Roll two dice. What is the probability that one die shows exactly two more than the other die? [closed]

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".

Total possible results: $6\times6=36$

Favorable results: $1-3,2-4,3-5,4-6$ and opposites, $8$.

Then the probability is $8/36=2/9$.

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.

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

Just for fun, I counted eight.