We deal 52 cards to 4 players, what is the probability that each player has an ace?

Question

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..

Answer 1

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...

Answer 2

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.

Answer 3

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.

$\dfrac{\binom{48}{12,12,12,12}\binom{4}{1,1,1,1}}{\binom{52}{13,13,13,13}}$