Roll five dice. What's the chance of rolling exactly one pair?

To have exactly one pair (no two pairs, not three of a kind, not "full house", ...) you will have a total of exactly four distinct values, one of these occuring twice. So out of the $6^5$ total possible dice rolls, there are ${6\choose 4}{4\choose 1}{5\choose 2}3!$ good possibilities (pick the four values, pick the duplicate value, pick the dice making the pair, rearrange the remaining three dice). So the probability is $$ \frac{{6\choose 4}{4\choose 1}{5\choose 2}3!}{6^5}=\frac{25}{54}.$$

Another way to argue: You can choose the two paired dice in $\binom52$ ways, the value of the pair in 6 ways and the non-paired dice in $5\cdot4\cdot 3$ ways, leading to the same answer as Hagen's