Is there a simple way to compute the following enumeration problem? A friend of mine gave me the problem to solve: we have N=5000 unique bowls and we draw m=6000 of them with replacement. What is the probability to see at least once each one of the 5000 bowls in our 6000 draws?
I have made tedious computation to find the answer (validated with a Monte-Carlo simulation with smaller values of m and N).
I got the impression that I've missed something so my question is the following: is there a simpler way to compute this enumeration problem?
 A: This probability is related to the Stirling numbers of the second kind
defined as
$$
S(k,n) = \frac{1}{n!} \sum_{i=0}^n (-1)^i \binom{n}{i}(n-i)^k
$$
Then, if you have a set of $n$ items, when you draw $k$ of them with replacement, the probability of having each of these item in the sample is
$$
p=\frac{S(k,n) n!}{n^k}
$$
Here $n^k$ is the total number of possible results while $S(k,n) n!$ gives the number of results that contain each item.
A: I am not a mathematician and just love solving probability problems in my free-time, so don't 100% trust this:
I think this can be solved numerically! Just split the question in many different easy-to-solve questions: 
Go through all numbers of N drawings needed to fulfill the requirement N=5000 unique bowls with the last draw.
So if N = 5000, it's super easy: P = fact(5000)/5000**5000
Unfortunately for all N > 5000 it's more complicated! Here I define small n as the additional tries over 5000. (n = N-5000)
This is my suggestion: (use this formula for all n of 1-1000 and sum the results)

Where the left part represents the successes, the middle part the permutations and the right part the failures. Here you find this magic number 1117.63, which is the geometric mean probabilities of failure, linearly weighted by their probabilities of success.
I completely understand if you don't agree with this magic number; maybe you can do a simulation to determine it a better way! But my general recommendation stands: Try to add some numeric properties to your simulation and split it up in easier parts. I am sure you will find a solution this way if you invest enough time!
EDIT:
So the idea is that you realize that your formula has 2 really straight forward parts, and the right part can be evaluated with a less computationally demanding simulation!
