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What is the probability that a shuffled standard deck of 52 cards has no two cards of the same rank together ?

I am unable to get a handle on this problem, and wonder whether there is an analytical solution ?

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  • If you haven't done so already, please consider voting up Jair Taylor's answer. It's a favorite of mine, and I think it deserves recognition. –  Feb 22 '13 at 16:14

1 Answers1

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You can use Jair Taylor's answer here. Put $k=4$ and $n=13$ to find the number of arrangements where no two cards of the same rank are together. The number of such arrangements turns out to be $$4184920420968817245135211427730337964623328025600.$$ Dividing by $52!/(4!)^{13}$ gives the probability: $.045476$.

Added: If you are willing to use an approximation, the number $Z$ of cards of the same rank that are together is approximately a Poisson random variable with mean $3$. Therefore, $$P(Z=0)\approx e^{-3}= .049787.$$

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