= Solution
With rows and columns ordered Head, Tail, the <payoff matrix> of ordered pairs is
$$
\begin{array}{c|cc}
& H&T\\\hline
H&(-2,2)&(X,-X)\\
T&(Y,-Y)&(-2,2)
\end{array}.
$$
This is a <zero-sum game>. There is \b[no pure <Nash equilibrium>]: when the coins match, player 1 gains by switching to a mismatch; when they do not match, player 2 gains by making them match. Since $X,Y>0$, these improvements are strict.
Let $p$ and $q$ be the respective probabilities that players 1 and 2 choose Head. Player 1's payoffs from Head and Tail are $X-(X+2)q$ and $(Y+2)q-2$. Player 2 is indifferent when the corresponding player-1 payoffs against its two pure actions, $Y-(Y+2)p$ and $(X+2)p-2$, agree. The resulting unique <fully mixed Nash equilibrium> is
$$
\boxed{p=\frac{Y+2}{X+Y+4},\qquad q=\frac{X+2}{X+Y+4}.}
$$
Both numbers lie strictly between zero and one. If either player instead used a pure action, the other's unique <best response> would force the pure-action cycle above, so there are no additional partly mixed equilibria. Substitution gives player 1's <expected payoff> $(XY-4)/(X+Y+4)$ and its negative for player 2.
For $X=1,Y=3$, the <best response correspondences> are
$$
BR_1(q)=\begin{cases}\{1\},&q<3/8,\\{[0,1]},&q=3/8,\\\{0\},&q>3/8,\end{cases}\qquad
BR_2(p)=\begin{cases}\{0\},&p<5/8,\\{[0,1]},&p=5/8,\\\{1\},&p>5/8.\end{cases}
$$
Their intersection is \b[$(p,q)=(5/8,3/8)$], with <expected payoffs> $(-1/8,1/8)$.
\Image[/past-exam-of-the-computer-science-course-of-the-university-of-oxford/2018/computational-game-theory-best-responses.png]
{title=The two <best response correspondences> intersect at the unique mixed equilibrium}
{description=The horizontal axis is player 1's Head probability $p$, and the vertical axis is player 2's Head probability $q$. Whole intervals at indifference are included.}
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