Question

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?

Answer 1

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) =Στί + Ρ(ai) 1=1

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