Probability distribution for distances between randomly selected integers within an interval

Let $X_{(1)}, \ldots, X_{(N)}$ be the chosen integers in increasing order (the order statistics). For simplicity I'll suppose $A = 1$. Of course we must have $B \ge N$. Then I claim that all the "gaps" $X_{(j+1)} - X_{(j)}$ as well as $B+1 - X_{(N)}$ and $X_{(1)} - 0$ have expected value $(B+1)/(N+1)$.

Note that $E[X_{(1)} | X_{(2)}] = X_{(2)}/2$, because given $X_{(2)} = x$, $X_{(1)}$ is equally likely to be any of the integers 1 to $x-1$. Thus $E[X_{(1)}] = E[X_{(2)} - X_{(1)}]$. Similarly, given $X_{(j)} = x$ and $X_{(j+2)} = y$, $X_{(j+1)}$ is equally likely to be any of the integers $x+1$ to $y-1$, so $E[X_{(j+2)} - X_{(j+1)}] = E[X_{(j+1)} - X_{(j)}]$. Similarly, $E[B+1-X_{(N)}] = E[X_{(N)} - X_{(N-1)}]$. Thus all $N+1$ gaps have the same expected value, and since they add up to $B+1$ that expected value is $(B+1)/(N+1)$.

Edit The answer below addresses a different question than the original one. That was a mistake of mine, properly signaled by Matthew in a comment, so I deleted my answer. Later on, the OP added some so-called precisions to the question, which in fact change it completely. As a consequence of this modification of the question, my answer becomes relevant, miraculously (modulo the endpoints thing). Call this a manifestation of prescience if you want, anyway I repost my answer, and this is the end of my interventions on this page.

There are $N-1$ distances between nearest-neighbors amongst $N$ points so the mean distance (averaged over a given sample) is the span of the sample divided by $N-1$. The span is the maximum $M$ of the sample minus the minimum $m$ of the sample. By symmetry, $m$ is distributed like $B+A-M$ hence the mean distance (averaged over the samples) is $$ E(S)=\frac1{N-1}E(M-m)=\frac1{N-1}(2E(M)-(A+B)). $$ For each $n$ such that $N\le n\le B-A$, there are $n!/(n-N)!$ samples such that $M\le A+n$, hence $$ B+1-E(M)=\sum_{n=N}^{B-A}P(M\le A+n)=\frac{(B-A-N)!}{(B-A)!}\sum_{n=N}^{B-A}\frac{n!}{(n-N)!}. $$ Putting all this together should yield $E(S)$.

For some reason, all existing answers to this question, including the accepted one, only discuss either the expected value of the nearest-neighbour distances or the distribution of the shortest nearest-neighbour distance, but not the distribution of the nearest-neighbour distances. This was also the subject of the later question Distribution probability of elements and pair-wise differences in a sorted list, which is very similar to this one, though neither is an exact duplicate of the other.

My answer to that question, adapted to the notation of the present question, gives the probability

$$P(d_i=d)=\frac{\binom{B-A+1-d}{N-1}}{\binom{B-A+1}N}$$

that the $i$-th nearest-neighbour distance $d_i$ is $d$, independent of $i$.