In the game of roulette, a player can place a exist7 bet on the number 13 and ha

Question

In the game of roulette, a player can place a exist7 bet on the number 13 and have a 1/38 probability of winning. If the metal ball lands on 13, the player gets to keep the exist7 paid to play the game and the player is awarded an additional exist245. Otherwise, the player is awarded nothing and the casino takes the players exist7. 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? Note that the expected value is the amount, on average, one would expect to gain or lose each game. The expected value is exist (Round to the nearest cent as needed.) The player would expect to lose about exist (Round to the nearest cent as needed.)

Explanation / Answer

E(x) = 210(1/38) + (-6)(37/38) =-0.3684

If you played the game 1000 times, how much would you expect to lose?

1000(-0.3684) = $ 368.4

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