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December 15th, 2007, 09:56 PM   #1
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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*(1-p)^(n-k))] 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?
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December 16th, 2007, 04:11 AM   #2
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Originally Posted by XTTX
Is there anything wrong with my thinking or calculations?
Yes, since "23" is correct. Your figure for the associated probability is incorrect.
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December 18th, 2007, 01:50 PM   #3
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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*..*(365-n+1)
It is easy to verify that P(B)<=0,5 for n>=23.
Richard André-Jeannin is offline  
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