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Brian Whatcott writes:
For the record, I wonder how David accounts for the probability which his
formula gives for d=365, n=365?
(Where each person is carefully chosen to have a different birthdate...)
...or n=366 ?
... or n=367?
What's to account for? For d=365, n=365 the formula gives a value
*very* close to 1 (I get p = 1 - 1.45 x 10^-157) as expected. For
n=366, 367, ... it gives a value of exactly 1, again as expected.
John