2004 AIME II Problem 2

They match only if both draw two reds, both draw two blues, or both draw one candy of each color

Mary draws from what Terry left: for two reds each, the probability is $\frac{\binom{10}{2}}{\binom{20}{2}} \cdot \frac{\binom{8}{2}}{\binom{18}{2}}$

The combinations match exactly when both draw two reds, both draw two blues, or both draw one candy of each color. The probability that Terry draws two reds is $\frac{\binom{10}{2}}{\binom{20}{2}} = \frac{45}{190} = \frac{9}{38},$ after which $8$ reds and $10$ blues remain, so Mary draws two reds with probability $\frac{\binom{8}{2}}{\binom{18}{2}} = \frac{28}{153}.$ That case has probability $\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323},$ and by symmetry two blues each is also $\frac{14}{323}.$

For mixed draws, Terry succeeds with probability $\frac{10 \cdot 10}{\binom{20}{2}} = \frac{10}{19},$ leaving $9$ of each color, and Mary with probability $\frac{9 \cdot 9}{\binom{18}{2}} = \frac{9}{17},$ for $\frac{10}{19} \cdot \frac{9}{17} = \frac{90}{323}.$

The total is $\frac{14 + 14 + 90}{323} = \frac{118}{323}.$ Since $118 = 2 \cdot 59$ and $323 = 17 \cdot 19,$ the fraction is in lowest terms, and $m + n = 118 + 323 = 441.$