Subject:
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Re: Birthday Mathematics (generalized)
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Newsgroups:
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lugnet.people
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Date:
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Mon, 25 Jun 2001 13:25:47 GMT
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Reply-To:
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ERIC@avoidspamERICHARSHBARGER.ORG
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Viewed:
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1456 times
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Also, correct me if I'm wrong here... my days of formal mathematics are
sometime in the past and I am just scribbling notes at my side as I type
this:
To Generalize The Problem, instead of finding 2 people with the same
birthdates in a crowd of N, if one wishes to find the probability that M
people in a crowd of N share a common birthdate the formula becomes (a
bit more complicated):
1 - [(365 ^ [M-1] - 1)/(365 ^ [M-1])] ^ [(N!)/(M!)(N!-M!)]
The final exponent indicates the 'N choose M' part of the formula while
the numbers in the first set of brackets reflects the probability that
given M people, they will NOT all share a common birthdate (there's only
1 chance in [365 ^ (M-1)] chance that they WILL share one...)
eric
Frank Filz wrote:
>
> blessing wrote:
> > > And, obviously, if your crowd has 365 people of more, you MUST have a
> > > common birthday somewhere.
> >
> > Now that makes sence.
>
> Except there's an "off by one error" (actually, it's an off by two, the
> year has 365 or 366 days, so therefore to guarantee an overlap you must
> have 367 people, unless you count Feb 29 the same as Mar 1, in which
> case you only need 366 people).
>
> --
> Frank Filz
>
> -----------------------------
> Work: mailto:ffilz@us.ibm.com (business only please)
> Home: mailto:ffilz@mindspring.com
--
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Eric Harshbarger / eric@ericharshbarger.org / www.ericharshbarger.org
"Oh please, if people are going to start telling the truth around
here... I'm going to bed." - Jackie-O (Parker Posey, THE HOUSE OF YES)
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Message is in Reply To:
 | | Re: Birthday Mathematics
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| (...) Except there's an "off by one error" (actually, it's an off by two, the year has 365 or 366 days, so therefore to guarantee an overlap you must have 367 people, unless you count Feb 29 the same as Mar 1, in which case you only need 366 (...) (23 years ago, 25-Jun-01, to lugnet.people)
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