[Math] Probability that a random 13-card hand contains at least 3 cards of every suit

card-gamescombinatoricsprobability

A random 13-card hand is dealt from a standard deck of cards. What is the probability
that the hand contains at least 3 cards of every suit? (Introduction to Probability, p.36)

My solution:

  • There are $\binom{52}{13}$ possible hands.
  • Because there are 13 cards for the hand, to obtain at least three cards of one suit per hand, we need to have exactly three cards of one suit per hand plus one additional card of any suit, thus $\binom{13}{3}^4 * 4 \binom{10}{1}$
  • Result: $\frac{40*\binom{13}{3}^4}{\binom{52}{13}} = 0.4214$

However, simulating it in R yields:

deck <- rep(1:4, 13)
out <- replicate(1e5,{
  hand <- sample(deck, size=13, replace=FALSE)
  all(table(hand) >= 3)
})
mean(out)
> 0.14387

Can anybody tell me what is wrong?

EDIT

I'm afraid, the correct code should be.

deck <- rep(1:4, 13)
out <- replicate(1e5,{
  hand <- sample(deck, size=13, replace=FALSE)
  length(table(hand))==4 & all(table(hand) >= 3 )
})
mean(out)
> 0.10639

Best Answer

Dominik, your answer was off by a factor of 4. This happened because you counted a hand containing J,K,Q,A of spades (for example) 4 times: (JQK)(A), (QKA)(J), (KAJ)(Q), and (JAQ)(K)

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