combinatorics - Probability that a random 13-card hand contains at least 3 cards of every suit?

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

Answer

We count the "favourables," the 4-3-3-3 hands. The suit in which we have $4$ cards can be chosen in $\binom{4}{1}$ ways. For each of these ways, the actual $4$ cards in that suit can be chosen in $\binom{13}{4}$ ways. For each of these ways, the cards in the other three suits can be chosen in $\binom{13}{3}^3$ ways, for a total of $\binom{4}{1}\binom{13}{4}\binom{13}{3}^3$.

Remark: Your counting procedure results in multiple counting. Think, for example, of the 4-3-3-3 hand that has the K, Q, J, 10 of hearts, and some specific cards for the rest of the hand. Your calculation views, for example, K, Q, J of hearts, and later 10 of hearts, as different from K, J, 10 of hearts, and then Q of hearts.