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.)

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
$$\frac{2}{6}\times\frac{1}{6} = \frac{2}{36}$$
