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 04:08:59 GMT
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Viewed:
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1408 times
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Eric Harshbarger <eric@ericharshbarger.org> wrote in message
news:3B354B3E.D79739E5@ericharshbarger.org...
>
> 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 realize that the shared date can be ANY day. We are NOT
> arbitrarily picking a date beforehan and asking if two people share THAT
> birthdate (which would be a much lower chance)... rather we are simply
> asking if two people share a date... whatever that day of the year is.
> Thus, with two individuals, we ask what the first person's birthdate is.
> The chance that the second person's date matches the first person is
> simply 1 in 365.
>
> Another way to think of it is that the probability is [1 - (the chance
> that the two people DON'T share the same birthday)] which would be:
>
> (1 - 364/365) == 1/365
ok I get it until this point.
> Now, if there are 3 people involved, the chance that two individiduals
> share a birthdate will be [1 - CHANCE THAT NO ONE SHARES A BIRTHDATE]
>
> The CHANCE THAT NONE OF THE THREE SHARE A BIRTHDATE is
>
> [364/365] ^ (3) ... meaning raised to the third power.
>
> This is because there are three ways to compare different pairs of
> individuals (AB, BC, AC) and for each pair the chance that they don't
> share a birthdate is, again, 364/365.
>
> Note that this exponent of 3 is, in probability terms, a '3 choose 2'
> calculation, or (3!)/(2! * 1!).
>
> So the probability we sought is 1 minus that value or:
>
> 1 - [364/365] ^ (3) ~= 0.8% (still a small chance)
>
> Generalizing the above formula to N individuals, one gets this:
>
> Probability of 2 people out of a group of N people sharing a common
> birthdate equals:
>
> 1 - [364/365] ^ (N!/(2!)(N-2)!)
>
> Note that when N == 23 (23 people in the group), the probability comes
> out to just above 50%, so if you are ever amongst a crowd of 23 people,
> you have better odds than not of two people in that crowd sharing a
> birthday.
Ouch! your givin' me one heck of a headache.
> And, obviously, if your crowd has 365 people of more, you MUST have a
> common birthday somewhere.
Now that makes sence.
Gary
Where did I leave that algebra book.......
*leaving to find a bottle of aspirin (and posibly a can of pepsi)*
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Message has 1 Reply:
 | | 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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Message is in Reply To:
| | Birthday Mathematics
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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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