In the game of roulette a player can place a 55 bet on the n

In the game of roulette, a player can place a $55 bet on the number 1414 and have a StartFraction 1 Over 38 EndFraction138 probability of winning. If the metal ball lands on 1414, the player gets to keep the $55 paid to play the game and the player is awarded an additional $175175. Otherwise, the player is awarded nothing and the casino takes the player\'s $55. 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?

Solution

I guess it is 14 and 175 and not 1414 and 175175. But I am solving this problem in a generlised way. So, put values and get the answer yourself also.

N1= 14; p= 1/38;

P[X= N1=14]= 1/38 ; P(X not equal to 14 )= 37/238

E[X] = (1/38)*(175) + (37/38) *(-55) = -48.9473

This is the expected loss that played, would be inflicted with.

if you have 1000 games then,

1000*-48.9473 = -48947.3