I have assumed that the easiest way would be to calculate the chance of a player drawing no aces. Would this not be the compliment of both players having at least one?
My first try was $$1-\frac{C(48, 22)}{C(52, 26)}.$$ I am not sure how to take into account the permutations, though. $1-\dfrac{P(48, 22)}{P(52, 26)}$ seems to be incorrect as does $1-\dfrac{C(48, 22)}{P(52, 26)}$.
I'm thinking there are $26$ from $48$ ways for a player to get half the deck with no aces which is the same number of ways for the other player's half deck to have all $4$ aces. The probability of this is therefore:
$$P(0) = \frac{\binom{48}{26}}{\binom{52}{26}}$$
This can be reversed so the probability of a player getting either no aces or all four aces is:
$$P(0\ \text{or}\ 4) = 2\cdot \frac{\binom{48}{26}}{\binom{52}{26}}$$
The compliment probability, both players get at least one ace is:
$$P(1\ \text{to}\ 3) = 1 - 2\cdot \frac{\binom{48}{26}}{\binom{52}{26}} = .889556$$