In the game of roulette a player can place a 99 bet on the n

In the game of roulette, a player can place a $99 bet on the number 1313 and have a StartFraction 1 Over 38 EndFraction 1 38 probability of winning. If the metal ball lands on 1313, the player gets to keep the $99 paid to play the game and the player is awarded an additional $315315. Otherwise, the player is awarded nothing and the casino takes the player\'s $99. 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 the game = 1/38

Probability of losing the game = 1 - Probability of winning the game = 1 - 1/38 = 37/38

E(x) = probability of winning * Winning amount - Probability of losing * losing amount

=> 1/38 * 315 - 37/38 * 99

=> -88.10$

Expected to lose in 1000 games = -88100$