Conditional Probability: Sheldon Ross Example 2h The following question comes from Example 2h, in Sheldon Ross's textbook A First Course in Probability on page 64. I got the same answer as the author through a different line of reasoning (given at the end of the solution). However, I would really like to understand the reasoning given in this solution, if someone could elaborate that would be great!
The question: An ordinary deck of 52 playing cards is randomly divided in 4 piles 13 cards in each pile. Compute the probability that each pile has an ace.
The solution:
We want to define four events for this problem :
\begin{align*}
E_{1} &= \{\text{The ace of spades in any one of the piles}\}\\
E_{2} &= \{\text{The ace of spades and the ace of hearts are in different piles}\}\\
E_{3} &= \{\text{The aces of spades, hearts, and diamonds are all in different piles}\}\\
E_{4} &= \{\text{All four of the aces are in different piles}\}
\end{align*}
The desired probability is $P(E_{1}E_{2}E_{3}E_{4})$ and by applying the multiplication rule,
$$P(E_{1}E_{2}E_{3}E_{4})=P(E_{1})P(E_{2}|E_{1})\cdots P(E_{4}|E_{1}E_{2}E_{3})$$
Now,
$$P(E_{1})=1$$
Since $E_1$ is in the sample space $S$. Also, 
$$P(E_{2}|E_{1})=\frac{39}{51}$$
Since the pile containing the pile contain the ace of spades will contain 12 of the remaining 51 card, and 
$$P(E_3 | E_1 E_2)=\frac{26}{50}$$
Since the piles containing the aces of spades and hearts will receive 24 of the remaining
50 cards. Finally,
and finally, 
$$P(E_4 | E_1 E_2 E_3)=\frac{13}{49}$$
Multiplying them all together we get 
$$P( E_1 E_2 E_3 E_4) \approx 0.105$$
My Confusion: I don't quite understand the reasoning; it is clear that $P(E_1)$ is 1, because the ace of spades has to end up some where. But the the conditional probability has me confused; given the the ace of spades is in one pile I have 51 cards to choose from to put in any of the piles. There are $\binom{51}{39}$  ways that I can choose the 39 cards for the piles that don't contain the ace of spades. If I want to assure that I get the ace of hearts in one of the 3 piles that doesn't contain the ace of spades, I can set it aside. Then I am left to choose 38 from 50, there are  $\binom{50}{38}$ ways to do this. Then I have 
$$\frac{\binom{50}{38}}{\binom{51}{39}}=\frac{39}{51}$$
And this same reasoning will lead to the same answer. I see benefit from not having to use the binomial coefficient, but I don't understand the soundness of the authors reasoning could someone elaborate on why his method works? 
 A: Imagine that each player has $13$ "slots," each of which will ultimately contain one card. We can assume that the Ace of $\spadesuit$, Ace of $\heartsuit$, Ace of $\diamondsuit$, and Ace of $\clubsuit$ are, in that order, the first four cards in the deck.
Once we have decided where the $\spadesuit$ Ace has gone, there are $51$ open slots, of which $39$ are not in the same group as the group that contains the $\spadesuit$ Ace. So the probability that the $\heartsuit$ Ace lands in one of these slots is $\frac{39}{51}$.
Similarly, when it comes to deal the $\diamondsuit$ Ace, there are $50$ open slots left, of which $26$ are "favourable." And given that the first three Aces have landed nicely, the probability the fourth lands nicely is $\frac{13}{49}$. 
A: (part 1)
We give a rationale for $P( E_2 | E_1 ) = 39/51$ since the pile containing the ace of spades will receive 12 of the remaining 51 cards.
An outcome is 4 (ordered) piles of 13 (unordered) cards from a deck of 52 cards.  Each of the 52 cards occurs in exactly one of the piles. An outcome $\omega$ is a 4-tuple and each item in the tuple is a set of  13 cards.  The sample space $\Omega$ is the set of all possible 4-tuples.
$E_1 = \{ \text{the ace of spades is in any one of the piles} \}$. $E_1 = \Omega$.  Therefore, $P( E_2 | E_1 ) = P( E_2 )$.
For a single outcome $\omega$, 12 cards different from the ace of spades are in the same pile as the ace of spades.  $51 - 12 = 39$.  The remaining 39 cards different from the ace of spades are not in the same pile as the ace of spades.  This is true for each outcome.
How can we use this information to show the proportion of outcomes in which the ace of hearts is not in the same pile as the ace of spades is 39/51?
$4 \cdot \dbinom{ 50 }{ 12 } \cdot \dbinom{ 39 }{ 13, 13, 13 }$
the number of outcomes in which the ace of hearts is not in the same pile as the ace of spades.  4 ways to choose a pile for the ace of spades, then choose 12 cards not the ace of spades and not the ace of hearts for the pile containing the ace of spades.
$4 \cdot \dbinom{ 50 }{ 11 } \cdot \dbinom{ 39 }{ 13, 13, 13 }$
the number of outcomes in which the ace of hearts is in the same pile as the ace of spades. 4 ways to choose a pile for both the ace of spades and the ace of hearts, then choose 11 cards not the ace of spaces and not the ace of hearts for the pile containing the ace of spades.
$\dbinom{ 50 }{ 12 } + \dbinom{ 50 }{ 11 } = \dbinom{ 51 }{ 12 }$
The proportion of outcomes in the sample space in which the ace of hearts is not in the same pile as the ace of spades is
$\dfrac{ 4 \cdot \dbinom{ 50 }{ 12 } \cdot \dbinom{ 39 }{ 13, 13, 13 } }
{ 4 \cdot \dbinom{ 50 }{ 12 } \cdot \dbinom{ 39 }{ 13, 13, 13 } +
4 \cdot \dbinom{ 50 }{ 11 } \cdot \dbinom{ 39 }{ 13, 13, 13 } } =
\dfrac{ \dbinom{ 50 }{ 12 } }{ \dbinom{ 51 }{ 12 } } = \dfrac{ 39 }{ 51 }$
Because the outcomes are equally likely, $P( E_2 )$ is the proportion.
$P( E_2 | E_1 ) = P( E_2 ) = | E_2 | / | \Omega | = 39/51$

You might say this is a lot of work when we could have just computed $| E_2 |$ and $| \Omega |$ directly.
$\dfrac{ | E_2 | }{ | \Omega | } =
 \dfrac{ 4 \cdot 3 \cdot \dbinom{ 50 }{ 12, 12, 13, 13 } }
 { \dbinom{ 52 }{ 13, 13, 13, 13 } } = \dfrac{ 39 }{ 51 }$
We have learned (verified) that counting the number of cards different from the ace of spades in the same pile as the ace of spades (12) is a shortcut ( 51 - 12 ) / 51 for the combinatorial calculation.
