Chance of each player getting an ace when the dealer has 12 cards? 
4 players are playing a card game wherein each of them gets 10 cards, and the dealer gets 12 cards. What is the probability that each player gets an ace?

I want the use the $p = \dfrac{n_A}{N}$ method, where $n_A$ equals the favourable outcomes and $N$ equals all possible outcomes.
Starting with $N$, I figured we could consider the dealer to be a fifth player, and considering we don't care about the order of the players we'd get: 
$$N = \dfrac{52!}{(10!)^4\times12!} \times \dfrac{1}{5!}$$
Now for $n_A$, the aces can be divided among the players in $4!$ ways, and each of the players would still get 9 other cards from a total of 48, with the dealer getting the remaining twelve, thus giving us: $$n_A = 4! \times \dfrac{48!}{(9!)^412!} \times \dfrac{1}{5!}$$
But if we calculate $p$ this way we get a probability of $\approx 3\% $, which is just intuitively orders of magnitude too large to be correct, so I am sure I made a mistake somewhere. Can anyone help me spot it and then explain what I did wrong?
 A: I wouldn't consider the dealer as a fifth player but instead, let me guide you through another way to  get the answer using combinatorics.
We should start by counting $N$ as ${52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10}$ for the number of ways the dealer can give ten cards from 52 to each of the four players.

$N = {52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10} \approx 971089585681469963688868551062400 $ 

Now for $n_A$ we will consider all four aces given in $4!$ and count for the number of ways nine cards from the remaining 48 can be given to each of the four players as ${48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9}$.

$n_A = 4!{48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9} \approx 35869963456698493441273194240000$

Hence 

$p = {n_A \over N} = {4!{48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9}\over {52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10}}  ={400 \over 10829} \approx 0.036$

A: Comment: Because 0.036 doesn't seem to match the answer you
anticipated, and because there was at least one false start towards
a combinatorial answer, I decided to simulate the 'deal' a million
times in R statistical software, and see what proportion of deals gave one ace to each player.
I got $0.037 \pm 0.0004.$ So I think @AlfredoLozano's method is
correct (but note that $400/10829 = 0.03693785 \approx 0.037).$
My deck has 1's for Aces and 0's for all other cards for simplicity
counting results, but the sample function treats each 'card'
as distinct. The m-vector each.1 is 'logical' with elements TRUE
and FALSE; its mean is the proportion of its TRUEs.
m = 10^6; each.1 = logical(m)
deck=c(1,1,1,1, rep(0,48))
for (i in 1:m) {
  d = sample(deck, 40)
  each.1[i] = (sum(d[1:10]==1)&sum(d[11:20]==1)&sum(d[21:30]==1)&sum(d[31:40]==1))  
}
mean(each.1)
## 0.037124

A: Here is another way to get the answer.  Imagine the dealer deals the first $10$ cards to player A, the next $10$ to player B, the next $10$ to player C, the next $10$ to player D, and keeps the rest for himself.  Clearly the cards can be shuffled in any of $52!$ ways.  The question becomes, in how many ways can the dealer "stack" the deck so that each player gets an Ace?
First, remove the Aces from the deck and shuffle the remaining $48$ cards, which can be done in $48!$ ways.  Then shuffle the $4$ Aces, in $4!$ ways.  Then insert the first Ace so it's one of the first $10$ cards in the deck, the second Ace so it's one of the next $10$ cards, and so forth.  All this can be done in $48!\times4!\times10^4$ ways.  So the the probability of each player getting an Ace is
$${48!\times4!\times10^4\over52!}={24\times10^4\over52\times51\times50\times49}=24\left(10\over52\right)\left(10\over51\right)\left(10\over50\right)\left(10\over49\right)\approx25\left(1\over5\right)^4=0.04\%$$
The exact answer, as a reduced fraction, is ${400\over10829}=0.0369378\ldots$
