= Solution
Strictly positive entries imply $v>0$. If $q$ is an optimal strategy, set $x=q/v$. Then $Ax\le\mathbf1$ and $\mathbf1^\mathsf Tx=1/v$. Conversely, a nonzero feasible $x$ with $s=\mathbf1^\mathsf Tx$ gives a strategy $q=x/s$ satisfying $\max_i(Aq)_i\le1/s$. Thus $v\le1/s$, or $s\le1/v$. The optimal objective is therefore $s_*=1/v$, and
$$
\boxed{v=1/s_*,\qquad q_*=x_*/s_*.}
$$
A nonzero feasible point exists by taking a sufficiently small positive multiple of a probability vector, and positivity of $A$ bounds the feasible objective, so the optimum is attained.
For general entries choose a constant $c$ so that $A'=A+c\mathbf1\mathbf1^\mathsf T$ is strictly positive. Every pair of strategies has its payoff increased by exactly $c$, so optimal strategies are unchanged. Solve the positive problem for $A'$, recover $v'=1/s_*'$, and then $\boxed{v=v'-c}$ for the original game.
Back to article page
