Problem source

A and B play a guessing game. Each simultaneously names one of the numbers $1,2,3.$ If the numbers differ by 2, whoever guessed the smaller pays the opponent £$2$. If the numbers differ by 1, whoever guessed the larger pays the opponent £$1.$
Otherwise no money changes hands. Many rounds of the game are played. 
\begin{questionparts}
\item If A says he will always guess the same number $N$, explain (for each value of $N$) how B can maximise his winnings. 
\item In an attempt to improve his play, A announces that he will guess each number at random with probability $\frac{1}{3},$ guesses on different rounds being independent. To counter this, B secretly decides
to guess $j$ with probability $b_{j}$ ($j=1,2,3,\, b_{1}+b_{2}+b_{3}=1$), guesses on different rounds being independent. Derive an expression for B's expected winnings on any round. How should the probabilities $b_{j}$ be chosen so as to maximize this expression? 
\item A now announces that he will guess $j$ with probability $a_{j}$ ($j=1,2,3,\, a_{1}+a_{2}+a_{3}=1$). If B guesses $j$ with probability $b_{j}$ ($j=1,2,3,\, b_{1}+b_{2}+b_{3}=1$), obtain an expression for his expected winnings in the form 
\[
Xa_{1}+Ya_{2}+Za_{3}.
\]
Show that he can choose $b_{1},b_{2}$ and $b_{3}$ such that $X,Y$ and $Z$ are all non-negative. Deduce that, whatever values for $a_{j}$ are chosen by A, B can ensure that in the long run he loses no money.
 \end{questionparts}

Solution source

\begin{questionparts}
\item Suppose A always plays $1$, then B should always play $2$ and every time they will win 1.
Suppose A always plays $2$ then B should always play $3$ and every time they will win 1.
If A always plays $3$ then B should always play $1$ and every time they will win 2.

\item 
\begin{array}{cccc}
& b_1 & b_2 & b_3 \\
\frac13 & (0, \frac{b_1}{3}) & (1, \frac{b_2}{3}) & (-2, \frac{b_3}{3}) \\
\frac13 & (-1, \frac{b_1}{3}) & (0, \frac{b_2}{3}) & (1, \frac{b_3}{3}) \\
\frac13 & (2, \frac{b_1}{3}) & (-1, \frac{b_2}{3}) & (0, \frac{b_3}{3}) \\
\end{array}

Therefore the expected value is: $\frac{b_1}{3} - \frac{b_3}{3}$ and to maximise this he should always guess $1$ (ie $b_1 = 1, b_2 = 0, b_3 = 0$.)

\item \begin{array}{cccc}
& b_1 & b_2 & b_3 \\
a_1 & (0, a_1b_1) & (1, a_1b_2) & (-2, a_1b_3) \\
a_2 & (-1, a_2b_1) & (0, a_2b_2) & (1, a_2b_3) \\
a_3 & (2, a_3b_1) & (-1, a_3b_2) & (0, a_3b_3) \\
\end{array}

Therefore the expected value is:

$(b_2-2b_3)a_1 + (b_3-b_1)a_2 + (2b_1-b_2)a_3$

We need $b_2 \geq 2b_3, b_3 \geq b_1, 2b_1 \geq b_2$ so $b_1 \leq b_3 \leq \frac12 b_2 \leq b_1$ so we could take $b_1 = b_3 = \frac12 b_2$ or $b_1 = b_3 = \frac14, b_2 = \frac12$ and all values would be $0$.

Therefore by choosing these values $B$ can guarantee his expected value is $0$ and therefore shouldn't expect to lose money in the long run.
\end{questionparts}