TITLE: Conditional Probability: Sheldon Ross Example 2h
QUESTION [10 upvotes]: 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?

REPLY [12 votes]: 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 $\frac{13}{49}$.