Expected number of close pairs Player A, picks $n_1$ integers $a_1$,...,$a_{n_1}$ uniformly at random from $1$..$N$, 
and player B picks $n_2$ integers $b_1$,...,$b_{n_2}$ the same way.
Given $N$, $n_1$, $n_2$, and $d$, what is the expected number of pairs ($a_i$,$b_j$) where |$a_i$-$b_j$| $=<$ $d$, 
A. if A and B are allowed to select duplicates in their lists.
B. if duplicate numbers are not allowed. 
Thanks, 
MG
 A: In both cases, the expected number is $n_1n_2p$ where $p$ is the probability that $|a_1-b_1|\le d$. There are $n(d)=N+2(N-1)+\cdots+2(N-d)$ such couples $(a,b)$ hence $p=n(d)/N^2$. Finally $n(d)=(2d+1)N-d(d+1)$, hence the expected number you ask for is 
$$n_1n_2[(2d+1)N-d(d+1)]/N^{2}.$$
A: If $d\ge N-1$, then all pairs must be within $d$ of each other, and the total number of close pairs must be exactly $n_1 n_2$.  Now assume $d < N-1$.  For any $i,j$, the random variables $a_i$ and $b_j$ are independent and uniformly distributed over $[N]=\{1,2,...,N\}$.  The points $(x,y)$ in $[N]^2$ with $|x-y|>d$ make up two right triangles with side length $N-1-d$, and the total number of such points is $(N-1-d)(N-d)$, or $(N-d)^2 - (N-d)$.  So the probability that a particular pair $(a_i,b_j)$ is close is given by
$$
\begin{eqnarray}
P\left[|a_i-b_j|\le d\right] &=& \frac{1}{N^2}\left(N^2 - (N-d)^2 + (N-d)\right) \\
&=& \frac{1}{N^2}\left(N(2d+1) - d(d+1)\right).
\end{eqnarray}
$$
The expected number of close pairs is just $n_1 n_2$ times this probability:
$$
E\left[\text{# pairs }(a_i, b_j)\text{ s.t. }|a_i-b_j|\le d\right] = \frac{n_1 n_2}{N^2}\left(N(2d+1) - d(d+1)\right).
$$
This is true whether or not $A$ and $B$ are allowed to have duplicates within their individual lists of integers.  It does, however, make a difference whether $A$ and $B$ may have duplicates between their lists.  If no duplicates are to be allowed even between $A$ and $B$, then $N$ points $(x,x)$ are removed from consideration for each pair $(a_i, b_j)$: the final result will have its $N^2$ denominator replaced by $N(N-1)$.
