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

Shyanne Chandler

Mathematics

9 June, 05:06

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?

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Answers (1)

Know the Answer?

Alani Kemp · Answer 1

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