We can consider only finite games if it makes a difference, but are there nash equilibria that can't be characterized as mixed equilibria?

  • 1
    $\begingroup$ I find this question either unclear or trivial. Please include your exact definitions. $\endgroup$
    – Giskard
    Mar 9 '18 at 20:00
  • $\begingroup$ The example that immediately comes to mind is prisoner's dilemma: the Nash equilibrium is for both players to pick "betray". $\endgroup$
    – alexgbelov
    Mar 14 '18 at 23:48

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 $1$.

  • 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. $\endgroup$
    – 123
    Mar 9 '18 at 21:33
  • $\begingroup$ @123 Good suggestion, the answer has been changed as recommended. $\endgroup$
    – user11305
    Mar 9 '18 at 23:02

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