Re: Another Boredom Filler
The thought I've been having lately is that if you were to translate the question into counting from 1-52 and matching the exact card instead of rank only, the problem simply becomes a Fixed Point calculation based on the permutations...and should tend to 1/e for no fixed points, I believe. Although, even that isn't a closed form solution. From what I've been able to read about this, as the sample size goes up, the probability of no matches tends toward 1/e^4.
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