# Probability question

#### buddingscientist

Suppose a room contains n people. Assuming that days of the year are equally likely to be birthdays for each person, calculate the probability that at least two of the people have a common birthday.

well I have the answer but I'm just curious as to the thought processes you go through to answer it.

what ive done (incorrectly) is:

if n = 2, P = 1/365^2
if n = 3, P = 1/365^2 (person 1, 2) + 1/365^2 (person 1, 3) + 1/365^2 (person 2, 3)

and generalized it to:
( 3 + (n sum i=3) (i)! / (i-1)! ) / 365^2

there's no question that it's wrong but thats the process I take

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#### matt grime

Homework Helper
work out the probability they all have different birthdays and subtract that from 1. Obviously if there are 366 people in the room the probabilit must be 1, mustn't it? (excludeing leap year things). Does n=366 give P=1 in your answer?

#### buddingscientist

probably not (npi), I know my answer is wrong, you recomend using conjugates (1 - P(all different)), I'll work on it some more, thanks for your response

#### OneSquared

Ignoring leap years:

P(bd2 = bd1) = 1/365
P(bd2 != bd1) = 364/365
P(bd3 = bd1) + P(bd3 = bd2) = 2/365, given that bd1 != bd2
P(bd3 != bd1) AND P(bd3 != bd2) = 363/365

Therefore P(bd1 != bd2 != bd3) = 1 - (363/365) * (364/365)

So for any n,
P(two birthdays are the same) =
1 - (((365-(n-1))*(365-(n-2))...(364))/(365^n))

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