Card EV

Solution

Answer: Yes

Let \(X\) be the payoff from drawing one card. A standard deck has \(52\) cards. Hearts: there are \(13\) hearts, so \[ P(\text{heart}) = \frac{13}{52}. \] These pay \(+\$10\). Face cards are \(J, Q, K\). There are \(3\) per suit, so \(12\) face cards total. Among these, \(3\) are hearts, which are already counted in the heart case. Thus the number of face cards that are not hearts is \[ 12 - 3 = 9. \] \[ P(\text{face card not heart}) = \frac{9}{52}. \] These pay \(+\$8\). All remaining cards lead to losing \$6. The number of such cards is \[ 52 - (13 + 9) = 30. \] \[ P(\text{other}) = \frac{30}{52}. \] These pay \(-\$6\). The expected value is \[ E(X) = 10\left(\frac{13}{52}\right) + 8\left(\frac{9}{52}\right) - 6\left(\frac{30}{52}\right). \] \[ E(X) = \frac{22}{52} \approx 0.423. \] So the expected gain is about \[ \$0.42. \] Since the expected value is positive, the game is favorable and you would want to play.