= Solution
In a two-player <zero-sum game>, the row player chooses a row, the column player chooses a column, and the row player's gain is the column player's loss. If their independent <mixed strategies> are probability vectors $x\in\mathbb R^m$, $y\in\mathbb R^n$, the row player's expected payoff is $x^TPy$.
The <optimal mixed strategies> solve
$$
x_*\in\operatorname*{arg\,max}_{x\ge0,\ \sum x_i=1}\min_j(x^TP)_j,\qquad
y_*\in\operatorname*{arg\,min}_{y\ge0,\ \sum y_j=1}\max_i(Py)_i.
$$
The <minimax theorem> equates these two guaranteed values. Equivalently, for a value $v$, $x_*^TP\ge v\mathbf1^T$ and $Py_*\le v\mathbf1$ componentwise.
For a row strategy $(p,1-p)$, the payoffs against the two pure columns are $2-6p$ and $-4+6p$. Their minimum is maximized at their intersection, $p=1/2$, where both are $-1$. Symmetrically the column player's maximum of the two row payoffs is minimized at $q=1/2$. Hence
$$
\boxed{x_*=(1/2,1/2),\qquad y_*=(1/2,1/2),\qquad v=-1.}
$$
These probabilities are unique: moving either probability away from $1/2$ worsens that player's guaranteed payoff.
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