223k views
4 votes
In the game of​ roulette, a player can place a ​$9 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 ​$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

User Labanino
by
5.2k points

1 Answer

3 votes

Answer:

- $0.47

Here, negative sign depicts the expected loss

Step-by-step explanation:

Data provided in the question:

Bet amount for number 24 = $9

Probability of winning on number 24 =
(1)/(38)

Award amount for number 24 = $315

Probability of losing on number 24 = 1 -
(1)/(38) =
(37)/(38)

Losing amount = $9

Now,

The expected value of the game to the player

= Winning amount × Probability of winning - Losing amount × Probability of losing

= ( $315 ×
(1)/(38) ) - ( $9 ×
(37)/(38) )

= 8.29 - 8.76

= - $0.47

Here, negative sign depicts the expected loss

User Strings
by
4.8k points