# Coupon collector's problem

Graph of number of coupons, n vs the expected number of tries needed to collect them all, E (T )

In probability theory, the coupon collector's problem describes the "collect all coupons and win" contests. It asks the following question: Suppose that there is an urn of n different coupons, from which coupons are being collected, equally likely, with replacement. What is the probability that more than t sample trials are needed to collect all n coupons? An alternative statement is: Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once? The mathematical analysis of the problem reveals that the expected number of trials needed grows as $\Theta(n\log(n))$.[1] For example, when n = 50 it takes about 225[2] trials to collect all 50 coupons.

## Understanding the problem

The key to solving the problem is understanding that it takes very little time to collect the first few coupons. On the other hand, it takes a long time to collect the last few coupons. In fact, for 50 coupons, it takes on average 50 trials to collect the very last coupon after the other 49 coupons have been collected. This is why the expected time to collect all coupons is much longer than 50. The idea now is to split the total time into 50 intervals where the expected time can be calculated.

The following table (click [show] to expand) gives the expected number of tries to get sets of 1 to 100 coupons.

## Solution

### Calculating the expectation

Let T be the time to collect all n coupons, and let ti be the time to collect the i-th coupon after i − 1 coupons have been collected. Think of T and ti as random variables. Observe that the probability of collecting a new coupon given i − 1 coupons is pi = (n − (i − 1))/n. Therefore, ti has geometric distribution with expectation 1/pi. By the linearity of expectations we have:

\begin{align} \operatorname{E}(T)& = \operatorname{E}(t_1) + \operatorname{E}(t_2) + \cdots + \operatorname{E}(t_n) = \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 \left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}\right) \, = \, n \cdot H_n. \end{align}

Here Hn is the harmonic number. Using the asymptotics of the harmonic numbers, we obtain:

$\operatorname{E}(T) = n \cdot H_n = n \log n + \gamma n + \frac1{2} + o(1), \ \ \text{as} \ n \to \infty,$

where $\gamma \approx 0.5772156649$ is the Euler–Mascheroni constant.

Now one can use the Markov inequality to bound the desired probability:

$\operatorname{P}(T \geq c \, n H_n) \le \frac1c.$

### Calculating the variance

Using the independence of random variables ti, we obtain:

\begin{align} \operatorname{Var}(T)& = \operatorname{Var}(t_1) + \operatorname{Var}(t_2) + \cdots + \operatorname{Var}(t_n) \\ & = \frac{1-p_1}{p_1^2} + \frac{1-p_2}{p_2^2} + \cdots + \frac{1-p_n}{p_n^2} \\ & = \left(\frac{n^2}{n^2} + \frac{n^2}{(n-1)^2} + \cdots + \frac{n^2}{1^2}\right) - \left(\frac{n}{n} + \frac{n}{n-1} + \cdots + \frac{n}{1}\right)\\ & = n^2 \cdot \left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} \right) - n \cdot H_n\\ & = n^2 \cdot \left(\frac{\pi^2}{6} - \frac{1}{n} + \frac{1}{2n^2}\right) - \left(n \log n + \gamma n + \frac{1}{2}\right) + o(1)\\ & < \frac{\pi^2}{6} n^2. \end{align}

The $\pi^2/6$ is a value of the Riemann zeta function (see Basel problem).

Now one can use the Chebyshev inequality to bound the desired probability:

$\operatorname{P}\left(|T- n H_n| \geq c\, n\right) \le \frac{\pi^2}{6c^2}.$

### Tail estimates

A different upper bound can be derived from the following observation. Let ${Z}_i^r$ denote the event that the $i$-th coupon was not picked in the first $r$ trials. Then:

\begin{align} P\left [ {Z}_i^r \right ] = \left(1-\frac{1}{n}\right)^r \le e^{-r / n} \end{align}

Thus, for $r = \beta n \log n$, we have $P\left [ {Z}_i^r \right ] \le e^{(-\beta n \log n ) / n} = n^{-\beta}$.

\begin{align} P\left [ T > \beta n \log n \right ] = P \left [ \bigcup_i {Z}_i^{\beta n \log n} \right ] \le n \cdot P [ {Z}_1^{\beta n \log n} ] \le n^{-\beta + 1} \end{align}

### Connection to probability generating functions

Another combinatorial technique can also be used to resolve the problem: see Coupon collector's problem (generating function approach).

## Extensions and generalizations

• Paul Erdős and Alfréd Rényi proved the limit theorem for the distribution of T. This result is a further extension of previous bounds.
$\operatorname{P}(T < n\log n + cn) \to e^{-e^{-c}}, \ \ \text{as} \ n \to \infty.$
• Donald J. Newman and Lawrence Shepp found a generalization of the coupon collector's problem when m copies of each coupon needs to be collected. Let Tm be the first time m copies of each coupon are collected. They showed that the expectation in this case satisfies:
$\operatorname{E}(T_m) = n \log n + (m-1) n \log\log n + O(n), \ \ \text{as} \ n \to \infty.$
Here m is fixed. When m = 1 we get the earlier formula for the expectation.
• Common generalization, also due to Erdős and Rényi:
$\operatorname{P}(T_k < n\log n + (k-1) n \log\log n + cn) \to e^{-e^{-c}/(k-1)!}, \ \ \text{as} \ n \to \infty.$
• In the general case of a nonuniform probability distribution, according to Philippe Flajolet,
$E(T)=\int_0^\infty \big(1-\prod_{i=1}^n(1-e^{-p_it})\big)dt$

2. ^ E(50) = 50*(1 + 1/2 + 1/3 + .. 1/50) = 224.9603, the expected number of trials to collect all 50 coupons. The approximation $n\log n+\gamma n+1/2$ for this expected number gives in this case $50\log 50+50\gamma+1/2 \approx 195.6011+28.8608+0.5\approx 224.9619$.