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?

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

