Each player dealt one king, queen and jack

Twelve cards (four each of kings, queens and jacks) are shuffled and dealt out so that each of $4$ players holds $3$ cards. What is the probability that every player ends up with exactly one king, one queen and one jack?

$\tfrac{72}{1925}$. The number of ways to deal the hands is $\binom{12}{3}\binom93\binom63=369600$. For every player to get one of each rank, the four kings, four queens and four jacks are each shared out one per player, in $(4!)^3=13824$ ways.