In the game of roulette, a player can place a $55 bet on the number 11 and have

Question

In the game of roulette, a player can place a $55 bet on the number 11 and have a StartFraction 1 Over 38 EndFraction 1 38 probability of winning. If the metal ball lands on 11, the player gets to keep the $55 paid to play the game and the player is awarded an additional $175. Otherwise, the player is awarded nothing and the casino takes the player's $55. 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.

Explanation / Answer

here expected value of game = expected gain -expected loss =(1/38)*175-(37/38)*5= -0.26316 ~ $-0.23

therefore expected loss for 1000 games =1000*0.26316 =$263.16

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