Probability that at least $2$ people will not receive any ace. 
I've a deck with 52 french cards ($13$ values for each of $4$ suits) and $4$ players. Randomly dealing out all cards, what's the probability that at least
  $2$ people will not receive any ace?

My try:
$$p=\frac{\frac{4!}{2!}\binom{48}{13,13,12,10}+\binom{4}{2}\binom{48}{13,13,11,11}+\frac{4!}{3!}\binom{48}{13,13,13,9}}{\binom{52}{13,13,13,13}}$$
Where:
$\binom{48}{13,13,13,9}$ is the case $A$ has all $4$ aces,
$\binom{48}{13,13,12,10}$ the case $A$ has $1$ ace $B$ has $3$ aces,
$\binom{48}{13,13,11,11}$ the $A$ has $2$ aces and same for $B$,
$\frac{4!}{2!}$ arrangements of $4$ people to be $A$ and $B$,
$\binom{4}{2}$ combination of $4$ people to be $A$ and $B$,
$\frac{4!}{3!}$ arrangements of $4$ people to be $A$
Am I right? If yes, is there a more elegant solution than mine?
 A: We will use a version of the Principle of Inclusion and Exclusion (PIE).  
What is the probability that some two of the players have no aces?  There are $\binom{4}{2}$ ways to pick the two players, so the total probability is
$$S_2 = \binom{4}{2} \frac{\binom{48}{26}} {\binom{52}{26}}$$
We have over-counted cases where three of the players have no aces, but hang on, we will compensate in a minute.
What is the probability that some three of the players have no aces?  There are $\binom{4}{3}$ ways to pick the three players, so the total probability is
$$S_3 = \binom{4}{3} \frac{\binom{48}{39}} {\binom{52}{39}}$$
The question is, how to compensate for the over-counting in $S_2$?  If there are actually three players who have no aces, the probability of this event has been counted three times in forming $S_2$.  We want to count it only once, so we must compensate by subtracting it twice.  Therefore the probability that at least two players have no aces is
$$S_2 - 2 S_3 \approx \boxed{0.310204}$$
Note: The usual statement of PIE shows how we may compute the probability of at least one of $n$ events.  But a modification of PIE allows us to compute the probability of at least $m$ events, and a simple case of that modified PIE has been applied above.  A full discussion may be found in Feller, An Introduction to Probability Theory and Its Applications, Third Edition, section IV.5(a), "The Realization of at Least $m$ Events".
A: With reference with  my last comment against your question.
Imagine  $4$ labelled rooms, each with $13$ labelled beds, to be occupied by $4$ travellers (aces)
As you have found, either $4$ are in one room, or $3-1 \;or\; 2-2$ in two rooms to satisfy the question's constraints
$Pr = \dfrac{4\binom{13}4 + (4\cdot3)\binom{13}3\binom{13}1 + \binom42\binom{13}2\binom{13}{2}}{\binom{52}4} = \dfrac{76}{245},\; \approx0.310204 $
Note that inclusion-exclusion has not been resorted to.  
A: There are $\binom42$ ways to choose two of the people who have 26
 available "slots" between them.
Consider only the distribution of aces, where the rest of the cards go doesn't matter.
The probability that all the aces fall in one or other of the 26
 slots of the chosen two is
$\binom42\cdot \frac{26}{52}\cdot\frac{25}{51}\cdot\frac{24}
{50}\cdot\frac{23}{49}$
There are $3$ groups of two containing A, say, viz. AB, AC and AD, so cases where A has all the aces have been counted three times, and similarly for the others, so to correct for this, subtract
$2\cdot4 \cdot \frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}$ 
The final answer comes out as  $\frac{76}{245}$
