Question

In the game of roulette, a player can place a $8 bet on the number 33 and have a 1/38 probability of winning. If the metal ball lands on 33, 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?The expected value is $? Round to the nearest cent as needed.The player would expect to lose about $? Round to the nearest cent as needed.

Answer 1

Answer:

• expected value: -$0.21
• loss on 1000 plays: $210.53

Step-by-step explanation:

The expected value is the sum of products of payoff and probability of that payoff:

  -$8(37/38) +$288·(1/38) = $(-296 +288)/38 = -$8/38 ≈ -$0.21

In 1000 plays, the expected loss is ...

  -$8000/38 ≈ $210.53