# Do Symmetric Games with Nash Equilibria always have a symmetric Equilbrium?

Define a game with S players to be Symmetric if all players have the same set of options and the payoff of a player depends only on the player's choice and the set of choices of all players. Equivalently A game is symmetric if applying a permutation to the options chosen by people induces the same permutation on the payoffs. For example if the original set of options chosen were 1,2,1,3 and the pay-offs were 6,0,6,100 respectively then if the game is symmetric the set of options 2,1,1,3 would have to lead to the pay-offs 0,6,6,100

Suppose a Symmetric Game S has at least 1 nash equilbrium, then must S have a symmetric Nash equilbrium i.e. a nash equilbrium where all players use the same strategy? If not under what conditions does there exist a nash equilbrium. If so is there a simple proof or a simple idea behind the proof?

Clearly this doesn't hold if we restrict to pure strategies the game with the following payoff matrix where all pure equilbria are asymmetric serves as a counter example, But I've yet to find a counterexample for impure strategies.

0/0 1/1
1/1 0/0

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The answer is yes for finite games and for zero-sum games. In general, however, the answer is no: http://www.rochester.edu/college/faculty/markfey/papers/SymmGame3.pdf

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The answer is yes for finite games and mixed strategies and this was already shown in the Ph.D thesis of John Nash, where it occurs as Theorem 4. Nash considered actually slightly more invariances in his theorem.

The proof amounts to the verification that one can do the usual fixed-point argument used for the proof that every finite game has a Nash equilibrium in mixed strategies, restricted to the set of symmetric strategy profiles and to symmetric best responses.

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