Question 530077: in the game of roulette, a player can plave a 6$ bet on the number 21 and have a 1/38 probability of winning. If the metal ball lands on 21, the player gets to keep the 6$ paid to play the game and the player is awarded 210$ . otherwise the player is awarded nothing and the casino takes the players 6$. What is the expected vaule of the game to the player? If you played the game 1000 times, how much would you expect to lose?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! in the game of roulette, a player can place a 6$ bet on the number 21 and have a 1/38 probability of winning. If the metal ball lands on 21, the player gets to keep the 6$ paid to play the game and the player is awarded 210$ . otherwise the player is awarded nothing and the casino takes the players 6$. What is the expected value of the game to the player?
E(x) = 210(1/38) + (-6)37/38) = (210-222)/38 = -12/38 = -$0.3158
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If you played the game 1000 times, how much would you expect to lose?
Ans: 1000(-0.3158) = $315.80
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cheers,
Stan H.
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