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.


Okay, let's try this. For the 3x2 game, there are two obvious pure strategy Nash equilibria, (8,8) and (6,6). Are there any mixed equilibria?

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.


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