coupon collector's problem

In probability theory, the coupon collector's problem describes "collect all coupons and win" contests. It asks the following question: If each box of a brand of cereals contains a coupon, and there are ''n'' different types of coupons, what is the probability that more than ''t'' boxes need to be bought 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 $\Theta(n\log(n))$. For example, when ''n'' = 50 it takes about 225E(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 $50\log 50+50\gamma+1/2 \approx 195.6011+28.8608+0.5\approx 224.9619$. trials on average to collect all 50 coupons.

Solution

Calculating the expectation

Let time ''T'' be the number of draws needed to collect all ''n'' coupons, and let ''t_i'' be the time to collect the ''i''-th coupon after ''i'' − 1 coupons have been collected. Then $T=t_1 + \cdots + t_n$. Think of ''T'' and ''t_i'' as random variables. Observe that the probability of collecting a coupon is $p_i = \frac = \frac$. Therefore, $t_i$ has geometric distribution with expectation $\frac = \frac$. By the linearity of expectations we have:

:$\begin \operatorname(T) & = \operatorname(t_1 + t_2 + \cdots + t_n) \\ & = \operatorname(t_1) + \operatorname(t_2) + \cdots + \operatorname(t_n) \\ & = \frac + \frac + \cdots + \frac \\ & = \frac + \frac + \cdots + \frac \\ & = n \cdot \left(\frac + \frac + \cdots + \frac\right) \\ & = n \cdot H_n. \end$

Here ''H_n'' is the ''n''-th harmonic number. Using the asymptotics of the harmonic numbers, we obtain:

:$\operatorname(T) = n \cdot H_n = n \log n + \gamma n + \frac + O(1/n),$
where $\gamma \approx 0.5772156649$ is the Euler–Mascheroni constant.

Using the Markov inequality to bound the desired probability:

:$\operatorname(T \geq cn H_n) \le \frac.$

The above can be modified slightly to handle the case when we've already collected some of the coupons. Let ''k'' be the number of coupons already collected, then:

:$\begin \operatorname(T_k) & = \operatorname(t_ + t_ + \cdots + t_n) \\ & = n \cdot \left(\frac + \frac + \cdots + \frac\right) \\ & = n \cdot H_ \end$

And when $k=0$ then we get the original result.

Calculating the variance

Using the independence of random variables ''t_i'', we obtain:

:$\begin \operatorname(T)& = \operatorname(t_1 + \cdots + t_n) \\ & = \operatorname(t_1) + \operatorname(t_2) + \cdots + \operatorname(t_n) \\ & = \frac + \frac + \cdots + \frac \\ & < \left(\frac + \frac + \cdots + \frac\right) \\ & = n^2 \cdot \left(\frac + \frac + \cdots + \frac \right) \\ & < \frac n^2 \end$

since $\frac6=\frac+\frac+\cdots+\frac+\cdots$ (see Basel problem).

Bound the desired probability using the Chebyshev inequality:

:$\operatorname\left(, T- n H_n, \geq cn\right) \le \frac.$

Tail estimates

A stronger tail estimate for the upper tail be obtained as follows. Let $_i^r$ denote the event that the $i$-th coupon was not picked in the first $r$ trials. Then
:
\begin
P\left _i^r \right = \left(1-\frac\right)^r \le e^.
\end

Thus, for $r = \beta n \log n$, we have P\left _i^r \right \le e^ = n^. Via a union bound over the $n$ coupons, we obtain
:
\begin
P\left T > \beta n \log n \right = P \left \bigcup_i _i^ \right \le n \cdot P _1^ \le n^.
\end

Extensions and generalizations

* Pierre-Simon Laplace, but also Paul Erdős and Alfréd Rényi, proved the limit theorem for the distribution of ''T''. This result is a further extension of previous bounds. A proof is found in.

::$\operatorname(T < n\log n + cn) \to e^, \text n \to \infty.$

* Donald J. Newman and Lawrence Shepp gave a generalization of the coupon collector's problem when ''m'' copies of each coupon need to be collected. Let ''T_m'' be the first time ''m'' copies of each coupon are collected. They showed that the expectation in this case satisfies:

::$\operatorname(T_m) = n \log n + (m-1) n \log\log n + O(n), \text n \to \infty.$
:Here ''m'' is fixed. When ''m'' = 1 we get the earlier formula for the expectation.

* Common generalization, also due to Erdős and Rényi:

::$\operatorname\left(T_m < n\log n + (m-1) n \log\log n + cn\right) \to e^, \text n \to \infty.$

* In the general case of a nonuniform probability distribution, according to Philippe Flajolet et al.
::$\operatorname(T)=\int_0^\infty \left(1 - \prod_^m \left(1-e^\right)\right)dt.$

:This is equal to

::$\operatorname(T)=\sum_^ (-1)^ \sum_ \frac,$

:where ''m'' denotes the number of coupons to be collected and ''P_J'' denotes the probability of getting any coupon in the set of coupons ''J''.

See also

* McDonald's Monopoly – an example of the coupon collector's problem that further increases the challenge by making some coupons of the set rarer
* Watterson estimator
* Birthday problem

Notes

References

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{{refend

External links

*
Coupon Collector Problem
by Ed Pegg, Jr., the Wolfram Demonstrations Project. Mathematica package.
*
How Many Singles, Doubles, Triples, Etc., Should The Coupon Collector Expect?
', a short note by Doron Zeilberger.

Articles containing proofs
Gambling mathematics
Probability theorems
Probability problems