Final answer:
When making a bet where drawing two threes in succession with replacement wins $60, and losing costs $10, the expected value per bet is approximately -$9.58. Over 20 games, the expected loss would be roughly $191.60.
Step-by-step explanation:
To calculate how much you would expect to win or lose from the card game mentioned, we need to use the concept of expected value. In this scenario, drawing two threes in succession from a standard deck of 52 cards with replacement is the winning condition. The probability of drawing a three for any given draw is 4/52 (since there are 4 threes in a deck of 52 cards). Since the draws are with replacement, the probability of getting a three on the second draw is also 4/52. The probability of both events happening in succession is their product:
(4/52) × (4/52) = (1/13) × (1/13) = 1/169.
Therefore, the expected value (EV) of a single bet is:
EV = (Probability of winning) × (Amount won) + (Probability of losing) × (Amount lost)
EV = (1/169) × $60 + (168/169) × (-$10)
EV =~ 0.00592 × $60 - 0.99408 × $10
EV =~ $0.36 - $9.94
EV =~ -$9.58
To find the expected outcome over 20 games, we multiply this expected value by the number of games:
20 × (-$9.58) = -$191.60
So, if you made the bet 20 times, you would expect to lose approximately $191.60, rounded to the nearest cent.