Suppose there is a left pile of 2 cards, a middle pile of 2 cards, and a right
pile of 2 cards.

It is not necessary to be concerned with which card is on top and which card is
on the bottom of each pile of 2.

Let's get the probability of the complement event:

The only unsuccessful ways are when both twos are in the same pile. We pick the
pile for the 2 twos to go in -- in 3 ways. 

Then the other two piles will consist of two aces each.

There are C(4,2) = 6 ways to choose the pair of aces for the leftmost of the
remaining two piles, and then, there is only 1 way left for the other two aces
to go in the one remaining pile.

So there are (3)(6)(1) = 18 unsuccessful ways.

Now we get the number of possible ways:

1. We select two cards to go in the left pile in C(6,2)=15 ways
2. We select two of the remaining 4 cards to go in the middle pile in C(4,2)=6
ways.
3. Then there is only 1 way left for the other two cards to go in the right pile.

So there are (15)(6)(1)=90 possible ways to arrange the cards.

So the probability of an unsuccessful event is 18/90 = 1/5.

So the probability of a successful event is 4/5.

Edwin