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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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3 Answers

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
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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
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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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