I am having a difficulty with the following multivariate hypergeometric distribution problem. The setting is as usual, an urn contains a total of $M$ balls of $K$ unique colors, with $N_1$ balls of color 1, $N_2$ balls of color 2, ..., $N_K$ balls of color $K$ s.t. $N_1+N_2...+N_K = M$. What is the probability that in a sample of size $n$ (without replacement), the ball drawn last has a color not sampled before. For simplicity, we can assume that $N_1=N_2=...=N_K=N$, i.e. $M=KN$.
I have been trying to look at particular cases with $K=2$ and $K=3$ (2 or 3 colors) with different values of sample size, $n$, hoping I could generalize the formulas for arbitrary $K$ and $n$. Thus, for example, for $K=2$ and any value of $n$, I showed that the probability in question could be found by $K \cdot \frac{{N_1 \choose n-1}{N_2 \choose 1}}{n {M \choose n}}$. For $K=3$, we may have two different cases: a) only 2 out of the available three colors are sampled (with $n-1$ balls of the same color and 1 ball of a second color). The desired probability is then $K(K-1)\frac{{N_1 \choose 2}{N_2 \choose 0}{N_3 \choose 1}}{n {M \choose 3}}$. And case b) all three colors are sampled (1 ball of each), then the desired probability is $K(K-1) \frac{{N_1 \choose 1}{N_2 \choose 1}{N_3 \choose 1}}{n{M \choose 3}}$ and the final answer is the sum of (a) and (b).
Does this logic seem reasonable ? Obviously by increasing the values of $k$ and $n$ the number of cases to keep track of will increase too but it seems that each new case may be simplified into (or represented by) a previously worked out scenario. In short, it seems that I may eventually be able to find some recursive relation but after some tedious work. Any ideas would be greatly appreciated. More specifically - is this a good route to go? If yes, are there any shortcuts that I can take ? Is there a completely different approach that I can try ? Thanks, in advance, Tamar