Question

(5.31) A roulette wheel has 38 slots, of which 18 are black, 18 are red, and 2 are green. When the wheel is spun, the ball is equally likely to come to rest in a any of the slots. Gamblers can place a number of different bets in roulette. One of the simplest wagers chooses red or black. A bet of $1 on red will pay off an additional dollar if the ball lands in a red slot. Otherwise, the player loses his dollar. When gamblers bet on red or black. The two green slots belong to the house.

(a) A gambler’s winnings on a $1 bet are either $1 or $1. Give the probabilities of these outcomes. Find the mean and standard deviation of the gambler’s winnings.

(b) Explain briefly what the law of large numbers tells us about what will happen if the gambler makes a large number of bets on red.

(c) The central limit theorem tells us the approximate distribution of the gambler’s mean winnings in 50 bets. What is this distribution? Use the 68-95-99.7 rule to give the range in which the mean winnings will fall 95% of the time. Multiply by 50 to get the middle 95% of the distribution of the gambler’s winnings on nights when he places 50 bets.

(d) What is the probability that the gambler will lose money if he makes 50 bets? (that is, the probability that the mean is less than 0.)

(e) The casino takes the other side of these bets. If 100,000 bets on red are placed in a week at the casino, what is the distribution of the mean winnings of the gamblers on these bets? What range covers the middle 95% of mean winnings in 100,000 bets? Multiply by 100,000 to get the range of gambler’s losses. (Gamblers’ losses are the casino’s winnings. Part (c) shows that a gambler gets excitement. Now we see that the casino has a business.)

Answer 1

(a)

Let X be the gambler’s winnings on a $1 bet.

P(X = 1) = 18/38 = 9/19

P(X = -1) = 20/38 = 10/19 (There will 18 black slots + 2 green slots = 20 slots where the gambler lose)

E(X) = (9/19) * 1 + (10/19) (-1) = -1/19 = -0.05263158

E(X²) = (9/19) * 1² + (10/19) (-1)² = 1

Var(X) = E(X²) - [E(X)]² = 1 - ( -1/19)²  = 0.99723

Standard deviation of winnings = $\sqrt{0.99723}$ = 0.998614

(b)

As, the expected value of winnings is negative, the law of large numbers tells that the gambler will lose money if the gambler makes a large number of bets on red.

(c)

The central limit theorem tells us the approximate distribution of the gambler’s mean winnings in 50 bets is Normal distribution.

By the 68-95-99.7 rule the range in which the mean winnings will fall 95% of the time is within 2 standard deviations from the mean.

For single bet, the range is (-0.05263158 - 2 * 0.998614, -0.05263158 + 2 * 0.998614) = (-2.04986, 1.944596)

For 50 bets, the range is (50 * -2.04986, 50 * 1.944596) = (-102.493, 97.2298)

(d)

Expected winnings for 50 bets = 50 * -0.05263158 = -2.631579

Variance of winnings for 50 bets = 50 * 0.99723 = 49.8615

Standard deviation = $\sqrt{49.8615}$ = 7.061268

Probability that the gambler will lose money if he makes 50 bets = Probability that the mean is less than 0

= P(Z < (0 - (-2.631579)) / 7.061268)

= P(Z < 0.3727)

= 0.6453

(e)

The distribution of the mean winnings of the gamblers on these bets is Normal distribution.

As, Gamblers’ losses are the casino’s winnings, the expected casino’s winnings on each bet is 0.05263158 with variance of 0.99723

For 100,000 bets, the mean is 100,000 * 0.05263158 = 5263.158 and variance 100,000 * 0.99723 = 99723

Standard deviation = $\sqrt{99723}$ = 315.7895

Range covers the middle 95% of mean winnings in 100,000 bets is,

(99723 - 2 * 315.7895, 99723 + 2 * 315.7895)

(99091.42 , 100354.6)