Balls in an urn, but they're all different and I want to see all of them.

Question

How would I calculate the number of times I need to pick one of N unique balls at random out of an urn with replacement until I'm 95% confident that I've seen every ball?

For example, let's say there are 16 balls in the urn, labeled uniquely from each other with the letters A thru P. I pick one ball out at random, write down the letter I got, then put the ball back. Repeat. My goal is to write down all 16 different letters (not being allowed to write down a letter until I've seen it on a ball I picked from the urn). Before starting, how many times should I expect to pull a ball to be 95% confident that I would have seen (and written down) each of the 16 letters at least once?

The example is not worth it. Instead, better edit your post to include your thoughts and attempts. Quick beginner guide for asking a well-received question + please avoid "no clue" questions. — Anne Bauval, Jun 28 '25 at 04:39
Given $n$ balls and $k$ draws, it can calculate the probability that it will be done in $k$ turns. But I don't think there is an easy way to compute a good $k$ in terms of $p$ and $n.$ probably the best you can do is get an approximation and then test the value $k$ near the approximation. — Thomas Andrews, Jun 28 '25 at 04:54
The expected number of draws it will take is $n(1+1/2+1/3+\dots +1/n).$ But that is surely less than $95%$ probability done. Still, it will be on the order of that. That value is approximately $n\log n$ and I bet you might reach 95% in at most $2n\log n$ draws, but that is just a guess. — Thomas Andrews, Jun 28 '25 at 05:05

ploosu2 · Answer 1 · 2025-06-28T06:14:27.270

1

The number of surjections from $[n]$ to $[N]$ is given by $N!{n \brace N} = \sum_{j=0}^{N-1}(-1)^j \binom{N}{N-j}(N-j)^n $ and hence the probability of having seen all after we've picked $n$ times is

$$ p(n) = \frac{1}{N^n}\sum_{j=0}^{N-1}(-1)^j \binom{N}{N-j}(N-j)^n $$

Then we solve the least $n$ such that $p(n)>0.95$. For $N=16$ we get $n=90$.

edited Jun 28 '25 at 06:14

answered Jun 28 '25 at 06:01

ploosu2

12,367

score 0 · Answer 2 · answered Jun 28 '25 at 05:29

A useful approximation is that if you are making enough draws to have a $95\%$ chance of seeing them all, you will certainly have seen at least $N-1$ of them. This is not true, you might have seen only $N-2$ of them, but if you follow the ideas here you will see the chance is very low. Having said that, if you make $k$ draws the chance you have never seen one is the chance you have chosen all the $k$ draws from $N-1$ of the balls. If you choose a specific ball to miss, the chance is $\frac {k(N-1)^k}{N^k}$ so the chance you miss any one is $N\frac {k(N-1)^k}{N^k}$ (Here we use the assumption you don't miss more than one as you would double count draws that miss two and so on.) Now you can numerically choose $k$ to get this below $0.05$

Alpha say for $N=16$ you need $169$ draws.

Balls in an urn, but they're all different and I want to see all of them.

2 Answers2