Solution:

There are $11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 = 10395$ ways to pair the letters. Each word begins with the smaller letter of its pair, so the last word alphabetically is the pair whose smaller letter is largest.

Case 1: $G$ is the smaller letter of the last word. Then $G$ pairs with one of $H, I, J, K, L$ ($5$ ways), and no two of the remaining four late letters may pair together (such a pair would start with a letter after $G$). Those four letters must take distinct partners from $\{A, \ldots, F\},$ in $6 \cdot 5 \cdot 4 \cdot 3 = 360$ ways, and the two leftover early letters pair with each other. That gives $5 \cdot 360 = 1800$ pairings. Case 2: $G$ is the larger letter, paired with some $x$ before $G.$ Then none of $H, \ldots, L$ may pair together, so all five take partners among the other five early letters; the six smaller letters are then exactly $A$ through $F,$ and the largest is $F.$ So the last word is $FG,$ and $H, \ldots, L$ match with $A, \ldots, E$ in $5! = 120$ ways.

The probability is $\frac{1800 + 120}{10395} = \frac{1920}{10395} = \frac{128}{693},$ so $m + n = 128 + 693 = 821.$