Let's consider the card labeled as follows:
Aces: \(A_1,A_2,A_3,A_4\)
Non-Aces: \(B_1,B_2 \)
The total number of ways to choose 2 cards out of 6 for the first pile is \(\binom{6}{2} \). After choosing the first pile, we choose 2 cards out of the remaining 4 for the second pile, which is \(\binom{4}{2} \). The remaining 2 cards automatically form the third pile, and this can be done in \(\binom{2}{2} \) ways.
Total ways is: \(\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}\)
But we have to divide 3!, since the order of the piles doesn't matter.
So we get \(\frac{\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}}{6}\) = 15
Next, we count the number of favorable outcomes where each pile contains exactly one Ace. Each pile must have one Ace and one non-Ace. There are 4 Aces (\(A_1,A_2,A_3,A_4\)) and 2 Non-Aces (\(B_1,B_2 \)).
The assignment of Aces to each pile can be done in 4 ways for first pile 3 ways for second 2 for final simplifying: \(4\times3\times 2\) = 24 ways
Every non-ace gets included ensuring simplified piles count as above:
\(\boxed{\frac{4}{15}}\)
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