2014 AMC 10A Problem 17

Note that if one die is the sum of the other two dice, then it is strictly greater than the other two dice. There are $3$ ways to choose which of the dice is the sum of the other two, which makes it the greatest. This die cannot be $1,$ since there is no way to sum two positive integers to get $1.$ There is a $\dfrac{1}{6}$ chance that this die is any of the other numbers. There is $1$ way to get a sum of $2,$ $2$ ways for $3,$ $3$ for $4,$ $4$ for $5,$ and $5$ for $6.$ We take these numbers of ways out of a total of $6^2 = 36$ possibilities. The desired probability is then $$3 \cdot \dfrac{1}{6} \cdot \dfrac{1 + 2 + 3 + 4 + 5}{36} = \dfrac{5}{24}.$$ Thus, D is the correct answer.