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Suppose a game is played in which you pay 59 to play, and will receive 530 if you win in addition to getting your 59 back) The casino has a 98% chance of winning, and you play the game 800 times. What is the expected value of the game to the casino? Round to the nearest dollar

User Sivabudh
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Final answer:

The expected value for the casino over 800 games of this probability-based game is $37,776 to the nearest dollar.

Step-by-step explanation:

The expected value for the casino in this game is calculated by considering the probability of the casino winning and the payouts for winning and losing. Since the casino has a 98% chance of winning, the expected value for the casino per game can be found by multiplying the probabilities and payouts accordingly. The casino gains $59 when they win and loses $530 (the $59 stake plus the $471 winnings) when they lose.

Calculation of Expected Value

The probability of the casino losing is therefore 2% or 0.02. The expected value (EV) for the casino per game is:

EV = (probability of winning × gain from a win) + (probability of losing × loss from a loss)

EV = (0.98 × $59) + (0.02 × -$530)

EV = $57.82 - $10.60

EV = $47.22

Thus, the expected value for the casino for each game is $47.22. To find the expected value over 800 games, we multiply this amount by 800:

Total EV for 800 games = $47.22 × 800 = $37,776

To the nearest dollar, the expected value for the casino over 800 games would be $37,776.

User Haoshu
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