This happens to be very similar to a problem that has recently
appeared here, namely at this MSE
link. We ask the
reader to consult this link for an introduction and additional
background material.
We start with the species of ordered set partitions with sets of more
than two elements marked. This is
$$\mathfrak{S}(\mathcal{U}\mathcal{Z}
+\mathcal{U}\mathcal{V}\mathfrak{P}_{\ge 2}(\mathcal{Z})).$$
We thus obtain the generating function
$$G(z, u, v) = \frac{1}{1-u(v\exp(z)-vz+z-1)}.$$
We then get for the probability that
$$P[T=m] = \frac{1}{n^m} {n\choose n-1}
(m-1)! [z^{m-1}] [u^{n-1}] G(z, u, v).$$
What happens here is very simple. We choose the $n-1$ coupons that go
into the prefix consisting of $m-1$ draws. Then we partition those
draws into sets, one for each type of coupon, containing the position
where it appeared. We mark sets of more than two elements. Doing the
extraction in $u$ we find
$$P[T=m] = \frac{1}{n^m} {n\choose n-1}
(m-1)! [z^{m-1}] (v\exp(z)-vz+z-1)^{n-1}.$$
Now to do the usual sanity check that we have a probability
distribution we remove the marking in $v$ and obtain
$$\sum_{m\ge 1} P[T=m]
= \sum_{m\ge 1} \frac{n!}{n^m}
(m-1)! [z^{m-1}] \frac{(\exp(z)-1)^{n-1}}{(n-1)!}.$$
This was evaluated at the cited link and the sanity check goes
through, more or less by inspection in fact. Continuing with the
expectation of coupons that were drawn mor than once we differentiate
with respect to $v$ and set $v=1$, getting
$$\frac{n! \times (m-1)!}{n^m}
[z^{m-1}] (n-1) \left.\frac{(v\exp(z)-vz+z-1)^{n-2}}{(n-1)!}
\times (\exp(z)-z)\right|_{v=1}
\\ = \frac{n! \times (m-1)!}{n^m}
[z^{m-1}] \frac{(\exp(z)-1)^{n-2}}{(n-2)!} \times (\exp(z)-z).$$
We write this in three pieces, namely
$$\frac{n! \times (m-1)!}{n^m}
[z^{m-1}] \frac{(\exp(z)-1)^{n-1}}{(n-2)!}
\\ - \frac{n! \times (m-1)!}{n^m}
[z^{m-2}] \frac{(\exp(z)-1)^{n-2}}{(n-2)!}
\\ + \frac{n! \times (m-1)!}{n^m}
[z^{m-1}] \frac{(\exp(z)-1)^{n-2}}{(n-2)!}.$$
Consulting the results from the main link we find for the first two pieces
$$n-1 - (H_n - 1) = n - H_n.$$
We then get for the third piece (recognizing the Stirling number EGF
and observing that the EGF morphs into an OGF)
$$\frac{n!}{n} \sum_{m\ge 1} \frac{1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-2} \frac{z}{1-qz}
= \frac{n!}{n} \prod_{q=1}^{n-2} \frac{1/n}{1-q/n}
\\ = \frac{n!}{n} \prod_{q=1}^{n-2} \frac{1}{n-q}
= \frac{n!}{n} \frac{1}{(n-1)!} = 1.$$
We thus have for the answer that the sibling collects $n+1-H_n$
coupons and hence is missing $H_n-1$ coupons probabilistically from
among the coupons collected in the prefix. Furthermore and
deterministically, the sibling never sees the last coupon collected
because it is always a singleton. Hence the sibling is missing
$$\bbox[5px,border:2px solid #00A000]{H_n}$$
coupons. We may add the halting singleton because it does not involve
any additional probability and is determined by the set partition of
the prefix.
What have we learned? On seeing this result it immediately becomes
evident that these two parameters (singletons and duplicates) are
prefectly additive on the level of generating functions and we could
have concluded by inspection, citing the result for singletons from
the link without any extra calculation.
