Exercise 6.3.5

• Payoff matrix 1: the optimal strategy for $X$ to pay the minimial value is to choose row 1. The optimal strategy for $Y$ to get paid max is to choose column 2. So the payoff from $X$ to $Y$ is 2.
• Payoff matrix 2:

If we follow the strategy of the previous payoff matrix. For $X$, on each row, we look at the worst payoff (i.e. maximum payoff), which are 4 and 8 respectively. To optimize $X$’s worst payoff, he will choose row 1 to minimize the maximum payoff, which is 4 here.

Similary, for $Y$, on each row, he’ll look at the worst payoff (i.e. minimum payoff), which are 1 and 2 for column 1 and column 2. To optimize $Y$’s worst payoff, he will choose column 2 to maximize the minimum payoff,which is 2 here.

We see that the optimal payoff for $X$ and $Y$ are different in this case, which is not in the first payoff matrix. So it indicates $X$ can’t always choose row 1, and $Y$ can’t always choose column 2. They need a probability distribution to choose different rows/columns each time.

Suppose $X$ choose the two rows with probabilities $x_1\ge 0,x_2\ge 0$, and $x_1+x_2=1$, then we have a mixed strategy with new row: $x_1+8x_2,4x_1+2x_2$, $X$ will choose $x_1,x_2$ to make the worst payoff as small as possible, i.e. the largest payoff. This will happen when the two largest payoffs are equal: $x_1+8x_2=4x_1+2x_2$, solve we have $x_1=\frac{2}{3},x_2=\frac{1}{3}$. Similarly, we have $y_1=\frac{2}{9},y_2=\frac{7}{9}$. So we see that the optimal payoff is $\frac{10}{3}$, notice this is smaller than 4, if $X$ chooses row 1 and $Y$ chooses column 2.