# Game Theory: Is there a Mixed Strategy Nash Equilibrium?

The game looks like this:

         a       b
A [(-12, 1) (8, 8)]
B [(15, 1), (8,-1)]


(15, 1) and (8,8) are Nash Equlibria. However, could you still mix between (8,8) and (15,1)? For example, for P2 (column player) to make P1 indifferent he could play b with a probability of 1. And player 1 could make player 2 indifferent with another probability mix.

However, can mixed strategies include weakly dominated strategies?

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In the usual formulation of the definition, yes. There are other equilibrium concepts in which they cannot, then $(B,b)$ is the unique equilibrium. –  Will Sawin Feb 24 '12 at 2:21
(B, b) can never be an equilibrium... –  user21641 Feb 25 '12 at 22:27

The answer given by Nixon is correct. If you label p the probability that player 1 chooses A, and q the probability that player 2 chooses a, then you have: $E(a)=E(b)$, so $1=8p-(1-p)$, that is $p=2/9$. On the other hand, you can see that $q=0$.