Coupon collector targeting a collection among many I am interested in the following problem:
We are given a universe $U$ of $n$ coupons, partitioned into $k$ collections, $C_1,\dots C_k$.
At each time step $t$, a coupon $X_t$ is selected uniformly at random in $U$.
I am interested in the expectation (and other properties) of the random variable $T$ which models the completion of at least one collection:
$$T = \min \{t > 0: \exists i, C_i\subseteq\{X_1,\dots,X_t\}\}$$
There are a few cases which have already been studied:

*

*If $k = 1$, this is just the usual coupon collector

*If $|C_i| = 1$, this is a geometric random variable of parameter $k/n$
Question: Has this problem been studied in the litterature ? (I haven't found any references)
Are there similar problems which might give insights on how to tackle this problem ?
Finally, I am also interested in an extension of the above problem where the $C_i$ are pairwise disjoint but do not necessarily cover $U$.
This version of the problem is somewhat easier to tackle since there is a recursive formula for $T$, considered as a function of $|C_1|,\dots,|C_k|$ (we move coupons that we have already collected to the part of $U$ that is not covered by the collection).
 A: Let us obtain an explicit expression for the distribution of $T$. This will be done for any family $(C_1,\dots,C_k)$ of subsets of $[n]:=\{1,\dots,n\}$. However, the formula will be rather complicated, expressed in terms of the Stirling numbers $S_t^{(j)}$ of the second kind, with $S_t^{(j)}$ defined as the number of ways to partition a set, say $A$, of cardinality $|A|=t$ into $j$ non-empty subsets. There is an explicit expression for $S_t^{(j)}$.
Consider the random set
\begin{equation*}
    S_t:=\{X_1,\dots, X_t\}. 
\end{equation*}
Let
\begin{equation*}
    p_{t,j}:=p_{n;t,j}:=P(|S_t|=j). 
\end{equation*}
Then for all integers $t\ge0$ and $j\ge0$
\begin{equation*}
    p_{0,j}=1(j=0),\quad p_{t,0}=1(t=0),\tag{1}
\end{equation*}
and, by conditioning on $|S_{t-1}|$, we get
\begin{equation*}
    p_{t,j}=p_{t-1,j}\,\frac jn+p_{t-1,j-1}\,\frac{n-j+1}n \tag{2}
\end{equation*}
if $t,j\ge1$.
Using the substitution
\begin{equation*}
    p_{t,j}=\frac{s_{t,j}}{n^t}\,\binom nj j!,
\end{equation*}
we rewrite (1)--(2) in an $n$-free way:
\begin{equation*}
    s_{0,j}=1(j=0),\quad s_{t,0}=1(t=0),\tag{1s}
\end{equation*}
for all integers $t\ge0$ and $j\ge0$ and
\begin{equation*}
    s_{t,j}=s_{t-1,j}\,j+s_{t-1,j-1} \tag{2s}
\end{equation*}
if $t,j\ge1$. Relations (1s) and (2s) are satisfied by, and hence determine, the Stirling numbers $S_t^{(j)}$ of the second kind. So, for all integers $t\ge0$ and $j\ge0$ we have $s_{t,j}=S_t^{(j)}$ and hence
\begin{equation*}
    p_{t,j}=\frac{S_t^{(j)}}{n^t}\,\binom nj j!. \tag{3}
\end{equation*}
Let now $R_j$ denote a set selected at random from the set $\binom{[n]}j$ of all subsets of $[n]$ of cardinality $j$. Then for any $C\subseteq[n]$
\begin{equation*}
    P(R_j\supseteq C)=\pi_{n,j,|C|}:=\binom{n-|C|}{j-|C|}\Big/\binom nj. \tag{4}
\end{equation*}
So, by conditioning on $|S_t|$ and then recalling (3) and using the mentioned explicit expression for $S_t^{(j)}$, we get
\begin{align*}
    P(S_t\supseteq C)&=\sum_{j=0}^n p_{t,j}\,\pi_{n,j,|C|} \\ 
    &=\frac1{n^t}\sum_{j=0}^n \binom{n-|C|}{j-|C|}\, j!\, S_t^{(j)} \\
    &=\frac1{n^t}\sum_{j=0}^n \binom{n-|C|}{j-|C|}\, \sum_{i=0}^j(-1)^{j-i}\binom ji i^t \\  
    &=\frac1{n^t} \sum_{i=0}^n i^t \sum_{j=i}^n (-1)^{j-i}\binom{n-|C|}{j-|C|}\,\binom ji  \\   
    &=\frac1{n^t} \sum_{i=0}^n i^t \binom{n-|C|}{i-|C|} \, _2F_1(i+1,i-n;i-|C|+1;1).    
    \tag{5}
\end{align*}
Finally, by the inclusion--exclusion formula,
\begin{align*}
    P(T\le t)&=P\Big(\bigcup_{i\in[k]}\{S_t\supseteq C_i\}\Big) \\ 
    &=\sum_{r\in[k]}(-1)^{r-1}\sum_{J\subseteq[k],\,|J|=r}P(S_t\supseteq C_J), 
\end{align*}
where $C_J:=\bigcup_{i\in J}C_i$, with $P(S_t\supseteq C_J)$ computed according to (5).
A: There a well known generating function  methods (the ''symbolic method'' and ''Poissonization'') which can be used to deal with this kind of question.
However, I am unable to point to a reference for this specific question (but it may well be out there).
Here is a self contained answer to get you started.
I rephrase your general question as an urn problem. Denote the cardinality of $C_i$ by $c_i$ and collect the useless coupons in class $C_{k+1}=U\setminus\bigcup_{i=1}^k C_i$, of cardinality $c_{k+1}=n-\sum_{i=1}^k c_i$.
Consider an urn which for $i\in\{1,\ldots k\}$ contains $c_i\geq 1$  distinguishable coupons of type $i$, and $c_{k+1}\geq 0$ further distinguishable coupons, altogether
$n:=c_1+\ldots+c_{k+1}$ coupons.
Coupons are drawn with replacement from this urn until for the first (random) time $T$ for an $i\in\{1,\ldots,k\}$ all different coupons
of type $i$ have been drawn at least once.
Let $Y_1(\ell),\ldots,Y_{k+1}(\ell)$ be the random variables  $Y_i(\ell):=$ "number of different coupons of type $i$'' that have been drawn
at "time" $\ell$.
I use generating functions and start from the following basic
Proposition
The generating function of (the joint distribution of)
$Y_1(\ell),\ldots,Y_{k+1}(\ell)$ is given by:
\begin{equation*}
\mathbb{E}\, t_1^{Y_1(\ell)}\ldots t_n^{Y_{k+1}(\ell)}=\frac{\ell!} {n^\ell}[t^\ell]  (1+(e^t-1)\,t_1)^{c_1}\cdot\ldots\cdot(1+(e^t-1)\,t_{k+1})^{c_{k+1}}
\end{equation*}
Proof Let $j_1+\ldots +j_{k+1}\leq\ell$.
Clearly
$$\mathbb{P}(Y_1(\ell)=j_1,\ldots,Y_{k+1}(\ell)=j_{k+1})=\frac{1}{n^\ell}\cdot {c_1 \choose j_1}\cdots {c_{k+1} \choose j_{k+1}}\cdot \mathrm{Sur}(\ell, j_1+\ldots+j_{k+1})$$
where $\mathrm{Sur}(\ell,r)$ denotes the number of surjective mappings from $\{1,\ldots,\ell\}$ onto $\{1,\ldots,r\}$. (Namely, once the (possibly empty) sets $J_i\subset C_i$ of cardinality $j_i$ of appearing coupons from class $C_i$ are chosen, the sequence of draws may be viewed as the value table of a surjective mapping from $\{1,\ldots,\ell\}$ to $\bigcup_{i=1}^{k+1} J_i$. )
It is known that
$\mathrm{Sur}(\ell,r)=\ell!\,[t^\ell]\,(e^t-1)^r$ (since a such a surjective mapping corresponds uniquely to an ordered partition of
$\{1,\ldots,\ell\}$ in $r$ into non-empty subsets, and  $e^t-1$ is the exponential generating function for non-empty sets). The assertion
about the g.f. follows.  End of proof
(I)  The distribution of $T$
Since $\{\,T > \ell\}=\{\,Y_1(\ell) <  c_1,\ldots,Y_{k}(\ell)<c_k\,\}$ the above gives
\begin{equation*}
\mathbb{P}(T>\ell) =\frac{\ell!} {n^\ell}[t^\ell]\, e^{c_{k+1}t} \prod_{i=1}^k \big(e^{tc_i} -(e^t-1)^{c_i})\big)
\end{equation*}
(II) The expectation of $T$
Finally, using $\mathbb{E} T=\sum_{\ell\geq 0} \mathbb{P}(T>\ell)$ and writing $\frac{\ell!}{n^\ell} =n\,\int_0^\infty\,s^\ell e^{-ns}\,ds$ leads to
$$\mathbb{E}(T)=n\int_0^\infty \prod_{i=1}^k\big(1-(1-e^{-s})^{c_i}\big)\,ds$$
ADDED Similar formulas for the expectation of related waiting times can be found here (Theorem 3.1) https://www.sciencedirect.com/science/article/pii/0166218X9290177C
A: You can modify a little the problem replacing the deterministe $t$ by a random Poisson variable $\hat{t}$ of expectation $t$. For large $t$, because $|\hat{t}-t|\lesssim\sqrt{t}$ (Central Limit Theorem), this new problem is (almost) equivalent to the initial one.
The good news is that now everything is completly explicite: Indeed the familly $$Y_u =\sum_{s\leq \hat{t}}1_{X_s=u} $$ (ie: the number of tickets of $u\in U$ after $\hat{t}$ drawing) are iid Poisson variable of parameter $\frac{t}{n}$. And then $$\{C_i\subset\{X_1,\cdots,X_{\hat{t}}\}\}=\{\forall u\in C_i:Y_u\geq 1\}$$
are independant (because $C_i$ are disjoint) Bernoulli of parameter $(1-\exp(-\frac{t}{n}))^{|C_i|}$.
