Two-Person Zero-Sum Game/Example/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 $B$.

This has an equilibrium point at $\left({A_2, B_3}\right)$.

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

Therefore by Elimination by Domination, $B_4$ can be eliminated.

$B_4$ 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, $\left({A_2, B_3}\right)$ is an equilibrium point.