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In the game of​ roulette, a player can place a ​$44 bet on the number 99 and have a StartFraction 1 Over 38 EndFraction 1 38 probability of winning. If the metal ball lands on 99​, the player gets to keep the ​$44 paid to play the game and the player is awarded an additional ​$140140. ​Otherwise, the player is awarded nothing and the casino takes the​ player's ​$44. 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.

Respuesta :

Answer:

Expected loss = $39,160

Step-by-step explanation:

Given:

Bet amount = $44

Probability of winning =  [tex]\frac{1}{38}[/tex]

Therefore,

Probability of losing =  [tex]1-\frac{1}{38}[/tex]=  [tex]\frac{37}{38}[/tex]

Award amount = $140

Number of times played = 1000

Now,

The expected value =  [tex]\$140\times\frac{1}{38}-\$44\times\frac{37}{38}[/tex]

or

The expected value = 3.68 - 42.84

or

The expected value = -$39.16 (Here negative sign mean the loss)

Therefore, for playing 1000 times the expected loss is 1000 × $39.16

= $39,160

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