In the game of roulette, a player can place a $8 bet on the number 17 and have a StartFraction 1 Over 38 EndFraction probability of winning. If the metal ball lands on 17, the player gets to keep the $8 paid to play the game and the player is awarded an additional $280. Otherwise, the player is awarded nothing and the casino takes the player's $8. 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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pedrocarratore pedrocarratore · Answer 1 · 2020-04-10T03:38:32+02:00

Answer:

Expected value of the game: -$0.421

Expected loss in 1000 games: $421

Step-by-step explanation:

There are two possible outcomes for the event:

- There is a 1 in 38 chance of winning $280

- There is a 37 in 38 chance of losing $8

The expected value for a single game is:

[tex]E(X) = \frac{1}{38}*\$280-\frac{37}{38}*\$8\\E(X)=-\$0.421[/tex]

The expected value of the game is -$0.421

In 1,000 plays, the expected loss is:

[tex]L = 1,000*-\$0.421\\L=-\$421[/tex]

You would expect to lose $421.