I shuffle my standard deck of 52 cards really, really well - a completely random shuffle. This generates some configuration of cards: 9Clubs, 4Spades, KHearts, etc...

What are the odds that this exact same configuration has been generated before? By before - I mean anytime before and anywhere - including all the poker games in all the world and all the magic tricks and all the kids just practicing and all their intermediate shuffles as well.


marked as duplicate by Rahul, user147263, Claude Leibovici, Jonas Meyer, TrueDefault Nov 5 '14 at 6:31

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  • $\begingroup$ Not a super hard or deep question, but I enjoyed thinking about it - made me feel special. $\endgroup$ – Mark McClure Nov 5 '14 at 4:47
  • $\begingroup$ So, I guess the question is mainly, how many shuffles have ever taken place in the world? A Fermi problem, I suppose. $\endgroup$ – Rahul Nov 5 '14 at 4:53
  • $\begingroup$ @Rahul Yes, to some extent. I think, though, that a reasonable upper bound (or even an unreasonable one) suffices to answer the question with near 100% certainty. How unreasonable you can be with your upper bound and how near to 100% is the fun part. $\endgroup$ – Mark McClure Nov 5 '14 at 4:56
  • $\begingroup$ @Rahul - Darn! I did search for such a thing for some time. I guess the closure police will be here soon. :) $\endgroup$ – Mark McClure Nov 5 '14 at 5:44

Lets just say 10 million poker games are played per day, and lets go back 20 years (I'm guessing before computers the number was neglegable). That's $2\cdot 10^9$ games. The probability of a single predetermined shuffle is $1/52!\approx 1.2\cdot 10^{-63}$. Thus the expected number of times your configuration has been played over the course of 20 years is around $4\cdot10^{-54}$.

These numbers (their inverses, rather) are on the order of magnitude of the number of atoms in the universe. As a thought experiment, the chances of you getting a predetermined shuffle on your first try is equivalent to finding the correct single atom in the entire universe. Even if humanity lives for another 10 billion years, you will never see a specific shuffle occur.

On the other hand, $(52*51*50*49*48*47)$ is about 14 billion, which corresponds to asking what the probability is of matching the first 6 cards of your predetermined shuffle, which is much more reasonable to occur.

  • $\begingroup$ Your last statement should be $\frac{1}{\,_{52}P_{12}}$ = $\frac{1}{\binom{52}{12} 12!}$ because you still need to choose the $12$ cards that you're organizing $12!$ different ways. $\endgroup$ – Axoren Nov 5 '14 at 5:28
  • $\begingroup$ Something like this, yeah. I might just accept it tomorrow. I've got a much more absurd upper bound. $\endgroup$ – Mark McClure Nov 5 '14 at 5:33
  • $\begingroup$ @Axoren: whoops! thanks! $\endgroup$ – Alex R. Nov 5 '14 at 5:39

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