An American roulette wheel contains 38 numbers. 18 are red, 18 are black, and 2

Question

An American roulette wheel contains 38 numbers. 18 are red, 18 are black, and 2 are green. When the roulette wheel is spun, the ball is equally likely to land on any of the 38 numbers. Suppose that a gambler bet $10 on red. If the ball lands on a red number, the gambler wins $10. Otherwise, the gambler loses $10. Let X be the amount the gambler wins on a $10 bet. Then the possible values for X are $10 or -$10. What's the expected value of the random variable X? On average how much will the gambler lose per play?

Explanation / Answer

E(X) = $10(18/38) + (-$10)(20/38) = -$0.53 On average a gambler will lose $0.53 per play.

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