2..twins dummy
2007-11-23 23:45:49
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answer #1
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answered by Anonymous
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2
2007-11-24 06:14:45
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answer #2
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answered by Layne M 3
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Just 2 for it to be possible.
I think you might mean how many people must share a room for it to be more likely than not that 2 of them share a birthday.
The answer to this is surprising low - 21 or 22 I think.
2007-11-24 10:25:51
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answer #3
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answered by a Real Truthseeker 7
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I'm not sure what all the numbers are for, but the answer is two. If two people are in a room, it is possible that those two people have the same birthday.
2007-11-23 23:47:54
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answer #4
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answered by marbledog 6
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To guarantee a matching birthday you would have to have 367 people sharing the room. 366 days in leap year so possible for them to all have different birthday, 367 means one of them will definitely share a birthday.
2007-11-23 23:48:40
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answer #5
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answered by john m 6
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Well considering it to be a normal year, and not a leap year, then it could theoretically means that there would have to be 366 people.
If every one of them had a different day then the 366th person must match one of the other people.
2007-11-23 23:47:04
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answer #6
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answered by ? 7
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When I was at school three of us in the class had the same birthday.
2007-11-24 03:00:22
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answer #7
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answered by ffsake 4
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Nobody can share a Birthday so it does'nt matter how many there are in the room. They can only share the anniversary of a birthday
How many of you twigged?
2007-11-23 23:51:05
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answer #8
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answered by Anonymous
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This is a well known problem.
It's less than you think (23)
Have a look at this web page.
http://www.mste.uiuc.edu/java/java/birthday/explanation.html
2007-11-24 02:14:05
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answer #9
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answered by efes_haze 5
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2 you said possible not probable
2007-11-23 23:57:08
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answer #10
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answered by Paddy 4
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2 - for example, twins.
1 person cannot share with anyone else.
2007-11-23 23:45:23
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answer #11
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answered by Anonymous
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