 Solution

Against the two pure column strategies, player one's expected payoffs are $L_1(p)=5-2p$ and $L_2(p)=2+(b-2)p$. A mixed strategy of the opponent cannot give a smaller expectation than their minimum. The zero-sum game optimization is thereforeThe first line decreases. If $0<b<2$, the second also decreases, and their minimum is maximized at $p=0$ with value $2$. If $b=2$, the second line is constantly $2$ and the first is at least $3$, so every $p$ is optimal. For $2<b\le 3$, the intersection $p=3/b$ lies at or beyond the right endpoint; the minimum increases throughout $[0,1]$ and is maximized at $p=1$ with value $b$. If $b>3$, the intersection lies inside the interval; the minimum increases before it and decreases afterwards. ThusFor the interior case the opponent can equalize row payoffs with first-column probability $(b-2)/b$, confirming the minimax theorem value.