Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 19H (Optimisation)

For probability vectors $p,q$, they are optimal with value $v$ exactly whenand $p^{T}Aq=v$. These are the mutual best-response and security-level conditions.

For every probability vector $x$, antisymmetry gives $x^{T}Ax=0$. Applying the minimax inequalities with the same strategy on both sides shows the lower value is at most zero and the upper value at least zero; transposition also changes the value to its negative. Hence the value is zero.

Starting from any optimal strategy, transfer all probability on row $i_0$ to row $i_1$. Every payoff against a pure column weakly increases, so the security level remains optimal. The resulting strategy has $p_{i_0}=0$.

The matrix is antisymmetric, so the value is zero. For a common candidate $p=(a,b,c,d{)}^T$, the column-player inequalities are $Ap\le 0$:The first three imply $a+d\le b\le c\le a-2d$, hence $d=0$ and $a=b=c$. Normalization givesIndeed $Ap=(0,0,0,-1{)}^T/3\le 0$, and antisymmetry gives the corresponding row inequalities.