Probability Question

I'm afraid I don't follow.
Where did 37.239 come from?

The odds that you get a new candy on the first try are 12/12, 2nd box is 11/12, etc... but what if you get the same candy over and over again, then box 3 could have a 11/12 chance of a new candy...

:x

Sorry, I didn't feel like I had to explain since I was lazy.

The probability of getting the first candy is 12/12. After you get the first one, the probability of the 2nd candy (one of the 11 kinds that isn't the first) is 11/12. The third would be 10/12, fourth would be 9/12, etc, until the last one which is 1/12.

So, to find the expected amount, you take the inverse of the probability. For example, if you roll a dice, the expected amount to roll a 1 would be 6 rolls, which is the inverse of 1/6.

*makes a table*

Candy #	Probability	Expected #
1st	12/12	12/12
2nd	11/12	12/11
3rd	10/12	12/10
4th	9/12	12/9
5th	8/12	12/8
6th	7/12	12/7
7th	6/12	12/6
8th	5/12	12/5
9th	4/12	12/4
10th	3/12	12/3
11th	2/12	12/2
12th	1/12	12/1

So to get the expected amount of attempts you would need to make, you have to get the sum of each one, so you add up the entire last column. (If you don't sum all of it, but instead the first two, that would be the expected amount of finding 2 unique candies.)

Make more sense?