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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