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Given a finite symmetric 2 player game with a strategy space $S$, a (mixed-strategy) symmetric equilibrium is a distribution $d\in \Delta(S)$ such that $(d,d)$ is a Nash equilibrium.

A known result is that in such game there always exists a symmetric equilibrium (see here).

Could there be multiple (extreme) symmetric equilibriums? that is two symmetric equilibriums $d_1,d_2$ such that $d=\frac{d_1+d_2}{2}$ is not an equilibrium?

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Yes. Consider the coordination game $(A,B) = (I,I)$, where $I$ is the $n \times n$ identity matrix.

For this game there are $2^{n}-1$ symmetric equilibria of the form $(x,x)$ where $x=(1/|S|,\ldots,1/|S|)$ for non-empty $S \subseteq \{1,\ldots,n\}$. I.e., the equilibria comprise all identical uniform mixtures. These are all equilibria.

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  • $\begingroup$ Nice, Thanks Rahul ! $\endgroup$
    – R B
    Nov 16, 2014 at 12:25
  • $\begingroup$ You're welcome. $\endgroup$ Nov 16, 2014 at 12:28

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