Distribution of the biggest gap Randomly select $n$ numbers from the universe $\{1,2\dots,m\}$ without replacement, and sort the numbers in ascending order. 
We can get a list of number $\{(a_1,a_2,\dots,a_n\)}$, and then we can get the difference between two consecutive numbers and get the gap list:
$\{(a_1, a_2-a_1,\dots ,a_n-a_{n-1})\}$
Among these gaps, there must be a biggest one, say $a_{\max}$.
So what is the distribution of $Pr(a_{\max}=k)$?
 A: This is the answer to a slightly modified version of 
the problem. I hope that it would also lead to a solution
of the original version.
As I point out in my answer to Math StackExchange question 66430
("What is the distribution of gaps?"),
if, in addition to the gaps $G_1=a_1$and $G_j:=a_j-a_{j-1}$ for $2\leq j\leq n$, 
you introduce final gap $G_{n+1}=(m+1)-a_n$,
 the random vector $(G_1,G_2,\dots, G_{n+1})$ gives a random composition
of the number $m+1$. That is, all outcomes $(g_1,g_2,\dots, g_{n+1})$
 with $$g_1+g_2+\cdots+g_{n+1}=m+1,\quad g_j\geq 1$$
are equally likely. There are $m\choose n$ such compositions, as 
found using stars and bars.  
Then $Pr(a_{\max}\leq k)$ (where my maximum includes the final gap)
is just the proportion of compositions using numbers from $1$ to $k$.
By inclusion-exclusion and stars and bars, this probability is
$$Pr(a_{\max}\leq k)={\sum_{x} (-1)^x {m-xk\choose n}{n+1\choose x}\over{m\choose n}}.$$
A: See
http://arxiv.org/pdf/cond-mat/0406116v2
for a more general version of the question (the 1-dim case is considered at length).
