space and games

March 19, 2010

When is an honest vote a cabal equilibrium?

Filed under: Voting — Peter de Blanc @ 2:11 pm

When I posted about cabal equilibria to election methods mailing list, Jameson Quinn asked me when honest voting is a cabal equilibrium. I have a partial answer.

Assume we are using an election method that satisfies the Condorcet criterion, and there exists a double Condorcet winner. Then an honest vote is a cabal equilibrium.

What is a double Condorcet winner, you ask? It is a candidate C such that, for every other pair of candidates A and B, there exists a majority of voters who each prefer C over both A and B.

Proof: Suppose that an honest vote is not a cabal equilibrium.

Then there must be some set S of voters that can improve the outcome for themselves by changing their votes. Let B be the new winner. Then no member of S may prefer C to B.

In the new strategy-profile, C is no longer a Condorcet winner (or else C would be elected). However, B is still ranked below C by a majority of voters. This is because only members of the set S changed their vote, and by assumption, members of S were already ranking B above C.

Thus, there must be some other candidate A who isn’t ranked below C (i.e. some members of S changed their vote by moving A above C).

Since all voters who prefer C to B are still voting honestly, the set of voters who prefer C to both B and A must not be a majority.

Thus C is not a double Condorcet winner.

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