In the game of roulette, a player can place a $9 bet on the number 25 and have a

Question

In the game of roulette, a player can place a $9 bet on the number 25 and have a 1/38 probability of winning. If the metal ball lands on 25, the player gets to keep the $9 paid to play the game and the player is awarded $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 $ (Round to the nearest cent as needed.) The player would expect to lose about $ (Round to the nearest cent as needed.)

Explanation / Answer

Solution:-
P(winning) = 1/38
P(losing) = 1-1/38 = 37/38
Expected value = 315(1/38)-9(37/38)
= (315-333)/38
= -18/38
= -0.4737$

When the game is played 1000 times, the amount expected to lose = 1000(-0.4737) = $473.70

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