probabilty problem on cards

[Brendan McKay]

A while back I posed this problem on the net:

If a standard 52-card pack is dealt out at random to four players,
what is the probability that each player has at least two cards
of each suit?

The reappearance of this problem recently (incorrectly stated) reminded
my that I hadn't posted the solution yet. 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, our
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.

Brendan McKay.
Computer Science Department, Australian National University,
GPO Box 4, Canberra, ACT 2601, Australia.

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[b...@anucsd.uucp]

In article <2...@anucsd.OZ>, b...@anucsd.OZ (Brendan McKay) writes:
> Mark Brady (Waterloo) provided the only exact solution, within a few
>

OOps, that should be Mark Brader. Sorry, Mark.

[Mark Brader]

Brendan McKay (b...@anucsd.OZ) writes:
> A while back I posed this problem on the net ...
>
> Mark Brady (Waterloo) provided the only exact solution ...

I don't usually complain publicly about people's spelling, but that should be
Mark Brader (Toronto).

----
Mark Brader (Toronto)

P.S. -- It WAS his turn. In the mail exchange when I gave the solution,
I called Canberra "Melbourne". However, I suspect "(Waterloo)" was
simply transferred from R.C.Read to me.