We deal 52 cards to 4 players, what is the probability that each player has an ace? I saw on internet the answer is : $$\frac{13^4}{{52}\choose{4}} $$
but I don't intuitively understand it. So I tried another one that is : $$\frac{{52}\choose{13 \ 13 \ 13 \ 13}}{52!} \cdot 4!$$
Because we will divide 52 cards into 4 groups of 13 cards, there is 52! ways to shuffle the deck and 4! ways to order the 4 aces. But this doesn't give me a correct result.. 
 A: Using multinomials. the answer is:
$$\frac{4!\cdot\binom{48}{12,12,12,12}}{\binom{52}{13,13,13,13}} = \frac{4!\,48!}{52!}\cdot \frac{13!\,13!\,13!\,13!}{12!\,12!\,12!\,12!} =\frac{13^4}{\binom{52}{4}}$$
Representing, as numerator on the left, a "pre-deal" of aces to each hand and the ways of distributing the remaining cards, and as denominator, the distribution without any pre-dealing.
The shuffling is not an extra factor, since the selection values already account for any possible ordering.
A: It would be good if you understand the internet answer, as the logic is simple.
Imagine $13$ slots for each player, and focus on the aces (where the remaining cards go is irrelevant)
Ignoring order, the aces need to be placed in${\binom{13}1}^4 = 13^4$ ways,
against $\binom{52}4$ total ways, and thus the result.
And if you prefer multinomials, you will get the correct numerator by considering the distribution of non-aces and aces in each hand, viz.
$\dfrac{\binom{48}{12,12,12,12}\binom{4}{1,1,1,1}}{\binom{52}{13,13,13,13}}$
A: There are $13$ possible shufflings (of aces) for each group so it must be $13\cdot13\cdot13\cdot13=13^4$ and then we have $4!$ ways to order $4$ aces, and all the situations in which we shuffle the cards are $52\cdot51\cdot50\cdot49$ (Because there are $4$ players and each time we give a card we have one fewer)  and if we continue we get; $$\frac{13^4\cdot4!}{52.51.50.49}=\frac{13^4}{52.51.50.49.\frac{1}{4!}}=\frac{13^4}{\binom{52}{4}}$$
I have also used the rule that $$\binom{n}{r}=\frac{n!}{(n-r)!\cdot r!}$$ in the case we plug in $n=52$ and $r=4$ we get $$\frac{52!}{48!\cdot4!}=\frac{52\cdot51\cdot50\cdot49\cdot48!}{48!\cdot4!}$$ doing the cancellations and...
A: The probability for finding one ace has been placed among the thirteen cards of each of the four hands, when the four aces have been randomly placed among the deck of fifty-two cards,* is:$$\dfrac{\dbinom{13}1^4}{\dbinom{52}4}$$

An alternate way to evaluate this would be to take ways each player receives one ace and twelve other cards, and divide by the total ways each player receives thirteen cards.
$$\dfrac{\dbinom{4}{1,1,1,1}\dbinom{48}{12,12,12,12}}{\dbinom{52}{13,13,13,13}}$$ 
Which can easily be seen to give the same result.
