Possible outcomes of rolling two dice When you roll two dice at the same time which look same, what's the number of possible cases? 
My friend thinks '21' because they are not distinguished, that is (1,2)=(2,1), (1,3)=(3,1), ... (5,6)=(6,5)
And I think the answer is 36 because they are different die even though they are not distinguished. 
I also think that 36 is right because if 21 is right, there will be a problem on computing probabilities of the events. 
I'm so confused
Can anybody tell me what's the correct answer of this problem and the reason? 
 A: For the first die, you have 6 possibilities. And the second die will also have 6 possibilities. So for every possibility of the first die, you got 6 possibilities for the second one. If we calculate it, it makes 6 possibilities of the first, times 6 of the second, which gives 36...
But if you talk about combinations, like (1;4) and such, you won’t get the same amount of possibilities, because (1;4) is the same as (4;1), if you see what I mean...
A: Case 1:
6 possible outcomes
Case 2:
6 possible outcomes
Total: 36 outcomes.
However, if you were to take this in certain contexts, for a problem concerning sums or something. Then it would be different because (3, 2) is the same as (2,3)
A: There are only $21$ cases which you can distinguish. However, these cases are not equally likely - 5-6 is twice as likely to occur as 6-6, because the 5 could be on either of the two dice.
So the right way to think of it is that there are $21$ possibilities that you can tell apart, but hidden behind these are actually $36$ possibilities (that you can't necessarily tell apart), and it's the $36$ you need to work with for computing probabilities, because they are the ones that are equally likely. 
It's a good idea to always approach questions about dice by pretending the dice have different colours and doing calculations on that basis.
