In the game of roulette, a player can place a $8 bet on the number 12 and have a

Question

In the game of roulette, a player can place a $8 bet on the number 12 and have a 1/36 probability of winning. If the metal ball lands on 12, 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.)

Explanation / Answer

Expected value

= 280 (1/38) - 8 (37/38)

= -0.42

Hence,

The expected value is $ -0.42

The player would expect to loose about $ 421.05

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