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This problem has been solved with an assumption that all years have 365 days, no leap years.
Instead of calculating the probability of at least 2 people in the group of N people having their birthday on the same day, we can calculate the probability that none of them have a birthday on the same day and then subtract that from 1.
Let's start with 2 people. The probability that they don't have a common birthday is (365/365)*(364/365). This gives the probability that they have the birthday on the same day as 1 - (365/365)*(364/365) = 1 - (365*364)/(365)^2
Next take 3 people. The probability that at least 2 share a common birthday is 1-(365*364*363)/(365)^3
Continuing, for N people the probability that at least have the birthday on the same day is:
1 - (365*363*363*...*(365 - n +1))/365^n
=> 1 - 365!/ [(365 - n)!*365^n]
The required probability that two people in a group of N have the birthday on the same day is 1 - 365!/ [(365 - n)!*365^n]
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