In the game of roulette, a player can place a $9 bet on the number 21 and have a

Question

In the game of roulette, a player can place a $9 bet on the number 21 and have a 33 probability of winning. If the metal ball lad on 21, the player gets to keep the S9 paid to play the game and the player is awarded an additional $315. Otherwise, the player is awarded nothing and the casino takes the player's S9. 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? The expected value is s (Round to the nearest cent as needed.) The player would expect to lose about s (Round to the nearest cent as needed.)

Explanation / Answer

Solution:-

p = 1/38

p = 0.02632

Amount recieved on winning the game = $315

Amonut lost on losing the game = $9

E(x) = P(won) × 315 - P(lost) × 9

E(x) = 0.02632 × 315 - (1 - 0.02632) × 9

E(x) = 8.2895 - 8.7632

E(x) = - $0.47

Player would expected to lose in 100 games = 0.4737 × 100 = $ 47.37

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