We can consider only finite games if it makes a difference, but are there nash equilibria that can't be characterized as mixed equilibria?
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1$\begingroup$ I find this question either unclear or trivial. Please include your exact definitions. $\endgroup$– GiskardCommented Mar 9, 2018 at 20:00
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$\begingroup$ The example that immediately comes to mind is prisoner's dilemma: the Nash equilibrium is for both players to pick "betray". $\endgroup$– alexgbelovCommented Mar 14, 2018 at 23:48
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Nope, every pure strategy equilibrium can be characterized as a degenerate mixed strategy equilibrium.
That is, it is a mixed strategy in which a pure strategy is played with probability $1$.
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2$\begingroup$ I might change "...is just..." to "...can be characterized as...". The current answer sort of muddies the waters, I think. There is a reason we define these as distinctly different equilibrium concepts. $\endgroup$– 123Commented Mar 9, 2018 at 21:33
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$\begingroup$ @123 Good suggestion, the answer has been changed as recommended. $\endgroup$– user11305Commented Mar 9, 2018 at 23:02