# Hypergeometric Distribution Question

In a hypergeometric distribution, you can compute the probability of drawing without replacement, which is useful in a number of studies and statistics. However, I'm having a problem that I've had for some time, but now it has become a more prevalent issue in my work.

If I were to use a hypergeometric distribution as the PMF for, say, drawing cards from a poker deck, and I wanted 3 Aces, and drew 5 cards total, the following would be a correct function for the draw probability:

$P= \frac{{4 \choose 3}{48 \choose 2}}{52 \choose 5}$

But what if I wanted to find the probability of drawing at least three aces, in addition to two other cards from the deck? One of the other two cards could be the last ace.

I've run computer simulations, and this does not appear to be the correct way to handle this scenario:

$P= \frac{{4 \choose 3}{49 \choose 2}}{52 \choose 5}$

The 48 was changed to a 49 because, instead of choosing any cards other than aces, I'm drawing two cards from whatever is remaining, and there's an extra ace in the deck. That means there are 49 cards still in the deck. To me, this seems like the function should spit out how likely I am to draw at least 3 aces in a five card hand. But it gives me weird results.

Specifically, I tested it with the following, much simpler problem:

I have a deck of 8 cards. There are only two kinds of cards in it; there are 4 of each in the deck. I want to draw 1 card of a specific type, say, "A", and one card from the deck that could be either "A" or "B". It's any card from the deck. This function tells me it's 100% likely to happen, which isn't possible:

$P= \frac{{4 \choose 1}{7 \choose 1}}{8 \choose 2}$

What am I doing wrong?

$$\mathbb{P}(\text{at least 3 aces})=\mathbb{P}(\text{3 aces})+\mathbb{P}(\text{4 aces})\\ =\dfrac{\dbinom{4}{3}\, \dbinom{48}{2}}{\dbinom{52}{5}}+\dfrac{\dbinom{4}{4}\, \dbinom{48}{1}}{\dbinom{52}{5}}$$
$P(\text{X = 3 or more})=1-P(X=0)-P(X=1)-P(X=2)$.