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03-18-2002, 12:37 PM
| #1 |
| Registered User Join Date: Feb 2002
Posts: 591
| Party question How many people does it need to be at a party for the chance for 2 people to have the same birthday to be 50%.? The correct answear acording to the person asking the question is 22. Somehow I have a mental block on this. Anyone out there know why? I might not ask the question exactly as intended. |
| Barjor is offline |
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03-18-2002, 12:40 PM
| #2 |
| Registered User Join Date: Jan 2002 Location: Cardiff
Posts: 2,219
| Well I come from plymouth and the only math our parties involve is: Code: Satisfaction == Volume of spliff * Percentage Weed || Girls |
| Brian is offline |
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03-18-2002, 12:52 PM
| #3 | |
| 5|-|1+|-|34|)  Join Date: Aug 2001
Posts: 4,429
| Re: Party question Quote:
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| ober is offline |
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03-18-2002, 12:55 PM
| #4 |
| Registered User Join Date: Feb 2002
Posts: 591
| Do you know about any houses for sale and well paid jobs in plymouth? Lets see if I can make it clearer. If I go to a party. How many people does it need to be on that party for the chance to be 50% that two people in that party have the same birthday. |
| Barjor is offline |
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03-18-2002, 12:59 PM
| #5 |
| 5|-|1+|-|34|) Join Date: Aug 2001
Posts: 4,429
| hmmm... and the answer was 22?? sounds like some fishy statistics stuff goin on... I didn't spend a whole lot of time paying attention in that class... |
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03-18-2002, 01:01 PM
| #6 |
| Registered User Join Date: Feb 2002
Posts: 591
| Well sqrt(365) is roughly 20 so maybee when the question been told acouple of times the answear changed alittle bit..Not sure |
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03-18-2002, 02:51 PM
| #7 |
| back?  Join Date: Oct 2001
Posts: 597
| I know, I know!!! Hey, I think I actually know this one! Who-hoo! ... Get to the answer, willya? Well, here it is: There are really only 2 possibilities here: either someone in the room shares a birthday with somebody else, or everyone has a different birthday. Now, the probability that you will get people that share a birthday is (100% - X) where X is the probablity that everyone has a DIFFERENT birthday. ...for some calculations... in order to find the probability for a same birthday, take the probability of all different ones, and subtract that from 100%. To find the chance of all different birthdays, it goes lke this: 1 person > 365 possible days a year 2 people > 365 for first, 364 for 2nd 3 people > 365 for first, 364 for 2nd, 363 for 3rd ... so say you got 10 people... what's the chance someone shares a birthday? Answer: 1 - ((365*364*363*362*361*360*35*358*357*356)/(365^10)) = 11.69% If any of this makes no sense, i could try to explain this differently. All this here is is probability... yeah. |
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03-18-2002, 03:39 PM
| #8 |
| Registered User Join Date: Feb 2002
Posts: 591
| Facinating..I should have known that but nevertheless you helped me from getting more grey hairs. Thanks |
| Barjor is offline |
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