I calculate a case that no Ace card was removed, so we have $5 \times \frac{10C4}{50C4} \approx 0.004559$. For every other cases, there is no way to get 4 Aces, so the probability is 0.
The correct answer is $\approx 0.00388$, what am I missing here? Thank you in advance!
On
We can just see how the aces must fall, and forget about the rest.
Imagine $50$ slots for the players + $2$ for discards
The first ace has to fall in one of the $50$ slots, and the rest must fall in the remaining slots of that player,
thus Pr = $\frac{50}{52}\frac{9}{51}\frac{8}{50}\frac{7}{49} = \frac{6}{1547}, \approx 0.00388$
This could not fit as a comment, so it is an answer. The counting problem is equivalent to the following one. Consider all permutations of the numbers $1,2,3,\dots,52$, we will write such a permutation as a "tuple", e.g. $(9,7,44,1,51,4,6,2,9,10,11,\dots,232)$. We distribute the cards to the five players in the order of the taken permuation, so the first player gets the first ten cards, in this example the cards $(9,7,44,1,51,4,6,2,9,10)$, the next player the next ten cards, and so on. There is an action of the cyclic group with five elements induced by cyclicly permuting the players (together with their cards), so applying this symmetry we need five times the probability that the first player gets the four aces. We may and do assume that the aces are the cards $1,2,3,4$. We start finding them a place in a permutation.
For the good cases $1$ has $10$ good places (in the hand of the first player), then depending on the place of $1$ there are each $9$ places for the $2$, then each $8$ places for the $3$, and finally $7$ places for the $4$. The remained cards can be distributed at the remained $48$ places, so there are $48!$ chances for this. So we have a total of $10\cdot 9\cdot 8\cdot 7\cdot 48!$ good permutations, when the first player gets the four aces.
The total number of cases is the number of all permutations, $52!=52\cdot 51\cdot 50\cdot 49\cdot 48!$, so in contrast to the above counting of the good cases the four aces have each $52$, then $51$, then $50$, then respectively $49$ independent possibilities.
Putting all together, the wanted probability is: $$ 5\times \frac {10\cdot 9\cdot 8\cdot 7\cdot 48!} {52\cdot 51\cdot 50\cdot 49\cdot 48!} = 5\times \frac {10\cdot 9\cdot 8\cdot 7} {52\cdot 51\cdot 50\cdot 49} = \frac {9\cdot 8\cdot 7} {52\cdot 51\cdot 49} = \frac {9\cdot 8} {52\cdot 51\cdot 7 } = \frac6{1547} \ . $$ This is approximatively $0.00387847446670976082740788623\dots $ .