Question 1147351: In the game of roulette, a player can place a $8 bet on the number 14 and have a 1/38 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 players $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?
Answer by VFBundy(438)   (Show Source):
You can put this solution on YOUR website! 1/38 * +280 = +140/19
37/38 * -8 = -148/19
Add these two values together to get -8/19. This is the expected value of the game. (About -$0.42.)
If you play the game 1000 times, the you could expect to lose: 1000*(-8/19) dollars...or, -8000/19 dollars. That's about $421.05.
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