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The original problem is as follows:

Find the probability $p$ that in a bridge game the players North, East, South, West have $a, b, c, d$ spades, respectively (which implies $a+b+c+d=13$).

And after simple counting, you'll get the answer, which can be rewritten as: $$p=\frac{\binom{13}{a}\,\binom{13}{b}\,\binom{13}{c}\,\binom{13}{d}}{\binom{52}{13}}$$

And it seems like this result can be directly obtained by interpreting the numerator and denominator respectively, but I'm stuck with how to do this. Any ideas?

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  • $\begingroup$ Don't think your answer is correct. Have a look at math.stackexchange.com/q/1416902/321264. $\endgroup$ – StubbornAtom Aug 19 '16 at 11:08
  • $\begingroup$ @StubbornAtom Actually they are the same, and you can rewrite it as above. $\endgroup$ – Daniel Aug 19 '16 at 11:13
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Finally I figured it out:

Suppose that there are 52 positions in a row where you can put 13 spades into different positions there. So you'll have $\binom{52}{13}$ choices in total. And now you want there to be $a$ spades in the first 13 positions, $b$ spades in the second 13 positions, $c$ spades in the third 13 positions, and $d$ spades in the fourth 13 positions, where $a+b+c+d=13$. So in this case you'll have $\binom{13}{a}\,\binom{13}{b}\,\binom{13}{c}\,\binom{13}{d}$ choices in total. Thus the result follows.

It's worth mentioning that in the above interpretation actually all cards are split into only two kinds, one kind is spades, and the other one is non-spades. And you can treat all spades as the same and all non-spades as the same.

To some extent, this seemingly simple interpretation can really provide some insights into this kind of problems.

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  • $\begingroup$ This is a really nice way to think about this problem - very natural and simple, yet not at all obvious (to me, at least). $\endgroup$ – Shagnik Aug 19 '16 at 18:56
  • $\begingroup$ This is quite a bit simpler than the solution linked to in the comments. (I changed the wording slightly, but feel free to change it back if you prefer.) $\endgroup$ – user84413 Aug 19 '16 at 20:53

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