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 1/38 probability of winning. 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.

(Round to the nearest cent as needed.)

Explanation / Answer

The expected value of the game to the player is

E(X) = 175(1/38) + (-5) 37/38

= 175/38 - 185/38

= -10 / 38

= -0.2632

E(X) = $ -0.2632

Negative expected value mean expected loss in game.

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

1000( -0.2632) = $263.2

We take this value positive because we are expecting to lose.

If you played the game 1000 times you would expect to lose $263.2

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