Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 21H / c / Solution

In this consecutive-number antisymmetric game, with the row player's number denoted $i$, the payoff matrix isIt is an antisymmetric matrix, so the matrix game has value zero for every $n\ge 3$. For $n=3$,satisfies $Ap=0$ and $p^{T}A=0$, proving optimality for both players.

For any larger $n$, extend the same strategy by zero probabilities:Indeed, the first three coordinates of $Ap$ are zero. For $i=4$, $(Ap{)}_4=-2/4-2/2+1/4=-5/4$; for $i\ge 5$, all three supported numbers are at least two below $i$, so $(Ap{)}_i=-2$. Hence $Ap\le 0$ and, by antisymmetry, $p^{T}A\ge 0$. The column player using $p$ prevents positive payoff, while the row player using $p$ prevents negative payoff, a mixed-strategy optimality certificate for a matrix game. This explicitly checks all the additional strategies rather than assuming the three-number subgame remains optimal.