Is there an alternative intuition for solving the probability of having one ace card in every bridge player's hand? I am trying to get to know probability a little better since it's a weak point for me and I was wondering what other ways there were to intuitively think about the problem of finding the probability that 4 bridge players each have exactly one ace after being dealt a deck of 52 cards.
My solution started with that there are $4!$ ways of distributing the aces initially, and then subsequently there are $48 \choose 12$ ways of distributing the rest of the first hand, $36 \choose 12$ ways of distributing the rest of the second hand, and $24 \choose 12$ ways of distributing the third hand, and one way to distribute the last hand. Thus there are $4! *$ $48 \choose 12$ $*$ $36 \choose 12$ $*$ $24 \choose 12$ ways of the players having exactly one ace. 
The total sample space, however, is counted for by first giving the first player his hand ($52 \choose 13$), the second player has hand from the remaining deck, ($39 \choose 13$), etc.; so the sample space is $ 52\choose 13$ $*$ $39 \choose 13$ $*$ $26 \choose 13$. 
The final answer after simplifying is that the probability that each player receives exactly one ace is $\frac{4!48!13^{3}}{52!}$
I think this is correct (correct me if my solution is wrong), but what are the other ways of thinking of this problem?
 A: An alternative intuition is to see it as 4 groups of 13 with  4 friends all in different groups.
The first can be in any group, the 2nd has 39 permissible slots out of 51, and so on.
Thus Pr =  $\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$
A: The solution is arguably simpler if you distinguish between permutations within the hands. There are $4!$ ways of distributing the aces over the $4$ players, and $13^4$ possible positions for the aces in the hands. That leaves $48!$ possibilities for filling the remaining $48$ slots. The size of the sample space in this perspective is just $52!$, so the desired probability is
$$\frac{4!\cdot13^4\cdot48!}{52!}\;.$$
A: Rather than trying to rephrase, I think a great choice is the explanation by  F. Scholz here:

A: I always thought it was easier to think in terms of the multinomial coefficient : you define it for $k_1,\cdots,k_d \ge 0, k_1 + \cdots + k_d = n$ and then you let
$$
\binom n{k_1,\cdots,k_d} \overset{def}= \frac{n!}{k_1! \cdots k_d!}. 
$$
It counts the number of ways to have $n$ distinguishable balls and put $k_1$ of them in the first box, $k_2$ in the second box, $\cdots$, $k_d$ of them in the $d^{\text{th}}$ box (the boxes are also distinguishable). The $k_i!$ factor in the denominator takes off "all the multiple-counts in $n!$" due to the fact that the $k_i$ balls you put in one box can be permuted amongst themselves, giving the same result. 
Using this new intuition, your count becomes
$$
\frac{\binom 4{1,1,1,1} \binom{48}{12,12,12,12}}{\binom{52}{13,13,13,13}}. 
$$
The numerator is the number of ways to put one ace in each hand and then $12$ of the remaining cards in each hand. The denominator is just the number of ways to put $13$ cards in each hand. Notice that 
$$
\binom{n}{k_1,\cdots,k_d} = \binom{n-k_d}{k_1,\cdots,k_{d-1}} \binom n{k_d}
$$
which you can check either combinatorically, or just notice that the $(n-k_d)!$ in both sides cancel out. If you set $k_i = k$ for $1 \le i \le d$, you get the formula
$$
\binom{dk}{k,k,\cdots,k} = \binom dk \binom{(d-1)k}k \cdots \binom {2k}k \binom kk = \prod_{j=1}^d \binom{jk}k,
$$
which is essentially what happens when you think about putting $12$ cards in the first hand $\binom{4 \cdot 12}{12}$, then $12$ cards in the second hand $\binom{3 \cdot 12}{12}$, etc.
In particular, your count was correct. (But for some reason, the number you explicitly compute at the end is off by a factor of $13$ ; check your computations. You clearly see $13^4$ in the numerator if you use my formula, they come from the $52!/(13!)^4$ in the denominator.)
Hope that helps,
A: In a deck of cards, for the aces to be equidistribute they need to be located in each of the modular positions.
That is to say we need an ace at each of $0\mod4, 1\mod4, 2\mod4$ and $3\mod4$.
If we consider a pack of cards with only 2 types of cards - 4 of type A and 48 of type , we can see that there are $\binom{52}{4}$ ways to deal the cards, and we need the hands with exactly 1 of type A in each modular slot. But this sum doesn't look easy.
Instead, consider how many ways there are to distribute 4 aces into 4 slots, namely $4^4$, but only $4!=24$ of these are any use to us. Once we have done this, each slot has $13$ possibilities, and so the distribution here is $4!.13^4$.
We have 48 cards remaining, giving the probability as:
$$ \text{P(each player gets an ace)}=\dfrac{4!.13^4.48!}{52!}=\dfrac{13^4}{\binom{52}{4}}$$
