Wednesday, February 20, 2013

non-Smith winner of approval voting equilibrium

Suppose I have 3 voters voting for 3 candidate outcomes.  Their preferences are A>C>B, B>C>A, and, since C stands for Condorcet, voter 3's first choice is C.  If preferences are strict and common knowledge, a voting scheme in which voters vote first between C and not C and then, in the event of not C, vote between A and B, will result in C's winning the first round, regardless of how you fill out 3's preferences.  If voter 3 is exactly indifferent between the bottom two choices, and 1 and 2 are approximately (in a von Neumann-Morgenstern sense) indifferent between their bottom two choices, then "not C" wins the first round; each of the first two voters is willing to take the chance that 3's coin comes up their way.  (They are hoping for opposite outcomes, obviously).

Now, in a post about half a year ago, I conjectured that, when there is a Condorcet winner, that candidate will (almost always) win in any approval voting equilibrium in the environment in which I generally think about voting systems: a lot of voters, maximizing expected utility, who know enough about everyone else to predict the vote outcomes with high relative precision but low absolute precision (i.e. if a candidate is getting about 1,000,000 votes, voters' best guesses will be off by more than 10 votes but less than 10,000 votes). The first paragraph scales up to a counter-example to that conjecture; if you have 1,000,000 voters, with about one third of them of each of the preferences given, and each voter believes that A and B have equal chances of winning but that C has a negligible chance of winning, then A and B each get (about) 500,000 votes and C gets 333,000.  There's another equilibrium in which the three practically tie, and each wins with a 1/3 probability.[1]  There's also an equilibrium that lies "in between" them, where C has just enough chance of winning to induce most C voters not to vote for their second choice, but a low enough chance that a few of them will vote for A and B, which thereby get slightly more votes than C; this equilibrium is unstable in the sense that a slightly different belief would lead to one of the other equilibrium outcomes instead. Both of the other equilibria are robust in this sense.

From a normative standpoint, I have to say, I don't think these are bad equilibria, especially ex ante.  I probably want to think more about this in light of some of the nice properties of Condorcet winners, many of which properties are described in terms that suggest that preference order is the only thing about preference that matters[2] — which suggests, perhaps, that the Condorcet solution concept is best ex post, but that something like a continuous approval vote might give better results ex ante in some contexts.

[1] In both cases the marginal probability of winning is practically equal to the probability conditional on there being a close race, which is what really matters, since they're both basically tied. I use the term "probability" here for conciseness.
[2] Note that the von Neumann-Morganstern utilities in the example given are essential; if the voters prefer their second choices to a coin toss between the other two — say, for definiteness, that half of the C voters prefer A to B and the other half prefer B to A — the construction in the first paragraph fails, and I'm pretty sure that now the only equilibrium is in fact the one in which every voter votes for two choices, giving C 1,000,000 votes and A and B 500,000 each.

1 comment:

dWj said...

If a Condorcet winner (strictly) beats every (50/50) pairwise lottery, then any equilibrium in which that candidate doesn't (almost always) win, second and third place must be (practically, asymptotically) tied: if second and third are not tied in some equilibrium, each voter votes for candidates it prefers to a 50/50 lottery over the top two candidates and against those whom it disprefers to that lottery, so the Condorcet winner gets votes from more than half of the voters and the second place candidate gets votes from no more than half of the voters.