Answer: Tuesday Teaser #34
This puzzle is pretty tough: the best way to think about it is to generalize what happens if a company hires one more person than it already has.
Let’s let D(n) be the expected number of working days for a company with n employees. If the company has 1 employee, it is clear that there is always 1 unique birthday. So D(1) = 364.
Now, inductively, adding an extra employee will have a D(n)/365 chance of reducing the day count by 1, and a [365-D(n)] chance of staying the same.
So: D(n+1) = D(n)*[365-D(n)]/365 + [D(n)+1]*D(n)/365 = D(n) - D(n)/365
Using our initial starting point of D(1) = 364, we can derive that D(n) = 365 * (364/365)^n
So the total expected number of man days worked by n employees in a year is then: E(n) = n * 365 * (364/365)^n
And the maximum E(n) is given when the derivative is equal to zero. Here, that occurs when n = 365. It’s a nice little coincidence that the number of days in the year coincides with the maximal number of employees.
For a little bit more detail around this solution, there is a comprehensive spreadsheet available on the web that you can have a play around with.
