# Facebook Friends' Birthdays: A Probability Question

• Jan 10th 2010, 06:56 AM
probabilitydunce
Facebook Friends' Birthdays: A Probability Question
Hey guys,

I have 2,000 Facebook friends and noticed that not one of them has a birthday today (January 10th). I quickly calculated the probability of this as (364/365) ^ 2000 = 0.4% (assuming, for the purposes of this calculation, that each birthday is equally likely).

Here's my question, though: how would I calculate the probability of nobody (out of my 2,000 friends) having a birthday on at least one of the 365 calendar days? In other words, what are the odds that I don't have a birthday on each of the 365 days in the year?

I realize it would be 1-P(birthdays on all 365 days); that said, I have no idea how to calculate P(birthdays on all 365 days) either. :p

Please assume, for the purposes, of this calculation that there are no leap days and that each birthday is equally likely.
• Jan 10th 2010, 07:59 AM
Laurent
Quote:

Originally Posted by probabilitydunce
Here's my question, though: how would I calculate the probability of nobody (out of my 2,000 friends) having a birthday on at least one of the 365 calendar days? In other words, what are the odds that I don't have a birthday on each of the 365 days in the year?

Hi,
I think there is no simple formula for that one. Let $N=365$ (number of days) and $n=2000$ (number of friends). If $A_i$ denotes the event "day number $i$ is nobody's birthday", then we have:

$\{\text{not everyday is a birthday}\}=A_1\cup\cdots\cup A_N$.

Therefore, using inclusion-exclusion formula and the fact that $P(A_1\cap\cdots\cap A_k)=\left(\frac{N-k}{N}\right)^n$, we get:

$P(\text{not everyday is a birthday})=\sum_{k=1}^N (-1)^{k+1}{N\choose k}\left(\frac{N-k}{N}\right)^n$.

In your case, I think the probability is approximately 0.783881.
• Jan 10th 2010, 01:39 PM
matheagle
better guestion, who would have 2000 'friends' at facebook?