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December 5th, 2016, 11:28 AM   #1
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From: kljlçjkçlkjç

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Probability to be born on a specific day

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Quote:
As an example, consider the scenario in which a teacher with a class of 30 students asks for everybody's birthday (for simplicity, ignore leap years), to determine whether any two students have the same birthday (corresponding to a hash collision as described further). Intuitively, this chance may seem small. If the teacher picked a specific day (say September 16), then the chance that at least one student was born on that specific day is 1 - (364/365)^{30}, about 7.9%. However, the probability that at least one student has the same birthday as any other student is around 70% for n = 30, from the formula 1-365!/((365-n)! * 365^n).
https://en.wikipedia.org/wiki/Birthday_attack

I don't understand the two formulas (for getting 7.9% and for 70%).
Could someone explain me, please?
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December 5th, 2016, 02:12 PM   #2
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Prob. not a specified date (Sep. 16) for one person is 364/365. Birthdays are assumed independent, so prob. that n people's birthday is not on specified date is $\displaystyle (\frac{364}{365})^n$, based on general rule that the probability of n independent events all happening is the product of the individual probabilities.

For the second question: The probability that two people have different birthdays is 364/365. For 3 people it is (364/365)(363/365). etc.
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December 6th, 2016, 02:47 AM   #3
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ty very much mathman.
the wikipedia link has a reference explaining the formula that I had not seen:
http://mathforum.org/dr.math/faq/faq.birthdayprob.html
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