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 $Pr(a\_{\max}=k)$?



Distribution of the biggest gap
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](https://math.stackexchange.com/questions/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
$$Pr(a\_{\max}\leq k)={\sum\_{x} (-1)^x {m-xk\choose n}{n+1\choose x}\over{m\choose n}}.$$


