# Coupon Collecting Problem: Find the Expectation of Boxes to Collect All Toys

## Problem 750

A box of some snacks includes one of five toys. The chances of getting any of the toys are equally likely and independent of the previous results.

**(a)** Suppose that you buy the box until you complete all the five toys. Find the expected number of boxes that you need to buy.

**(b)** Find the variance and the standard deviation of the event in part (a).

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## Solution.

### Solution of (a)

Let $X$ be the number of boxes that you need to buy until you complete all the five toys. Our goal is to compute the expected value $E[X]$.

To achieve this, we consider the next random variables. Let $X_i$ be the number of boxes you need to buy to get $i$th toy after getting $i-1$ toys. Then it is clear from definition that

\[X = X_1 + X_2 + X_3 + X_4 + X_5.\]
For example, $X_1$ is the number of boxes you need to buy to get the first toy. Since whenever you open the first box, it is guaranteed that you get a new toy, we have $X_1 = 1$.

Also, to get the second toy after the first one, there are $4/5$ chance of getting new toy and $1/5$ chance of getting the same toy as the first one. Thus, $X_2$ is a geometric random variable with parameter $4/5$. We denote this as $X \sim G_{4/5}$. Similarly, we get

\[X_3 \sim G_{3/5}, \quad X_4 \sim G_{2/5}, \quad \text{ and } X_5 \sim G_{1/5}.\]

By the linearity of expectation, we have

\begin{align*}

E[X] &= E[X_1 + X_2 + X_3 + X_4 + X_5]\\

&= E[X_1] + E[X_2] + E[X_3] + E[X_4] + E[X_5]\\

&= E[1] + E[G_{4/5}] + E[G_{3/5}] + E[G_{2/5}] + E[G_{1/5}]
\end{align*}

Now, the expected value of a geometric random variable $G_p$ is given by

\[E[G_p] = \frac{1}{p}.\]
It follows that

\begin{align*}

E[X] &= 1 + \frac{5}{4} +\frac{5}{3} + \frac{5}{2} + \frac{5}{1}\\

&= 5\left(1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}\right)\\

&\approx 11.41667

\end{align*}

Thus, the expected number of boxes that you need to buy to complete all the five toys is 11.41667.

### Solution of (b)

Now we compute the variance of $X$. Recall that the variance of a geometric random variable $G_p$ is given by

\[V(G_p) = \frac{1-p}{p^2}.\]
As we have seen that

\begin{align*}

X &= X_1 + X_2 + X_3 + X_4 + X_5\\

&\sim 1 + G_{4/5} + G_{3/5} + G_{2/5} + G_{1/5}

\end{align*}

and as each random variable $X_i$ is independent of each other, we obtain

\begin{align*}

V(X) &= V(1 + G_{4/5} + G_{3/5} + G_{2/5} + G_{1/5})\\[6pt]
&= V(1) + V(G_{4/5}) + V(G_{3/5}) + V(G_{2/5}) + V(G_{1/5})\\[6pt]
&= 0 + \frac{1-\frac{4}{5}}{\left(\frac{4}{5}\right)^2} + \frac{1-\frac{3}{5}}{\left(\frac{3}{5}\right)^2} + \frac{1-\frac{2}{5}}{\left(\frac{2}{5}\right)^2} + \frac{1-\frac{1}{5}}{\left(\frac{1}{5}\right)^2}\\[6pt]
&\approx 25.17361

\end{align*}

Thus, the variance is $V(X) = 25.17361$.

The standard deviation is the square root of the variance. Hence, we obtain

\[\sigma(X) \approx \sqrt{25.17361} \approx 5.01733.\]

## Remark.

This type of problems is called a **Coupon Collecting Problem**.

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