In the game of roulette, a player can place a $ 9 9 bet on the number 14 14 and have a StartFraction 1 Over 38 EndFraction 1 38 probability of winning. If the metal ball lands on 14 14, the player gets to keep the $ 9 9 paid to play the game and the player is awarded $ 315 315. Otherwise, the player is awarded nothing and the casino takes the player's $ 9 9. What is the expected value of the game to the player? If you played the game 1000times, how much would you expect to lose?

Question

In the game of​ roulette, a player can place a ​$ 9 9 bet on the number 14 14 and have a StartFraction 1 Over 38 EndFraction 1 38 probability of winning. If the metal ball lands on 14 14​, the player gets to keep the ​$ 9 9 paid to play the game and the player is awarded ​$ 315 315. ​ Otherwise, the player is awarded nothing and the casino takes the​ player&#039;s ​$ 9 9. 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?

Savanna Smith · Accepted Answer

If you play 38 games you expect to win once on average 38 games cost&#039;s  =  38  ×  9.9  =  376.2      $  the one time you win you get a total of $ 315 back  376.2  −  315  =  61.2-61.2 every 38 games.Divide by 38 to get the expected Value per game       61.2    38    =  1.610=1.61 loss per gameSo you respect to lose on average of 161 $ per game on averse multiplying by 1000 givens   =  1.6  ×  1000  =  1610

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