TITLE: Finding the probability that an ace is found in every pile when a deck of cards is split into 4
QUESTION [7 upvotes]: I'm trying to answer this question and you are supposed to use the multiplication rule to solve it: 
A deck of 52 playing cards is randomly divided into four piles of 13 cards each. Compute the probability that each pile has exactly 1 ace.
I started off by defining following 4 events: $A_{1}, A_{2}, A_{3}$ and $A_{4}$ where $A_{i}$ denotes the event that exactly one ace is found in the i$^{th}$ pile - so to find the probability I need to find the probability of the intersection of all these events which is where I can use the multiplication rule.
The multiplication rule says that $$\mathbb{P}(A_{1} \cap A_{2} \cap A_{3} \cap A_{4})= \mathbb{P}(A_{1}) \mathbb{P}(A_{2}|A_{1}) \mathbb{P}(A_{3}|A_{2} \cap A_{1}) \mathbb{P}(A_{4}|A_{3} \cap A_{2} \cap A_{1})$$
to find each of the probabilities on the RHS I looked compared the possible combinations allowed for each situation:
The number of possible card combinations such that $A_{1}$ holds is $\binom{48}{12}$ and the total number of possible card combinations for the first pile is $\binom{52}{13}$. It then follows that $$\mathbb{P}(A_{1})= \frac{\binom{48}{12}}{\binom{52}{13}}=\frac{1406}{4165}$$
When moving on to the second pile, it follows that we now have 39 cards remaining so in order for the pile 2 to have exactly one ace, it leads to $\binom{36}{12}$ possible combinations out of the $\binom{39}{13}$ total number of combinations and so we get that
$$\mathbb{P}(A_{2}|A_{1}) = \frac{\binom{36}{12}} {\binom{39}{13}} =\frac{225}{703}$$.
Continuing on in this way I got that 
$$\mathbb{P}(A_{3}|A_{2} \cap A_{1}) = \frac{\binom{24}{12}}{\binom{26}{13}}=\frac{13}{50}$$
and by the way I have defined my events, it means that 
$$\mathbb{P}(A_{4}|A_{3} \cap A_{2} \cap A_{1}) = 1$$
so by the multiplication rule I get that $$\mathbb{P}(A_{1} \cap A_{2} \cap A_{3} \cap A_{4}) = \frac{1406}{4165} \frac{225}{703} \frac{13}{50} \approx 0.0281$$
However the answer I am given says it should be $\approx 0.105$. Can anyone help me to see where I have gone wrong? Would it perhaps be that defining the events differently lead to different probabilities? Thanks!

REPLY [2 votes]: Choose $12$ non-ace cards for the first pile, $12$ non-ace cards for the second pile, $12$ non-ace cards for the third pile and the remaining $12$ non-ace cards for the fourth pile.
Place the four aces in the four piles, one in each pile.
This can be done in $${48 \choose 12}{36 \choose 12}{24 \choose 12}{12 \choose 12}\cdot 4!$$
ways.
This has to be divided with the number of all possibilities: choose $13$ cards for the first pile, $13$ for the second, $13$ for the third and the remaining $13$ for the fourth. This can be done in 
$${52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}$$
ways.
The result is
$$\frac{{48 \choose 12}{36 \choose 12}{24 \choose 12}{12 \choose 12}\cdot 4!}{{52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}} = 0.1055$$