In the game of roulette, a player can place a bet on the number 2 and have 1/38

Question

In the game of roulette, a player can place a bet on the number 2 and have 1/38 probability of winning. If the metal ball lands on 2, the player gets to keep the $6 paid to play the game and the player is awarded an additional $210. Other wise the player is awarded nothing and the casino takes the payers $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? ROUND TO NEAREST CENT

Expected value-

How much the player would expect to lose-

Explanation / Answer

In the game of roulette, a player can place a bet on the number 2 and have 1/38 probability of winning. If the metal ball lands on 2, the player gets to keep the $6 paid to play the game and the player is awarded an additional $210. Other wise the player is awarded nothing and the casino takes the payers $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? ROUND TO NEAREST CENT

Expected value

E(x) = 210(1/38) + (-6)37/38) = (210-222)/38 = -12/38 = -$0.3158

If you played the game 1000 times, how much would you expect to lose?

Amount expect to lose = 1000(-0.3158) = $315.80

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