Tim_Lou Posted December 8, 2004 Report Posted December 8, 2004 somethings have been bothering me.im thinking about the probability of at least 2 people sharing the same birthday in a sample. assume that equal amount of people are born each day, and the sample is picked totally random. well, excluding feb 29th, i figured out that it is = to 1- probability of not sharing birthday:which i come up with1-(nPr (365,n) / 365^n) but, as feb 29th is included, things get tricky, since the probably of a person having feb 29th as birthday is less than normal.. something like 1/1461 i guess.this would make the birthday probability a bit less.. ......my brain is all messed up, how am i suppose to account this in and get a final forumla?
Tim_Lou Posted December 9, 2004 Author Report Posted December 9, 2004 what do you mean? in 1-(nPr (365,n) / 365^n), im assuming 365 days in a year, but once feb 29th adds up...is it 366? no, since this day has a lower probability to be selected. i did try to solve it...i put 1-(nPr (365,n) / 365^n) times probability that a sample does not contain feb 29th, which=[1-(nPr (365,n) / 365^n)]*(1460^n / 1461^n) + (1 - ?) (1 - 1460^n / 1461^n)but i fail to figure out the probability of birthdays without repeating (including feb 29th), which is the ? thing. the problem is, what should i put for the total arrangement? simply 366^n? but the 366th day has a different probability to be picked...??!???!? ouch, my brain hurts...(correct me if anything i show is wrong....)
GAHD Posted December 9, 2004 Report Posted December 9, 2004 Since feb 29th occurs every 4 years, why not list it as fractional or decimal? Ex 365.25 (365.2422 if you want to have more precision, but that gets more into rotations than 'days') Edit: you could also do the formula twice, weighting the 365 day year as .75 and the 366 day formula as .25 and then add the results together.
Freethinker Posted December 9, 2004 Report Posted December 9, 2004 The probablity of "same birthday" is extremely different depending on how you approach it. Same Birthday based on a specific Birthday is a large number. But determining at what group size the chance of two people having same (random) birthday is small. At roughly 23 people it becomes 50%. http://mathforum.org/dr.math/faq/faq.birthdayprob.html It actually has a name, "Birthday Paradox". Or for a more detailed formula based explanation http://en.wikipedia.org/wiki/Birthday_paradox
Tim_Lou Posted December 11, 2004 Author Report Posted December 11, 2004 hmm, i can use decimals in factorials?but... what does it mean?365.25*(365.25-1)(365.25-2)....*?is it related to the decimal power binomial expansion? whats the last number (the ?) then, 1? or ?? what?...........well, 365.25 comes out as 1.09775179238*10^779......in my calculator, weird. and the 2nd site freethinker given... is way too advanced, er, im only in basic calculus ;)anyway, it doesnt seem to address the feb 29th problem.....
Tim_Lou Posted December 11, 2004 Author Report Posted December 11, 2004 and yes, the probability of at least 2 people having the same birthday is not tiny,excluding feb 29th:people probability2 .0027410 .1169520 .4114430 .7063240 .8912350 .97037and so on...notice that there is great jump between 20 people and 30 people..... hehe, looks like we got some people sharing same birthday in hypography..... lets see, post your birthday here!dont lie man! mine is July 2nd (should this thread be moved to watercooler?)
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