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December 15th, 2007, 09:56 PM  #1 
Newbie Joined: Dec 2007 Posts: 1 Thanks: 0  Birthday problem [statistics]
Well, I was just googling some math puzzles (how nerdy,) but anyways... I came across this one: http://mathforum.org/dr.math/faq/faq.birthdayprob.html Now, the question poses something along the lines of "how many students would need to be in a class in order for the probability of at least 2 to have the same birthdays be >= 50%." Now, I view this as a binomial distribution. let X = the event that 2 or more students have the same birthday. P(X >= 2) = ∑[(n over k)(p^k*(1p)^(nk))] where k = {2,3,4,...n) and n = class size. Formula can be found here: http://en.wikipedia.org/wiki/Binomial_distribution. First find the probability 2 or more students have the same birthday of a single day, say January 1st. Suppose p = the probability of a student having a birthday on a single day of the year: 1/365 ~ .0027 n = class size = unknown variable k = {2,3,4...n) In order to consider all days of the year, we multiply the P(X >= 2  day is January 1st) * 365. Now if we just brute force this, plugging in 23 for class size (as suggested) gives you a probability of .6671, much higher than 50%. The get the closest value, 20 should be defined as n, giving you a probability of .5037. Is there anything wrong with my thinking or calculations? 
December 16th, 2007, 04:11 AM  #2  
Global Moderator Joined: Dec 2006 Posts: 20,096 Thanks: 1905  Quote:
 
December 18th, 2007, 01:50 PM  #3 
Senior Member Joined: Oct 2007 From: France Posts: 121 Thanks: 1 
let A= the event that 2 or more students have the same birthday Let B be the complementary event (all the birthdays are different). We want P(B)<=0,5. We have P(B)=A(365,n)/365^n, where A(365,n) is the number of arrangements of 365 days taken n at a time, namely: A(365,n)=365*364*..*(365n+1) It is easy to verify that P(B)<=0,5 for n>=23. 

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