Two-Person Zero-Sum Game/Examples/Abstract 4
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 | $\begin{array}{r {{|}} c {{|}} c {{|}} c {{|}} }
& B_1 & B_2 & B_3 \\ \hline A_1 & 2 & -5 & -2 \\ \hline A_2 & 3 & -1 & -1 \\ \hline A_3 & -3 & 4 & -4 \\ \hline \end{array}$ |
This has an equilibrium point at $\tuple {A_2, B_3}$ with a payoff to $A$ of $-1$.
Proof
By inspection, strategy $B_3$ dominates strategy $B_1$.
Therefore by Elimination by Domination, $B_1$ can be eliminated.
$B_1$ having been eliminated, it is seen by inspection that $A_2$ dominates $A_1$.
$A_1$ having been eliminated, it is seen by inspection that $B_3$ dominates $B_2$.
$B_2$ 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 $-1$.
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
- 1983: Morton D. Davis: Game Theory (revised ed.) ... (previous) ... (next): $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: Introductory Problems: Figure $2.4$