Probability Question

In Mabi, there's this "Wish Upon a star" or something event.
Simplifying it a little, you get a box, open it, and get a candy. To get the prize, you need all 12 types of candy.
Assuming each candy is equally likely to be in a box, what is the expected number of boxes which need to be opened before all 12 candies are obtained?
If that doesn't work for one reason or another, how would you graph the probability distribution of how likely you are to have all 12 after n boxes?

Assume equal distribution and that the probabilities are constant. That is the same as placing 12 candies within a box, and pick up one. Then you place it back again, and after that scramble the box, and pick up another one at random. How many picks on average do you need in order to have picked up all?

First of all, the probability of getting a new candy when you have 0, 1, 2 ... different candies is the following:

(So for P(12), or when you have 12 new candies to obtain, you have 100% chance of getting a new candy. That's obvious!)

Now, if you've picked the first candy, there's chance of getting a new candy on next attempt. That also means that there's chance of not obtaining a new candy. And what happens if you don't obtain a new candy? You have to try until you get one!

The estimate for candy number n will then be:


If you get a new candy, then then the remaining amount of boxes to open will be E(n - 1). Because we've opened a new box right now however, we need to add 1 to our calculation. This happens with P(n) probability. If not, we don't get a new candy: Then we've used one box, and the average amount of boxes to left open is E(n). In total, (1 + E(n)). This is multiplied by the probability of not opening a new box.

By substituting P:


Then, we get E(12) = 37.4.

Trying to find a certain probability, or for a set amount of boxes, insert these values:



Noah