# Are there Nash Equilibria that aren't mixed strategies?

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

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

## 1 Answer

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 $1$.

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