in the game of roulette, a player can place a $9 bet on the number 36 and have a

Question

in the game of roulette, a player can place a $9 bet on the number 36 and have a 1/38 probability of winning. If the metal ball lands on 36, the player gets to keep the $9 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 $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? The expected value is $___ The player would expect to lose about $__

Explanation / Answer

Consider:

Thus,  
  
E(x) = Expected value = mean = Sum(xP(x)) = -0.473684211 [ANSWER]

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Hence,if you play 1000 times, you expect to lose

E(1000x) = 1000E(x) = -47.3684211 [ANSWER]

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