Coupon collector's problem explained

In probability theory, the coupon collector's problem refers to mathematical analysis of "collect all coupons and win" contests. It asks the following question: if each box of a given product (e.g., breakfast cereals) contains a coupon, and there are different types of coupons, what is the probability that more than boxes need to be bought to collect all coupons? An alternative statement is: given 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

. For example, when it takes about 225^[1] trials on average to collect all 50 coupons. Sometimes the problem is instead expressed in terms of an -sided die.

Solution

Calculating the expectation

Let time be the number of draws needed to collect all coupons, and let be the time to collect the -th coupon after coupons have been collected. Then

. Think of and as random variables. Observe that the probability of collecting the -th coupon is . Therefore, has geometric distribution with expectation . By the linearity of expectations we have:

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

where is the Euler–Mascheroni constant.

Using the Markov inequality to bound the desired probability:

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

And when

then we get the original result.

Calculating the variance

Using the independence of random variables, we obtain:

since

(see Basel problem).

Bound the desired probability using the Chebyshev inequality:

Stirling numbers

Let the random variable be the number of dice rolls performed before all faces have occurred.

The subpower is defined

}=a!\left\, where is a Stirling number of the second kind.^[2]

Sequences of

die rolls are functions counted by , while surjections (that land on each face at least once) are counted by }, so the probability that all faces were landed on within the -th throw is }. By the recurrence relation of the Stirling numbers, the probability that exactly rolls are needed is }-\frac=\frac

Generating functions

Replacing

with in the probability generating function produces the o.g.f. for . Using the partial fraction decomposition , we can take the expansion revealing that for , Given an o.g.f., since , a variation of the binomial transform is . (Specifically, if , .)

Rewriting the binomial coefficient via the gamma function and expanding as the

of the polygamma series (in terms of generalised harmonic numbers), we find , so which can also be written with the falling factorial and Lah numbers as The raw moments of the distribution can be obtained from the falling moments via a Stirling transform; due to the identity , this provides

Tail estimates

A stronger tail estimate for the upper tail can be obtained as follows. Let

denote the event that the -th coupon was not picked in the first trials. Then

Thus, for

, we have . Via a union bound over the coupons, we obtain

Extensions and generalizations

• Pierre-Simon Laplace, but also Paul Erdős and Alfréd Rényi, proved the limit theorem for the distribution of . This result is a further extension of previous bounds. A proof is found in.^[3]

which is a Gumbel distribution. A simple proof by martingales is in .^[4] .

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

Here is fixed. When we get the earlier formula for the expectation.

• Common generalization, also due to Erdős and Rényi:

• In the general case of a nonuniform probability distribution, according to Philippe Flajolet et al.

This is equal to

where denotes the number of coupons to be collected and denotes the probability of getting any coupon in the set of coupons .

See also

• Cover time – a generalization of the coupon collector's problem to Markov chains and random walks on graphs
• 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

References

Bibliography

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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.

Notes and References

1. E(50) = 50(1 + 1/2 + 1/3 + ... + 1/50) = 224.9603, the expected number of trials to collect all 50 coupons. The approximation

   nlogn+\gamman+1/2

   for this expected number gives in this case

   50log50+50\gamma+1/2 ≈ 195.6011+28.8608+0.5 ≈ 224.9619

   .
2. Rus. Mircea Dan. Mircea Dan Rus. 15 January 2025. Yet another note on notation. 2501.08762. math.NT.
3. Book: Mitzenmacher, Michael . Probability and computing : randomization and probabilistic techniques in algorithms and data analysis . 2017 . Eli Upfal . 978-1-107-15488-9 . 2nd . Cambridge, United Kingdom . Theorem 5.13 . 960841613.
4. Kan . N. D. . 2005-05-01 . Martingale approach to the coupon collection problem . Journal of Mathematical Sciences . en . 127 . 1 . 1737–1744 . 10.1007/s10958-005-0134-y . 1573-8795. subscription .