SOLUTION: In the game of roulette, a player can place a $6 bet on the number 9 and have a 1\38 probability of winning. If the metal ball lands on 9, the player gets to keep the $6 paid

Question 1109723: In the game of roulette, a player can place a $6 bet on the number 9 and have a
1\38 probability of winning. If the metal ball lands on 9, 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. 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 jorel1380(3719) (Show Source): You can put this solution on YOUR website!
The expected outcome would be:
(37/38)(-6)+(1/38)(210)=-0.316, or a loss of 31 cents per spin. Over 1000 spins, the player can expect to lose $315.79
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