Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 42 / 4 / b

Use the PDF's entry $P_{12}=4$, rather than the TeX aid's $2$. To check nondegeneracy of a bimatrix game, every opponent strategy of support size $k$ must have at most $k$ pure best responses. Against each pure column, the unique best rows are respectively $3,1,2$. Against each pure row, the unique best columns are respectively $1,2,2$.

The only mixture making all three rows indifferent solves $-y_1+2y_3=0$ and $-2y_1+2y_2-3y_3=0$, giving $y=(4,7,2)/13$, which has support size three. Thus no two-column-support mixture has three best rows. For the other player, column $1$ strictly dominates column $3$, because their payoff difference is $(1,1,2)$, so three best columns are impossible. These observations cover all support sizes, proving the game is nondegenerate.

For the Lemke-Howson algorithm, use unnormalized strategy vectors $u,v$ in the polytopesA row label $i\in \{1,2,3\}$ occurs at $u_i=0$ in $\mathcal{X}$ or at $(Pv{)}_i=1$ in $\mathcal{Y}$. A column label $3+j$ occurs at $(Q^{T}u{)}_j=1$ in $\mathcal{X}$ or at $v_j=0$ in $\mathcal{Y}$. The origin pair has all six labels.

Drop row label $3$ by increasing $u_3$. Column $2$ is the first binding payoff constraint, at $u_3=1/4$, so label $5$ becomes duplicated. Drop $5$ in $\mathcal{Y}$ by increasing $v_2$; row $1$ binds first, at $v_2=1/4$, duplicating label $1$. Drop $1$ in $\mathcal{X}$ by increasing $u_1$ while keeping $u_2=0$ and the column-$2$ constraint binding. Since that constraint does not involve $u_1$, $u_3$ stays $1/4$. Column $1$ binds at $u_1=1/4$, duplicating label $4$.

Now drop $4$ in $\mathcal{Y}$ by increasing $v_1$. Keep $v_3=0$ and row $1$ binding, so $v_2=1/4$. Row $3$ reaches payoff $1$ at $v_1=1/6$, earlier than row $2$, which would bind at $v_1=1/4$. The missing label $3$ returns, terminating the path. The successive vertex-label pairs areThe final pair is completely labelled and nonzero. Normalize each vector by its own sum to obtainThe row payoffs against $y$ are $(12/5,2,12/5)$ and the column payoffs against $x$ are $(2,2,1/2)$. Both supported actions are best responses, so this is a Nash equilibrium, with payoffs $\boxed{(12/5,2)}$.