R. Wainright asks:
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Recently, my daughter asked the following question: After shuffling an
ordinary deck of cards, what is the probability of having none of the cards in
their proper place with respect to rank only? That is, ace through king, ace
through king, etc. (four times). In this problem the suit of each card does not
matter.
Any help or direction to a reference would be appreciated. Thank you.
>>
For a deck of n cards, this probability is well-known to approach
1/e (where e = 2.7218281828... is the base of natural logarithms)
as n -> oo. So this is already a pretty good approximation for n = 52.
1/e = .3628794411714423 to 16 decimal places.
The answer turns out to be exactly 1 - 1/2! + 1/3! - 1/4! + ... - 1/52!
where k! denotes k(k-1)(k-1)...3.2.1. According to my calculation,
(which may be wrong in the last decimal place) this comes out to
.3678794411714424 to 16 decimal places (agreeing with 1/e at least
through the 15th decimal place) though with more work
this probability can be expressed precisely as a fraction.
--Dan Asimov
