10 cards out - chance of 2 aces out Poker - 52 cards
4 suits of 13 cards each
A deck will have 4 aces  
Let say 5 player - 10 cards pre flop  
Poker probability
What is the chance exactly 2 aces are out (in the 10 cards dealt)?
Is this correct?
$$\binom{4}{2} \binom{12}{8} \binom{4}{1}^8 \over \binom{52}{10}$$
(aces)(other)(suits)/(possible)     
What are the chances 2 or more aces are out (dealt)?
Another way to say that is at least 2.
Or 2+. 
Is this even close?
$$\binom{4}{2} \binom{50}{8} \over \binom{52}{10}$$
(aces)(number of random hands left) / (possible)
If this should be two separate questions then fine. 
 A: Let's look at the first question. You want exactly 2 aces and 8 other cards. The probability is given by $${{4 \choose{2} }{48 \choose {8}}} \over {{52} \choose {10}}$$ 
Do you see now how to get the answer to your second question?
As JohnK noted, question 2 requires finding the probability of getting 3 and 4 aces and then adding those probabilities to the answer to question 1. 
So, for 3 aces, $${{4 \choose{3} }{48 \choose {7}}} \over {{52} \choose {10}}$$
Following this approach and continuing on by finding the probability for 4 aces and summing up over 2, 3, and 4, we find the probability of having 2 or more aces out in 10 cards to be 0.1625
A: @Frisbee, the probability of having at least 2 is the probability of having 2, 3, or 4. By the exact same logic as in my previous answer, the probability of having three aces is 
$$ \frac{\binom{4}{3} \binom{48}{7}}{\binom{52}{10}} $$ 
and the probability of having four aces is 
$$ \frac{\binom{4}{4} \binom{48}{6}}{\binom{52}{10}} $$ 
These are disjoint events, so add them all up to get the total probability....
A: After you question, I tried reproducing the frequency figures that can be calculated with combinations, as posted in Wikipedia by simply computer-simulating each hand in R, and repeating $5$-card draws millions of times.
I don't know if, strictly speaking, this is an "answer" to your question, but I think it is nice to confirm that combinatorics does indeed reflect perfectly what you would encounter if you were to just tally the outcomes of millions of random draws.
Here is the code, and the results so that you don't even have to run it.
A: @Frisbee, the "it" you keep asking about is obviously referring to your proposed answer. As the other commenter said, your answer is wrong because you double counted two aces in the group of cards you're choosing from (and no idea where the $\binom{4}{1}^8$ part came from). 
Once you choose exactly 2 out of the 4 aces, of which there are  $\binom{4}{2}$ (read "4 choose 2") possible ways to do, there are 48 cards left and you choose 8 of them, and there are $\binom{48}{8}$ ways to do that. So the number of possible ways both can happen is $\binom{4}{2} \binom{48}{8}$
