$A$ 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 $3/4$ 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 $2$.
What value of $p$ maximizes the minimum possible value of $B$’s expected gain, and what is this maximin value? (Note that $B$’s expected gain depends not only on $p$, but also on what $A$ does.)
Consider now player $A$. Suppose that she also randomizes her decision, writing down number $1$ with probability $q$. What is $A$’s expected loss if
$(c)$ $B$ chooses number $1$?
$(d)$ $B$ chooses number $2$?
What value of $q$ minimizes $A$’s maximum expected loss? Show that the minimum of $A$’s maximum expected loss is equal to the maximum of $B$’s minimum expected gain.
My attempt (just the results)
$(a)$: $\frac{1}{4}(7p-3)$.
$(b)$: $2-\frac{11}{4}p$.
$(c)$: $\frac{1}{4}(3-7q)$.
$(d)$: $2-\frac{5}{4}q$.
I don't understand the what the "maximize minimum" and "minimize maximum" part of the question means.
Maximizing the minimum possible value of B's expected gain means that A either chooses 1 or 2, and you take the one that is less favorable to B (minimum expected gain). Then you choose the p for which this minimum will be of maximal value.
Minimax means that both players act perfectly rationally, so your opponent always makes the best move (min for you), and you need to choose your best move accordingly (max for you).
I think the last question is about finding the Nash equilibrium.
(I wanted to comment but I didn't have 50 reputation.)