A dealer splits a standard 52-card deck of cards into 5 piles. The dealer tells you that one of the piles consists of all the face cards and aces.

Question

Each of the remaining piles contains a single suit (so, one of the piles contains all the hearts with face values 2–10, another pile

contains all the spades with face values 2–10, etc.). The dealer asks you to pick a pile at

random and then pick a card from that pile. The dealer looks at the card and tells you that

the suit is a heart. What is the probability that the card is a face card or ace?

Answer 1

This is much like the Monty Hall problem. The probability that it is a face card or ace starts off as 1/5, and this is not altered by the fact that a heart is shown. You can also argue that P(face card or ace|heart) = P(face card or ace|spade) = P(face card or ace|club) = P(face card or ace|diamond). As these four are all equal, they have to equal P(face card) (think Baye's theorem).

Another way to show this.

P(heart and not a face card or ace) = 1/5

P(heart and face card or ace) = P(face card or ace)P(heart|face card or ace) = 1/5 * 1/4 = 1/20

P(face card or ace |heart) = P(face card or ace and heart)/P(heart) = 1/20/(1/20 + 1/5) = 1/20/5/20 = 1/5