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 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 ${\displaystyle \Theta (n\log(n))}$. For example, when n = 50 it takes about 225 trials on average to collect all 50 coupons. Sometimes the problem is instead expressed in terms of an n-sided die.

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 ${\displaystyle T=t_{1}+\cdots +t_{n}}$. Think of T and t_i as random variables. Observe that the probability of collecting the i-th new coupon is ${\displaystyle p_{i}={\frac {n-(i-1)}{n}}={\frac {n-i+1}{n}}}$. Therefore, ${\displaystyle t_{i}}$ has geometric distribution with expectation ${\displaystyle {\frac {1}{p_{i}}}={\frac {n}{n-i+1}}}$. 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.

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 k be the number of coupons already collected, then:

And when ${\displaystyle k=0}$ then we get the original result.

Using the independence of random variables t_i, we obtain: