In the game of roulette, a player can place a $6 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 $6 paid to play the game and the player is awarded an additional $210. Otherwise, the player is awarded nothing and the casino takes the player's $6. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(X) is usually negative. This value gives the average amount per game the player can expect to lose. The expected value is $

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Answer:

The expected value to lose is $0.32

Step-by-step explanation:

First you find the average value from the winnings.

You do this by multiplying the probablity of victory by how much you would win:    1/38 * 210  == $5.526

Then you add that to the average value of losing times the probablity of losing:    37/38 * -6       ==    $ -5.842     (it is important to have the negative there because you are losing money)

Then you add those together and you get your anwser:

$5.526 - $5.842 = $0.316 which rounds up to -$0.32

The expected value to lose is $0.32

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