Determine Nash equilibrium

Question

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.

Answer 1

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.

Answer 2

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 $(e,0)$

see battle of the sexes type game