space and games

March 4, 2009

A Stronger Type of Equilibrium

Filed under: Decision Theory, Voting — Peter de Blanc @ 5:57 pm

In game theory, a Nash equilibrium is an assignment of strategies to players, such that no player can benefit from changing vis own strategy (assuming the other players’ strategies remain the same). For example, in the one-shot Prisoner’s Dilemma, mutual defection is the only Nash equilibrium. There are several other ideas for how to define an “equilibrium” of a game, many of which are “refinements” of Nash Equilibrium, in the sense that the set of equilibria is smaller.

Here’s another one: a “cabal equilibrium” is an assignment of strategies to players, such that no subset (or “cabal”) K of players can simultaneously change strategies in such a way that some players in K benefit and no players in K are worse off. In other words, each cabal of players is playing a strategy which is Pareto optimal for the players within that cabal. Note that any subset of players is considered a cabal, so if there are N players, then there are 2N cabals.

Any cabal equilibrium is also a Nash equilibrium. A single player is just a cabal of size one, so if no cabal of players wishes to change strategies, then single no player wishes to change strategy. So this definition of equilibrium is “stronger” than the Nash equilibrium, in the sense that it is a more stringent test.

For example, the Prisoner’s Dilemma has no cabal equilibrium. First observe that mutual defection is the only Nash equilibrium, so there is at most one cabal equilibrium. But mutual defection is not an cabal equilibrium, because both players would prefer to simultaneously switch their strategy to cooperation.

Stag hunt has two Nash equilibria but only one cabal equilibrium. Battle of the Sexes has two Nash equilibria, both of which are cabal equilibria.

Mike Dobbins and I talked about cabal equilibria in voting. Because the outcome of an election depends on the voters in such a coarse way, the Nash equilibrium does not seem like a strong enough definition to apply to voting situations. For example, any election in which the winner beats the first loser by more than one vote is a Nash equilibrium, because no single voter can change the outcome by changing vis strategy. So not only are there multiple Nash equilibria, but in fact almost any election is a Nash equilibrium.

So let’s consider cabal equilibria in a plurality or approval vote. Suppose there is an equilibrium, and this equilibrium elects candidate A.

Suppose that more than half of the voters preferred B over A. Then they could all vote for B (and disapprove of A in an approval vote), resulting in a better outcome from their point of view. Then the original election was not a cabal equilibrium: contradiction! So there is no candidate that is preferred over A by a majority.

So if we ignore the possibility that exactly half the voters prefer some other candidate over A, this would mean that A is a Condorcet winner!

Now let’s consider the converse. Suppose that there is no cabal equilibrium. Then, in particular, it is not a cabal equilibrium for everyone to vote for the same candidate A. Then there exists some cabal K of voters which could make a Pareto improvement by changing votes. In order for the new strategy to be a Pareto improvement, it must affect the outcome of the election somehow; thus it must cause some other candidate B to be elected instead of A. In order to change the outcome of the election, we must have changed at least half of the votes (since originally everyone was voting for A), so K must contain at least half of the voters. Since no member of K prefers B over A, we know A is not preferred over B by a majority, and thus is not a Condorcet winner.

So if there is no cabal equilibrium, then A is not a Condorcet winner. Since the same argument works for any other candidate (not just candidate A), this means that no cabal equilibrium implies no Condorcet winner. Taking the contrapositive, this means that in an election with a Condorcet winner, there is a cabal equilibrium!

So we’ve just proven a theorem: A plurality or approval vote with an odd number of voters has a cabal equilibrium iff there is a Condorcet winner, and a cabal equilibrium always elects the Condorcet winner.

Powered by WordPress