Captive Friends

Jack and Tony must be in two different pairs to be both set free. The chance of this happening is $\frac{18}{19}$ because Tony (for example) could be paired with 19 different people, and one of them is Jack.

Then they have to win in their pairs, which happens with probability $\frac{1}{2}\cdot\frac{1}{2} =\frac{1}{4}.$

Thus, the total probability is $\frac{18}{19}\cdot \frac{1}{4} = \frac{9}{38},$ so $A+B=47.$

The answer is the same, by the way, as if he had randomly chosen the ten men to be set free: $\frac{\binom{18}{8}}{\binom{20}{10}} = \frac9{38}.$ (Or if you like: $\frac12 \cdot \frac9{19}$ .)

I think this makes intuitive sense.

To generalize for $k$ friends and $2n$ captives where $k \leq n$ :

1. The probability that none are in the same pair is: $\displaystyle\prod_{i=1}^{k} \frac{2n-(i+k-1)}{2n-(2i-1)}$

2. The probability that all of them are the chosen one in their pair: $\frac{1}{2^{k}}$

The probability that they are all free is the product of the two probability.

There are going to be 10 people set free in the end. We can chose a Jack and a Tony from the freed group in $\binom{10}{2}$ ways.

There are a total of $\binom{20}{2}$ ways to chose Jack and Tony from a group of 20 people, so the final answer is $\frac{\binom{10}{2}}{\binom{20}{2}}$ = $\frac{10 \cdot 9}{20 \cdot 9}$ = $\frac{9}{38}$ $\rightarrow$ $\boxed{47}.$