Collecting cards The following problem is from Princeton University Mathematics competition.
Kelvin and Quinn are collecting trading cards; there are 6 distinct cards that could appear
in a pack. Each pack contains exactly one card, and each card is equally likely. Kelvin buys
packs until he has at least one copy of every card, then he stops buying packs. If Quinn is
missing exactly one card, the probability that Kelvin has at least two copies of the card Quinn
is missing is expressible as $\frac{m}{n}$ for coprime positive integers $m,n$. Determine $m + n.$
Assuming that Kelvin stops after getting $N$ cards, in which he has exactly one copy of the card, say A,  Quinn is missing, the computation of the probability of this event is a bit messy. The official solution is given below and appears incomplete:
However, we also have the probabilities for each of the new cards that appear. This
is $\frac{5}{6} \cdot \frac{4}{6} \cdots \frac{1}{6} \cdot \frac{1}{6}$ since we are fixing when A appears so we have two copies of $\frac{1}{6}$, one for A
and one for the last card that isn't A, Thus, in total, the probability is $\frac{6^5}{5!\cdot (6-n)} \cdot \frac{5!}{6^6} = \frac{1}{(6-n)\cdot 6}$
We now need to sum this over all possible n, giving us
$$ \sum_{n=0}^6 \frac{1}{6(n-6)} = \frac{49}{120} $$
Since we computed the complement, the probability we want is $1 - \frac{49}{120} = \frac{71}{120}$ and $m+n = 191$.
Some text appears to be missing at the beginning of the official solution.
 A: I answered it here: Probability of exactly 1 given everything appears. Here is a solution:
"Consider pulling cards until you pull  twice. Note that the order of  is equally likely to be the first, second, ..., sixth. Now, condition on the order of , let it be the th unique card. Let us assume that your position  is not equal to 6. Now,  is again equally likely to be the last card among the remaining (6−) cards, giving a probability of  being last 1/(6−+1).
For  to be drawn once in your scenario,  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)∑5=11/(6−+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."
