= Solution
Let $p$ be the row player's <mixed strategy> and $q$ the column player's <mixed strategy>. The fourth row is strictly dominated by the first. A useful candidate mixes the first two rows so that their first and third column payoffs agree. If the first row has probability $t$, these payoffs are $7t-2$ and $2-7t$, giving $t=2/7$ and common value zero.
To prove that these are <optimal mixed strategies>, use the two explicit <mixed strategies>
$$
\boxed{p=(2/7,5/7,0,0)^T,\qquad q=(1/2,0,1/2)^T,\qquad v=0.}
$$
Their payoff vectors satisfy
$$
p^TA=(0,11/7,0),\qquad Aq=(0,0,-1/2,-1)^T.
$$
Thus the row player guarantees a payoff at least zero against every column <mixed strategy>, and the column player guarantees a payoff at most zero against every row <mixed strategy>. This matching pair proves that \b[the value of the zero-sum game is zero].
These <optimal mixed strategies> are also unique. Since $Aq$ is strictly negative in rows three and four, any optimal row strategy gives them probability zero. Against columns one and three it must then satisfy both $7t-2\geq0$ and $2-7t\geq0$, so $t=2/7$. Similarly, $p^TA$ is strictly positive in column two, forcing an optimal column strategy to avoid that column; its constraints against rows one and two then force equal weights on columns one and three.
Back to article page
