# Collecting Coupons

## Problem

Coupons in cereal boxes are numbered $$1$$ to $$5$$, and a set of one of each is required for a prize. With one coupon per box, how many boxes on the average are required to make a complete set?

## Solution

Let $$X_{1}$$, $$X_{2}$$, …, $$X_{5}$$ be random variables representing the number of boxes opened until receiving the $$1$$st, $$2$$nd, …, and $$5$$th unique coupon, respectively, where the number of boxes is counted since the last box containing a unique coupon was opened. For example, $$X_{3}$$ represents the number of boxes opened after the $$2$$nd unique coupon was found up until the $$3$$3rd unique coupon was found.

Each $$X_{i}$$ is a geometric random variable associated with a different probability of success. For example, $$X_{1}$$ is associated with $$p_{1} = 1$$ because the $$1$$st box will obviously contain a unique coupon. $$X_{2}$$ is associated with $$p_{2} = \frac{4}{5}$$ because only $$4$$ of the $$5$$ coupons are unique at that point. Probabilities for the remaining $$X_{i}$$ can be calculated in a similar fashion. For geometric random variables, we also know that $$E[X_{i}] = \frac{1}{p_{i}}$$.

To solve the problem, we can apply the Linearity of Expectation property as follows:

$\begin{split} E \left[\sum_{i=1}^5 X_{i} \right] &= \sum_{i=1}^5 E[X_{i}] \\ &= \sum_{i=1}^5 \frac{1}{p_{i}} \\ &= 5 \left( \frac{1}{5} + \frac{1}{4} + \frac{1}{3} + \frac{1}{2} + \frac{1}{1} \right) \\ &= \frac{137}{12} \approx 11.42 \end{split}$

## Notes

When generalized to $$n$$ coupons, the solution to this problem can be written as:

$C(n) = n \sum_{i=1}^n \frac{1}{i} = nH_{n}$

In the above, $$H_{n}$$ represents the $$n$$th Harmonic Number, which is the sum of the reciprocals of integers from $$1$$ to $$n$$.