Exercise 6.3.6

In an antisymmetric payoff matrix, its diagonal elements are all zero, and for all entries, it has $a_{\mathit{ij}}=-a_{\mathit{ji}}$. For $X$, he’s going to look at each row, and find the maximum payoff from each row (his worst payoff) and choose the minimum among them. Suppose for a row $i$, we have its maximum of $S_i=\max <!--\; FUNCTION\; APPLICATION\; -->(a_{i,:})$. So we have maximums from each row of $S_1,S_2,\dots ,S_n$, assume the minimum of them is $S_m=\min <!--\; FUNCTION\; APPLICATION\; -->(S_1,S_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,S_n)$, that means $X$ will pick row $m$ as his optimal row.

Now we look at $Y$. For $Y$, he needs to look at the worst payoff (minimum payoff) on each column, for a column $a_{:,j}$, since it’s antisymmetric matrix, we know that $a_{:,j}=-a_{j,:}$, so $\min <!--\; FUNCTION\; APPLICATION\; -->(a_{:,j})=\min <!--\; FUNCTION\; APPLICATION\; -->(-a_{j,:})=-\max <!--\; FUNCTION\; APPLICATION\; -->(a_{j,:})=-S_j$. So for each column, $Y$ gets minimums of $-S_1,-S_2,\dots ,-S_n$. Then $Y$ needs to look at the maximum of them, which is $\max <!--\; FUNCTION\; APPLICATION\; -->(-S_1,-S_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,-S_n)=-\min <!--\; FUNCTION\; APPLICATION\; -->(S_1,S_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,S_n)=-S_m$.

We see that both $X$ and $Y$ are going to pick the row and column with the same index. So their fair payoff will always be on the diagonal, which is always 0.