Kevin Venzke's brilliantly simple demonstration that "favorite betrayal" is sometimes "strategically forced" in every Condorcet voting method (whether rank-equalities are allowed or not)

Assumptions. We shall only consider "Condorcet voting methods" which are "anonymous" and "neutral" and in which a ballot has the form FrGsH where each relator r and s is selected from the set {=,>} and F,G,H is a permutation of the three candidates. (As we shall see, the anonymity and neutrality assumptions may be dropped. Also in two places we shall depend on certain additional "reasonableness assumptions.")

A method is "Condorcet" if it elects the Condorcet winner whenever one exists. We shall allow randomized ("lottery") voting methods in our proof. "Favorite betrayal" means dishonestly ranking your favorite candidate strictly below top in your vote, and it being "strategically forced" means that if you ranked your favorite top or coequal-top, then a strictly worse election result would always happen (from your point of view).

Proof: The election-table below shows an election with 15 voters. Because this is a A>C>B>A cycle and the scenario is symmetrical, any anonymous and neutral method must elect each candidate with 1/3 probability.

#voters Their Vote
3 A=B>C
3 C=A>B
3 B=C>A
2 A>C>B
2 B>A>C
2 C>B>A

Now suppose the A=B>C voters change their vote to A>B>C, thus betraying their true co-favorite B. This turns A into the Condorcet winner, who would have to win with 100% probability since it is a Condorcet method. Votes A=B>C and A=B=C which do not betray B, cause either the original 3-way-tie scenario, or cause C to be the Condorcet winner which is even worse from this voter's point of view. So: betrayal is strategically forced. Q.E.D.

Proof ultimately based on an electorama web post by Kevin Venzke in June 2005. I would like to thank Abd ulRahman Lomax and Kevin Venzke for some combination of confusing, harassing, and unconfusing me about this until I got it right. This proof has been criticized because it depends for its operation on a symmetric election-tie; the critic would prefer a proof involving no ties. See below for some remarks about that... we can (nearly) fully satisfy this criticism...

Objection: This proof involved 3 identical voters simultaneously changing their vote. I want only one voter to change his vote.

Response: Consider the 3 voters changing their votes one at a time. Focus on the first voter whose vote-change actually has a beneficial effect, and the scenarios immediately before and after he does his change. That gives a favorite-betrayal election example with only a single voter changing. (This same trick is useable all over voting theory, incidentally.) WARNING: This response only is valid under the reasonableness assumption that this critical voter could not get the same or better effect by instead changing his A=B>C vote to A=B=C, or that the vote "A=B=C" is forbidden, or defining any time to need to vote "A=B=C" to be a favorite-betrayal. However, at least part of this extra assumption can be dropped thanks to the following argument: if it were really true that the change to A=B=C had a beneficial effect for this honestly-A=B>C voter, then the inverse change would have had a beneficial effect for a voter who honestly felt C>A=B, which would have been a single-voter favorite-betrayal scenario of a different kind.

First extension of the proof: The demand that the voting method be anonymous & neutral may be avoided by a 3-case argument:

  1. Suppose that, due to privileged voters or candidates, the winner is A. Then the B=C>A voters have incentive to rank B above C.
  2. Suppose that the winner is B. Then the C=A>B voters have incentive to rank C above A.
  3. If we suppose the winner is C, then we have the original situation, in which the A=B>C voters have incentive to rank A above B.

Second extension of the proof: If equalities are disallowed in rankings, then this even simpler election yields a 3-way tie (1/3 probability of each of A,B,C winning):

#voters Their Vote
1 A>B>C
1 C>A>B
1 B>C>A

Now if the A>B>C voter changes her vote to B>A>C, betraying A, then B becomes the Condorcet winner, with 100% victory probability. This is a superior outcome from this voter's point of view if the utilities are UA=12, UB=9, UC=0 since it increases the expected utility from 7 to 9. The possible votes which did not betray A, namely the original A>B>C and A>C>B, both are worse from this voter's point of view (the latter turns C into the Condorcet winner – utility=0) so that the betrayal is strategically forced. (And we can similarly make a 3-case argument extending this proof to avoid having to assume anonymity & neutrality.) Q.E.D.

Third extension of the proof: The criticism that the scenarios above are highly symmetric ties may be avoided in the following way to get a proof not involving a tied-scenario, or at least not really depending essentially on needing ties. (There are many variants of the following "random perturbation" trick; we only exhibit the simplest.)

#voters Their Vote
30 A=B>C
30 C=A>B
30 B=C>A
20 A>C>B
20 B>A>C
20 C>B>A
1 unknown

In this 150-voter election (which is just 10 times the original one), we have a symmetrical 3-way tie, except for the 151st voter, called the "tiebreaker," whose vote is unknown to us until after the election is over. (In realistic elections, one usually does not have full knowledge of other voters' votes until afterwards.) We therefore model that extra voter as possibly casting any of the 6 possible nontrivial equality-containing votes A=B>C, A=C>B, C=B>A, A>B=C, B>A=C, C>A=B with equal likelihood. (Also, one could merely assume the last three of these six to get another proof that would also work...) Then the election generally will not be tied. However, from the point of view of any particular among the 150 voters working from the incomplete information available to him, the winners A,B,C (due to symmetry & anonymity) are each equally likely to be elected.

Now (as before) we consider the 30 A=B>C voters. They are motivated to betray their true co-favorite B by changing their vote to A>B>C. That turns A into the Condorcet winner among the first 150 voters. That causes A to have a greater than 1/3 probability of winning the 151-voter full election – in fact the probability is at least 2/3 because in 4 out of the 6 cases A is the Condorcet winner. This situation is at least as good, and if there is a nonzero win probability for A or B in the 2 cases where A is not the Condorcet winner, better, in the eyes of these 30 voters. Personally I find it "unreasonable" for a Condorcet method not to give A the victory with at least some nonzero probability in at least one of those 2 cases. That proves that this betrayal is strategically forced, in this kind of partial information scenario, if the Condorcet method obeys this "reasonableness" condition. Q.E.D.

Again it is probably possible to drop this extra assumption, at least partially, if some extra arguments are put in. You would try to argue that any "unreasonable" Condorcet method would have to exhibit favorite betrayal of another sort. For example, clearly in each of the two unreasonable C-always-wins scenarios, a single B>A>C voter could betray B by voting A>B>C to make A the Condorcet Winner, which would be an improvement from his point of view.

Another Venzke Proof


And if we forbid ranking-equalities

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