Ask Question
5 November, 16:00

In a large casino, the house wins on one of its games with a probability of 50.6%. All bets are 1 :1. If you win, you gain the amount you bet; if you lose, you lose the amount you bet. If you play 180 games in an evening, betting $1 each time, how much should you expect to win or lose?

+4
Answers (1)
  1. 5 November, 16:14
    0
    E (x) = - $ 2.16

    Step-by-step explanation:

    If the house wins on one of its games with a probability of 50.6 % (0.506), then you lose 180*0.506 games, it means you lose in 91.08 games, how your bet by game $1, then you lose $91.08.

    You win in 180-91.08 = 88.92 games; how you gain the amount you bet, then you gain $88.92.

    Thus the expect value of your gain is:

    E (x) = 88.92 - 91.08

    E (x) = - $ 2.16
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “In a large casino, the house wins on one of its games with a probability of 50.6%. All bets are 1 :1. If you win, you gain the amount you ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers