Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

I'm just a little confused on this. I'm pretty sure that I need to use indicators for this but I'm not sure how I find the probability. The question goes like this:

A company puts five different types of prizes into their cereal boxes, one in each box and in equal proportions. If a customer decides to collect all five prizes, what is the expected number of boxes of cereals that he or she should buy?

I have seen something like this before and I feel that I'm close, I'm just stuck on the probability. So far I have said that $$X_i=\begin{cases}1 & \text{if the $i^{th}$ box contains a new prize}\\ 0 & \text{if no new prize is obtained} \end{cases}$$ I know that the probability of a new prize after the first box is $\frac45$ (because obviously the person would get a new prize with the first box) and then the probability of a new prize after the second prize is obtained is $\frac35$, and so on and so forth until the fifth prize is obtained. What am I doing wrong?! Or "what am I missing?!" would be the more appropriate question.

share|cite|improve this question

marked as duplicate by joriki probability Jun 25 at 15:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 1 down vote accepted

As the expected number of tries to obtain a success with probability $p$ is $\frac{1}{p}$, you get the expect number : $$1+\frac{5}{4}+\frac{5}{3}+\frac{5}{2}+5=\frac{12+15+20+30+60}{12}=\frac{137}{12}\approx 11.41$$

share|cite|improve this answer

Look for phrases like the coupon collector's problem.

The point is that once you have found $k$ prizes, the probability you find a new prize in the next box is $\frac{n-k}{n}$ so the expected number of extra boxes needed to find a $k+1$th prize is $\frac{n}{n-k}$ and so the total number of boxes needed is $$\frac{n}{n}+\frac{n}{n-1}+\frac{n}{n-2}+\cdots+\frac{n}{1}$$ which is $n$ times the $n$th harmonic number.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.