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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 ${\displaystyle \Theta (n\log(n))}$. For example, when n = 50 it takes about 225 trials on average to collect all 50 coupons.

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Solution

Calculating the expectation

Let T be the time 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. Think of T and t_i as random variables. Observe that the probability of collecting a new coupon is p_i = (n - (i - 1))/n. Therefore, t_i has geometric distribution with expectation 1/p_i. By the linearity of expectations we have:

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

where ${\displaystyle \gamma \approx 0.5772156649}$ is the Euler-Mascheroni constant.

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

Calculating the variance

Using the independence of random variables t_i, we obtain:

since ${\displaystyle {\frac {\pi ^{2}}{6}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{n^{2}}}+\cdots }$ (see Basel problem).

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

Tail estimates

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

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

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Extensions and generalizations

• Paul Erd?s and Alfréd Rényi proved the limit theorem for the distribution of T. This result is a further extension of previous bounds.

• Donald J. Newman and Lawrence Shepp found 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:

Here m is fixed. When m = 1 we get the earlier formula for the expectation.

• Common generalization, also due to Erd?s and Rényi:

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

Source of the article : Wikipedia