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?

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 

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

