Random but very interesting probability question!

Question

There are 100 types of cards, each type with a number from 1 to 100 on it. Each minute I am given a random card. Random means that the number of my card has an equal chance of being 1,2,3,...,100. I stop when I have cards of each number, that is, I stop collecting once I have each number on at least one card. How many minutes is this process expected to take? That is, find the expected value of the number of minutes I have to wait.

E.g. I could potentially get duplicates so this method will take a lot more than 100 minutes. Once I get 99 different cards, I will stop as soon as I get the other one which I haven't got yet.

Thanks for helping everyone!

Why not start small, with 1, 2, 3,... cards, so as to get some insight into what is happening ? — true blue anil, Oct 01 '15 at 07:00
Does the source of the cards have infinite capacity, i.e. say if the first $100$ cards are marked $1$, then for the rest of the time can we see any more card marked $1$? — Samrat Mukhopadhyay, Oct 01 '15 at 07:10
This is called the coupon collector's problem and has already many duplicates on the site. — Did, Oct 01 '15 at 07:35

score -1 · Answer 1 · answered Oct 01 '15 at 06:57

-1

Well the first card 1 minutes, the second card $\frac{100}{99}$ minutes, the third card $\frac{100}{98}$ minutes, and so on. So it is expected to take $(1 + \frac{100}{99} + \frac{100}{98} + ... + \frac{100}{1})$ which is around 519 minutes.

answered Oct 01 '15 at 06:57

cr001

12,778

Random but very interesting probability question!

1 Answers1