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

 A: 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)  
A: 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.
A: I doubt, that it is still interesting for you, but I want to check myself.I think either you or I misunderstand the question. Because it asks to find the probability of the hand contains AT LEAST 3 cards of every suit and you calculated the probability of EXACTLY 3 cards of every suit.
From my point of view answer must be:


*

*for 0 cards of every suit there is no point to calculate, since
nothing happened 

*for 1 cards of every suit (choose from
13-1)^3(choose from 13-remainig 10)(choose from 4-1) and divide all
this to (choose from 52-13) 

*for 2 cards of every suit (choose from 13-2)^3(choose from 13-remainig 7)(choose from 4-1) and divide all this to (choose from 52-13)


Then just substract from 1 the calculated probability and you will get P(>=3)
A: Another solution that's more mechanical and generalizable to other arrangements like two pair or full house:
We know we must have a 4-3-3-3 hand. You can pick any card for the first suit: 52. The next three cards must match the suit of the first so there are 12 x 11 x 10 possibilities for those. Now for the 4th card, we can pick any card not in the 1st suit: 39. Using the same reasoning as before, for the next two cards there are 12 x 11 possibilities. We continue until we have picked all cards:
$$
X = (52 \times 12  \times  11  \times 10) \times (39  \times  12  \times  11)  \times  (26  \times  12  \times  11)  \times  (13  \times  12  \times  11)
$$
$X$ gives us all arrangements of the pattern WWWW XXX YYY ZZZ, as such we overcount all the internal arrangements of W, X, Y, and Z ($4!3!^3$). We also overcount that X, Y, and Z can be interchanged ($3!$).
So the answer is $X /(4!3!^4) / {52 \choose 13}$.
