In the game of roulette a player can place a 9 bet on the nu

In the game of roulette, a player can place a $9 bet on the number 11 and have a 1/38 probability of winning. If the metal ball lands on 11, the player gets to keep the $9 paid to play the game and the player is awarded an additional $315 Otherwise, the player is awarded nothing and the casino takes the player\'s $9. What is the expected value of the game to the player? If you played the game 1000 times, how much would you expect to lose?

Solution

Probability of winning = 1/38 Probability of losing = 1-1/38 = 37/38 Expected value = [175 x 1/38] -[ 5 x 37/38 ] = -5/19 Therefore player\'s expected value is losing $ 5/19 therefore when he plays for 1000 times, amount expected to lose = 1000 x 5/19 = 263.15 $