Probability of no void in a bridge table 52 cards, 13 of four different suits are randomly distributed to 4 people, each person receiving exactly 13 cards. What is the probability that all 4 persons get at least one card of each suit? (That is there is no void anywhere in the table?)
Note: The question is not about one particular person having a void but about no void in the bridge table. It is possible to work out the solution of the above problem from this by considering case by case break-up. But this is very tedious. Is there any elegant solution?  
 A: The following solution gives an answer which is relatively concise, but requires a fair amount of computing by a computer algebra system (a few minutes on my aging PC).
There are $\binom{52}{13 \; 13 \; 13 \; 13}=53644737765488792839237440000$ (a multinomial coefficient) ways to deal four bridge hands, all of which we assume are equally likely.  We want to count the number of deals in which no player is void in any suit.  
To that end, we will find a generating function for the answer to a more general problem: How many ways are there to deal $i$ cards to player 1, $j$ cards to player 2, $k$ cards to player three, and $\ell$ cards to player 4 with no player having a void in any suit?  Let's say that number is $a_{ijk \ell}$, and define $f(w,x,y,z)$ to be the four-variable generating function of $a_{ijk \ell}$, i.e.
$$f(w,x,y,z) = \sum_{i,j,k,\ell} a_{ijk \ell} w^i x^j y^k z^{\ell}$$
Now define
$$S_0(w,x,y,z) = (1+w+x+y+z)^{13} $$
$$S_1(w,x,y,z) = (1+x+y+z)^{13} +(1+w+y+z)^{13}+ (1+w+x+z)^{13}+(1+w+x+y)^{13}$$
$$S_2(w,x,y,z) = (1+w+x)^{13}+(1+w+y)^{13}+(1+w+z)^{13}+(1+x+y)^{13}+(1+x+z)^{13}+(1+y+z)^{13}$$
$$S_3(w,x,y,z) = (1+w)^{13}+(1+x)^{13}+(1+y)^{13}+(1+z)^{13}$$
and
$$S_4 = 1$$
The point of these definitions is that $S_0-S_1+S_2-S_3+S_4$ consists of the terms in $(1+w+x+y+z)^{13}$ of the form $w^i x^j y^k z^{\ell}$ with $i,j,k$ and $\ell$ all positive.
Then
$$f(w,x,y,z) = [S_0(w,x,y,z)-S_1(w,x,y,z)+S_2(w,x,y,z)-S_3(w,x,y,z)+S_4(w,x,y,z)]^4$$
A computer algebra system can expand $f(w,x,y,z)$ and find the coefficient of $w^{13} x^{13} y^{13} z^{13}$, which is the number of bridge deals in which no player has a void in any suit.  Courtesy of Mathematica, we have
$$[w^{13} x^{13} y^{13} z^{13}] f(x,w,y,z) = 43786641333457372963248979200$$
So the probability that no player has a void in any suit is 
$$\frac{[w^{13} x^{13} y^{13} z^{13}] f(x,w,y,z)}{\binom{52}{13 \; 13 \; 13 \; 13}} \approx \boxed{0.8162336728}$$
Readers unfamiliar with generating functions may find the answers to this question interesting: How can I learn about generating functions?
