Two-Person Zero-Sum Game/Examples/Abstract 3

Example of Two-Person Zero-Sum Game

The two players are $A$ and $B$.

Player $A$ has $m$ strategies: $A_1, A_2, \ldots, A_m$.

Player $B$ has $n$ strategies: $B_1, B_2, \ldots, B_n$.

The game is zero-sum in that a payoff of $p$ to $A$ corresponds to a payoff of $-p$ to $B$.

The following table specifies the payoff to $A$ for each strategy of $A$ and $B$.

		$B$'s strategy
$A$'s strategy		$\begin{array}{r {{\|}} c {{\|}} c {{\|}} c {{\|}} } & B_1 & B_2 & B_3 & B_4 \\ \hline A_1 & -3 & 17 & -5 & 21 \\ \hline A_2 & 7 & 9 & 5 & 7 \\ \hline A_3 & 3 & -7 & 1 & 13 \\ \hline A_4 & 1 & -19 & 3 & 11 \\ \hline \end{array}$

This has an equilibrium point at $\tuple {A_2, B_3}$ with a payoff to $A$ of $5$.

By inspection, strategy $B_1$ dominates strategy $B_4$.

$B_4$ having been eliminated, it is seen by inspection that $A_2$ dominates both $A_3$ and $A_4$.

$A_3$ and $A_4$ having been eliminated, it is seen by inspection that $B_3$ dominates both $B_1$ and $B_2$.

$B_1$ and $B_2$ having been eliminated, it is seen by inspection that $A_2$ dominates $A_1$.

The optimum strategies are therefore $A_2$ and $B_3$, and the payoff to $A$ is $5$.

Given $B_3$, every other strategy adopted by $A$ will result in a smaller payoff to $A$.

Given $A_2$, every other strategy adopted by $B$ will result in a larger payoff to $A$.

Hence, by definition, $\tuple {A_2, B_3}$ is an equilibrium point.

$\blacksquare$

1983: Morton D. Davis: Game Theory (revised ed.) ... (previous) ... (next): $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: Introductory Problems: Figure $2.3$