# Richard Pavlicek's combinatorial problem

In the game of bridge, a standard deck is dealt to four players, 13 cards each. That gives a total of $\binom{52}{13,13,13,13}$ distinct deals.

How many distinct deals can be dealt if all spot cards - $2$ through $9$ - are considered equal in rank (but retain their suits.)

This question originates on Richard Pavlicek's bridge site.

He asks a combinatorial question. I am also interested in a solution but unfortunately, I have no plan how the formula can be constructed.

Can anyone help ?

• You could at least state the problem yourself. – Robert Israel Apr 14 '15 at 21:23
• @Peter Done. It was odd to see Pavlicek's name on a math site. I'm so used to seeing his name on bridge sites. – Thomas Andrews Apr 14 '15 at 21:38
• @thomas if you are a bridge freak, you might be interested in this question : boardgames.stackexchange.com/questions/23788/… – Peter Apr 14 '15 at 21:44
• I think I invented the phrase "par zero deal," actually. @Peter At least, I hadn't seen it anywhere before I wrote up a list of them. If you google "Par Zero Deal" my collection is the first thing that comes up. – Thomas Andrews Apr 14 '15 at 21:45
• @thomas Did you collect some more par-zero deals ? If yes, you might create a site with all par-zero-deals you know. I am very interested in this topic and I would like to see surprising or even crazy par-zero-deals. – Peter Apr 14 '15 at 21:47

This problem can be solved by the Polya Enumeration Theorem. Ignore the no spot cards for the moment as they only contribute trivial symmetries. The setup here is that we have $8\times 4$ slots into which we distribute assignments to players $A,B,C$ and $D.$ We have the symmetric group $S_8$ acting on each block of spot cards from the same suit. This gives the cycle index $$Z(Q) =Z(S_8)^4.$$

It follows that the number of deals where player $A$ receives $a$ spot cards, player $B$ receives $b$ spot cards and so on is given by $$[A^a B^b C^c D^d] Z(S_8)^4(A+B+C+D).$$

If all four degrees are at most thirteen we can combine such an assignment with an assigment of the no spot cards of the remaining cards, which is given by a simple multinomial coefficient, for a contribution of $${20\choose 13-a,13-b,13-c,13-d} [A^a B^b C^c D^d] Z(S_8)^4(A+B+C+D).$$ It remains to sum these terms from $Z(Q)$ to get the answer, which is $$800827437699287808.$$

Observe that all of these terms have $a+b+c+d=32.$ Note also that the multinomial corresponds to multiplying $Z(Q)$ by $a_1^{20},$ representing twenty fixed points for the no spot cards which are not being permuted.

Here the computation features the recurrence by Lovasz for the cycle index $Z(S_n)$, which is $$Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l Z(S_{n-l}) \quad\text{where}\quad Z(S_0) = 1.$$

This was the Maple code that I used.

with(combinat);
with(numtheory);

pet_cycleind_symm :=
proc(n)
local p, s;
option remember;

if n=0 then return 1; fi;

end;

pet_cycleind_idspots := pet_cycleind_symm(8)^4;

pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;

res := ind;

polyvars := indets(poly);
indvars := indets(ind);

for v in indvars do
pot := op(1, v);

subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];

subs2 := [v=subs(subs1, poly)];

res := subs(subs2, res);
od;

res;
end;

count :=
proc()
option remember;
local sind, res, term, Ad, Bd, Cd, Dd;

sind := pet_varinto_cind(A+B+C+D, pet_cycleind_idspots);
res := 0;

for term in expand(sind) do
Bd := degree(term, B);
Cd := degree(term, C);
Dd := degree(term, D);

if Ad<=13 and Bd<=13 and Cd<=13 and Dd<= 13 then
fi;
od;

res;
end;


The output of the Maple program is as follows. The timing here was less than one tenth of a second.

> count();
memory used=37195.2MB, alloc=8.3MB, time=436.90
memory used=37197.7MB, alloc=8.3MB, time=436.94
800827437699287808


This matches the value presented at the linked web site.

A somewhat more advanced computation involving suits being distributed is at this MSE link.

• It would be neat to see what this algorithm we could get with all cards except the ace and king treated as equal, so we could possibly test it. – Thomas Andrews Apr 15 '15 at 16:41
• If I have not made a mistake (which is unfortunately not that unlikely) the answer I get is $$630343600320,$$ which it seems is still beyond reach of brute force enumeration. – Marko Riedel Apr 15 '15 at 18:57
• Well, there are steps between brute force and your algorithm. :) – Thomas Andrews Apr 15 '15 at 18:59
• Thanks for pointing that out, I should have reflected on my choice of terms here before commenting. In fact the algorithm on the linked web page should not be difficult to adapt to this new scenario. – Marko Riedel Apr 15 '15 at 19:01