I am trying to find the formula for the probability distribution of the coupon collector problem, but considering existing draws.
Let's say there are $m$ coupons to collect each of which having an equal odds of being selected. Let's call $N$ the random variable associated with the number of coupons one has to buy to complete the set (at least 1 of each $m$ coupons).
Literature shows that the general formula is: $$P(N=n)=\frac{m!}{m^n}{n-1 \brace m-1}$$ where ${n \brace k}$ is a Stirling number of the second kind.
Now let's say one already has drawed $k$ unique coupons at least once, with $0 \le k \le m$. I have been trying to find the formula, without much success so far. It's trivial that when $k=0$ the above formula should be found, and when $k=m-1$ then it's a simple geometric distribution with probability $p=\frac{1}{m}$. Anything in between, I am at loss.
Let $N_k$ be the number of additional coupons you need to draw, where you already have found $k$ coupons.
First, I will prove that, for all $n\in \{0,1,2,\dots\}$, $$ P(N_k\le n)=\sum_{i=0}^{m-k}(-1)^i\binom {m-k}i \left(1-\frac im\right)^n $$ This fully answers your question, since $P(N_k=n)=P(N_k\le n)-P(N_k\le n-1)$.
In order for $\{N_k\le n\}$ to occur, the next $n$ coupons you would draw need to contain all $m-k$ of the unseen coupons. The formula above computes the probability that $n$ coupons suffice using the principle of inclusion exclusion. For each $i\in \{1,\dots,m-k\}$, let $E_i$ be the event that that the $i^\text{th}$ unseen coupon was not present in the $n$ coupons drawn. Then $$ \begin{align} P(N_k\le n) &= P(\overline {E_1}\cap \dots \cap \overline{E_n}) \\&= \sum_{i=0}^{m-k}(-1)^i\sum_{1\le j_1 <\dots <j_i\le m-k}P(E_{j_1}\cap \dots \cap E_{j_i}) \\&= \sum_{i=0}^{m-k}(-1)^i\binom {m-k}iP(E_{1}\cap E_2\cap \dots \cap E_{i}) \end{align} $$ Finally, in order for $E_1\cap \dots \cap E_i$ to occur, the first $i$ coupons must not be chosen. This clearly means that $P(E_1\cap \dots \cap E_i)=(1-i/m)^n$, which leads to the formula advertised at the beginning.
Alternatively, you can also express this probability in terms of the Stirling numbers of the second kind. When you start with $k$ seen coupons, and draw $n$ additional coupons, let $W$ be the number of the additional coupons which were not among the $k$ seen. $$ \begin{align} P(N_k= n) &=\sum_{w=0}^{n} P(N_k= n\mid W=w)P(W=w) \\ &=\sum_{w=m-k}^{n} \frac{(m-k)!}{(m-k)^w}{w-1\brace m-k-1} \cdot \binom{n}{w}(1-k/m)^w(k/m)^{n-w} \end{align} $$