<s>[INST] 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?
 [/INST] 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.  
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