Why is minmax value equal to maxmin value in two-player game?

Question

In two-player games, a player's minmax value is always equal to his maxmin value.

This proposition is from Essentials of Game Theory. The minmax value is $\min_{s_{-i}}\max_{s_{i}} u_{i}(s_i,s_{-i})$ while the maxmin value is $\max_{s_{i}}\min_{s_{-i}} u_{i}(s_i,s_{-i})$, I believe.

I only understand that $\max\min\le\max\min$ because we have $\min\le \max$ at first and then applying $\max$ and $\min$ on both sides respectively changes nothing. But I really don't know why they are equal in two-player game as mentioned in the book.

(I know that this is true in zero-sum game for von Neumann's theorem. Maybe that proposition is a typo of the book?)

Thanks in advance.

Answer 1

When you consider only Player's $i$ payoff, it is as if they play a zero-sum game with only these payoffs. Player $i$ tried to maximize it, the other player to minimize it and indeed, the minimax=maximin (in mixed actions only. See @Herr K.'s counterexample when only considering pure).

Answer 2

The statement appears to be wrong. For a counterexample, consider player 1's (row player's) payoffs in the following two-player game: \begin{array}{|c|c|c|} \hline &C&D\\\hline A&2,\;\cdot&1,\;\cdot\\\hline B&0,\;\cdot&3,\;\cdot\\\hline \end{array} It can be verified that \begin{equation} \min_{s_2}\max_{s_1}u_1(s_1,s_2)=u_1(A,C)=2 \end{equation} and \begin{equation} \max_{s_1}\min_{s_2}u_1(s_1,s_2)=u_1(A,D)=1. \end{equation} The two values are not equal.