Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) = P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725

There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways to select exactly 3 aces is:

$4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192$

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

$P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}$

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

$P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052$