= 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 therefore
$$
\boxed{\max_{0\le p\le1}\min\{5-2p,\ 2+(b-2)p\}}.
$$
The 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 $2<b\le3$, 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. Thus
$$
\boxed{\begin{array}{c|c|c}
b&\text{optimal }p&\text{game value}\\\hline
0<b<2&0&2\\
b=2&\text{any }p\in[0,1]&2\\
2<b\le3&1&b\\
b>3&3/b&5-6/b
\end{array}}.
$$
For the interior case the opponent can equalize row payoffs with first-column probability $(b-2)/b$, confirming the <minimax theorem> value.
Back to article page
