Question

24. A gambler decides to bet on the first 12 numbers on a roulette table. There are 12 chances in 38 to win. This bet pays 2 to 1. The gambler bets $3 on each play and plays 190 times. (a) (4 points) Construct a box model for one bet. (b) (2 points) Compute the expected value for the sum of all 190 plays (c) (3 points) Compute the standard error for the sum of all 190 plays. (d) (5 points) What is the probability that the gambler leaves the roulette table after his 190 plays with more money that when he arrived? (Hint: The sum of 190 plays > 0)

Answer 1

(a)

The gambler bets $3 on 12 numbers. He wins $6 if that for those 12 numbers or lose $3 for remaining 26 numbers.

There are 38 objects in the box; 12 are labelled 6 and the 26 are labelled -3.

(b)

The expected value of the box for one bet = (1/38) [ 12 * 6 - 26 * 3] = -0.158

Expected value of sum for 190 plays = 190 * (-0.158) = -30.02

(c)

Variance of sum for one bet = (1/38) [ 12 * 6² + 26 * 3²] - (-0.158)²

= 17.50135

Variance of sum for 190 bet = 190 * 17.50135 = 3325.256

Standard deviation of sum for 190 bet = sqrt(3325.256) = 57.665

Standard error of sum for 190 bet = 57.66503 / sqrt(190) = 4.183

(c)

P(Sum of 190 days > 0) = P[Z > (0 - (-30.02)) / 57.665] = P(Z > 0.5206) = 0.3013