In the game of roulette, a player can place a S8 bet on the number 19 and have a

Question

In the game of roulette, a player can place a S8 bet on the number 19 and have a 1/38 probability of winning. If the metal ball lands on 19, the player gets to keep the $8 do 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

The player loses 8 dollars with probability 37/38 and wins 280 dollars with probability 1/38.

So expected value is -8*37/38 + 280/38 = -8/19 = -0.42 dollars

In 1000 games the player is expected to lose 1000*0.42105 dollars

= 421.05 dollars

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