The coupon collector’s problem describes the following question: there are $n$ different types of coupons and the collector will receive a random coupon every time, what is the expected trials that the collector can collect all $n$ types of coupons? We can conclude that the expected trails grows as $\mathcal{O}(n\log{n})$.

The Distribution of Trails for a new Coupon

Let $T$ be the time to collect all $n$ types of coupons, and $t_i$ be the time to collect a new coupon after $i-1$ coupons have been collected. We can see that the probability of $t_i$ satisfies geometric distribution with $p_i = \frac{n-(i-1)}{n}$, then we conclude

\[E(t_i) = \frac{1}{p_i} = \frac{n}{n-(i-1)}\]

We also obviously know that $t_i$ and $t_j$ are independent. Thus

\[\begin{aligned}E(T) &= \sum{E(t_i)} \\ &= \frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_n} \\ &= \frac{n}{n} + \frac{n}{n-1} + \cdots + \frac{n}{1} \\ &= n \cdot (\frac{1}{n} + \frac{n}{n-1} + \cdots + \frac{1}{1}) \\ &\simeq n\log{n} + \gamma n + \frac{1}{2} + O(\frac{1}{n}) \end{aligned}\]

The variance of the random variable $T$ can be calculated as (since $t_i$ are independent)

\[\begin{aligned}Var(T) &= \sum{Var(t_i)} \\ &= \frac{1-p_1}{p_1^2} + \frac{1-p_2}{p_2^2} + \cdots + \frac{1-p_n}{p_n^2} \\ &<= \frac{n^2}{n^2} + \frac{n^2}{(n-1)^2} + \cdots + \frac{n^2}{1^2} \\ &<= n^2 \cdot (\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2}) \\ &= n^2 \cdot \frac{\pi^2}{6} \end{aligned}\]

A classical puzzle that can be modeled as Coupon Collector’s problem is that, for a cubic dice, we are expected to get all six points after how many trails? The answer for the question should be

\[\sum_1^6{\frac{6}{i}} \simeq 14.7 \simeq 15\]


The Coupon Collector’s Problem has been generalized as the expected trails of collect $m$ copy of each $n$ coupons. When $m=2$, the problem is also called The Double Dixie Cup Problem1. The expectation of $T_m$ satisfies

\[E(T_m) = n\log{n} + (m-1) n \log{\log{n}} + O(n), \texttt{as}\, n \to \infty\]

In the more general case, when $p_i$ is nonuniform, Philippe Flajolet gives2

\[E(T) = \int_0^\infty{(1-\prod_{i=1}^n{1-e^{-p_i \cdot t}})}\,dt\]

If the collector receives $d$ coupons each run3, in the asymptotic case the $m$ copy of of each $n$ coupons will be collected expected within time

\[E(T_m) = n\log{n}/d + (m-1) (n/d) \log{\log{n}} + O(n \cdot m)\]

The Coupon Collector’s Problem can also be generalized as stochastic process456. The paper from MAT7 also gives exhaustive analysis of this problem. The expectation can also be expressed using the Stirling numbers89.