Consider a 2-player symmetric game given by a payoff matrix $A\in [0,1]^{n,n}$ for the row player (i.e. the column player matrix is $A^t$).

Let $s=<s_1,s_2>$ be a (possibly mixed-strategies, not-necessarily symmetric) equilibrium for the game.

Define $Sup(s)=Sup(s_1)\cup Sup(s_2)$, where $Sup(s_i)$ is the set of strategies in $s_i$ that has a strictly positive probability.

Is there always a symmetric equilibrium $<s',s'>$ such that $Sup(s')\subseteq Sup(s)$?

(note that there always exist a symmetric equilibrium for the game, but I'm interested in knowing whether any equilibrium imply a symmetric equilibrium with subset support).

For example (yes, kinda pointless if $n=2$, but yet), consider the following simple game:

$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $

And the column player profit, given by $A^t$ is:

$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $

There exists exactly two equilibriums here (up to symmetry), both of which are in pure-strategies, in the first both players play strategy 1, and in the second some player plays 1 and the other plays 0.

The support of the symmetric equilibrium is $\{1\}$ which is a subset of the support of the asymmetric equilibrium which as $\{1,2\}$ as it's support.


1 Answer 1


Here is a counterexample (from p.76 of my thesis):

$$ A = A^T = \begin{bmatrix} 0 & 3 & 2 \\ 3 & 0 & 2 \\ 2 & 2 & 3 \end{bmatrix}. $$

Labeling the strategies in order as $a$, $b$, and $c$, there are asymmetric Nash equilibria $(a,b)$ and $(b,a)$ with support $\{a,b\}$, but there is no symmetric Nash equilibrium supported on this set. If the players play the same mixed strategy supported on this set then at least half of the time they will receive zero payoff, so their expected payoff can be at most $\frac{3}{2}$. But then they are better off deviating to $c$.


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