In the game of roulette, a player can place a $10 bet on the number 16 and have a 1/38 chance of probability chance of winning. If the metal balls lands on 16, the player gets to keep the $10 paid to play the game and the player is awarded an additional $350. Otherwise, the player is awarded nothing and the casino takes the takes the player's $10. What is the expected value of the game to the player? If you played the game 1,000 times, how much would you expect to lose?
Win/ lose (X) | Probability(X) | x* P(x) |
350 | (1/38)= 0.026 | 9.211 |
-10 | (37/ 38) = 0.974 | -9.737 |
Total | 1.00 | -0.526 |
If the ball lands on 16, player gains =350
If the ball lands on any other no, player loses = - 10
E(X) = (350 * 1/38) + (-10 * 37/38) = 9.211 - 9.737 = - 0.526
E(X) = - 0.526
Expected loss for 1000 games = 1000 * -0.526 = 526.32
Expected loss for 1000 games = 526.32
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