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Is the following statement always true:

if there is a mixed-strategy Nash equilibria then it is unique.

I know that there can be several pure strategy Nash equilibrias.

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migrated from cstheory.stackexchange.com Jun 12 '12 at 18:05

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No. The easiest way to prove this is by counterexample. Expanding matching pennies, we can create a game with two mixed strategies (players either mix between A and B or between C and D):

Dual Matching Pennies

What might be true about that statement is for a mix of particular pure strategies for two players, there is a unique mixed strategy equilibrium. This presumes that there is a threshold after which you switch from using strategy $0$ to strategy $1$. At this threshold, your opponent is ambivalent between her strategies. As depicted below, the intersection (and hence mix) will be unique.

Reaction Correspondence Matching Pennies

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