Problem:
Prove that if in a matrix game 2x2 if the player 1 has a pure optimal strategy, so has player 2
Attempt:
Given:
We know that player 1 has pure optimal strategy, meaning: $$P(x, \overline{y}) \leq P(\overline{x}, \overline{y}) \leq P(\overline{x}, y) $$ Do we actually know that?
I thought about expressing $P(x,y)=-Q(x,y)$ but nowhere it's said that it's an antagonistic game, but then I found this theorem:
If we knew that there is a saddle point, we would know that, there is a price and that: $$P(x, \overline{y}) \leq v \leq P(\overline{x},y)$$ Meaning that there is dual solution:
$$\begin{cases} v -> min\\ P(i,y) \leq v \\ y \in Y \end{cases} $$
$$\begin{cases} v -> max\\ P(x,j) \geq v \\ x \in X \end{cases} $$
Question:
The Problmem is ... do we actually know that we have antagonistic game or a seddle point, just by knowing that Player 1 has pure strategy? How can we prove this ?