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. For example, when n = 50 it takes about 225 trials on average to collect all 50 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 coupon is. Therefore, has geometric distribution with expectation. By the linearity of expectations we have:
Here Hn is the n-th harmonic number. Using the asymptotics of the harmonic numbers, we obtain:
where 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 ti, we obtain:
since .
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 denote the event that the -th coupon was not picked in the first trials. Then:
Thus, for, we have.

Extensions and generalizations