TITLE: Rolling $2$ dice: NOT using $36$ as the base?
QUESTION [15 upvotes]: I apologize for such a simple question. It has been a while since I took math classes. 
When you roll $2$ dice, there are $36$ possibilities. However, there are only $21$ combinations, if order does not matter. Rolling a $(4,2)$ = rolling a $(2,4)$.
Let's say in a game, rolling a $(1,1)$ makes you lose. The odds of rolling this is a $1/36$. But why can't you say the probability is a $1/21$, assuming you roll both dice at the same time? There's only one combination that makes you lose, so why can't you use $21$ as the denominator?
I have tried searching on this topic, but have not found a good answer. (Most likely because my thinking is fallacious.)

REPLY [5 votes]: Simply because, no matter what, you have 36 possibilities! $6*6 \qquad possibilities = 36$

Well, this makes it even more obvious compared to other explanations, and you can see that combinations occur once or more. But there are only 21 combinations: $[1,1],[2,2],[3,3],[4,4],[5,5],[6,6],[1,2],[1,3],[1,4],[1,5],[1,6],[2,3],[2,4],[2,5],[2,6],[3,4],[3,5],[3,6],[4,5],[4,6],[5,6]$. So you are right about there being 21 combinations. But, you also have $[3,1], [3,2]$ etc. Therefore, because $[a,b]=[b,a]$, the possibilities for $[a,b]$, where $ a \neq b$ is $\frac{2}{36} = \frac{1}{18}$ You have 6 doubles, and 15 combinations, leading to $2*15 + 1*6 = 36$ combinations! (Where in $[a,b]$, $a \neq b$, will count as 2 combinations! 
I know that it is a bit repetitive ,so if you think anything is missing, please comment below!