Four bridge hands with no two people having 8 or more cards of the same suit between them. I am trying to solve this problem:

A 52-card deck is dealt out to 4 people (13 to each). What is the probability that no two people have 8 or more cards of the same suit between them?

It seems to me that there are only 4! ways for this to be possible (when each player has four of a different suit and three of each of the others). This seems to be too small but I'm unsure of how to count all the possible satisfying combinations there are.  
I also can't quite figure out the denominator. Is it ${52 \choose 13}{39 \choose 13}{26 \choose 13}$? Or $\frac{52!}{13!^4}$?
 A: First, your two choices for the denominator are the same since
$${52\choose13}{39\choose13}{26\choose13}=\frac{52!}{39!13!}\frac{39!}{26!13!}\frac{26!}{13!13!}=\frac{52!}{(13!)^4}$$
Second, you are right that everyone needs to have a 4333 hand, but there are many more than $4!$ ways to do it. We will make 4 hands with a specific 4 cards suit, then distribute the hands to the four players.
North hand with 4 spades
$${13\choose4}{13\choose3}{13\choose3}{13\choose3}$$
East hand with 4 hearts, from remaining card.
$${9\choose3}{10\choose4}{10\choose3}{10\choose3}$$
South hand with 4 diamonds, from remaining cards
$${6\choose3}{6\choose3}{7\choose4}{7\choose3}$$
Finally, West hand with 4 clubs, from remaining cards
$${3\choose3}{3\choose3}{3\choose3}{4\choose4}$$
The number of ways that no team has a 8 cards suit between them is
$$4!{13\choose4}{13\choose3}{13\choose3}{13\choose3}{9\choose3}{10\choose4}{10\choose3}{10\choose3}{6\choose3}{6\choose3}{7\choose4}{7\choose3}{3\choose3}{3\choose3}{3\choose3}{4\choose4}$$
which could be simplify to
$$4!\left(\frac{13!}{4!(3!)^3}\right)^4$$
We could have found this expression directly. The interior of the parenthesis is the ways to distribute a suit evenly (to the fourth power for the four suits) and the $4!$ in front to decide who received each 4 cards suits.
The probability is
$$4!\left(\frac{13!}{4!(3!)^3}\right)^4\times\frac{(13!)^4}{52!}\approx0.000\ 931\ 419\ldots$$
A: For each suit, you have to choose the player having four cards, so that's $4!$ ways.
3 of the players choose 4 cards in their "long" suit, and 3 in each other, so there is
$$4!\times\left({13\choose4}{13\choose3}^3\right)^3$$
ways to deal the cards.
The denominator is ${52\choose13}\times{39\choose13}\times{26\choose13}$ (three players choose their 13 cards, the last one has no choice).
