I'm not a math major but I guess it makes sense. With two door aces out and you not having any aces, that leaves 42 unseen cards, two of which are aces.

The two hands with door aces have 4 other cards which could be aces, so the odds are:
2/42 + 2/41 + 2/40 + 2/39 = .198 or about 20%

With only one door ace out that leave 42 unseen cards, THREE of which could be aces. The hand with the door ace has two other cards that could be an ace, thus:

3/42 + 3/41 = .145 or about 15%

So the when there are two aces out, the odds of split aces being present is greater than when only one ace is out.