Algebra -> Probability-and-statistics-> SOLUTION: In the game of roulette, a player can place a $8 bet on the number 14 and have a StartFraction 1 Over 38 EndFraction probability of winning. If the metal ball lands Log On
Question 1110069: In the game of roulette, a player can place a $8 bet on the number 14 and have a StartFraction 1 Over 38 EndFraction
probability of winning. If the metal ball lands on 14, the player gets to keep the $8 paid to play the game and the player is awarded an additional $280. Otherwise, the player is awarded nothing and the casino takes the player's $8. 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.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
n the game of roulette, a player can place a $8 bet on the number 14 and have a StartFraction 1 Over 38 EndFraction
probability of winning. If the metal ball lands on 14, the player gets to keep the $8 paid to play the game and the player is awarded an additional $280. Otherwise, the player is awarded nothing and the casino takes the player's $8. 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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Random 'winnings':: -8...280
Probabilities:::::: 37/38..1/38
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Expected 'winnings' = (37/38)(-8) + (1/38)280
= (-296+280)/38
= -42 cents (amt. you can expect to lose on each play)
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If you play 1000 time expect to lose 42*1000 = $420
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Cheers,
Stan H.
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