Let's say that I have a one-dimensional line of finite length 'L' that I populate with a set of 'N' random points. I was wondering if there was a simple/straightforward method (not involving long chains of conditional probabilities) of deriving the probability 'p' that the minimum distance between any pair of these points is larger than some value 'k' -i.e. if the line was an array, there would be more than 'k' slots/positions between any two point. Well that, or an expression for the mean minimum distance (MMD) for a pair of points in the set – referring to the smallest distance between any two points that can be found, not the mean minimum/shortest distance between all possible pairs of points.
I was unable to find an answer to this question after a literature search, so I was hoping someone here might have an answer or point me in the right direction with a reference. This is for recreational purposes, but maybe someone will find it interesting. If not, apologies for the spam.
Best Answer
This can answered without any complicated maths.
It can be related to the following: Imagine you have $N$ marked cards in a pack of $m$ cards and shuffle them randomly. What is the probability that they are all at least distance $d$ apart? Consider dealing the cards out, one by one, from the top of the pack. Every time you deal a marked card from the top of the deck, you then deal $d$ cards from the bottom (or just deal out the remainder if there's less than $d$ of them). Once all the cards are dealt out, they are still completely random. The dealt out cards will have distance at least d between all the marked cards if (and only if) none of the marked cards were originally in the bottom $(N-1)d$. The probability that the marked cards are all distance d apart is the same as the probability that none are in the bottom $(N-1)d$.
The points uniformly distributed on a line segment is just the same (considering the limit as $m$$\rightarrow∞$). The probability that they are all at least a distance $d$ apart is the same as the probability that none are in the left section of length $(N-1)d$. This has probability $(1-\frac{(N-1)d}{L})^N$.
Integrating over $0$$\le$$d$$\le$$\frac{L}{(N-1)}$ gives the expected minimum distance of $\frac{L}{(N^2-1)}$.