Question

In the game of​ roulette, a player can place a ​$44 bet on the number 99 and have a StartFraction 1 Over 38 EndFraction 1 38 probability of winning. If the metal ball lands on 99​, the player gets to keep the ​$44 paid to play the game and the player is awarded an additional ​$140140. ​Otherwise, the player is awarded nothing and the casino takes the​ player's ​$44. 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? Note that the expected value is the​ amount, on​ average, one would expect to gain or lose each game.

Answer 1

Expected loss = $39,160

Explanation:

Given:

Bet amount = $44

Probability of winning =
`(1)/(38)`

Therefore,

Probability of losing =
`1-(1)/(38)`=
`(37)/(38)`

Award amount = $140

Number of times played = 1000

Now,

The expected value =
`\$140*(1)/(38)-\$44*(37)/(38)`

or

The expected value = 3.68 - 42.84

or

The expected value = -$39.16 (Here negative sign mean the loss)

Therefore, for playing 1000 times the expected loss is 1000 × $39.16

= $39,160