Expected number of hours that all employees take a break at the same time? This is derived from a real question my company has, so the wording might sound a bit weird.
There are 4 employees, each working 8 hours a day. Breaks are in 1 hour blocks and employees may only take breaks at the beginning of an hour (so there are 8 chunks to choose from). Each employee is entitled 2 breaks per day and are not required to use them. (EDIT) Each employee will choose uniformly choose among 8 break time slots available.
What is the expected number of hours that all 4 employees take a break at the same time?
Here's what I got so far.
The probability of everybody using 0 hours to break is _I_DONT_CARE_ because it doesn't affect the expected hours.
The probability of everybody using the same break, GIVEN THAT everybody used 1 hour break is
$$x = (\frac1{8 \choose 1})^4$$
The probability of everybody using the same breaks, GIVEN THAT everybody used 2 hour breaks is
$$y = (\frac1{8 \choose 2})^4$$
The expected number of hours that everybody takes a break at the same time is    
$$1x + 2y$$
Is my calculation right? Is there an easier way to look at the problem?
 A: Consider the probabilities $p$, $q$ and $r$ that any given employee chooses to use 0, 1 or 2 breaks during the day, hence $p+q+r=1$. Assume these probabilities are the same for each employee and that each employee chooses its break hour(s) uniformly and independently on the 3 others.
Then the probability a given employee chooses the first hour for a break is $0$ if the employee takes 0 break, $\frac18$ if the employee takes 1 break and $\frac14$ if the employee takes 2 breaks, hence this employee chooses the first hour for a break with probability $b=p\cdot0+q\cdot\frac18+r\cdot\frac14$. The probability that all the 4 employees choose the first hour for a break is $b^4$. 
The same is true for each one of the 7 other hours hence, by linearity of the expectation, the expected number $h$ of hours that all 4 employees spend on a break together is $h=8b^4=(q+2r)^4/256$.
Edit: If there are $n\geqslant1$ employees and $k\geqslant2$ possible break  periods, the expected number of break periods that all the employees spend together is $h=kb^n$ with $b=p\cdot0+q\cdot\frac1k+r\cdot\frac2k$, that is, $$
\color{red}{h=(q+2r)^n/k^{n-1}}.
$$
