In the game of roulette, a player can place a $8 bet on the number 33 and have a 1/38 probability of winning. If the metal ball lands on 33, the player gets to keep the $8 paid to play the game and the player is awarded an additional $280. Otherwise, the player is awarded nothing and the casino takes the player's $8. 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.

Question

Mathematics

Guest

11 October, 09:35

In the game of roulette, a player can place a $8 bet on the number 33 and have a 1/38 probability of winning. If the metal ball lands on 33, the player gets to keep the $8 paid to play the game and the player is awarded an additional $280. Otherwise, the player is awarded nothing and the casino takes the player's $8. 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.

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Answers (1)

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Armani · Answer 1

expected value: - $0.21 loss on 1000 plays: $210.53

Step-by-step explanation:

The expected value is the sum of products of payoff and probability of that payoff:

-$8 (37/38) + $288· (1/38) = $ (-296 + 288) / 38 = - $8/38 ≈ - $0.21

In 1000 plays, the expected loss is ...

-$8000/38 ≈ $210.53