Subject:
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Birthday Mathematics
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Newsgroups:
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lugnet.people
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Date:
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Sun, 24 Jun 2001 02:06:54 GMT
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Reply-To:
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ERIC@ERICHARSHBARGERantispam.ORG
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Viewed:
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1232 times
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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 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
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.
And, obviously, if your crowd has 365 people of more, you MUST have a
common birthday somewhere.
So, a little math-geekiness to interrupt the LEGO-geekiness.
eric harshbarger
ps: oh... my b'day is 22 March.
Shiri Dori wrote:
>
> XFUT lugnet.people
>
> Hi all!
>
> Well, I'm sure the geeks of us all know that in a room with 30 people,
> there's a fairly large chance of having two people with the same birthday.
> (I can explain to the non-geeks if you really want... not now though.)
>
> Since we are well over 1000 people (1), there have to be tons of overlaps on
> birthdays. So here's my question and challange:
>
> Does *everyone* on lugnet have someone else born on their birthday? I
> predict that yes, everyone can find someone with a matching birthday.
> 'Course this is hard to prove, and I bet not everyone will participate in
> this thread, but it's worth a shot. I'll try to keep records of this.
>
> So if you wanna see if I'm right or if you wanna prove me wrong - reply to
> this post with your birth date. Year not required, but that's up to you.
> Since there are both Europeans and Americans here, please type out the name
> of the month, so as to avoid confusion of day/month numbers.
>
> ----
>
> I'll start off with my own:
>
> June 25th (1984)
>
> ----
>
> Out of courtesy, please snip the rest of this message!...
>
> Thanks, and hoping I'm right,
> -Shiri
>
> (1) 1000+ members already.
--
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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 has 1 Reply:
 | | Re: Birthday Mathematics
|
| 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)
|
Message is in Reply To:
| | Find your Birthday Buddy
|
| XFUT lugnet.people Hi all! Well, I'm sure the geeks of us all know that in a room with 30 people, there's a fairly large chance of having two people with the same birthday. (I can explain to the non-geeks if you really want... not now though.) Since (...) (23 years ago, 23-Jun-01, to lugnet.general, lugnet.off-topic.geek, lugnet.people) !!
|
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