Probability of cards when dealing an entire deck What the probability of dealing an entire 13-card suit to EACH of four players when dealing an entire 52-card deck at random?
 A: Deal the cards to players East, West, North, and South. There are 
$${52!\over(13!)^4}$$
ways to do this.   Suppose you have four packs, each of just one suit.  There are $4!$ possibilities here. Now divide.
A: There are
$$
\frac{52!}{(13!)^4}=53644737765488792839237440000
$$
equally likely ways to form four $13$-card hands from a $52$-card deck.  Only $4!$ of these give everybody a super-duper-flush.  So the probability is
$$
\frac{4!(13!)^4}{52!} \approx 4.4739 \times 10^{-28}.
$$
A: Imagine that the deck is thoroughly shuffled, and the cards are dealt unconventionally, first $13$ to West, then $13$ to North, and so on. There are $\binom{52}{13}$ equally likely hands for West, of which $4$ are $1$-suiters.
For every hand that West gets, there are $\binom{39}{13}$ equally likely hands for North. Given that West got a $1$-suiter, $3$ of these hands are $1$-suiters. Now continue on to East, and, to make things look nice, to West, though this doesn't matter, since if the others get $1$-suiters, then so does she.  Our probability is 
$$\frac{4}{\binom{52}{13}}\cdot\frac{3}{\binom{39}{13}}\cdot\frac{2}{\binom{26}{13}}\cdot\frac{1}{\binom{13}{13}}.$$ 
