Back to Probability Theory

Problem

• Three cards are randomly selected from a standard deck.  What’s the probability they’re all hearts given that the cards are selected:
  • Consecutively with replacement
    • that is, putting the selected card back in the deck
  • Consecutively without replacement
    • that is, holding onto the selected card
  • Simultaneously

Solution

Selecting three hearts consecutively with replacement

• Picking three cards consecutively with replacement means selecting the first card, returning it to the deck, selecting the second card, returning it, and selecting the third card.  Since each selection is independent, not affecting the probability of the others, the Special Conjunction Rule applies:
  • P(H1&H2&H3) = P(H1) x P(H2) x P(H3) =   ¼ x ¼ x ¼ = 1/64

Selecting three hearts consecutively without replacement

• Picking three cards consecutively without replacement means selecting the first card and holding it; selecting the second card and holding; and selecting the third card.  Unlike with-replacement, the probabilities change with each selection
• The probability of a heart on the first pick is ¼, as earlier.  But for the second pick, assuming a heart was selected on the first draw, the deck has changed, having 51 cards with 12 hearts; so the probability of selecting a heart is 12/51.  For the third pick, assuming the first two were hearts, 50 cards remain with 11 hearts; so the probability of a heart is 11/50.  The probability of three hearts is therefore:
  • 13/52 x 12/51 x 11/50 = 1716 / 132600 = 0.01294.
• The principle underlying the calculation is the General Conjunction Rule (GCR):
  • P(A&B) = P(A|B) x P(B).
• P(A|B), read the probability of A given B, expresses conditional probability, the probability that something is true if something else is true.  The formal definition is:
  • P(A|B) = P(A&B) / P(B), where P(B) > 0.
• The General Conjunction Rule, which derives from the definition of conditional probability, is like the Special Conjunction Rule except that P(A) is replaced by P(A|B).  The idea is that the probability of P(A&B) = the probability of A multiplied by the probability of A given B, i.e. P(A|B).  A corollary of GCR is that P(A|B) x P(B) = P(B|A) x P(A), since P(A&B) = P(B&A).
• For three hearts with no replacement the General Conjunction Rule yields:
  • P(H1&H2&H3) =  P(H1) x P(H2| H1) x P(H3| (H1&H2)) =  13/52 x 12/51 x 11/50 = 1716 / 132600 = 0.01294.
• P(H2|H1) is the probability of picking a heart on the second draw given that a heart was picked on the first.  Thus to say P(H2|H1)  = 12/51 is to say if a heart is selected on the first pick, the probability of a heart on the second is 12/51.  
• Similarly, P(H3|(H1&H2)) is the probability of picking a heart on the third draw given that hearts were picked on the first two.  To say P(H3| (H1&H2)) = 11/50 is to say if hearts are selected on the first two picks, the probability of a heart on the third is 11/50.

Selecting three hearts simultaneously

• Selecting three cards simultaneously doesn’t differ materially from selecting them without replacement.  As long as the cards are not returned to the deck, it doesn’t matter whether they’re picked simultaneously or consecutively; picking a heart still alters the probability of picking another heart. The probability of selecting three hearts simultaneously is therefore 0.01294.