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.

Answer

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) \\ & < n^2 \cdot \left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} \right) \\ & < n^2 \cdot \left(\frac{\pi^2}{6} - \frac{1}{n} + \frac{1}{2n^2}\right) \\ & < \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}

Extensions and generalizations

\operatorname{P}(T < n\log n + cn) \to e^{-e^{-c}}, \ \ \text{as} \ n \to \infty.
\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.
\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.
E(T)=\int_0^\infty \big(1-\prod_{i=1}^n(1-e^{-p_it})\big)dt

See also

Notes

  1. ↑ Here and throughout this article, "log" refers to the natural logarithm rather than a logarithm to some other base. The use of Θ here invokes big O notation.
  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.

References

External links