Doubt regarding the maximin value in this question $A$ 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. 
 A: 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.
