For known $k$ ,whats the probability of picking $2$ numbers-$x,y$ between $1$ and $n$ which $x-y=k$ if its known that $x>y$ For known $0<k<n$,what's the probability of picking $2$ numbers, $x,y$ between $1,2,3,...,n$ for which $x-y=k$ if its known that $x>y$, $k$ is between $1$ to $n$. Let $A - x>y$ and let $B - x-y=k$.
 A: We need x=k+i and y=i, where i can be any integer from 1 to n-k. Since x > y let's assume x is randomly chosen from 2,...,n and y is subseqquently chosen from 1,...,x. Then the probablility of getting x=k+i and y=i is (1/(n-1))*(1/(k+i-1)). So the answer is to sum that quantity from i=1 to n-k.
You may need to be more clear about how we are picking the numbers. They cannot be randomly chosen simultaneoulsy or you might end up with x=y.
A: The answer depends on how we're picking these numbers.

First solution:
Let's pick $x$ first uniformly at random in $[2,n-1]$, then pick $y$ to satisfy $1\le y<x\le n$. 
For each $x$ in the range $[k+1,n]$, there is exactly one $y$ which will work.  For $x$ in the range $[2,k]$, there are no $y$ that will work.  Hence your desired probability is 
$$\frac{1}{n-1}\sum_{x=k+1}^n \frac{1}{x-1}=\frac{1}{n-1}\left(\frac{1}{k}+\frac{1}{k+1}+\cdots+\frac{1}{n-1}\right)=\frac{1}{n-1}(H_{n-1}-H_{k-1})$$
Where $H_t$ denotes the Harmonic number $\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{t}$.  There's no nice closed form  for these, but one approximation is $H_t\approx \ln t-\gamma$, where $\gamma\approx 0.577$ is the Euler-Mascheroni constant.  Hence your desired probability is approximately $$\frac{\ln (n-1)-\ln k}{n-1}$$

Alternate solution:
Let's pick $x,y$ uniformly at random from $[1,n]$, and discard any cases where $x\le y$.  There are $1+2+\cdots+(n-1)=\frac{n(n-1)}{2}$ equally likely cases remaining, of which exactly $n-k$ are ones where the desired relationship holds.  Hence the desired answer is $$\frac{2(n-k)}{n(n-1)}$$
