= Solution
Player A maximizes the <payoff>, while B minimizes it. The fourth row is strictly worse for A than the first in every column, so remove $A_4$ by <dominated strategy elimination>. On the remaining rows, column $B_3$ is strictly better for B than $B_2$, so remove $B_2$. Now $A_3$ is strictly worse than $A_2$ in the two remaining columns. The reduced <zero-sum game> has <payoff matrix>
$$
\begin{pmatrix}4&-5\\-2&3\end{pmatrix},
$$
with rows $A_1,A_2$ and columns $B_1,B_3$.
If A uses <mixed strategy> $(p,1-p)$, the expected <payoffs> against these columns are $-2+6p$ and $3-8p$. The first increases and the second decreases, so their minimum is maximized where they are equal: $14p=5$. Thus
$$
\boxed{A:\ (5/14,9/14,0,0),\qquad v=1/7.}
$$
For an optimality certificate, B uses <mixed strategy> $(q,0,1-q)$, making A's first two row <payoffs> $-5+9q$ and $3-5q$. Equality gives $q=4/7$. Against this B strategy the four row <payoffs> are $(1/7,1/7,-6/7,-6/7)$, while against the displayed A strategy the three column <payoffs> are $(1/7,13/7,1/7)$. Therefore A guarantees $1/7$ and B holds A to $1/7$, verifying the <value of a zero-sum game> and the original, unreduced <mixed strategy>.
Back to article page
