A deck of $52$ cards is divided into four piles of $13$ cards. What is the probability that each pile has one ace? 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 one ace.
The answer provided is is $(39*26*13)/(51*50*49) \approx 0.105$
The above answer uses conditional probability, but I would like to know what's wrong with my reasoning:

Call the four piles of partitions $1$, $2$, $3$, and $4$.


*

*For partition $1$, there are ${4 \choose 1}$ ways to choose which ace the partition will contain. Then, there are ${48\choose 12}$ ways to choose the remaining $12$ cards, as we cannot choose any other aces.

*For partition $2$, there are ${3\choose 1}$ ways to choose which ace the partition will contain. Then, there are ${36\choose 12}$ ways to choose the remaining $12$ cards, as there are only $36$ non-ace cards left.


Following similar reasoning for partitions $3$ and $4$, we find that there are ${2\choose 1}{24 \choose 12}$ and ${1\choose 1}{12 \choose 12} = 1$ ways to form those hands.
Therefore, my probability is given by
$$\frac{4 \cdot {48\choose 12} + 3\cdot{36\choose 12} + 2\cdot{24\choose 12}}{{52\choose 13}{39\choose 13}{26\choose 13}} \not \approx 0.105$$
The denominator is the number of ways to choose the cards in each hand without any constraints.
I am not sure what is wrong with my computation.
 A: You were close. You just have to multiply the possibilities in the numerator:
$$\frac{4 \cdot {48\choose 12} \cdot 3\cdot{36\choose 12} \cdot 2\cdot{24\choose 12}}{{52\choose 13}{39\choose 13}{26\choose 13}} \approx 0.105$$
A: Simpler way: There are $52$ places where an ace might be put ($13$ places in each of $4$ piles). We don't care which ace is which, so we can count $\binom{52}{4}$ different ways to choose the four places where the aces will be put; each of these sets of four places is equally likely.
But to get exactly one ace in each pile, we have to choose one of the $13$ places in the first pile,  one of the $13$ places in the second pile,  one of the $13$ places in the third pile, and one of the $13$ places in the fourth pile.
There are $13^4$ ways to do that, so $13^4$ of the sets of four places satisfy the condition. The probability is therefore
$$ \frac{13^4}{\binom{52}{4}} \approx 0.1055.$$
A: I like to think of this problem a little more explicitly, in terms of conditional probabilities (conditioning the sample space).
You can define events ${E_1,E_2,E_3,E_4}$ as ${E_i=}$ {the probability pile i has exactly 1 ace}.
Then, we have ${P(E_1E_2E_3E_4)=P(E_1)P(E_2|E_1)P(E_3|E_1E_2)P(E_4|E_1E_2E_3)}$.
${P(E_1) = \frac{{4 \choose 1}{52-4 \choose 12}}{{52 \choose 13}}}$. Since, we start with a deck of 52 cards, 4 of which are aces and 48 non-aces. From these, we choose 1 ace and 12 non-aces.
Now given ${E_1}$ occurred, we are left with a deck of cards that contains 39 cards. 3 of which are aces and 36 of which are non-aces. Hence, given ${E_1}$ we can compute ${P(E_2)}$ in the same manner as we did for ${P(E_1)}$ except that our sample space is no longer the traditional "deck" of cards.
${P(E_2 | E_1)=\frac{{3 \choose 1}{36 \choose 12}}{39 \choose 13}}$.
Next, given ${E_1}$ and ${E_2}$ occurred, our sample space is reduced to a deck of 26 cards, 2 of which are aces and 24 of which are non-aces. From here, the solution is clear.
