Birthday Problem - Company Stats Strange or Average?

Question

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!

Answer 1

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 $.

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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.

Answer 2

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.

Answer 3

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.