In the game of roulette, a player can place a $5 bet on the number 11 and have a startfraction 1 over 38 endfraction probability of winning. if the metal ball lands on 11, the player gets to keep the $5 paid to play the game and the player is awarded an additional $175. otherwise, the player is awarded nothing and the casino takes the player's $5. 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?

Question

Answer:

we know that
P(winning) = 1/38
P(losing) = 1-1/38 = 37/38
Expected value = 175*1/38 - 5*37/38
175/38 - 185/38 = - 10/38 =-$0.2632

the answer part A) is
The player's expected value is -$0.2632 ( is losing)

b) if you played the game 1000 times, how much would you expect to lose?
-$0.2632*1000=-$263.20

the answer Part B) is $263.20

solved

general 10 months ago 7180