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A reddit thread recently asked this question: What is the probability a shuffled deck of cards will have no number-wise repetitions? That is, what is the probability that your shuffled deck will not have a two next to a two, or an ace next to an ace, but suit-wise repetitions are allowed.

One user proposed that the answer is $(48/51)^{51} \approx 4.542\%$ (corresponding to a 48/51 probability of not matching for 51 pairs of cards), and several users ran simulations which agreed with this value, but another user pointed out that if you take the 8-card case, this formula is only an approximation, and that it underestimates the true probability found by considering every combination by hand.

How can one find an exact probability for this problem, and how far off is the estimate above?

user2530102
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  • Messy. The only easy thing is the expected number of repetitions. – André Nicolas Jun 04 '15 at 18:28
  • It does seem deceptively difficult, but given all the research that has gone into card shuffles, I thought there may be someone who has explored this issue. Calculating it out by hand, it seems like there should be some structure to it, but like you said it gets messy fast. – user2530102 Jun 04 '15 at 18:39
  • The exact probability is here: http://math.stackexchange.com/questions/310971/no-two-identical-ranks-together-in-a-standard-deck-of-cards/311007#311007 –  Oct 03 '16 at 12:42

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