In the game of roulette, a player can place a $4 bet on the number 25 and have a

Question

In the game of roulette, a player can place a $4 bet on the number 25 and have a 1/38 probability of winning. If the metal ball lands on 25, the player gets to keep the $4 paid to play the game and the player is awarded $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?

a) The expected value is $........

(Round to the nearest cent as needed.)

b) The player would expect to lose about $.....

(Round to the nearest cent as needed.)

Explanation / Answer

A) The expected value is given as Sum[x.p(x)]

For Winning, x = $140, p(x) = 1/38

For Losing, x = -$4, p(x) = 1 - 1/38 = 37/38

Therefore Expected Value = 140 * (1/38) - 4 * (37/38) = (140-148)/38 = -8/38 =

-4/19 = 0.21cents loss per game.

(b) If 1000 games are played, the expected loss = 1000 * (-4/19) = -4000/19 = $210.53 Loss

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.