Is There a Closed Form for This Recursive Probability Function?

[CRGreathouse]

I'm not sure if I should be posting here or in General, but here goes.

I have a probability problem and I'm trying to get a closed form, or something resembling it, for a tricky recursive formulation. This problem is the 'simple' base case for a much more complicated problem, but I think I can solve it (in recursive terms) if I can somehow simplify this special case.

The variables (a, b, c, d) represent a partition of a whole, thus all must be nonnegative.

Base cases:

\mathfrak{f}(0, b, c, d) = 0

\mathfrak{f}(a, 0, c, d) = 0

\mathfrak{f}(a, b, c, d) = 0 if any of {a, b, c, d} are negative

Recursive case:
\binom{a+b+c+d}{2}\mathfrak{f}(a,b,c,d)=
ab+ac\mathfrak{f}(a-1,b,c-1,d)+bc\mathfrak{f}(a, b-1,c-1,d)<br /> +\binom{c}{2}\mathfrak{f}(a,b,c-2,d)+cd\mathfrak{f}(a,b,c-1,d-1)

Any ideas for me? Does this resemble some famous (hopefully solved) problem? Should I just give up and use a computer algebra system for the whole messy expression (there are several layers of recursion to go for the general solution!)?

 

[CRGreathouse]

Well, I've started to work on the problem with Maxima to get a handle on the simplest cases of this special case. (The full problem is in 5 or 6 variables and seems quite intractable.)

Trivially, we have the special case for c = 0:
f(a,b,0,d)=\frac{ab}{\binom{a+b+d}{2}}

Unexpectedly, c = 1 simplifies to the same thing (!) even though the denominators are initially different. This was not clear to Maxima, but some factoring by hand showed it to be correct:
f(a,b,1,d)=\frac{ab}{\binom{a+b+d}{2}}

(c = 3?) I made mistakes calculating this the first time, but it seems now to be the same as the others... if that's true in general then the problem is solved in this special case! I'm going to check this over carefully since I don't believe it.