In this <consecutive-number antisymmetric game>, with the row player's number denoted $i$, the <payoff matrix> is
$$
a_{ij}=\begin{cases}
0,&i=j,\\
1,&i=j+1,\\
-1,&j=i+1,\\
-2,&i\ge j+2,\\
2,&j\ge i+2.
\end{cases}
$$
It is an <antisymmetric matrix>, so the <matrix game> has value zero for every $n\ge3$. For $n=3$,
$$
A=\begin{pmatrix}0&-1&2\\1&0&-1\\-2&1&0\end{pmatrix},\qquad
p=q=\left(\frac14,\frac12,\frac14\right)
$$
satisfies $Ap=0$ and $p^TA=0$, proving optimality for both players.

For any larger $n$, extend the same strategy by zero probabilities:
$$
\boxed{v=0,\qquad p=q=\left(\frac14,\frac12,\frac14,0,\ldots,0\right).}
$$
Indeed, the first three coordinates of $Ap$ are zero. For $i=4$, $(Ap)_4=-2/4-2/2+1/4=-5/4$; for $i\ge5$, all three supported numbers are at least two below $i$, so $(Ap)_i=-2$. Hence $Ap\le0$ and, by antisymmetry, $p^TA\ge0$. 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.

Solution

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

In this game the higher of two chosen integers wins one unit if they are consecutive, but pays two units if their difference is at least two. Ties pay zero. For any choices $1,\ldots,n$ with $n\ge3$, both players can optimally choose $1,2,3$ with probabilities $1/4,1/2,1/4$ and assign zero probability to larger numbers. If $p$ is this vector, $Ap$ is zero in its first three entries, $-5/4$ in its fourth when present, and $-2$ in all later entries. Thus $Ap\le0$ and $p^TA\ge0$, proving the zero value and both security guarantees.


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 Incoming links: Consecutive-number antisymmetric game