= Solution
Put $p=\Pr(T)$ and $q=\Pr(L)$, so the complementary <probabilities> are $1-p$ and $1-q$. All four expressions are the row player's <expected payoff>, including those indexed by a column action. Direct <expected value> calculations give
$$
\boxed{\begin{aligned}
\pi_1(p,L)&=2p-(1-p)=3p-1,\\
\pi_1(p,R)&=-3p+(1-p)=1-4p,\\
\pi_1(T,q)&=2q-3(1-q)=5q-3,\\
\pi_1(B,q)&=-q+(1-q)=1-2q.
\end{aligned}}
$$
The column player's <expected payoff> is the negative of the row player's because this is a <zero-sum game>.
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