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I had a birthday problem question that I'm really interested in knowing the answer for:

In a group of 2,000 people, what is the probability of one day during the year that no one has that particular day of birth? What about the probability of having 10 days not having a birthday within them?

For the real life story, I work for a company which we have just discovered that there are 10 particular days which no one has a birthday on. We thought it strange, but we were wondering if this was actually an expected outcome. I didn't know where else to turn. Thanks in advance for your input!

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Also note that many of the solutions we are coming up with are based on assumption of distribution of birthdays which may not be true of reality. It may be true that there is a higher probability of seeing birthdays in spring than winter which means a uniform discrete distribution may not be a good fit – Kamster Jul 14 '14 at 22:53
Though actually a study like this makes it seem that uniform may be reasonable assumption – Kamster Jul 14 '14 at 22:55
I would posit that the uniformity of the distribution of birthdays depends highly on the uniformity of the distribution of ages. If all 2000 people were born in the same year, it wouldn't strike me as terribly odd if there were 25~50 days in the year that had no births. Doctors have lives, too, and since it was invented in the 1960's, labor inducement has been on the rise; Holidays and Weekends would show a dearth of births for people 50 and under (on a rotating cycle by the year of birth, as alluded to). – Patrick M Jul 15 '14 at 5:25
Thanks everyone for your answers! Very interesting! – EternalSeeker Jul 15 '14 at 17:33

If there are 2000 birthdays randomly (uniformly for simplification) distributed over 365 days (we'll ignore leap year), this gives an average of $\frac{2000}{365}\approx 5.48$ birthdays per day. Then the number of birthdays on any particular day would be well approximated by a Poisson random variable with $\lambda = 5.48$. Thus the probability of $k$ birthdays on a given day would be $e^{-5.48}\cdot\frac{5.48^k}{k!}$. So the probability of a particular day having no birthdays is about $e^{-5.48}\approx .00417$. Since there are 365 days in the year, the number of birthday-free days would be expected to be about $365(.00417)\approx 1.5$. So $10$ seems rather high for the number of birthday free days.

P.S. I wrote some Python code to simulate the situation. I ran it 5000 times. The most number of missed birthdays I saw was 7. The average number of missed birthdays was 1.508.

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I like how your statistical analysis arrives at the same 0.0041~ish figure. It's ... Reassuring :-) +1 – Patrick M Jul 15 '14 at 5:20

If everyone's birthdays are independent and uniformly distributed across the year, each employee has a chance of $\frac{364}{365}$ of avoiding that day. (I'm ignoring leap years for simplicity).

The chance that all 2000 people avoid the chosen day is $(\frac{364}{365})^{2000}\approx 0.00414 $.

How unlikely is it that there are ten days that everyone avoids? If ten particular days are fixed, the answer is obviously $(\frac{355}{365})^{2000}\approx 10^{-24}$.

However, this calculation is not straightforward to generalize to "any set of ten days", because we then have to leave the assumption of independence behind -- "nobody has a birthday on January 1 through 10" is not independent of "nobody has a birthday on January 11 throuh 21".

If we cheat and ignore that problem, we can multiply our $10^{-24}$ with the number of possible 10-day subsets $\binom{365}{10}\approx 10^{19}$ to get an upper bound of about 1/132,000 for the probability of having at least 10 non-birthdays in a year.

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Is that the relevant computation, though? I don't think one wants the probability of missing a specific date, but I think what matters is the probability of any day being missed. – Semiclassical Jul 14 '14 at 22:41
@Semiclassical: It's how I understand the first question being asked. But we agree that it is probably meant just as a stepping stone to "the probability that there is any set of 10 days such that ...". – Henning Makholm Jul 14 '14 at 22:44
My main reason for concern is that, as @YannHamdau 's answer notes, the probability of 10 specific days being missed is miniscule. So I doubt thats the correct measure for how unlikely the real-life scenario was. – Semiclassical Jul 14 '14 at 22:50
@Semiclassical: You're right that the probability of 10 specific days being missed is not in the question. That's an intermediate result I chose to show in order to explain the (not rigorous) way I ended up with 1/132,000 at the end. – Henning Makholm Jul 14 '14 at 22:53
@HenningMakholm Your mock probability is an upper bound obtained by a "union bound" or "Boole's inequality". – Byron Schmuland Jul 15 '14 at 2:05

As has been said, any specific day is without birthdays with probability $p=(1-\frac1{365})^n\approx 0.0041$. If we repeat an experiment with such a success probability $365$ times, the expected number of successes is $np\approx1.51$ and the standard deviation is $\sqrt{np(1-p)}\approx1.23$. Then ten birthday-free days are almost $7\sigma$ above the mean, a highly unlikely event.

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But the 365 experiments are not independent. (Most dramatically, it's impossible for all of them to succeed). Intuitively approximating with a binomial distribution is probably not too far off, but how do we quantify that? – Henning Makholm Jul 15 '14 at 12:56
@HenningMakholm Well, they are negatively correlated so the true standard deviation will be even smaller. – Byron Schmuland Jul 15 '14 at 14:24

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