If John is the talles in his group, what are the odds he is the tallest of the whole University? There is a University with $r$ groups. Each group has $n$ students, students are randomly distributed within groups. From one of those groups, John is the tallest kid. Which is the expected ranking of John according to height in the whole University?
In general, if a kid is ranked the $k-$th tallest kid in his group, in which position will he be in the whole University rank?
For example, with two groups and two students, brute force calculations show that if John is the tallest in one group he will be in the 1.66 position in the general ranking.
Thanks.
 A: The tallest in the university is certainly the tallest in their group. So the question is what is the probability that the tallest in the university was in the group we're looking at, which is $1/r$.
If someone is ranked $k$th within his group then there are $k-1$ students definitely taller, and $n-k$ who are definitely shorter (just looking at those in the same group). There are $(r-1)n$ other students. Each of these other students might be taller than the first person in the group, between the first and second, and so on, down to shorter than the $n$th person. There are $n+1$ possibilities, and each is equally likely. So each other person has a probability of $\frac{k}{n+1}$ of being taller.
Unfortunately these are not independent: if the first person you look at is taller than the whole group, that suggests the group is shorter than average, and it is more likely that other people will be taller than the whole group too. But you can still get the average number of people who are taller than the person you are interested in, simply because expectation is linear. So on average there will be $\frac{k}{n+1}(r-1)n$ taller people in the other groups, and the expected rank is $k+\frac{k}{n+1}(r-1)n$.
A: The first part is relatively easy 
There are $r$ groups of equal size so each is equally likely to have the tallest student overall, making the probability that John is the tallest student overall is  $\dfrac{1}{r}$
In fact you can extend this: if there are $N$ students overall then the probability that the tallest of a randomly selected set of $n$ of them is also the tallest overall is $\dfrac{n}{N}$; in the original question $N=nr$
In the second part you get a distribution, essentially binomial 
Each of the $N-n$ people not in John's group could be in the gaps above John, between John and the second tallest person in John's group, and so on down to being below the shortest person in John's group.  All of these are equally likely so with probability $\dfrac{1}{n+1}$, so the probability of being taller than the $k$th person in John's group is $\dfrac{k}{n+1}$
That makes the probability that the $k$th tallest person in John's group is $m$th tallest overall is $${N-n \choose m-k}\frac{k^{m-k}(n+1-k)^{N-n-m+k}}{(n+1)^{N-n}}$$ with expected position $k+(N-n)\dfrac{k}{n+1}=k\dfrac{N+1}{n+1}$ and again for the original question you can use $N=nr$  
