In the game of roulette, a player can place a $8 bet on the number 6 and have a

Question

In the game of roulette, a player can place a $8 bet on the number 6 and have a 1/38 probability of winning. If the metal ball lands on 6, 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 players $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

Gain = 280 + 8 = 288

Probability of winning = 1/38

Expected value for the player = 288(1/38) - 8 = $-0.42

If the game is played 1000 times, the expected loss = 0.42 * 1000 = $420

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