Probability of getting a certain sum of two dice; confusion about order If you roll two six-sided dice, the probability of obtaining a $7$ (as a sum) is $6/36$.
Here is what is confusing me. Aren't $(5,2)$ and $(2,5)$ the same thing? So we shouldn't really double count?
Thus by that logic, wouldn't the actual answer be $3/21$ instead?
EDIT: My $21$ possibilities came from $\{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,2), (2,3), \dots, (5,6), (6,6) \}$
 A: We can perfectly well decide that the outcomes are double $1$, a $1$ and a $2$, double $2$, and so on, as in your proposal. That would give us $21$ different outcomes, not $36$.
However, these $21$ outcomes are not all equally likely. So although they are a legitimate collection of outcomes, they are not easy to work with when we are computing probabilities.
By way of contrast, if we imagine that we are tossing a red die and a blue die, and record as an ordered pair (result on red, result on blue) then, with a fair die fairly tossed, all outcomes are equally likely. Equivalently, we can imagine tossing one die, then the other, and record the results as an ordered pair.
You can compute probabilities using your collection of outcomes, if you keep in mind that for example double $1$ is half as likely as a $1$ and a $2$. The answers will be the same, the computations more messy, and more subject to error. 
A: The problem with that logic is that the 36 possible options take into account treating the dice as individual rolls. There are indeed fewer than 36 possible outcomes if you just consider pairings of the dice, but that isn't where the probability of the rolls come from. The probability comes from allowing each die to vary individually, and the fact that one of the dice rolling a two doesn't exclude the chance that the other die could still roll a five and the sum would be seven, the same being true for the first die rolling a five not excluding the chance of the second rolling a two. So we take this into account by paying attention to both possibilities when count the possible outcomes.
A: 
Here is what is confusing me. Aren't (5,2)(5,2) and (2,5)(2,5) the
  same thing? So we shouldn't really double count?

That assumption is where the issue lies. 5,2 and 2,5 are not exactly the same thing, because the dice aren't physically the same objects. 
Your logic would work if the numbers on the dice were sports teams playing at a fixed stadium, so Team 5 playing Team 2 and Team 2 playing team 5 would be the same game.
But if we introduce the difference in stadiums (i.e. different dice), we get 2 different games: we have Team 5 Playing Team 2 on 5's home stadium (i.e. die #1) and Team 2 Playing Team 5  on 2's home stadium (i.e. die #2)
Another way to think about this is:
How can I get a 2,5 combo? Either first die shows 5 and second shows 2, or the first shows 2 and the other shows 5. Even though both dice are the "same" in terms of appearance, they are not physically the same objects.
How can I get a 3,3, combo? Only one: both dice show 3. (This doesn't really fit in our sports analogy, maybe if a team is practicing with their second-tier side at their home stadium)
So let's say 3,3 has a "probability weight" of 1, while 2,5 has a "probability weight of 2"
Out of your 21 arrangements, 6 are doubles (1,1 2,2 3,3 .. 6,6) which only have a probability weight of 1.
So 15 have a "probability weight" of 2 while 6 have a "probability weight" of 1.
Which gives us 15*2 + 6 = 36 "outcomes".
So for 7, you have 1-6 2-5 3-4 4-3 5-2 6-1 = 6 out of the total 36 "outcomes".
P.S. Another way to think about it is to consider the second die a reroll of the first die. In this case the dice are physically the same, but the order in which you roll a 2,5 matters because "rolling a 5 on my first and a 2 on my second" is different from "rolling a 2 on my first and a 5 on my second".
