In the game of roulette, a player can place a $10 bet on the number 5 and have a

Question

In the game of roulette, a player can place a $10 bet on the number 5 and have a 1/38 probability of winning If he metal ball lands on 5, the player gets to keep the $10 paid to play the game and the player nothing and the casino takes the player's $10. hat s the expected value of the game to the player? If you played the game 1000 times, is awarded an additional $350. Otherwise, the player is awarded how much would you expect to lose? The expected value is $ Round to the nearest cent as needed.) The player would expect to lose about $ (Round to the nearest cent as needed.)

Explanation / Answer

Expected value = Sum( P(x) * x) = 1/38 *(350) + 37/38*(-10) = -20/38=-10/19 = -0.526

In 1000 times, the player should expect to lose about 1000*0.526 or 526 dollars

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