When we get a new equilibrium by crossing strategies of two other equilibria?

Question

Let consider a 2-player, zero-sum game.

Are there (quite) general conditions under which "if $(a, b)$ and $(a', b')$ are Nash equilibria, then even $(a, b')$ and $(a', b)$ are Nash equilibria"?

Thanks in advance!

Answer 1

Equilibria in a zero-sum game are always interchangeable in this way.

All equilibria in a zero-sum game have the same payoff $v$ (to player 1, $-v$ to player 2), called the value of the game. In equilibrium, any security strategy of player 1 that ensures that he gets payoff at least $v$ can be played against any security strategy of player 2 that ensures she pays no more than $v$.

To prove this, for example, you can first prove that the payoffs of $(a,b)$, $(a',b')$, $(a',b)$, and $(a,b')$ are all the same, and then use this fact to show that the equilibrium condition holds for $(a',b)$ (and symmetrically $(a,b')$).