Probability question, not sure if I'm doing this right... at least vs exactly. So I'm trying to figure a few things out with probability/counting. These a probability questions, but my understanding of the counting behind them is a little fuzzy still.  
For example, The probability of getting 3 of a kind out of a 5 card poker hand is:
COMBIN(13, 1) to choose 1 kind of out the 4 kinds
COMBIN(4, 3) to choose 3 cards out of the 1 kind
COMBIN(12, 2) to choose the 2 kinds for the last 2 cards (the first card took 1 kind already, so it's 12 instead of 13)
(COMBIN(4,1))^2 to choose the 2 cards out the the last kind(s)
So, the answer is: (COMBIN(13,1)*COMBIN(4,3)COMBIN(12,2)(COMBIN(4,1))^2)/(COMBIN(52,2))
How would this change if I were to look for AT LEAST 3 cards of the same kind instead of EXACTLY 3 cards of the same kind?
 A: The probability of at least $3$ is the probability of exactly $3$ plus the probability of exactly $4$. (I am assuming that you do not want to count the full house hands in your "at least $3$ of a kind. If you do, the answer changes. And if you mean a hand that is $3$ of a kind or better, in the poker sense, then the answer changes again.) 
You did the probability of exactly $3$ correctly. Counting the $4$ of a kind hands is even easier. By the reasoning that you used, it is $\binom{13}{1}\binom{48}{1}$. We need to choose the kind. Ater we have done that, we have $\binom{48}{1}$ ways to choose the (almost) useless card. 
Remark: You did not ask for an explanation of the "$3$ of a kind" calculation, so I assume it is understood.  
The subtlety there is that we must avoid a full house, which is a much better hand. So after the kind and the $3$ cards are chosen, we must choose two different kinds for our remaining cards. 
A: There are 3 kinds of poker hands that have "at least" 3 of the same rank: 3 of a kind, 4 of a kind, full house.
One way to count how to get exactly 3 cards of the same rank: 13 * (4 C 3) * (48 C 2) and add this to the ways to get exactly 4 cards of the same rank: 13 * 48.
Note also that getting exactly 3 of the same rank is 3 of a kind AND full house, thus there are more ways (higher probability) to do this than getting what the poker world calls "3 of a kind."
