= Solution
An $m\times n$ <matrix game> with payoff matrix $A$ is a two-player <zero-sum game>: player I chooses a row and receives $A_{ij}$, while player II chooses a column and loses the same amount. For <mixed strategy>[mixed strategies] $p$ and $q$, the expected payoff to player I is
$$
p^TAq.
$$
Player I's <optimal mixed strategy> maximizes the payoff guaranteed against every $q$, while player II's minimizes the largest payoff obtainable by any $p$. Thus
$$
\max_p\min_qp^TAq
=v
=\min_q\max_pp^TAq
$$
by the <minimax theorem>. Equivalently, optimal strategies satisfy
$$
p^TA\geq v\mathbf1^T,
\qquad
Aq\leq v\mathbf1.
$$
Here $A^T=-A$, so this is an <antisymmetric zero-sum game>. For every probability vector $r$,
$$
r^TAr=0.
$$
It follows that the row player's guaranteed payoff cannot exceed zero and the column player's worst loss cannot be below zero. Minimax therefore gives
$$
\boxed{v=0}.
$$
If $p$ is optimal for player I, then
$$
p^TA\geq0.
$$
Transposing and using $A^T=-A$ gives
$$
Ap\leq0,
$$
which is precisely the optimality condition for player II. Thus every optimal strategy for player I is also optimal for player II.
The condition $Ap\leq0$ explicitly reads
$$
\begin{aligned}
p_2+p_3-4p_4&\leq0,\\
-p_1+2p_3+2p_4&\leq0,\\
-p_1-2p_2+3p_4&\leq0,\\
4p_1-2p_2-3p_3&\leq0.
\end{aligned}
$$
The probability vector
$$
\boxed{p=\frac17(2,4,0,1)^T}
$$
satisfies
$$
Ap=(0,0,-1,0)^T\leq0,
$$
so it is optimal for both players.
To prove uniqueness, let $q$ be any optimal strategy. Since $p$ is optimal for player II and $q$ is optimal for player I,
$$
0\leq q^TAp=-q_3,
$$
and hence $q_3=0$. Writing $q=(a,b,0,d)^T$, the first, second, and fourth inequalities in $Aq\leq0$ give
$$
b\leq4d,\qquad a\geq2d,\qquad b\geq2a.
$$
Consequently
$$
4d\geq b\geq2a\geq4d,
$$
so equality holds throughout: $a=2d$ and $b=4d$. The normalization $a+b+d=1$ yields $d=1/7$. Therefore the displayed $p$ is the unique optimal strategy.
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