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.
If A says he will always guess the same number \(N\), explain (for each value of \(N\)) how B can maximise his winnings.
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?
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.
Solution:
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.
\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 \(b_1 = 1, b_2 = 0, b_3 = 0\).)
\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.