In the game of roulette, a player can place a $10 bet on the number 19 and have a StartFraction 1 Over 38 EndFraction

probability of winning. If the metal ball lands on 19, the player gets to keep the $10 paid to play the game and the player is awarded an additional $350. Otherwise, the player is awarded nothing and the casino takes the player's $10. 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.

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Newton9022 Newton9022 · Answer 1 · 2020-07-06T03:42:58+02:00

Answer:

(a)-$0.53

(b)$530

Step-by-step explanation:

If the player wins the game, he gets a profit of $350.

If the player losses the game, he gets a profit of -$10.

Probability of Winning [tex]=\dfrac{1}{38}[/tex]

Probability of Loosing [tex]=\dfrac{37}{38}[/tex]

(a)Expected Value of the game

[tex]E(x)=\left(\dfrac{1}{38} \times 350\right) + \left(\dfrac{37}{38} \times -10\right)\\=-\$0.53[/tex]

The expected value of the game to the player is -$0.53.

(b)If the game is played 1000 times

Expected Loss = 0.53 X 1000

=$530

The player should expect to lose approximately $530 if he plays the game 1000 times.