# Setting up a probability problem - What is the best way to interpret rolling two dice?

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.

• 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. – fleablood Dec 28 '15 at 4:13
• 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. – Ross Millikan Dec 28 '15 at 4:18