Rolling $2$ dice: NOT using $36$ as the base? 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.)
 A: Because to get $(1,1)$, both dice must show a $1$. To get a $1$ and a $2$, it could be either $(1,2)$ or $(2,1)$.

Here's another way to look at it ...

The probability of getting $(1,1)$ is
$$\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}$$
Explanation: The dice are independent and each die has probability $\large{\frac{1}{6}}$ of showing a $1$.
To get a $1$ and a $2$, the first die must show either $1$ or $2$,
and the second die must show whichever of $1,2$ did not show on the first die. Hence the probability of getting a $1$ and a $2$ is
$$\frac{2}{6}\times\frac{1}{6} = \frac{2}{36}$$
A: In order to throw (1, 1), the values of both dice (let's call them A and B) need to equal 1. Since these events are statistically independent of one another:
$$Prob[(1, 1)] = Prob[A = 1\,and\,B = 1] = Prob[A=1] Prob[B=1] = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$
If you specify that $(1, 2) = (2, 1)$, it essentially does not make a difference whether $A=1$ and $B=2$, or $A=2$ and $B=1$. In this case, the probability of throwing (1, 2) comes down to:
$$Prob[(1, 2)] = Prob[A = 1\,and\,B = 2] + Prob[A = 2\,and\,B = 1] = \frac{1}{36} + \frac{1}{36} = \frac{2}{36}$$
As you can see, the events $(1, 1)$ and $(1, 2)$ are not equiprobable: the former has a probability of 1/36, while the latter has a probability of 1/18.
There are six types of throws with equal numbers $(1, 1), (2, 2), \ldots, (6, 6)$ and fifteen types of throws with different numbers $(1, 2), (1, 3), \ldots, (5, 6)$. Since these events cover the total sample space, the sum of the probabilities equals:
$$6 \cdot \frac{1}{36} + 15 \cdot \frac{2}{36} = \frac{36}{36} = 1$$
A: 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!
A: The main problem is that the answer of $\frac1{21}$ doesn't fit with empirical data, and so must be wrong.
Throw the dice $1,000,000$ times and count the number of occurrences of $(1,1)$. It will be around $27,778$ (from $\frac1{36}$), whereas if the probability was $\frac1{21}$ we would expect an answer around $47,619$.
So we must assume $\frac1{21}$ is the wrong answer, and this is because, whether the dice are identical (shape, size, colour, etc..) or not, there are still $36$ possibilities.
A: The key point is that if you distinguish the two dice all $36$ possibilities are equally likely. That is the only thing that allows you to convert number of possibilities to probability.
If you don't distinguish the two dice then there are only $21$ possibilities, but some of them are more likely than others -- and this issue gets more complicated the more dice you have. So knowing that there are $21$ possibilities doesn't give you a probability of $1/21$. 
To extend your approach to a point where it more clearly doesn't work, suppose I change all the numbers other than $1$ to $2$s (so each die has $1,2,2,2,2,2$ on it). This clearly doesn't affect the probability of double-$1$, but now there are only three possibilities...
A: Paint the dice different colors, say red and blue.  Now (red 1, blue 2) is clearly different from (red 2, blue 1).  But the dice don't know they're painted, because they're dice.
