A question for those with more mathematical talent than I

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Nyarlathotep
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A question for those with more mathematical talent than I

Post by Nyarlathotep » Wed Aug 04, 2004 5:28 pm

Which is most people, unfortunately.

My workplace has 13 people. Out of those 13 we have two sets of people who share the same birthday. What are the odds of that? I want to say it's 1 in 13,225 (365*365) but I seem to recall reading soemwhere that the odds of two people being born the same day are NOT 1 in 365. Anyone have any ideas?

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Sundog
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Post by Sundog » Wed Aug 04, 2004 5:40 pm

They say it better than I could...

http://www.benbest.com/science/theodds.html

Additionally, some coincidences are more probable than we might expect due to our lack of appreciation of actual probabilities. The probability of two people not having the same birthday is 364/365 (multilpying the 364 days remaining for the second person times the 1/365 probability of the birthday of the first person). 1 - 364/365 => 0.275% chance of having the same birthday. But for 23 people there is a greater than 50% chance that at least two of them will have the same birthday because

P(same birthday) ;= 1 - (364/365)(363/365)...(343/365) > 0.5

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Nyarlathotep
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Post by Nyarlathotep » Wed Aug 04, 2004 6:47 pm

Sundog wrote:They say it better than I could...

http://www.benbest.com/science/theodds.html

Additionally, some coincidences are more probable than we might expect due to our lack of appreciation of actual probabilities. The probability of two people not having the same birthday is 364/365 (multilpying the 364 days remaining for the second person times the 1/365 probability of the birthday of the first person). 1 - 364/365 => 0.275% chance of having the same birthday. But for 23 people there is a greater than 50% chance that at least two of them will have the same birthday because

P(same birthday) ;= 1 - (364/365)(363/365)...(343/365) > 0.5
So if I do the math right, the odds of two sets of identical birthdays in a group of 13 people is about 1 in 20 (.049 is the nuber I get). Rare but not spectacularly so.

Thanks, Sundog.