We are counting the probability of at least one pair, meaning that we are allowing "three of a kind" or "two pairs."
How many no pair hands are there? We must choose $6$ different denominations from the $13$ denominations available. There are $\binom{13}{6}$ ways to do this.
For every choice of denominations, there are $4^6$ ways to choose the actual cards.
So the number of no pair hands is $\binom{13}{6}4^{6}$.
To find the probability of a no pair hand, divide by $\binom{52}{6}$. So the probability of getting at least one pair is $1-\frac{\binom{13}{6}4^6}{\binom{52}{6}}.$
Remark: To count the number of hands that have exactly one pair, do this. The denomination can be chosen in $\binom{13}{1}$ ways. For each choice, the actual card can be chosen in $\binom{4}{2}$ ways. Now choose the denominations we will have singles in. This can be done in $\binom{12}{4}$ ways. And now the actual singletons can be chosen in $4^4$ ways, for a total of $\binom{13}{1}\binom{4}{2}\binom{12}{4}4^4.$ For the probability we have precisely one pair, divide by $\binom{52}{6}$.