# 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?

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Feb 24 '13 at 21:02

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.

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So, I got 5.3644 e 28. Does that mean 1 in 5 followed by 28 zeros? – Jon Feb 24 '13 at 21:06

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 $$\frac{4!(13!)^4}{52!} \approx 4.4739 \times 10^{-28}.$$

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As a rough idea of magnitude, $10^{28}$ is larger than the number of stars in the visible universe. These odds are "truly" astronomical – Tim Seguine Feb 24 '13 at 20:52
How much further does the probably "worsen" to have the cards dealt IN ORDER of ace to 2 to each individual? – Jon Feb 24 '13 at 21:10

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 $$\frac{4}{\binom{52}{13}}\cdot\frac{3}{\binom{39}{13}}\cdot\frac{2}{\binom{26}{13}}\cdot\frac{1}{\binom{13}{13}}.$$

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