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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?

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Do these 4 employees work the same shift, or different shifts ? Why do you assume that each employee will choose uniformly choose among 8 break time slots available ? The answer will strongly depend on their preferences. Say, each will be more likely to take a break on 3rd and 5th hours. –  Sasha Mar 30 '12 at 17:08
    
"The probability of everybody using 1 hours and break at the same time" should be better stated: "The probability of everybody using the same break, GIVEN THAT everybody used 1 hour break", and so on. –  leonbloy Mar 30 '12 at 17:12
    
You should care very much about the probability of everybody using 0 hours to break. If this probability is 1, all 4 employees never take a break at the same time. –  Did Mar 30 '12 at 17:13
    
@Sasha, Great questions! That's what I forgot to mention. Each employee is a robot and the chances of choosing an hour to break are all equal. They all work the same shifts. Please don't follow up saying robots don't need to take breaks. :) –  Russell Mar 30 '12 at 17:13
    
@Didier, good point. If that's the case then I thought x and y would be 0 anyway. Or am I missing something. –  Russell Mar 30 '12 at 17:14
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1 Answer

up vote 1 down vote accepted

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, $$ \color{red}{h=(q+2r)^n/k^{n-1}}. $$

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Wow, this is great. I forgot about the p,q,r relationship. Thank you! –  Russell Mar 30 '12 at 17:29
    
and now it's time for a break. –  draks ... Mar 30 '12 at 19:57
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