Question

A roulette wheel has 38 slots: 18 red, 18 black, and 2 green. A ball is tossed into the wheel and eventually settles in a slot at random. You play many games, betting $1 on red each time. If the ball lands in a red slot you win $1, otherwise you lose the dollar you bet. After n games, what is the probability that you have more money than you started with? Give (approximate) numerical answers for n = 100 and n = 1000. How would the answer change if there were no green slots?

Answer 1

Expected winnings from one game = $1*18/38 - $1*20/38

= $-1/19

Variance of winnings from one game =

=

For n = 100 and n = 1000, the expected winnings will follow Normal distribution (From Central Limit Theorem)

For n = 100, expected winnings = 100*$-1/19 = $-100/19

Variance of winnings = 360/361*100

Thus, standard deviation of winnings = $9.986

The probability that you have more money than you started with = P(winnings > 0)

Using correction of continuity, the required probability = P(winnings > 0.5)

= P[Z > {0.5 - (-100/19)}/9.986]

= P(Z > 1.03) = 0.1515

For n = 1000,

Mean = $-1000/19, Standard deviation = $31.58

The required probability = P(Z > 1.68) = 0.0465

If there were no green slots, expected winnings from one game = $1*18/36 - $1*18/36 = 0

Variance of winning = 1

In this case, probability that you have more money than you started with

For n = 100 is P{Z > 0.5/10} = P(Z > 0.05) = 0.4801

For n = 1000 is P{Z > 0.5/√1000} = P(Z > 0.0158)

= 0.4937