TITLE: Counting card distributions when cards are duplicated
QUESTION [6 upvotes]: If we have a deck of 48 different cards and 4 players each get 12 cards, it is well known how to calculate the number of possible distributions: take fac(48) and divide 4 times by fac(12).
In a german card came (Doppelkopf) there are 24 different card types, but 2 copies of each type, that is 48 cards at all.
How many distributions when there are 4 players? My first approach was to assume there are 48 different cards, and then to divide by 2 and by 2 (... 24 times). But this underestimates the real number of distributions. Because: Say player 1 has both spade kings. This has been ruled out already by dividing by fak(12). But if player 1 and player 2 both have a spade king, then we must divide by 2.
How to calculate the number of different distibutions?
Is there a closed formula as in the single card deck case?

REPLY [6 votes]: Here is an elementary approach which works. It is based on Tony Huynh's method of dealing out the hands one at a time, but using transition matrices/dynamic programming instead of sums. 
Let the state of the deck be $(p,s)$ where $p$ is the number of pairs it contains and $s$ is the number of singletons. Dealing out a hand changes the state of the deck, e.g., from the starting deck $(24,0)$ we can go to $7$ possible states: $(18,0),$ $(17,2)$, ..., $(12,12)$. 
The number of ways to go from $(p_1,s_1)$ to $(p_2,s_2)$ by dealing one hand only depends on the labels, and is a single sum over the number of pairs $r$ in the new hand: $\sum_r {p_1 \choose r,p_2,p_1-r-p_2}{s_1 \choose 12-r-p_1+p_2}.$
This gives us transition matrices which are $1\times 7$, $7\times 13$, $13 \times 7$, and $7 \times 1$, whose product in the proper order, as a scalar, is the number of distinct deals, $2248575441654260591964$. 
Here is some Mathematica code which implements this:
transitions[p1_, s1_, p2_, s2_] := 
   Sum[Multinomial[r, p2, p1-p2-r] Binomial[s1, 12+p2-r-p1], {r,0, p1-p2}]
M1 = Table[transitions[24-i, 2i, 18-j, 2j], {i, 0, 0}, {j, 0, 6}]
M2 = Table[transitions[18-i, 2i, 12-j, 2j], {i, 0, 6}, {j, 0, 12}] 
M3 = Table[transitions[12-i, 2i, 6-j, 2j], {i, 0, 12}, {j, 0, 6}]
M4 = Table[transitions[6-i, 2i, 0-j, 2j], {i, 0, 6}, {j, 0, 0}] 
(M1.M2.M3.M4)[[1, 1]]

Final output: 
2248575441654260591964