Question

For this project, you will analyze the famous casino game, Roulette. In the game of Roulette, there is a wheel consisting of 38 numbers: 18 black numbers, 18 red numbers, and 2 green numbers (0 and 00). The wheel is spun and a ball is released into the wheel. As the wheel comes to a stop, the ball will land on one of the numbers. A player will place their bet before the wheel is spun. A player can make the following bets (along with other bets):

Single (or multiple) number

1^st, 2^nd or 3^rd column (e.g. 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34)

1^st, 2^nd or 3^rd dozen (e.g. 1 through 12)

Odd / Even (does not include 0 or 00)

Red / Black

1 through 18 or 19 through 36

Please answer the following questions based on the game of Roulette:

What is the probability of betting on all odds and winning?

What is the probability of betting on the 2^nd dozen and winning?

A player bets on the number 5 in the first game; then bets on number 22 in the second game. Did the player’s odds of winning change from the first to the second game? Explain.

What is a better bet: all reds or all evens? Explain. (Hint: 0 and 00 are not included in an all even bet)

If a player always bets on 00, what is the expected winnings (or loses) when playing 20 games of Roulette given the player receives $100 if they win and owes $1 if they lose?

If a player always bets on 3 and 23, what is the expected winnings (or loses) when playing 75 games of Roulette given the player receives $120 if they win (on either number) and owes $3 if they lose?

If a player always bets on the 2^nd dozen (13 to 24), what is the expected winnings (or loses) when playing 25 games of Roulette given the player receives $32 if they win and owes $15 if they lose?

If you are playing Roulette, how would you bet? Would you play a lot of games or try to win a lot on one or two games?

Answer 1

1. Out of the 38 numbers when 0 and 00 are not included, there 36 numbers left and 18 of them are odd. So the probability of betting on all odds and winning is 18/38.

2. There are 12 numbers (13 to 24) in the second dozen. So the probability of winning on the second dozen is 12/38 or 6/19.

3. Switching the bet from 5 to 22 does not change the odds of winning. Because we are still betting on a single number the probability of which is 1/38.

4. 0 and 00 are green so in all reds we have the probability as 18/38 and the probability of getting all evens is also 18/38. Since the probability of both the events are the same no event is better than the other. So we have an equal chance of winning by placing on all reds or on all evens.

5. Probability of winning on 00 is 1/38. And probability of losing is 37/38. So the expected value of 1 game : 100/38 - 37/38 = 63/38. Expected value of 20 games : 20*63/38 = 33.16

6. Probability of getting 3 or 23 is 2/38 ie 1/19 and the probability of not getting it is 18/19. So expected value of 1 game : 120*1/19-3*18/19= 66/19. Expected value of 75 games= 66*75/19= 260.53

7. Probability of getting a value from 13 to 24 is 12/38 and probability of not getting it is 26/38. So expected value of one game is 32*12/38 - 15*26/38 = -6/38. So expected value of 25 games is -6*25/38 = - 3.95