Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 18H / c

Every payoff is positive, so every pair of mixed strategies has positive expected payoff and the game value satisfies $v>0$. Player 1 maximizes $v$ subject to $A^{T}p\ge ve$, $e^{T}p=1$, $p\ge 0$. Setting $x=p/v$ turns these intoMaximizing $v$ is therefore equivalent to the stated minimization of $e^{T}x$.