What is wrong with my solution? Probability that every bridge player gets an ace. Answers here and here do not answer my question.
We have $4$ players playing bridge. What is the probability that each player gets an ace? The solution provided with the problem first finds the total no. of ways of dealing the cards using the multinomial coefficient:
$$\frac{52!}{13! \ 13! \ 13! \ 13!}$$
For favourable outcomes, there are $4!$ ways of dealing out the aces ($1$ to each player) and the remaining cards can be dealt in (again, by multinomial coefficient)
$$\frac{48!}{12! \ 12! \ 12! \ 12!}$$
ways. Thus, the probability becomes:
$$4!\frac{48!(13!)^4}{52!(12!)^4}$$
Simplifying:
$$24\frac{13^4}{49\cdot 50\cdot 51\cdot 52}$$
$$\frac{13^3}{49\cdot 25\cdot 17} \approx 0.1055$$
The last calculation is still hard without a calculator, so I thought of an (apparently incorrect) approach.
My solution (what is the mistake here?):
Let us forget about the other cards, and concentrate on the aces. There are $4!$ ways that each player will get an ace. There are $4^4$ total ways of distributing the aces. So, the probability is:
$$\frac{4!}{4^4} = \frac{3!}{4^3} = \frac{3}{32} = \color{red}{0.09375}$$
Which is significantly off. So obviously there is a mistake in my argument. My guess is that the other $12$ (or $48$ cards) somehow influence the answer. Could you please point out the mistake in the answer? I would also be grateful if anyone could explain the mistake with a smaller example, which shows how what I've done and the original answer differ.
 A: Good question; on some level, your solution seems reasonable.  The problem is that it implicitly assumes all $4^4$ "ways of distributing the aces"... that is, of assigning the $4$ aces to the $4$ hands... are equally probable.  This is not the case; it's more likely that the aces are shared (in a particular way) than that one (particular) player gets them all, and so the true probability is (slightly) greater than the one you proposed.
Making the problem smaller makes this more clear.  In particular, suppose the deck consists only of kings and aces, and we deal two cards to each player.  Then the probability that each player gets an ace is $(4!)^2/\left({{8}\choose{2}}{{6}\choose{2}}{{4}\choose{2}}\right)=1/105$, larger than the $1/256$ from your solution.  (In the extreme limit, if the deck consists of only the four aces, then the probability that each player gets an ace is $1$; but this seems less informative somehow.)
A: We can obtain the good result, with simple calculs and only small numbers.
We deal the 52 cards to 4 players.
We ask : who has the ace of Clubs ?  We know which player has this card. What is the probability that this player don't have the Ace of Diamonds : 51 places, 39 are succes, probability is $\frac{39}{51}$.
Consider this step is a success, Ace of Diamond is not in same hand than Ace of clubs.
What is the probability that Ace of Hearth is in the 2 remaining hands : 50 places, 26 are success, probability is $\frac{26}{50}$
Last step, if we success also at this step, what is the probability that Ace of spade is in the remaining hand : $\frac{13}{49} $
Final result is $\dfrac{39}{51} \times \dfrac{26}{50} \times \dfrac{13}{49} $
