TITLE: Distribution of a maximum
QUESTION [16 upvotes]: Randomly select $n$ numbers from ${\{1,2,\dots,m\}}$ without replacement, and order the chosen elements increasingly: $X_1 < X_2 < \dots < X_n$
And we can view each $X_i$ as a random variable, and we can get $\mathbb{E}(X_i) = \frac{(m+1)i}{n+1}$
And we can define $Y_i=|X_i-\mathbb{E}(X_i)|$ which is the distance of each variable to its corresponding expectation. 
And we can also define $Z = \max_{1 \le i \le n} Y_i$
So what is the distribution of $Z$?
[Updated]
Any bound of $Z$ is helpful.
[Updated]
I have re-posted the question at Mathoverflow:
https://mathoverflow.net/questions/78822/distribution-of-a-maximum
People there gave some new ideas.

REPLY [3 votes]: This answer addresses your update "Any bound of $Z$ is helpful."  It does not give the probability distribution of $Z$.  
For fixed $m$ and $n$ with $n < m$, we have $$Z \leq n \left(\frac{m+1}{n+1}-1\right),$$ and for fixed $m$, where $n$ is allowed to range over all possible values from $1$ to $m$, we have $$Z \leq \left(\sqrt{m+1}-1\right)^2.$$

For the first inequality, let $m$ and $n$ be fixed with $n < m$.  So $\frac{m+1}{n+1} > 1$.  Thus the maximum possible value of $\left|X_i - \frac{(m+1)i}{n+1}\right|$ occurs for one of the extreme values in an extreme in the other direction sample.  In other words, the largest possible value of $Z$ occurs with $Y_n$ when the sample $\{1, 2, \ldots, n\}$ is chosen or with $Y_1$ when the sample $\{m-n+1, m-n+2, \ldots, m\}$ is chosen.  By symmetry, these two values should be equal, and that is the case: In the first sample, we have $Y_n = \left|n - \frac{(m+1)n}{n+1}\right| = n \left(\frac{m+1}{n+1}-1\right)$, and in the second sample we have $Y_1 = \left|m-n+1 - \frac{m+1}{n+1}\right| = (m+1)\left(1 - \frac{1}{n+1}\right) - n = n \left(\frac{m+1}{n+1}-1\right).$  Thus $Z \leq n \left(\frac{m+1}{n+1}-1\right).$
To see the second inequality, just use calculus to maximize the expression $f(n) = n\left(\frac{m+1}{n+1}-1\right)$.  We get that $f(n)$ is maximized when $n = \sqrt{m+1}-1$, and thus $Z \leq \left(\sqrt{m+1}-1\right)^2$.
As an illustration of this last result, consider the numerical work posted by Sasha.  The largest values of $Z$ occur on the plots for which $n$ is closest to $\sqrt{m+1}-1$.