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.)

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:
$$6 \cdot \frac{1}{36} + 15 \cdot \frac{2}{36} = \frac{36}{36} = 1$$
