Answer
$\frac{1}{193}$
Work Step by Step
There are 4 ways to choose three aces: (no heart, no spade, no club, no diamond).
Since you now have 3 cards, there are 49 cards left (1 ace, and all cards from 2 to King).
So there are 49 ways to choose the fourth card.
Of these 49 ways, 1 card is an ace. Having all four aces is $\frac{1}{270,725}$ of the possible four-set cards.
This leaves us with 48 ways of choosing a card that is not an ace (to go along with the three aces in hand).
So, there are 4 × 48 = 192 sets of cards where exactly three are aces.
That gives 193 (192 +1) sets where there are at least three aces. The conditional probability is (1/270,725) / (193/270,725) = 1/193.
