= Solution
\b[The unique <Nash equilibrium> uses row <probabilities> $(2/7,5/7)$ on $(T,B)$ and column <probabilities> $(4/7,3/7)$ on $(L,R)$. The <value of a zero-sum game> to the row player is $-1/7$; the column player receives $1/7$.]
The first graph solves the row player's max–min <optimization>: choose a <mixed strategy> guaranteeing the greatest possible worst-case <expected payoff>. The second solves the column player's min–max <optimization>: choose a <mixed strategy> placing the smallest upper bound on the row player's <expected payoff>. Their optimal heights agree, illustrating the <minimax theorem>.
There is also a direct <Nash equilibrium> certificate, so the conclusion does not depend on reading coordinates from a plot. At $p^*=2/7$, both opposing pure columns give row <expected payoff> $-1/7$; at $q^*=4/7$, both pure rows give row <expected payoff> $-1/7$. By linearity, every <unilateral deviation>, including an arbitrary <mixed strategy>, leaves the deviator's <expected payoff> unchanged against the fixed opposing equilibrium mixture. Neither player can improve it. This is a <mixed-strategy optimality certificate for a matrix game>.
The equilibrium outcome is a <probability distribution> over cells, rather than one deterministic cell. <Independent> randomization gives
$$
\boxed{\Pr(TL)=\frac8{49},\quad \Pr(TR)=\frac6{49},\quad
\Pr(BL)=\frac{20}{49},\quad \Pr(BR)=\frac{15}{49}.}
$$
The resulting row <expected payoff> is $(16-18-20+15)/49=-1/7$. Finally, each envelope has a unique optimum, so any <Nash equilibrium> must use the displayed <optimal mixed strategies>. In particular there is no <pure strategy> <Nash equilibrium>.
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