Confusion about a simple conditional probability problem

I try to solve the following problem:

5 cards are dealt from a shuffled deck.
What is the probability that the dealt hand contains exactly 2 aces,
given that we know it contains at least 1 ace?

I have arrived at two potential solutions to this problem, but the obtained results differ.
In the first way I use the Bayes' formula.
Let's #A: the number of aces in the hand.

$P(\#A=2|\#A \ge 1) = \frac{P(\#A \ge 1| \#A=2)P(\#A=2)}{P(\#A \ge 1)}=\frac{1 \cdot P(\#A=2)}{1-P(\#A = 0)}=\frac{\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}}{1-\frac{\binom{4}{0}\binom{48}{5}}{\binom{52}{5}}}\approx0.12$

In the second way, I assume that 1 ace is already in my hand, and I have to draw additional 4 cards from a deck of 51 cards (3 aces, and 48 non-aces) and then I calculate the probability of drawing exactly 1 additional ace (in order to have 2 aces in my hand in total).

$\frac{\binom{3}{1}\binom{48}{3}}{\binom{51}{4}}\approx0.21$

So I get a different number. I would like to know which solution is the correct one (or maybe they are both incorrect?) and I would also like to know why aren't they equivalent?
Thank you in advance for your help.

Solution 1:

Your first answer is correct given the question. As you can see, your work in the first approach only considers number of aces and non-aces and not the order.

If you were to find the conditional probability of being dealt exactly $1$ more ace in next $4$ cards given the first card was already an Ace, the second approach would be correct. I am adding a more elaborate work that would show you how your second answer is answering a different question that I just stated.

If $B$ is the event of first card being Ace and $A$ being the event of exactly two aces in five cards hand,

$P(A \cap B)= \displaystyle \frac{{4 \choose 1} {4 \choose 2} {48 \choose 3} \cdot 2! \cdot 3! }{ {52 \choose 5} \cdot 5!} = \frac{4324}{270725}$

(we know the first card is an Ace. We choose which of the next $4$ cards would be an Ace. We choose $2$ Aces and $3$ non-Aces and we permute Aces within and we permute non-Aces within.)

Also, $ \displaystyle P(B) = \frac{1}{13}$

$ \displaystyle \therefore P(A|B) = \frac{4324}{20825} \approx 0.21$

I hope this clarifies.