TITLE: Roll two dice. What is the probability that one die shows exactly two more than the other die?
QUESTION [15 upvotes]: Two fair six-sided dice are rolled. What is the probability that one die shows exactly two more than the other die (for example, rolling a $1$ and $3$, or rolling a $6$ and a $4$)?

I know how to calculate the probabilities of each event by itself, but I do not know how to proceed with this problem.

REPLY [2 votes]: If the first die is 1, the other can only be 3, probability = 1/6
If the first die is 2, the other can only be 4, probability = 1/6
If the first die is 5, the other can only be 3, probability = 1/6
If the first die is 6, the other can only be 4, probability = 1/6
If the first die is 3, the other can only be 1 or 5, probability = 2/6
If the first die is 4, the other can only be 2 or 6, probability = 2/6
Total probability is (1+1+1+1+2+2)/(6+6+6+6+6+6) = 8/36 = 2/9