# Game Theory - Mixed strategy Nash equilibria

I have the following games and have to find all of the Nash equilibria (mixed and pure strategies). See games here

The problem I have is:

Taking the first game (3x2) I tried to see if the strategy "c" is strictly dominated by a mixed strategy between "a" and "b", as "c" never is a best response for player 1. However, I saw "c" is not dominated by any mixed strategy between "a"and "b".

Thus, I'm left with a 3x2 game and have to find all of Nash equilibria in mixed strategies.

This is where I'm having issues, how can I find these equilibria? I tried calculating the expected payoffs to see which probabilities would constitute the equilibria but when I try to calculate the expected payoffs for player 2 I'm left with the result r=2p (with the probabilities for strategies a, b and c being p, r and 1-p-r respectively)

If anyone could tell me how to solve this issue I would greatly appreciate it. Thanks.

Start with player 2, the Column player, who chooses X with probability $$r$$ and Y with probability $$(1-r)$$, and player 1, the Row player, choose a with probability $$p$$, b with probability $$q$$, and c with probability $$(1-p-q)$$, so I've slightly changed OP's notation. Column's payoff is $$r[8p+2q+(1-p-q)5]+(1-r)[0p+6q+(1-p-q)5].$$ Taking the derivative with respect to Column's choice variable, $$r$$ yields $$8p-4q.$$ If we set that equal to zero, we get $$p=1/2q$$, but that is the wrong way to look at the problem. What the derivative really says is that if $$p<1/2 q$$ then $$r=0$$, because the derivative is negative and the maximum occurs at the lower endpoint, 0. Similarly, if $$p>1/2 q$$ then $$r=1$$ because the derivative is positive.
So the only hope for a mixed strategy equilibrium is if $$p=1/2 q$$. Will that work? Look at the problem from the viewpoint of player 1, who plays Row. Her expected value is $$pr8 +0 +qr2+q(1-r)6+(1-p-q)5.$$ Take the derivative with respect to $$p$$, simplify, and get $$8r=5$$, or $$r=5/8$$. As before, this says that if $$r<5/8$$ then $$p=0$$. Take the derivative with respect to $$q$$, and a similar approach shows that $$r<1/4$$ implies $$q=1.$$
So, the only places where $$p$$ and $$q$$ are not either zero or one, is $$p=5/8$$ and $$q=1/4.$$ But then we don't have $$p=1/2q$$ and so there is not mixed equilibrium. or so it seems to me.