# Determine Nash equilibrium

How can I determine the nash equilibria in the following matrix? $$\begin{pmatrix}-\pi,-\pi & e,0 \\ 0,e & -\pi,-\pi \end{pmatrix}$$

I know the definition of a Nash equilibrium, but because of the $-\pi$, it will always be the smallest value, so there wouldn't exist a Nash equilibrium....

I think I can still try something with minimax theorem but I dont know how to use it here. I also never seen the double value per matrixposition before... How do I interpret this? And how can I find the nash equilibria here?

Can someone help me here? Thanks in advance!

Edit: So my question is: how to find the nash equilbrium with mixed strategies.. (0,e) and (e,0) is obvious but I dont know how to find the mixed strategy nash equilibrium, thats why I mentioned minimax sorry for the unclear question.

Suppose that in the mixed strategy equilibrium player 1 is choosing the first option with probability $p$. It is a fact about Nash equilibria that player 2 is then indifferent between which of the two options she plays. For if she was not then she would only play the better strategy. But then player 1 would use the best response to that strategy, contradicting the fact that we were in a mixed strategy equilibrium. (This sort of reasoning is standard in game theory and I assume you are familiar with it.)

Thus $-p\pi+(1-p)e= p\cdot 0 -(1-p)\pi$. This implies that in equilibrium player 1 plays the first option with probability $\frac{e+\pi}{e+2\pi}$. The game is symmetric so player 2 does likewise.

• $\frac{e+\pi}{e+2\pi}$ is then the intersection of the two graphs of $-p\pi+(1-p)e$ and $(p-1)\pi$? So this means; we have 3 nash equilibria. Quite nice since it's just a 2x2 matrix I think, or is this normal? I'm just doing some exercises from the net to practise with, dont actually know a lot from the subject. – Bob Nov 19 '12 at 16:27
• Yes, it is the intersection of the graphs. I would say it is fairly common to have three equilibria in 2x2 games. – Johan Nov 20 '12 at 15:32
• there are a couple of theorems saying that in games with finite number of players and finite number of strategies, the number of nash equilibria is generically odd. – Cristian Nov 20 '12 at 19:25

Unless I'm wrong, the first value should be the payoff to player one and the second value should be the payoff to player two. We need to find where they are both best responding.

If player 1 plays the first option, player two's best response is the second. ($0>-\pi$) If player 1 plays the second option, player two's best response is the first. ($e>-\pi$) If player 2 plays the first options, player one's best response is the second. ($0>-\pi$) If player 2 plays the second option, player one's best response is the first. ($e>-\pi$)

Therefore there are two nash equilibrium at $(0,e$) and $(e,0)$

• Welcome! Please try to use LaTeX for the math in your posts. – Martin Argerami Nov 19 '12 at 5:47
• Ok, clear, but what I actually ment was how to find the nash equilbrium as in; with mixed strategies.. (0,e) and (e,0) is obvious but I dont know how to find the mixed strategy nash equilibrium, thats why I mentioned minimax, but sorry for the unclear question then... Do you know how to do this or should I start a new topic? – Bob Nov 19 '12 at 15:11