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Coupon collector's problem asks:

Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once?

The well-known solution is $E(T)=n \cdot H_n$, where T is the time to collect all n coupons(proof).

I am trying to approach another way, by calculating possible arrangements of coupons using inclusion-exclusion(Stirling's numbers of the second kind) and that one coupon should only be collected at last and other coupons should be collected at least once: $$P(T=k)=\frac{n!\cdot{k-1\brace n-1}}{n^k}\\ =\frac{\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot i^{k-1}}{n^{k-1}}\\ E(T)=\sum\limits_{k=n}^{\infty}k\cdot P(T=k)\\ =\sum\limits_{k=n}^{\infty}k\cdot\frac{\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot i^{k-1}}{n^{k-1}}\\ =\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot\sum\limits_{k=n}^{\infty}k\cdot (\frac i n)^{k-1}\\ =\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot(\frac i n)^{n-1}\cdot(\frac 1 {1-\frac i n})\cdot(n-1+\frac 1 {1-\frac i n})$$

Calculation of first 170 terms yields same results.

Are two formulas same?

ylem
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1 Answers1

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By way of enrichment here is a proof using Stirling numbers of the second kind which encapsulates inclusion-exclusion in the generating function of these numbers.

First let us verify that we indeed have a probability distribution here. We have for the number $T$ of coupons being $m$ draws that

$$P[T=m] = \frac{1}{n^m} \times n\times {m-1\brace n-1} \times (n-1)!.$$

Recall the OGF of the Stirling numbers of the second kind which says that

$${n\brace k} = [z^n] \prod_{q=1}^k \frac{z}{1-qz}.$$

This gives for the sum of the probabilities

$$\sum_{m\ge 1} P[T=m] = (n-1)! \sum_{m\ge 1} \frac{1}{n^{m-1}} {m-1\brace n-1} \\ = (n-1)! \sum_{m\ge 1} \frac{1}{n^{m-1}} [z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz} \\ = (n-1)! \prod_{q=1}^{n-1} \frac{1/n}{1-q/n} = (n-1)! \prod_{q=1}^{n-1} \frac{1}{n-q} = 1.$$

This confirms it being a probability distribution.

We then get for the expectation that $$\sum_{m\ge 1} m\times P[T=m] = (n-1)! \sum_{m\ge 1} \frac{m}{n^{m-1}} {m-1\brace n-1} \\ = (n-1)! \sum_{m\ge 1} \frac{m}{n^{m-1}} [z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz} \\ = 1 + (n-1)! \sum_{m\ge 1} \frac{m-1}{n^{m-1}} [z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz} \\ = 1 + (n-1)! \sum_{m\ge 2} \frac{m-1}{n^{m-1}} [z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz} \\ = 1 + \frac{1}{n} (n-1)! \sum_{m\ge 2} \frac{1}{n^{m-2}} [z^{m-2}] \left(\prod_{q=1}^{n-1} \frac{z}{1-qz}\right)' \\ = 1 + \frac{1}{n} (n-1)! \left.\left(\prod_{q=1}^{n-1} \frac{z}{1-qz}\right)'\right|_{z=1/n} \\ = 1 + \frac{1}{n} (n-1)! \left. \left(\prod_{q=1}^{n-1} \frac{z}{1-qz} \sum_{p=1}^{n-1} \frac{1-pz}{z} \frac{1}{(1-pz)^2} \right)\right|_{z=1/n} \\ = 1 + \frac{1}{n} (n-1)! \prod_{q=1}^{n-1} \frac{1/n}{1-q/n} \left. \sum_{p=1}^{n-1} \frac{1}{z} \frac{1}{1-pz} \right|_{z=1/n} \\ = 1 + \frac{1}{n} (n-1)! \prod_{q=1}^{n-1} \frac{1}{n-q} \sum_{p=1}^{n-1} \frac{n}{1-p/n} \\ = 1 + \frac{1}{n} \sum_{p=1}^{n-1} \frac{n^2}{n-p} = 1 + n H_{n-1} = n \times H_n.$$

What we have here are in fact two annihilated coefficient extractors (ACE) more of which may be found at this MSE link. Admittedly the EGF better represents inclusion-exclusion than the OGF and could indeed be used here where the initial coefficient extractor would then transform it into the OGF.

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Marko Riedel
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