Question 1110069
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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