Does the value of a pure strategy Nash equilibrium(if exists) equal the value of the mix strategy Nash equilibrium in two-person zero-sum game?

Given a two-person zero-sum game, a mixed strategy Nash equilibrium always exists and all such equilibria have the same value. A pure strategy Nash equilibrium, however, may not exist.

My question is: suppose that a pure strategy Nash equilibrium exists, does the value of it equal the value of the mixed strategy Nash equilibrium?

Pure strategy Nash equilibria are a subset of mixed strategy Nash equilibria, so as long as your statement

a mixed strategy Nash equilibrium always exists and all such equilibria have the same value

is true, all pure strategy Nash equilibria will have the same value as well.

• It seems that I was asking a dumb question =_= Jun 9, 2020 at 7:32

Just to be careful, the above equivalence need not hold in general games. Consider the following game (which is a modified matching pennies):

$$\begin{array}{c|c|c|c|} & \text{H} & \text{T} & \text{P}\\ \hline \text{H} & (1,-1) & (-1,1) & (-2,-2) \\ \hline \text{T} & (-1,1) & (1,-1) & (-2,-2) \\ \hline \text{P} & (-2,-2) & (-2,-2) & (-2,-2) \\ \hline \end{array}$$

This game has a pure strategy nash equilibrium $$(P,P)$$ which yields -2 to each player.

However, a mixed strategy of the game involves $$\big\{\frac{1}{2}\circ H; \frac{1}{2}\circ T; 0 \circ P\big\}$$ for each player (i.e. each player randomizes uniformly between H and T). This mixed strategy yields a value of 0 to each player. The issue arises because there may be Nash Equilibria in (weakly) dominated strategies.