In the game of roulette, a player can place a $9 bet on the number 9 and have a

Question

In the game of roulette, a player can place a $9 bet on the number 9 and have a 1/38 probability of winning. If the metal ball lands on 9, 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? Note that the expected value is the amount, on average, one would expect to gain or lose each game. The expected value is $. The player would expect to lose about $.

Explanation / Answer

Expected value:

E(x) = 315(1/38) + (-9)37/38 = (315 - 333)/38 = -12/38 = -$0.4737 or 47 cents

Expected to lose:

1000 * (-0.4737) = $473.70

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