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}$$= \frac{39}{51}$  In the end, I found the right answer for $P(E_2)$, but I went through a whole line of reasoning that was not trivial. In the other hand, the book apparently deduces it in a trivial way:     since the pile containing the ace of spades will receive 12 of the remaining 51 cards   What I'm not getting?
Suppose that a deck of 52 cards containing four aces is shuffled thoroughly and the cards are then distributed among four players so that each player receives 13 cards. Determine the probability that each player will receive one ace.   The answer to this is given as$$\frac{13^4}{\binom {52}4}$$  My doubt is the following:   The book justifies ${\binom {52}{4}}$ as number of possible different combinations of the four positions in the deck occupied by 4 aces. That sounds like a case of arrangements to me, so shouldn't we think about permutations and not combinations if we are concerned about how the aces are to be arranged in the deck ?. Shouldn't the denominator be ${\binom {52}{13}}$ since you are choosing 13 cards for 4 people.
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