n the game of roulette, a player can place a $9 bet on the number 3 and have a 1
ID: 2656296 • Letter: N
Question
n the game of roulette, a player can place a $9 bet on the number 3 and have a 1/38 probability of winning. If the metal ball lands on 3, 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?
The expected value is $___
The player would expect to lose about $__
Explanation / Answer
The expected value = Probability of winning * Pay off in Winning + Probability of losing * Payoff in losing =
(1/38)* 315 + (37/38)* *(-9) = -0.47368 or -0.47
The Expected amount to lose = 1000 * Exected Value = 1000 * -0.473684 = -4736.84
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