I have a probability question if anyone is willing to help. In a $7$ card hand without replacement what is the probability of getting at least a $3$ of a kind (standard deck A-K, $4$ suits). I was trying to look into it and the working equation I have so far is:
$$\frac{{13\choose 1} \times {4\choose 3} \times {13\choose 4} \times {4\choose 1} \times {4\choose 1} \times {4\choose 1} \times {4\choose 1}}{52\choose 7} $$
I can't even really explain how I got that I was just trying numbers to try to get close to the value I arrived at experimentally which seems to be closer to $7.69\%$ instead of $7.11\%$ which is what the combination formula gives me.
Any help would be appreciated I just really want to get a better understanding of this specific case without having to go through a whole probability course.