Probability that there exist at least one day without any Birthday

Question

Probability that a given day do not have any birthday among N people is: (364/365)^N

However, what will be the probability that there exist at least one day in a year that have no birthday?

This question popped into head while watching atleast someone celebrating their birthday every day. Upon some exploration, understood how the above equation works for one day. But I am clueless how to approach about at least one day in a year.

Answer 1

As 1mdlrjcmed said in a deleted answer, this is more the coupon collector's problem than the birthday problem. The expected number until you first see all days is $365H_{365} \approx 2364.646$ using harmonic numbers.

As Sassatelli Giulio said in a comment, you can in theory calculate your probability of a free day using Stirling numbers of the second kind, and you are looking for $1 - \frac{365!\, S_2(N,365)}{365^N}$.

I would suggest instead using recursion, where $p(N, m)$ is the probability of $m$ days taken with $N$ people. You want $1-p(N,365)$ and you can use $$p(N,m) = \frac{m}{365}p(N-1,m)+\frac{366-m}{365}p(N-1,m-1)$$ starting with $p(0,0)=1$ and $p(N,0)=p(0,m)=0$ when $N$ or $m$ are non-zero.

Clearly you will get $1-p(N,365)=1$ when $N<365$. You will get

• about $1-1.5\times 10^{-157}$ when $N=365$,
• about $0.99$ when $N=1606$
• about $0.9$ when $N=1853$
• about $0.5$ when $N=2287$
• about $0.1$ when $N=2971$
• about $0.01$ when $N=3827$

which if graphed looks like this