Interesting Probability Question - Birthday Problem Variation 
Suppose at a hipster eatery they make craft pickles. At this eatery they have $n$ pickle makers (picklers). Every day each pickler makes $10$ jars of pickles. Whenever any pickler has a birthday, there is a holiday (the eatery is closed, they do not work at all). There are no other holidays provided even weekends. Under these conditions, how many picklers have to be employed, if they want to maximize the expected number of jars of pickles produced in a year (i.e. maximize person-hours)?

Hi, this is a very interesting problem I've come across. It seems to be a variation of the birthday problem. I tried let $y$ be the number of holidays and $f(y)$ be the pdf of the number holidays with respect to $n$. This is the pdf I came up with but it does not seem to be normalized.
$$f(y) = \frac{365! \cdot y^{n-y}}{(365-y)!365^n}$$
Any suggestions?
 A: The pickle factory can produce arbitrarily many pickles by hiring only people born on a particular day. 
To make a problem in the Birthday Problem tradition, we will assume that birthdays of pickle makers are independent uniformly distributed, and that a year has $365$ days. . Suppose we hire $n$ people.
For $i=1$ to $365$, let $Y_i=1$ if no one has a birthday on Day $i$, and let $Y_i=0$ otherwise. Then the pickle production on Day $i$ is $10nY_i$.  The yearly production is $\sum_1^{365}10nY_i$, and the expected yearly production, by the linearity of expectation, is $\sum_1^{365}10nE(Y_i)$.
The probability no one has a birthday on Day $i$ is $(364/365)^n$. So we want to maximize
$$3650n(364/365)^n.$$
This is a standard calculus problem, except that we will have to produce an integer answer. Use the calculus to maximize $te^{-kt}$ where $k=\ln(365/364)$.
Remark: We used the linearity of expectation to sidestep the more complicated problem of finding the distribution of the number of pickle jars produced. That approach leaves us with a complicated expression for the expectation. Indicator random variables such as our $Y_i$ can be very useful.
A: I know this has already been answered mathematically, but for what it's worth, here is a histogram of average values gotten from random trials.

You can clearly see a maximum somewhere around 365 employees.
A: Let $y$ denote the number of days in a year.  Let $m$ denote the number of employees.
Assume that every day of the year is equally likely to be a given employee's birthday. Equivalently, every way of arranging $m$ birthdays in the year is equally likely. The event space is the set of arrangements of $m$ birthdays in a year of $y$ days (keeping in mind that one day can have many birthdays). The size of the event space is
$\left(\negthickspace\begin{pmatrix} y\\m\end{pmatrix}\negthickspace\right) = \begin{pmatrix} y+m-1\\m\end{pmatrix}$.
$\left(\negthickspace\begin{pmatrix} y\\m\end{pmatrix}\negthickspace\right)$ denotes the multiset coefficient.
To get the probability mass function for the number of holidays $h$, I need to count the number of events where the number of holidays is $h$. This occurs when there are $h$ days with at least one birthday, and all birthdays occur during those days.
The number of ways to arrange $h$ holidays in a year of $y$ days is $\begin{pmatrix} y\\h\end{pmatrix}$. This is equivalent to taking $h$ employees and counting the number of ways they can have birthdays such that no two birthdays coincide. For every arrangement of $h$ holidays, I need to count the number of ways the remaining $m-h$ employees' birthdays can be arranged, such that each birthday coincides with one of the previously chosen $h$ holidays. This number is 
$\left(\negthickspace\begin{pmatrix} h\\m-h\end{pmatrix}\negthickspace\right) = \begin{pmatrix} m-1\\m-h\end{pmatrix} = \begin{pmatrix} m-1\\h-1\end{pmatrix}$. 
So the number of ways to arrange $m$ birthdays such that there are $h$ holidays is
$\begin{pmatrix} y\\h \end{pmatrix}\begin{pmatrix}m-1\\h-1\end{pmatrix}$. Finally, the probability mass function is this number divided by the size of the event space.
$$
P\{H=h\} = \frac{\begin{pmatrix} y\\h \end{pmatrix}\begin{pmatrix}m-1\\h-1\end{pmatrix}}{\begin{pmatrix} y+m-1\\m\end{pmatrix}}
$$
I don't know how to prove this, but you can verify on WolframAlpha that this is indeed normalised. Sum the numerator for $h = 1$ to m or $h = 1$ to y depending on which is smaller, and compare to the denominator.
