jmc

Since a multiple of 196 must have 2 factors of 2 and 2 factors of 7, we can count the pairs by focusing on the factors of 7. For one thing, 98 can be paired with any even number as it has 1 factor of 2, since $98=2\cdot 7^2$ takes care of all the other primes. So, 98 can be paired with 2, 4, 12, 14, and 28, for 5 pairs. Then, 28 can be paired with (excluding 98 which we already counted) 21 and 14, both of which have the necessary factor of 7, giving us 2 more pairs. There are no remaining pairs of numbers 21 and smaller that are multiples of 196, because the only pair with two factors of 7, $\{14,21\}$, has a factor of 2 but not 4. So, there are $5+2=7$ pairs. And in total, there are $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ possible pairs, giving us a probability of $\frac{7}{21}=\boxed{\frac{1}{3}}$.