2 identical candies are given to 2 distinct children at random. What is the Probability that each children has at least one candy? OK i have two different ways to approach the problem and both give different answers, it would be great if someone could clarify which of them is correct(if they are).
Method 1:
Since the candies are identical; either child 1 has both the candies, 1 child gets 1 each or child 2 gets both the candies. Hence 1/3; or: $\frac{\dbinom{2-1}{2-1}}{\dbinom{2+2-1}{2-1}}=\frac{1}{3}$
Method 2:
A candy has $\frac{1}{2}$ chance of getting given to child 1 or child 2. Since there are two candies and that we can give child 1 first then child 2, we have $2\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{2}$.
I can understand why method 2 is preferred/correct, because one can argue the the outcomes are not equally likely so method 1 fails.
But does it matter if the candies are identical or not? If they are distinct, i think it is clear that method 2 is correct, but is there any difference when the candies are set to be identical?
 A: You are facing the problem that the probabilities are not exactly given by simply saying that "indistinguishable candies are given at random to starving children"
Hypothesis No. 1: 
You declare that the following possibilities are equally likely
$$\{(cc,0),(c,c),(0,cc)\}.$$ 
This choice is logically possible. In this case, however, you do not have to do any calculation. The result is given by the hypothesis. The only question is if this hypothesis corresponds to your actual acting in reality. We'll come back to this issue later.
Hypothesis No. 2:
You forget about indistiguishability for a moment and consider the following possibilities to be equally likely 
 $$\{(c,c'),(c',c),(cc',0),(0,cc')\}.$$ 
In this case the probability that each child receivs one candiy is $$P((c,c'))+P((c',c))=\frac14 +\frac14=\frac12$$
This is another logically possible choice. In this case you made the candies indistinguishable by pulling together two outcomes: $(c,c')$ and $(c',c)$.
In this second method you explicitly assume that the candies are distinguishable.
Reality:
In reality the candies are made distinguishable by saying that it was possible to take and give the first candy and then to take and give the second candy.
So, the use of the word "indistinguishable" may be very confuzing. In the case of your problem (with touchable real candies) "indistinguishability" meant only that you did not want to distinguish between two otherwise distinguishable cases.
