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The classic coupon collector problem (or double dixie problem) is the following (refs: https://www.renyi.hu/~p_erdos/1961-09.pdf) . There are $n$ classes of different coupons, and you can buy one coupon at the time at random. The classes are labelled with integer $1,2,...,n$. Let's $\xi_j$ be equal to $k$ if the $j-th$ coupon is of the $k-$th class. These random variables $\xi_j$ are in the classic coupon problem independent, and such that each coupon class is treated equally, i.e. $P(\xi_j=k)=\frac{1}{n}$ for each class $k=1,2...,n$ and time $j=1,2...$. We ask the average number of coupons needed to have at least one coupon for each class. If we call $\nu(n)$ this random number, its average is: $$ \mathbb{E}[\nu(n)]=n\log(n)+O(n). $$ I want now to ask how this average number changes in a more difficult setting. Let's suppose that there is a finite number $M$ of coupons per class. Therefore, the variables $\xi_j$ are not independent: the probability that the extracted $j-$th coupon belongs to a given class $k$ depends on the classes of the previous $j-1$ extracted coupons. In this setting, which is the average number $\mathbb{E}[\nu(n)]$ of samples needed to have at least one coupon per each one of the $n$ classes?

Umberto Tomasini
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    Surely that $o(n)$ should be $O(n)$ (we can even be explicit about what the constant is). As for your question, it certainly depends on the relative sizes of $M$ and $n$: for $M$ bounded there are only $Mn$ total coupons and so the answer is $c \cdot n$ for some constant $c$ that depends on $M$, but if $M$ grows fast enough relative to $n$ then you'll end up back with the $M = \infty$ answer. – JBL Oct 31 '22 at 15:40
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    In my linked answer, I give the exact answer $$\sum_{i=1}^n (-1)^{i-1}\binom{n}{i} \frac{nM+1}{iM+1}, $$ plus this approximation which works well as long as $M$ is large enough:$$nH_n -\frac1M\left[n\left(\frac{(H_n)^2+H_{n,2}}2\right)-H_n\right] +O\left({1\over M^2}\right)$$Here $H_n=1+\frac12+\dots+\frac1n\approx \log n +\gamma$, and $H_{n,2}=\frac1{1^2}+\frac1{2^2}+\dots+\frac1{n^2}$. For large $n$, $H_{n,2}$ is approximately equal to $\pi^2/6$. – Mike Earnest Oct 31 '22 at 19:21
  • Thank you really a lot! Greatly appreciated :) – Umberto Tomasini Oct 31 '22 at 20:17

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