 Solution

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 $p^{\ast }=2/7$, both opposing pure columns give row expected payoff $-1/7$; at $q^{\ast }=4/7$, 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 givesThe 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.