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Re: Marilyn, again



(This little note seemed to have a problem getting
to the list, when I first sent it...)

At 22:58 8/6/97 -0800, John Mallinckrodt wrote:

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

Unfortunately, the probability of 366 people having the same
birthday is not exactly one.

That's what to account for....

Regards
Brian