Subject:
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Re: Birthday Mathematics
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Newsgroups:
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lugnet.people
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Date:
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Mon, 25 Jun 2001 12:52:53 GMT
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Viewed:
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1435 times
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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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Message has 2 Replies:
 | | Re: Birthday Mathematics
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| Correct, Frank. I slipped up by 1 there at the end. 366 to guarantee a non-leap-year overlap. sorry for the slip, eric (...) (23 years ago, 25-Jun-01, to lugnet.people)
| | | Re: Birthday Mathematics (generalized)
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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, (...) (23 years ago, 25-Jun-01, to lugnet.people)
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Message is in Reply To:
| | Re: Birthday Mathematics
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| Eric Harshbarger <eric@ericharshbarger.org> wrote in message news:3B354B3E.D79739...ger.org... (...) ok I get it until this point. (...) Ouch! your givin' me one heck of a headache. (...) Now that makes sence. Gary Where did I leave that algebra (...) (23 years ago, 25-Jun-01, to lugnet.people)
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