A>B>C could always switch his reported preference to B>C>A. We then have the previous situation, which I just argued must elect A or B to satisfy strong FBC. Also, if one of those voters switches his preference to A>C>B, C now beats B (the B vs. C pairwise contest has a margin of 1) and is the CW. By induction, if we say that A or B wins in the situation n A>B>C m-n+1 B>C>A m C>A>B then A or B must win in the situation n+1 A>B>C m-n B>C>A m C>A>B Otherwise one of the A>B>C voters would have an incentive to betray his favorite by switching to B>C>A, creating a situation which elects A or B. Also, if the A>B>C voter tries to use an alternate non-favorite-betrayal strategy, switching to A>C>B, C is then the Condorcet winner, since in both cases B beats C by a single vote. So, A or B must win in the situation m A>B>C 1 B>C>A m C>A>B But if A wins then the B>C>A voter has an incentive to betray his favorite by switching to C>A>B, causing C to become the CW and win. Also, if that voter switches to B>A>C then A will beat C and become the CW, since the A vs. B contest had a margin of one vote. So B must win. Now, use the same process, moving voters from C>A>B to B>C>A. In each case, B must still win by the logic of the first example. Otherwise one of the elect C by switching to C>A>B. And in each case, the A vs. C contest has a margin of one vote, so switching to B>A>C will elect A, the voter's least favorite. Therefore, B must win in the case m A>B>C m B>C>A 1 C>A>B But in this case, the C>A>B voter can switch his vote to A>B>C and elect A, so he has an incentive for favorite betrayal. If he tries to accomplish the same result without betraying his favorite, by switching to C>B>A, then B (his least favorite) will be the CW since the A vs. B contest has a margin of one vote. So while our arguments above based on strong FBC require that B win in this case, strong FBC prohibits B from winning in this case. We have a contradiction, and hence we cannot simultaneously elect the CW whenever he exists and still satisfy strong FBC. Weaknesses of this argument: 1) It ignores truncation, an important strategy to consider when analyzing various Condorcet methods. 2) It is specific to 3 candidates. Strong FBC is satisfied if a voter can obtain the same (superior) result by either ranking somebody else above his true favorite OR by changing the relative rankings of candidates lower on his list. With 3 candidates I only have to consider one possible insincere ranking of the lower choices. With 4 candidates I'd have to consider 5 such rankings. It isn't clear to me how to invoke the Condorcet criterion when there is a cycle A>B>C>A and I'm considering a swtich from A>B>C>D to A>B>D>C (assume all candidates defeat D). Anyway, that's all for now. Alex ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
