Two-Person Zero-Sum Game/Examples/Abstract 1

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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 \\ \hline A_1 & 5 & -2 & 1 \\ \hline A_2 & 6 & 4 & 2 \\ \hline A_3 & 0 & 7 & -1 \\ \hline \end{array}$

This has an equilibrium point at $\tuple {A_2, B_3}$ with a payoff to $A$ of $2$.


Proof

By inspection, strategy $A_2$ dominates strategy $A_1$.

Therefore by Elimination by Domination, $A_1$ can be eliminated.

$A_1$ 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 $A_3$.


The optimum strategies are therefore $A_2$ and $B_3$, and the payoff to $A$ is $2$.


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.

$\blacksquare$


Sources