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.
2026-03-29 14:55:50.1774796150
Mixed-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