If a standard 52-card pack is dealt out at random to four players, what is the probability that each player...

The answer is exactly

        44164536626535800904
       ---------------------
       214099368476567659799

      = 0.20628 05553 36478 89789 ...

Mark Brady (Waterloo) provided the only exact solution, within a few
hours of the original posting.  Kevin Schmidt (NYU) gave a solution to
14D.  Carl Witthoft (AOA) found it boring.

Here's an outline of one solution (found with help from R.C.Read):

We determine the total number of deals (out of 52!) with the required
properties.  Suppose each player (denoted W,X,Y,Z) has one box for each
suit (denoted S,H,C,D).  Each of the sixteen boxes on the table
contains between two and seven cards.  For the W-S box, for example,
that gives us an exponential generating function of F(WS), where
  F(t) = t^2/2! + t^3/3! + ... + t^7/7!.
The state of the whole table is thus described by the sixteen term
product F(WS)F(WH)...F(ZD), subject to the constraints that there are
13 cards of each suit and 13 cards with each player.  These constraints
correspond to extracting the coefficient of W^13 X^13 ... D^13.  You
then multiply by (13!)^8 to get the required number of deals.

The coefficient extraction can be simplified a little to
Coeff of W^13 X^13 Y^13 Z^13 in [ Coeff of t^13 in F(Wt)F(Xt)F(Yt)F(Zt) ]^4.
From here you can apply your favourite symbolic algebra package, or
write a custom program.  Sorry, I don't know a human-easy way.  [Do
you?]  Mark Brader managed to avoid multi-precise arithmetic except for
a few clean-up operations near the end.

[ Editor: the solver is called "Mark Brady" above and "Mark Brader"
here; the original does not say which spelling is right. ]

-- Brendan McKay <bdm@anucsd.anu.oz>, Computer Science Department,
   Australian National University