# Expected return on a roulette bet

Consider the following problem, from Henk Tijms's Understanding Probability:

In European roulette the ball lands on one of the numbers 0, 1, . . . , 36 in every spin of the wheel. A gambler offers at even odds the bet that the house number 0 will come up once in every 25 spins of the wheel. What is the gambler’s expected profit per dollar bet?

Call $X$ the event that the house number shows up once in 25 spins. The probability that the house number shows up once in 25 spins is: $$P(X=1) = \binom{25}{1}\frac{1}{37}\left(\frac{36}{37}\right)^{24}\simeq 0.370$$ Consequently, the gambler's expected profit is: $$E(\text{profit})=1\cdot P(X=1)-1\cdot P(X=0) = 0.370-0.630 = -0.260,$$ or he loses 26 cents on every dollar.

Is this correct?

Edit: if the problem is interpreted as "the house number shows up at least once in 25 spins", the probability would be $$P(X=1)=1-\left(\frac{36}{37}\right)^{25}\simeq 0.496,$$ which gives the expected profit: $$E(\text{profit})=0.496 - 0.504 = -0.008,$$ or a loss of 0.8 cents per dollar.

• Does X mean exactly once or at least once? – herb steinberg Jan 9 '18 at 17:23
• @herbsteinberg the text does not specify, so I guessed it means exactly once. – J. D. Jan 9 '18 at 17:39
• I would accept the second reading, that it shows up at least once. There is the old story of the gambler betting on double sixes in 24 rolls and losing and this is very close. – Ross Millikan Jan 9 '18 at 18:12