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In a card-and-coin game, a face card with heads wins $6 (probability 6/52), a face card with tails wins $2 (probability 6/52), and any non-face card loses $2 no matter the coin (probability 40/52). What does the solved expected-value table say this game is worth, per play?
AAbout -$0.62, an average loss per game
BAbout +$0.62, an average gain per game
CExactly -$2.00 every single play
DAbout -$6.00, an average loss per game
Answer & Solution
Correct answer: A. About -$0.62, an average loss per game
1. Multiply each payoff by its probability: (6)(6/52) + (2)(6/52) + (-2)(40/52) = 36/52 + 12/52 - 80/52.
2. Add those three fractions: (36 + 12 - 80)/52 = -32/52.
3. -32/52 rounds to about -$0.62, an average loss of 62 cents per game.
4. Option D applies only the largest single loss amount and ignores the weighting, so it overstates the true average loss.
_Source: OpenStax Introductory Statistics (CC BY 4.0), Ch 4 "Discrete Random Variables", section 4.2 Mean or Expected Value and Standard Deviation_
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