 | | Birthday Mathematics —Eric Harshbarger
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| | | So here's an explanation of the probability of two people in a crowd sharing a birthdate (independent of the birth year -- and disregarding leap year birthdays): The chance of two people sharing a birthday is 1/365. This should be clear if you (...) (23 years ago, 24-Jun-01, to lugnet.people)
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| | | | | Re: Birthday Mathematics —Gary E. Blessing
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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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| | | | | | Re: Birthday Mathematics —Frank Filz
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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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| | | | | | Re: Birthday Mathematics —Eric Harshbarger
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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)
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| | | | | | Re: Birthday Mathematics (generalized) —Eric Harshbarger
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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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