Is there a closed form solution to the following recurrence relation? $$P(i,j) = \frac{i^{5}}{5i(5i-1)(5i-2)(5i-3)(5i-4)}\sum\limits_{k=0}^{j-5}(1-P(i-1,k))$$ where $P(i,j)=0$ for $j<5$.
The above recurrence is the solution I obtained to a probability problem, and I've been trying to simplify it even further by obtaining a closed formula. The problem:
If we have $5i$ colored balls, $i$ of each color. Let's call the colors $1,2,3,4$ and $5$. We pick the balls without replacement until we obtain colors $1,2,3,4$ and $5$ in sequence.Then $P(i,j)$ is the probability that we will pick colors $1,2,3,4$ and $5$ in sequence in at most $j$ trials.
Note that, using a counting argument, one can show that $$P(n,5n)= \frac{{n!}^{5}}{(5n)!}\sum\limits_{i=1}^n \frac{{(5n-4i)!}(-1)^{i+1}}{(n-i)!i!}$$