Expected number of trials before all cards are collected

[xcrunner448]

Homework Statement

A cereal box company started a promotion in which they put one of five different baseball cards in each box of cereal. All cards occur with equal probability. What is the expected number of cereal boxes you have to buy before you collect all five cards?

The Attempt at a Solution

The expected value is the sum from k=5 to infinity of k*(probability of taking exactly k boxes to collect all 5 cards). However, I cannot find a general expression that gives that probability. For k=5, it would be (0.2^5)*5! = 0.0384, because the probability of a certain card in each box is 0.2, and there are 5!=120 different ways to collect all five cards. I cannot figure out how to calculate this probability for k>5. I have tried the same method as I used for k=5, that is, (0.2^k) times the number of ways of collecting all five cards, but because some cards must be repeated 1 or more times I am having difficulty finding out how to calculate that.

 

[tiny-tim]

welcome to pf!

hi xcrunner448! welcome to pf!

hint: what is P(not ending at k) ? ​

 

[xcrunner448]

Hi tiny-tim, thanks for your response! I also had that idea, but I could not figure out a general expression for that either. For k=5, the probability of not getting, say, card #1 by the fifth box is (0.8)^5. I thought I'd multiply that by 5 to cover the all 5 possible cards, but that adds to greater than 1. I realized that I would then have to subtract the overlap (the probability of not getting cards #1 and 2, or the probability of not getting cards #2,3, and 4, and so on). But that got very complicated very quickly. There may be a much better way of finding it, but I could not think of it.

In the meantime, however, I found "[URL ,[/URL] which describes the problem perfectly. It is an interesting derivation, based on the expected number of boxes to find the i^th card after finding (i-1) cards. It gave me the answer of 11.416666=137/12, which I believe is correct.