# Mixed-strategy Nash equilibria

I didn't find in books, so I'm asking - Mixed-strategy Nash equilibria is always only one or doesn't exist for the one certain game? And I know that there can be several(and can not be at all) pure strategy Nash equilibria.

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Pure strategies can be seen as special cases of mixed strategies, in which some strategy is played with probability $1$. In a finite game, there is always at least one mixed strategy Nash equilibrium. This has been proven by John Nash[1].

There can be more than one mixed (or pure) strategy Nash equilibrium and in degenerate cases, it is possible that there are infinitely many. In a well-defined sense (open and dense in payoff-space), almost every finite game has a finite and odd number of mixed strategy Nash equilibria.

A typical example of a game with more than one equilibrium is Battle of Sexes, which has two pure strategy equilibria and one completely mixed equilibrium, meaning every strategy is played with positive probability.

[1]: J.Nash. Non-Cooperative Games. http://www.cs.upc.edu/~ia/nash51.pdf

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yes, I meant that real mixed strategy, with probabilities, can be only one? Or there can be several probabilities? I know that pure strategy NE is degenerate cases of mixed. – alec May 13 '12 at 17:30
There can be several, but not in degenerate $2\times 2$-bimatrix-games. – Michael Greinecker May 13 '12 at 17:33
And then how can I find this several probabilities? I know how to calculate only one variant of probabilities. – alec May 13 '12 at 17:36
The usual way consists essentially of the following. You take for each player a nonempty subset of pure strategies. In a completely mixed equilibrium, every player is indifferent between all her strategies. So you use this condition to find completely mixed equilibria in this restricted games. You then check whether they are also equilibria in the unrestricted game. With more than two players, this can become quite complicated. – Michael Greinecker May 13 '12 at 17:41