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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.

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    $\begingroup$ Yes. And red and black die is a good way to see it. There are 3 ways. red can be 4,5,6 and the black will have to be 6,5,4. It's counter intuitive but intuition is fundamently wrong. $\endgroup$ – fleablood Dec 28 '15 at 4:13
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    $\begingroup$ Your red and black is a good approach. The fact that you don't care which is red and which is black is what makes $4+6$ twice as likely as $5+5$. There are two ways to do the first and only one to do the second. $\endgroup$ – Ross Millikan Dec 28 '15 at 4:18
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I'm not sure what you are looking for from a solution (the comments seem to resolve your concerns). Keep in mind with probability, you don't care just about the outcomes, but how likely they are to occur. Because rolling (4,6) is twice as likely, you should count it twice in the sum.

Alternatively, create precise definitions of events, so you don't have to guess based on intuition. For example, the sample space is all the combinations of two dice -- that is (1,2,3,4,5,6) x (1,2,3,4,5,6) -- and there are 3 outcomes that are acceptable here, thus the answer is 3/36.

There is no reason to count dice rolls by unordered pairs, since you'll just have to complicate things later on.

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