I'm trying to figure out the following question, and I'm stuck! It comes from an old PUMaC problem with a partial solution (there's a typo in the one provided on Stack Exchange as well):
Stack Post:
Original Problem and Solution (problem 4):
If I'm not mistaken, I think what the problem is saying is that if you have cards $\{A,B,C,D,E,F\}$, drawn independently and with equal probability 1/6 (uniformly), what is (1 minus) the probability that you obtain exactly one A, given that you stop drawing when you've drawn every card? (We can choose A w.l.o.g. by law of total probability).
e.g. $(A,B,C,D,E,E,F)$, $(B,B,C,C,D,F,F,E,E,A)$ <-- these draws work.
According to their partial solution, it seems not so much of a coincidence that we see the harmonic series as would be in the case if we found E(number of draws until all 6 appear) using linearity of expectation. But, beyond that I'm not sure how they achieved this answer.
When I tried to go about it, it seemed like it would take significant case work (dependent on the number of draws and whether or not A is the last to appear).
Consider pulling cards until you pull $A$ twice. Note that the order of $A$ is equally likely to be the first, second, ..., sixth. Now, condition on the order of $A$, let it be the $n$th unique card. Let us assume that your position $n$ is not equal to $6$. Now, $A$ is again equally likely to be the last card among the remaining $(6 - n)$ cards, giving a probability of $A$ being last $1/(6-n + 1)$.
For $A$ to be drawn once in your scenario, $A$ had to be the last card in the first or the second draw (do you see why?), giving the final answer $(1/6) + (1/6) \sum_{n=1}^5 1/(6-n + 1)$, which matches the formula they gave. I believe they used a different counting argument, but since their solution is incomplete, and this one is simpler, you might as well use this.
As an important side note, I think the better way to think about this problem to understand all the symmetries and independences is by considering a Poisson splitting (that is how I solved it). It is a hard problem without Poisson processes for sure.