A roulette wheel has 38 slots. Thirty-six slots are numbered from 1 to 36; the remaining two slots are numbered 0 and 00. Suppose the number 00 is considered not to be even, but the number 0 is still even. In a
1 to play. If the ball lands in an even numbered slot, he receives back the dollar he bets plus an additional dollar. If the ball does not land on an even numbered slot, he loses his dollar. Let X denote the bet gain to the bettor on one play of the game. Construct the probability distribution X. Compute the expected value E(X) of X, and explain why this game is not offered in a casino (where 0 is not considered even). Compute the standard deviation as well.
The Correct Answer and Explanation is:
Let random variable XXX represent the bettor’s net gain on one play of the game. The roulette wheel has 38 slots: numbers 1–36, 0, and 00.
Defining outcomes:
- Even numbers: 2, 4, 6, …, 36, plus 0 (which is considered even here).
- Total even slots: 18 (from 2 to 36) + 1 (0) = 19.
- Total not even slots: 38 total slots − 19 even slots = 19.
Values of XXX:
- If the ball lands on an even number: bettor gains $1.
- If the ball lands on a not even number: bettor loses $1.
Probability distribution of XXX:
| XXX | Probability P(X)P(X)P(X) |
|---|---|
| 1 | 1938\frac{19}{38}3819 |
| -1 | 1938\frac{19}{38}3819 |
Expected value E(X)E(X)E(X):
E(X)=(1)×1938+(−1)×1938E(X) = (1) \times \frac{19}{38} + (-1) \times \frac{19}{38} E(X)=(1)×3819+(−1)×3819E(X)=1938−1938=0E(X) = \frac{19}{38} – \frac{19}{38} = 0E(X)=3819−3819=0
Thus, the expected value is 0. This means, on average, the bettor neither gains nor loses money over many plays. This is a fair game, unlike real casino roulette where 0 and 00 are both treated as not even, shifting the odds in favor of the house.
In actual casinos, the house edge exists because the probability of winning on even is 1838\frac{18}{38}3818, making the expected value negative for the player. Therefore, under these modified rules, casinos would not offer this version since there is no advantage.
Variance and standard deviation:
First, compute E(X2)E(X^2)E(X2):E(X2)=(12)×1938+(−12)×1938E(X^2) = (1^2) \times \frac{19}{38} + (-1^2) \times \frac{19}{38}E(X2)=(12)×3819+(−12)×3819E(X2)=1938+1938=1E(X^2) = \frac{19}{38} + \frac{19}{38} = 1E(X2)=3819+3819=1
Variance σ2\sigma^2σ2:σ2=E(X2)−(E(X))2=1−0=1\sigma^2 = E(X^2) – (E(X))^2 = 1 – 0 = 1σ2=E(X2)−(E(X))2=1−0=1
Standard deviation σ\sigmaσ:σ=1=1\sigma = \sqrt{1} = 1σ=1=1
Conclusion:
The game is perfectly fair with an expected value of 0 and a standard deviation of 1. Casinos typically avoid offering fair games since they do not yield profit over time.
