Dividing a deck in four stacks, aces must be in different piles (Conditional Prob.)

Question

I'm really stuck in this fairly simple example of conditional probability, I don't understand the book reasoning:

An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 ace.

Solution. Define events $E_i, i = 1, 2, 3, 4$, as follows:

$E_1$ = {the ace of spades is in any one of the piles}

$E_2$ = {the ace of spades and the ace of hearts are in different piles}

$E_3$ = {the aces of spades, hearts, and diamonds are all in different piles}

$E_4$ = {all 4 aces are in different piles}

The desired probability is $P(E_1E_2E_3E_4)$, and by the multiplication rule,

$P(E_1E_2E_3E_4) = P(E_1)P(E_2|E_1)P(E_3|E_1E_2)P(E_4|E_1E_2E_3)$

Now, $P(E_1) = 1$ since $E_1$ is the sample space S. Also, $P(E_2|E_1) = \frac{39}{51}$ since the pile containing the ace of spades will receive 12 of the remaining 51 cards (...)

I was lost with $P(E_2|E_1)$, I don't understand why it's $\frac{39}{51}$. I tried to think like this:

by definition, $P(E_2|E_1) = P(E_1E_2)/P(E_1)$ and, since $P(E_1) = 1$, $P(E_2|E_1) = P(E_1E_2)$. But it's obvious that $E_2 \subset E_1 \Rightarrow E_1 \cap E_2 = E_2 \Rightarrow (E_2|E_1) = P(E_2)$.

So I tried to calculate $P(E_2)$ to see if it matched the answer of the book. By definition, $E_2$ is the event where the ace of spades and the ace of hearts are in different piles. So the sample space is $52\choose13,13,13,13$. Now, suppose you take out the ace of hearts and the ace of spades of your deck, now you have 50 cards and there are $50\choose12,12,13,13$ ways of dividing the deck and $4\choose2$ ways of deciding which piles receive 12 cards and which receives 13. After making this division, there are $2!$ ways, for each result, to put back the 2 aces you took off (each one in one of the piles with 12 cards). So

$P(E_2) =$ ${50}\choose{12,12,13,13}$$ {4}\choose{2}$$2!$$/$${52}\choose{13,13,13,13}$$= \frac{39}{51}$

In the end, I found the right answer for $P(E_2)$, but I went through a whole line of reasoning that was not trivial. In the other hand, the book apparently deduces it in a trivial way:

since the pile containing the ace of spades will receive 12 of the remaining 51 cards

What I'm not getting?

Answer 1

I just found the answer in another post, it goes like this:

Imagine that there are 52 slots to be filled with the cards, each pile has 13 slots. After you place the first ace, you have 51 slots remaining to be filled, but for $E_2|E_1$ event to happen, just 39 slots of the 51 are suitable to receive the second ace, so $P(E_2|E_1) = \frac{39}{51}$. After the second ace was placed, you have 50 slots remaining and now just 26 are suitable to receive the third ace for $E_3|E_1E_2$ to happen, so $P(E_3|E_1E_2) = \frac{26}{50}$. Following the same reasoning, for the fourth card, there're 13 slots suitable from the 49 slots remaining, so $P(E_4|E_1E_2E_3) = \frac{13}{49}$

Answer 2

I would say, what is the chance that the first pile has exactly 1 ace.

$\frac{{4\choose 1}{48\choose 12}}{{52\choose 13}}$

What is the chance that the second pile has exactly 1 ace, given that the first pile holds one ace?

$\frac{{3\choose 1}{36\choose 12}}{{39\choose 13}}$

And one ace in the 3rd pile given that there are 2 aces in the first two piles

$\frac{{2\choose 1}{24\choose 12}}{{26\choose 13}}$

And the last pile takes care of itself.

$\frac{{4\choose 1}{48\choose 12}}{{52\choose 13}}\frac{{3\choose 1}{36\choose 12}}{{39\choose 13}}\frac{{2\choose 1}{24\choose 12}}{{26\choose 13}} = \frac{4!{48\choose 12,12,12,12}}{{52\choose 13,13,13,13}} = \frac {4! 13^4}{52\cdot 51 \cdot 50\cdot 49}$