Question

In the game of​ roulette, a player can place a ​$4 bet on the number 24 and have a StartFraction 1 Over 38 EndFraction probability of winning. If the metal ball lands on 24​, the player gets to keep the ​$4 paid to play the game and the player is awarded an additional ​$140. ​ Otherwise, the player is awarded nothing and the casino takes the​ player's ​$4. 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?

Answer 1

Answer :

given data :-

the game of​ roulette, a player can place =  ​4 bet on the number 24

the player gets to keep the ​ paid to play the = 4

the game and the player is awarded an additional ​= 140.

the game times = 1000

here,

probability of wiining =1/38 as 24 is one of slot of 38

probability of losing =37/38

therefore expected value of the game to the? player

=140(1/38)-4(37/38)

= 140*0.0263 - 4*0.973

= 3.684 - 3.892

= -0.208

expected lose for 1000 games = -1000(140(1/38)-4*(37/38))

= -1000*-0.208

= 208

the  expect to​ lose is = $ 208