Random but very interesting probability 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!
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.
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.