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?

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  $\frac{76}{245}$
