AMC 12 Mock Exam 2 Problem 13

If one number equals the sum of the other two, it is strictly larger than each of them, so it is the unique largest number; call it $m$, where $2\le m\le6$. For each $m$, the largest die can be any of the $3$ dice, and the other two dice show an ordered pair of positive integers with sum $m$, of which there are $m-1$. Thus the number of favorable outcomes is $$3\left(1+2+3+4+5\right)=45,$$ and the probability is $\frac{45}{216}=\boxed{\tfrac{5}{24}}$.