Consider a recursive function P(D,S,A,N) of four paremeters: D, the size of the deck, S, the number you select from the deck, A, the number of aces in the deck, and N, the number of aces that must be found to declare "success". P is itself the probability of success from a given starting condition.

P is defined as follows:

• P == 1 if N == 0 because if you don't need to find more aces, you've succeeded.
• P == 0 if A, S, or D are zero; in any of these cases it should be evident that no more aces may be found within the limits of the problem description.
• Otherwise, P is defined as follows: P(D,S,A,N) = (A/D) * P(D-1, S-1, A-1, N-1) + (1 - A/D) * P(D-1, S-1, A, N)
  
  The probability of success from this position has contributions from two possibilities; either the next card is an ace (with probability A/D, in which case the posterior probability of success is that of a position in which we are one step closer to the end, the deck has had one ace removed, and the number of aces to find is one left -- or the next card is not an ace (probability 1 - A/D) in which case the new situation is one in which we are one step closer to the end, the deck has one non-ace to remove, and the number of aces needed remains the same.

So.... after writing a function to compute all that, I get P(52, 26, 4, 2) ~~ 0.6951. I've backed this up with a monte-carlo simulation that shuffles and cuts decks, and counts the number in which there are at least two aces; successive runs found at least two aces in 695,229; 694,539; and 695,174 trials out of a million.