To encourage John’s promising chess career, his coach offers him a prize if he wins (at least) two games in a row in a three-game series to be played with his coach and Garry Kasparov alternately: coach-kasparov-coach or kasparov-coach-kasparov, according to John’s choice. Kasparov is (evidently) a better player than John’s coach. Which series should John choose? 1
Let \(p_C\) be the probability of winning the coach and \(p_K\) the probability of winning Kasparov, where \(p_C < p_K\). The winning outcomes for the first scenario (C-K-C) are:
| outcome | probability |
|---|---|
| $C_0K_1C_1$ | $p_C~p_K~(1-p_C)$ |
| $C_1K_1C_0$ | $p_C~p_K~(1-p_C)$ |
| $C_1K_1C_1$ | $p_C~p_K~p_C$ |
And thus $p_C~p_K~(1-p_C) + p_C~p_K~(1-p_C) + p_C~p_K~p_C = p_K~p_C~(2 - p_C)$. For the second scenario (K-C-K):
| outcome | probability |
|---|---|
| $K_0C_1K_1$ | $p_K~p_C~(1-p_K)$ |
| $K_1C_1K_0$ | $p_K~p_C~(1-p_K)$ |
| $K_1C_1K_1$ | $p_K~p_C~p_K$ |
Where $p_K~p_C~(1-p_K) + p_K~p_C~(1-p_K) + p_K~p_K~p_C = p_K~p_C~(2 - p_K)$. Hence:
\[\begin{align*} p_K~p_C~(2 - p_K) & \stackrel{?}{=} p_K~p_C~(2 - p_K) \\ 2 - p_C & \stackrel{?}{=} 2 - p_K \\ p_C & < p_K \end{align*}\]-
This is adapted from problem 2 of Frederick Mosteller’s “Fifty Challenging Problems in Probability”. ↩︎