Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 212 / 4 / Solution

Let $u,v$ be the row and column mixed strategies, so their entries are nonnegative and sum to one. The payoffs are $u^{\mathsf{T}}Pv$ and $u^{\mathsf{T}}{P}^{\mathsf{T}}v$. A Nash equilibrium is a pair in which each strategy is a best response to the other: neither player can improve its expected payoff by a unilateral change of mixed strategy. Equivalently, every positive-probability pure strategy attains that player's maximum pure-strategy payoff against the opposing mixture.

For the complementarity construction of a symmetric Nash equilibrium put $s={\sum }_i{x}_i>0$ and $u=x/s$. The equation $Px+z=1$ gives $Px\le 1$. Nonnegativity and $x^{\mathsf{T}}z=0$ imply $x_i>0\Rightarrow z_i=0\Rightarrow (Px{)}_i=1$. Therefore every used pure strategy has payoff $1/s$ against $u$, while all other pure strategies have payoff at most $1/s$. The same vector of pure-strategy payoffs applies to the column player in this symmetric bimatrix game. Thus a symmetric Nash equilibrium is

For a genuine Lemke-Howson algorithm path, use two separate vectors, $\xi$ for the row player and $\eta$ for the column player, with slacksGive labels $i=1,2,3$ to ${\xi }_i=0$ or $r_i=0$, and labels $3+j$ to ${\eta }_j=0$ or $s_j=0$. A nonzero completely labeled pair has ${\xi }_i{r}_i=0$ and ${\eta }_j{s}_j=0$, and normalization yields mutual best responses. At the artificial pair $(\xi ,\eta )=(0,0)$ all six labels are present.

Drop label $3$ by increasing ${\xi }_3$. The two limiting rows of $P\xi \le 1$ are $3{\xi }_3\le 1$ and $2{\xi }_3\le 1$, so the minimum-ratio pivot stops at ${\xi }_3=1/3$, where $s_2=0$. Label $5$ is now duplicated, since ${\eta }_2=0$ also carries it. Increase ${\eta }_2$ to remove that duplicate. The limiting inequalities are $3{\eta }_2\le 1$ and $4{\eta }_2\le 1$, so the next pivot stops at ${\eta }_2=1/4$, where $r_3=0$ restores the dropped label $3$. The endpoint isIt has all six labels and satisfies both complementarity conditions. After normalization, the Lemke-Howson algorithm returnsAgainst column strategy $2$, row strategy $3$ pays $4$, exceeding $3$ and $0$. Against row strategy $3$, column strategy $2$ pays $3$, exceeding $0$ and $2$. This verifies the endpoint directly.

To find every other Nash equilibrium, observe that row strategy $3$ strictly dominates row strategy $1$: the payoff differences against the three columns are $(2,1,2)$, all positive. By symmetry, column strategy $1$ is also strictly dominated. Neither can appear in an equilibrium. The reduced strategies $2,3$ have row payoff matrixThe two pure Nash equilibria are $(e_2,e_3)$ and $(e_3,e_2)$. If the opponent uses strategy $2$ with probability $q$, the payoffs of strategies $2,3$ are $3(1-q)$ and $2+2q$. Indifference requires $q=1/5$. The same computation applies to the other player. A player mixing both strategies forces this exact opposing mixture; a player playing a pure strategy has a unique opposing best response, producing one of the two pure equilibria. Hence there are exactly three Nash equilibria:For the symmetric mixed equilibrium, the original complementarity construction can use $x=(0,1/12,1/3)$ and $z=(3/4,0,0)$; it gives payoff $12/5$ to each player after normalization.