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I've been trying to find all the Nash equilibria for a 2-player game with 2 actions either, where the payoffs were dependent of variables. I'm finished, except for one special case (I had to divide by something and pretend it's not zero). Now I want to consider that case.

If I do, I get a game where the outcomes for both players and both strategies is just 0.

Does that mean that there infinitely many Nash equilibria? So if I called my strategies $p^\ast = (p,1-p)$ and $q^\ast = (q,1-q)$, will there be a Nash equilibrium for all $(p,q) \in [0,1]^2$? Wouldn't that also mean that all pure strategies would be pure Nash equilibria?

Thanks in advance for the answer.

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If each player has the same payoffs in the four cells of a $2 \times 2$ game, any strategy profile (pure or mixed) is a Nash equilibrium. The answer to each of your questions is yes.

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