# 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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