I got .0753 = ((13)(52C9))/(52C13) although i suspect that there are some repetitious combinations ie. 4,4,4,4,5,6,7,7,7,7,5,6,2 and 7,7,7,7,5,6,4,4,4,4,5,6,2.
You haven't eliminated repetitions like getting two or three sets of four of a kind, but that's okay because you said you wanted the probability of at least one four of a kind, not exactly one four of a kind.
"You haven't eliminated repetitions like getting two or three sets of four of a kind"
In calculating the combinations for any particular four-of-a-kind, for example for the combinations of the four-of-a-kinds for cards 4 and 7, there will be one combination that is counted twice: 4,4,4,4,5,6,7,7,7,7,5,6,2 and 7,7,7,7,5,6,4,4,4,4,5,6,2.
[tex]\{\texttt{hands with three four-of-a-kinds}\} \subset \{\texttt{hands with at least two four-of-a-kinds}\}[/tex]
[tex]\subset \{\texttt{hands with at least one four-of-a-kind}\}[/tex]