To solve the problem, we need to make some assumptions about the process of choosing. We assume that $108$ numbers are chosen "at random," one at a time, with replacement, from a set of $256$ numbers.
Imagine writing down the chosen numbers as a sequence of length $108$. There are $256^{108}$ such sequences, all equally likely. This number will be the denominator of our probability.
Now we calculate the number of "good" sequences, that is, sequences that have exactly $6$ distinct pairs, and $96$ singles. The types of the pairs can be chosen in $\binom{256}{6}$ ways. Once we have settled on our $6$ types, imagine ranking the chosen types in some arbitrary way, say numerical, but it doesn't matter.
Decide on the locations of the two numbers of the highest ranked type. This can be done in $\binom{108}{2}$ ways. For each of these ways, the locations of the two numbers of the next highest ranked type can be chosen in $\binom{106}{2}$ ways, and for each of these the locations of the two numbers of the next type can be chosen in $\binom{104}{2}$ ways, and so on. So the full task of dealing with pairs can be done in $D=\binom{256}{6}\binom{108}{2}\binom{106}{2}\binom{104}{2}\binom{102}{2}\binom{100}{2}\binom{98}{2}$ ways.
Now we have $96$ empty slots left. The first can be filled in $250$ ways. For each of these ways, the second empty slot can be filled in $249$ ways, and so on. Finally, the $96$th empty slot can be filled in $250-96+1$ ways. So our task of filling the $96$ slots can be done in $S=(250)(249)(248)\cdots(156)(155)$ ways.
So the number of ways we can have $6$ distinct pairs and the rest singles is $DS$. For the probability, divide $DS$ by $256^{108}$.