Roll two dice. What is the probability that one die shows exactly two more than the other die? 
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. 
 A: Any result will do as long as the other die can score the same number plus two, that gets us with n-2 per die (n being number of sides). This gets us 2(n-2) posible results over n^2 (as we have two identical dice)
then the probability is: 2(n-2)/n^2 
A: Just for fun, I counted eight.

A: To get yourself started, you could draw a table. The rows could be one roll, and the columns could be the other roll. Then the checkmark shows where the rolls are "two away" from each other.
\begin{array}{r|c|c|c|c|c|c}
&1&2&3&4&5&6\\\hline
1&&&\checkmark&&&\\\hline
2&&&&\checkmark&&\\\hline
3&\checkmark&&&&\checkmark&\\\hline
4&&\checkmark&&&&\checkmark\\\hline
5&&&\checkmark&&&\\\hline
6&&&&\checkmark&&
\end{array}
Notice that, since all pairs are equally likely, we have a $8/36 = 2/9$ chance of being "two away".
A: Could use the multiplication rule:
The probability of Die 1 landing on 1-4 is 4/6.
The probability of the Die 2 landing on the number that's Die1+2 is then 1/6. 
(4/6) * (1/6) = 4/36
We multiply this by 2 to account the scenario where Die 2 is the 1-4 die, and then Die 1 is two higher than Die 2. So, 8/36. 
A: Total possible results: $6\times6=36$
Favorable results: $1-3,2-4,3-5,4-6$ and opposites, $8$.
Then the probability is $8/36=2/9$.
A: 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
A: The probability of rolling a 1 and 3 is 1/18. Same for the probability of 2&4, 3&5, and 4&6.
So the overall probability of the dice being two apart equals 4/18 = 2/9.
