Probability in cards that $4$ people each get queen and king?

Question

A $52$ card deck is shuffled and then dealt out to $4$ people $($each person gets $13$ cards$)$. What is the chance that each person gets an a Queen & a King?

My attempt: I know there's $\binom{52}{13}$ different hands that can be shuffled out. The chances of getting a Queen is $4/52$ and the chances of getting a King is $4/52$. This makes the chance of having both $16/(52*52)$ = $16/2704$. Since there are $4$ people, does that mean the overall chance of each person having a Queen and King in their $13$ card deck $(\frac{16}{2704})^4$?

Answer 1

There are $\binom{52}{13}\binom{52-13}{13}\binom{52-13-13}{13}$ ways of dealing 4 hands of 13 cards each. Given that each hand has to have one king and queen, there are $\binom{52-8}{13-2}\binom{52-13-2-8}{13-2}\binom{52-13-2-13-2-8}{13-2}$ for the remaining cards, and $(4\cdot 3\cdot 2)\cdot (4\cdot 3\cdot 2)$ ways for the king and queen to go, So probability is $\frac{24\cdot 24\cdot \binom{52-8}{13-2}\binom{52-13-2-8}{13-2}\binom{52-13-2-13-2-8}{13-2}}{\binom{52}{13}\binom{52-13}{13}\binom{52-13-13}{13}}$

Answer 2

Despite the fact that (for good reason) this is not how the dealing is done, let us assume A gets the first $13$ cards, and B gets the next $13$, and so on.

There are $\binom{52}{8}$ equally likely ways to choose the locations of the places where the $8$ cards we are interested in will go, and for each of these ways there are $8!$ ways to permute the cards.

There are $13^4$ ways to choose a Queen location in each group of $13$. For each of these, there are $12^4$ ways to choose a King location in each group. The chosen Queen locations can be filled with Queens in $4!$ ways. For each of these, the chosen King locations can be filled with Kings in $4!$ ways.

Thus the number of "favourables" is $13^412^44!4!$, and the required probability is $$\frac{13^412^4 4!4!}{\binom{52}{8}8!}.$$