Probability that the aces of spades, hearts, and diamonds are all in different piles $ \textbf{Question:} $ A deck of $ 52 $ cards are randomly divided into $ 4 $ piles of $ 13 $ cards each. Let $ E_{1} $ be the event that the ace of spades is in one of the piles, $ E_{2} $ be the event that the ace of spades and hearts are in different piles, and $ E_{3} $ be the event that the ace of spades and hearts and diamonds are all in different piles, compute $ P(E_{3} | E_{1}E_{2}). $
There is a simple way to approach this problem by saying that after the ace of spades and hearts have been placed into $ 2 $ piles, we have the remaining $ 50 $ slots left, of which $ 26 $ is suitable to place the ace of diamonds, so the desire probability is $ \displaystyle \frac{26}{50}. $
I approach using a different way and I currently got stuck so any help would be appreciated.
I have $ \displaystyle P(E_{3} | E_{1}E_{2}) = \frac{P(E_{1}E_{2} | E_{3})P(E_{3})}{P(E_{1}E_{2})}. $ 
Now $ P(E_{1}E_{2} | E_{3}) = 1 $ and $ P(E_{1}E_{2}) = P(E_{2}|E_{1})P(E_{1}) = P(E_{2}|E_{1}) $ since $ P(E_{1}) = 1. $ So at this point I only have to compute $ P(E_{3}) $ and $ P(E_{1}E_{2}). $ 
To compute $ P(E_{1}E_{2}), $ I compute the probability that the ace of spades and hearts are in the $ \textit{same} $ piles, which is $ \displaystyle \binom{4}{1}.\frac{\binom{50}{11}}{\binom{52}{13}} = \frac{4}{17}, $ so $ \displaystyle P(E_{1}E_{2}) = \frac{13}{17}. $     
To compute $ P(E_{3}), $ since there are $ \displaystyle \binom{4}{3} $ ways to put the $ 3 $ aces of spades and hearts and diamonds into $ 3 $ different piles, I got $ \displaystyle P(E_{3}) = \binom{4}{3}.\frac{\binom{49}{12}\binom{37}{12}\binom{25}{12}\binom{13}{13}}{\binom{52}{13, 13, 13, 13}}. $ However, when I put everything together, $ \displaystyle P(E_{3} | E_{1}E_{2}) = \frac{P(E_{1}E_{2} | E_{3})P(E_{3})}{P(E_{1}E_{2})} \neq \frac{26}{50} $ and I don't know where I got the computation wrong.  
 A: To calculate $P(E_3\mid E_1,E_2)=\frac{P(E_1,E_2,E_3)}{P(E_1,E_2)}$ Let us approach via direct counting.  (as an aside, the event $E_1$ is redundant... the ace of spades is always in a pile, so $E_1$ is in fact the universal event)
As we are working in an equiprobable sample space, we may instead use $P(E_3\mid E_1,E_2) = \frac{|E_1\cap E_2\cap E_3|}{|E_1\cap E_2|}$
The number of ways in which our three aces are in all different piles:


*

*Pick which pile gets which ace (spades, hearts, diamonds) and which has none of those three.  $4!$ choices

*Pick the remaining twelve cards in the ace of spades' pile from the remaining $49$.  $\binom{49}{12}$ choices

*Pick the remaining twelve cards in the ace of hearts' pile.  $\binom{37}{12}$ choices

*Pick the remaining twelve cards in the ace of diamonds' pile $\binom{25}{12}$ choices

*The remaining thirteen cards go into the as of yet unused pile.  $\binom{13}{13}$ choices (redundant to include)



The number of ways in which the ace of spades and hearts are in different piles:


*

*Pick the pile for the ace of spades.  Pick the pile for the ace of hearts.  $4\cdot 3$ choices

*From left to right (or clockwise starting from north if arranged circularly) pick the cards to go into the first unused pile.  $\binom{50}{13}$ choices

*Pick the cards to go into the second unused pile.  $\binom{37}{13}$

*Pick the remaining cards to go into the ace of spades' pile.  $\binom{24}{12}$

*Pick the remaining cards to go into the ace of hearts' pile.  $\binom{12}{12}$ (redundant)


We have as a result:
$$Pr(E_3\mid E_1,E_2) = \frac{4!\binom{49}{12}\binom{37}{12}\binom{25}{12}\binom{13}{13}}{4\cdot 3\binom{50}{13}\binom{37}{13}\binom{24}{12}\binom{12}{12}}=\frac{13}{25}=\frac{26}{50}$$

That is how I personally think.  As for your work, you seem to have done it all rather well.  You made the mistake in calculating $P(E_3)$ to have a $\binom{4}{3}$ instead of a $4!$.  By fixing that one mistake, you get the correct answer:
$$\frac{\color{red}{4!}\binom{49}{12}\binom{27}{12}\binom{25}{12}}{\binom{52}{13,13,13,13}} / \frac{13}{17}=\frac{13}{25}=\frac{26}{50}$$
