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?