I have a zero-sum game as described in the table below:
- Using minimax = maximin approach I get that the first player should play A with $p^*=\frac{5}{12}$ and B with $p^*=\frac{7}{12}$.
- The second player should then mix A and B with $q^*=\frac{1}{2}$.
- I found the value of the game to be: $v=0.5$.
The question is asking to determine the winner. And since, the game is zero-sum, I know that one player wins and the other loses (what is one player's gain is the other's loss etc.). But how does this apply to this situation? Basically, every time they play the same action, player 2 wins and when they play different actions - player 1 wins. But how to determine the overall winner? I think it is based on the value of the game. My guess is that when the value of the game, v is positive, then the first player is the winner? Is that correct?
$$\begin{array}{c|c|c|c|} & \text{A} & \text{B} & \text{min} \\ \hline \text{A} & -3 & 4 & -3\\ \hline \text{B} & 3 & -2 & -2 \\ \hline \text{max} & 3 & 4 \\ \hline \end{array}$$
There is no such thing as an "overall winner". Value $v=0.5$, means that (when the two players use the specified "optimal" strategies) then the first player's winnings average $0.5$ per play. Because of the random nature of the play, he wins some and loses some. If one player does not use the optimal strategy, then that $0.5$ will change in the other player's favor.