Expected number of coupons occurring at most once As a variation on the coupon collector problem, I would like to find on average how many coupons ($n$ types of coupons) occur in $k$ trials are picked at most once.
I have approached this by summing the probability for a coupon occurring once (binomial) and the probability of it occurring 0 times, this gives me the probability of a single coupon occurring at most once.
However, it has occurred to me that the subsequent coupons don't seem to be independent of this probability, and hence I can't really just multiply this probability by $k$
I do not necessarily need to find an exact expression, and an approximation is acceptable. The value of $k$ I am interested in is $n\ln{n}$ but I am not sure if and how it will help with approximating. I don't know how to evaluate if pretending the probabilities are independent and simply summing them is an acceptable approximation.
I've found something similar here which seems relevant but I am struggling to understand it completely as it seems quite general.
Probability distribution of number of unique coupons after multiple draws
 A: As JMoravitz already stated in comments, linearity of expectation doesn't require independence. The probability that a given coupon type is drawn at most once in $k$ trials is
$$
n^{-k}\left((n-1)^k+k(n-1)^{k-1}\right)\;,
$$
and by linearity of expectation the expected number of coupon types drawn at most once in $k$ trials is
$$
n\cdot n^{-k}\left((n-1)^k+k(n-1)^{k-1}\right)=(n-1+k)\left(1-\frac1n\right)^{k-1}\;.
$$
A: Confirming the  accepted answer  we have  a very  simple combinatorial
class here:
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SEQ}_{=n}(\mathcal{U}+\mathcal{U}\times\mathcal{Z}
+ \textsc{SET}_{\ge 2}(\mathcal{Z}))$$
which gives the generating function
$$G(z, u) = (u+uz+\exp(z)-1-z)^n.$$
The desired statistic is
$$\frac{1}{n^k} k! [z^k] \left. \frac{\partial}{\partial u}
G(z, u) \right|_{u=1}
\\ = \frac{1}{n^k} k! [z^k] \left.
n (u+uz+\exp(z)-1-z)^{n-1} (1+z) \right|_{u=1}
\\ = \frac{1}{n^{k-1}} k! [z^k] (1+z) \exp((n-1)z)
\\ = \frac{1}{n^{k-1}} ((n-1)^k + k (n-1)^{k-1})
= (n-1+k) \frac{1}{n^{k-1}} (n-1)^{k-1}
\\ = (n-1+k) \left(1-\frac{1}{n}\right)^{k-1}.$$
