$A$ and $B$ play the following game: $A$ writes down either number $1$ or number $2$, and $B$ must guess which one. If the number that $A$ has written down is $i$ and $B$ has guessed correctly, $B$ receives $i$ units from $A$. If $B$ makes a wrong guess, $B$ pays $\frac{3}{4}$ unit to $A$. If $B$ randomizes his decision by guessing $1$ with probability $p$ and $2$ with probability $1-p$, determine his expected gain if
$(a)$ $A$ has written down number $1$
$(b)$ $A$ has written down number $2$
$(c)$ What value of $p$ maximizes the minimum possible value of $B$'s expected gain, and what is this maximin value?
Now as far as $(a)$ and $(b)$ go, I am quite confident that the answers are $\frac{7}{4}p-\frac3{4}$ and $-\frac{11}{4}p+2$ respectively.
It is $(c)$ that troubles me. I assumed $A$ is equally likely to write $1$ or $2$. Therefore, if a Random Variable $X$ represents the gains of $B$, the expectation of $X$ can be given by: $$E[X]=1\bigg(\frac{1}{2}\cdot p\bigg)-\frac3{4}\bigg(\frac{1}{2}\cdot (1-p)\bigg)-\frac3{4}\bigg(\frac{1}{2}\cdot p\bigg)+2\bigg(\frac{1}{2}\cdot (1-p)\bigg)=-\frac{p}{2}+\frac5{8}$$
Now, the minimum possible value of $B$'s expected gain would be $\frac{5}{8}$, which is attained when $p=0$.
But this approach is apparently incorrect as per the solution manual. The way it has been solved in the manual is by equating the answers of $(a)$ and $(b)$ and solving for $p=\frac{11}{18}$.
So what exactly is the question asking here in $(c)$? Why does equating the individual expected gains give the probability that we're after?
If someone could clear things up here, I'd really appreciate it.
The easiest way to see what's happening is to look at the graphs of the two functions $\ v = \frac{7}{4}p-\frac{3}{4}\ $, shown in the graph below as line $\mathrm{L1}$, and $\ v = 2-\frac{11}{4}p\ $, shown as line $\mathrm{L2}$, which intersect at the point $\ \left(\frac{11}{18}, \frac{23}{72}\right)\ $, labelled $\mathrm{F}$ in the diagram.
The graph of the function $\ v=\min\left(\frac{7}{4}p-\frac{3}{4},2-\frac{11}{4}p\right)\ $ is shown by the red line. When $\ p\le \frac{11}{18}\ $—that is, to the left of the point $\mathrm{F}$—the first of the two functions is smaller, and therefore coincides with the minimum of the two, whereas the second of the two functions does this when $\ p > \frac{11}{18}\ $—to the right of the point $\mathrm{F}$.
It's clear from the graph, that the maximum of the function represented by the red line, $\ v=\min\left(\frac{7}{4}p-\frac{3}{4},2-\frac{11}{4}p\right)\ $, occurs at the point of intersection $\mathrm{F}$, where $\ p= \frac{11}{18}\ $
The problem with your analysis is the assumption that A will choose the two numbers $1$ and $2$ with equal probability. If he did do that, then B would indeed be able to increase his expected payoff to $\ \frac{5}{8}\ $. However, if A chooses to write $1$ with probability $\ \frac{11}{18}\ $, and $2$ with probability $\frac{7}{18}\ $, then B's expected payoff remains $\ \frac{23}{72}\ $, no matter what he does.