Majamin, once more, thank you very much for taking the time to propose a solution for my problem, it is much appreciated. If nothing else, it helped me formulate the mathematical problem more precisely.

In the meantime I got a solution from another source, I have tried it on some data and so far it gives logical probabilities.

It is more complex than we both thought - no way I would have come to that formula myself. I will give it here by courtesy and for future users who might find it useful:

A: m draws, r unique outcomes

B: n draws, s unique outcomes

What we want is the the simple probability that m draws from A and n draws from B (all with replacement) produce disjoint sets of outcomes, regardless of what r and s happen to be.

Let's call this P_{disjoint}(N,m,n) (with N=20, "number of balls").

This assesses the likelihood of a certain 'null hypothesis', that urns A and B do in fact contain all the integers [1,20] in equal numbers, against a hypothesis that better fits the data, that urns A and B actually contain disjoint sets of numbers.

The most straightforward way to find P_{disjoint}(N,m,n) is to sum probablities over all possible r:

(sorry for the format, I can't seem to be able to paste the equation from Word...)

P_{disjoint}(N,m,n) = ( SUM(r=0, N) (combine(N, r)*{m,r}*r!*(N-r)^{n}) / N^{m+n}

Here,

combine(N, r) is the number of ways to choose r of the N numbers to be the outcomes of the drawings from urn A,

{m,r} is a Stirling number of the second kind and counts the number of ways that the m outcomes can take the r chosen values assuring that each value is chosen at least once (multiplied by r! because order matters),

(N-r)^{n}is the number of ways to chose the n balls from urn B so that they are disjoint from the numbers drawn form urn A, and

N^{m+n}is the number of ways to choose m balls from urn A and n balls from urn B without regard for disjointness.

Simplifying this expression is not possible due to the nature of Stirling numbers, and it is a real pain to implement in a programming language (java) as the numbers involved are huge (namely, the factorials) and cannot be handled by primitive types (special methods exist for that but there are still a lot of different number types to take into account).

The calculated probability is - oh, by magic - the same whether we take P_{disjoint}(N,m,n) or P_{disjoint}(N,n,m), however the result is different for m1 != m2, n1 != n2, with m1+n1=m2+n2 ('!=' stands for different). Hence the cumulative number of draws is not determinant, the specific numbers of draws from A and B are.

There we go...