Two piles of 26 cards There is a 52-card deck. The deck is split into two piles of 26 cards each. 
I need to find the following probabilities:
$1)$ Each pile contains two aces 
$2)$ The first pile contains more hearts than the second 
$3)$ There are more hearts in the first pile than clubs in the second
My solutions:
$1)$ $$\frac{{4\choose 2}{48\choose 24}}{52\choose 26}$$
$2)$ $$\frac{\sum_{k=7}^{13} {13\choose k}{39\choose 26-k}}{52\choose 26}$$
$3)$ $$\frac{\sum_{k=1}^{13}\sum_{l=0}^{k-1}{13\choose k}{13\choose 13-l}{26\choose 13-k+l}}{52\choose 26}$$
Are my answers correct?
 A: Your set up is correct but $(2)$ and $(3)$ can be simplified.
For $(2)$, notice there are $13$ cards of hearts. So one of the piles will have more hearts than the other and each of the two outcomes is equally likely. So the probability that the first pile has more hearts than the second is $\frac{1}{2}$.
For $(3)$, here is what I would do to simplify. Let's first find the probability that number of hearts in the first pile is equal to the number of clubs in the second ( that would also mean that the number of clubs in the first is equal to the number of hearts in the second). So from $26$ clubs and hearts together, we select $13$ cards for pile $1$ and assign the rest to pile $2$. Take an example. Say we chose $k$ hearts and $(13-k)$ clubs from $26$ cards in the first pile. So the second pile will obviously have $k$ clubs which is equal to the number of hearts in the first pile. We then choose rest $13$ cards from remaining $26$ diamonds and spades.
So, $ \small \displaystyle P(H1 = C2) = {26 \choose 13} {26 \choose 13} / {52 \choose 26} = \frac{16232365000}{74417546961}$
Now subtracting from $1$ gives us the probability that the number of hearts in pile $1$ is not equal to the number of clubs in pile $2$. Dividing by $2$ gives the probability that we have more hearts in pile $1$ than clubs in pile $2$. The desired probability comes to,
$ \displaystyle \small \frac{58185181961}{148835093922} \approx 0.3909$
