This is the solution to problem 7 from Fifty Challenging Problems in Probability by Frederick Mosteller (1987). The problem is paraphrased below; for reference, it is inspired by the original book.
Mr. Brown is playing roulette. He always bets one dollar on 13. If the 38-number wheel lands on it, then he wins $35 and the stake back. His friend bets him $20 that he will be behind after 36 plays.
What is Mr. Brown’s expected net change after 36 plays?
Let’s dive in!
Expected net change without the friend’s bet
Let \(\delta\) be the net change after one play. The probability of winning $35 is \(1/38\), while the probability of losing $1 is \(37/38\). Therefore, \(P(\delta = -1) = 37/38\) and \(P(\delta = 35) = 1/38\). The expected net change after one play is thus \(\mathbb{E}[\delta] = -1 \cdot 37/38 + 35 \cdot 1/38 = -2/38\). After 36 plays, this totals \(\mathbb{E}[\Delta] = 36 \cdot \mathbb{E}[\delta] = -2/38 \cdot 36 \approx -1.895\).
Expected net change including the friend’s bet
What’s the probability of being net-negative after 36 plays, ignoring his friend’s bet? If Mr. Brown loses every time, he stands at -36 dollars. If he wins once, he’s at \(-35 + 35 = 0\). If he wins twice, he’s at \(-34 + 2\cdot 35 =56\). If he wins more often, he’s always net-positive. So the probability of being net-negative is the same as the probability of losing every time: \(P_{\mathrm{net-neg}} = P(\delta = -1)^{36} \approx 0.383\). This is the same as the probability of losing $20 in the bet.
The expected reward from this bet \(B\) is thus \(\mathbb{E}[B] \approx -20 \cdot 0.383 + 20 \cdot 0.617 = 4.68\). If we add this to the expected net change from the base roullete game, we get the total expected change: \(\mathbb{E}[\Delta] + \mathbb{E}[B] \approx -1.895 + 4.68 = 2.785\).
Without the friend’s bet, Mr. Brown would be losing money. Now, he’s gaining money!
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