Consider the following problem: If I roll two fair six-sided dice, what is the probability of me rolling two numbers that add up to $10$?
Obviously, the only unordered ways to do this is to roll $(4, 6)$ or $(5, 5)$. My inclination is to say the probability is $3/36 = 1/12$ on the basis that of the $36$ ordered rolls, the only ways to get the roll to sum to $10$ is if it's $(4, 6), (5, 5)$, or $(6, 4)$. To justify the ordered interpretation, I consider the "thought-experiment" that I roll, but immediately cover one die. Looking at the first (unconvered) die, I know I have a $3 / 6 = 1 / 2$ chance of rolling something I want, i.e. I must roll a $4, 5$, or $6$. But then I know that if I rolled a $4, 5$, or $6$, then there is only a $1 / 6$ chance that the second die (i.e. the one that has until now been covered) is what it must be. That is, if I rolled a $4$ then there is only a $1 / 6$ chance that the second die is the $6$ I need, and similarly for the other options. Thus I have a $1 / 36 + 1 / 36 + 1 /36 = 1 / 12$ chance.
My trouble with this is that it also seems a bit counter-intuitive because when I roll, I could not care less which die said which; that is, if I had a red die and a black die in my hand, then after I roll I have no interest in whether the red was a $4$ and the black a $6$ or vice versa. So my question: Is this interpretation of the question in terms of ordered pairs correct? That is, is it correct to say that the unordered roll $(5, 5)$ is only half as likely as the unordered roll $(4, 6) = (6, 4)$? Thanks in advance.