In the game of roulette, a player can place a $5 bet on the number 16 and have a

Question

In the game of roulette, a player can place a $5 bet on the number 16 and have a probability of winning 38 if the metal ball lands on 16, the player gets to keep the $5 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 $5. 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 $ 131.58 (Round to the nearest cent as needed.)

Explanation / Answer

Please check below on how to solve the question:

P of winning = 1/38
P of loosing = 1- P of winning = 37/38
E(X) = (1/38)(5+175) - (37/38)*(5) = -$.13158

If it happens for 1000 times then the expected value is -$131.58, which means that if you play 1000 times you are bound to loose $131.6, on average.

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