Probability of any random n points on a line being within a given distance

Let $m$ be the ratio between the minimum distance and the length of the segment. Up to a rescaling, we can equivalently compute the probability that the minimum distance among $n$ points on a unitary segment, that is, $n$ real numbers in $[0,1]$, is at least $m$.

Let $x_1, x_2, \dots, x_n$ be these numbers. The probability of them being sorted is $1/n!$, so, up to a factor $n!$, we can assume $x_1 < x_2 < \dots < x_n$. We want to compute the measure of the set \[ S = \{ (x_1, \dots, x_n) \in [0,1]^n \mid x_i + m < x_{i+1} \text{ for } 1 \leq i \leq n-1 \}. \] Consider the map $f \colon S \rightarrow [0,1]^n$ given by \[ (x_1, \dots, x_n) \mapsto (x_1, x_2 - m, x_3 -2m, \dots, x_n - (n-1)m). \] This map is clearly injective, and the image is the set \[ f(S) = \{ (x_1, \dots, x_n) \in [0, 1 - (n-1)m]^n \mid x_i < x_{i+1} \text{ for } 1 \leq i \leq n-1 \}; \] indeed, for each $i$, $f(x_i) = x_i - (i-1)m < x_{i+1} - m - (i-1)m = x_{i+1} - i \cdot m = f(x_{i+1})$, and $f(x_n) = x_n - (n-1)m \leq 1 - (n-1)m$, and it admits an inverse $g$ given by \[ (x_1, \dots, x_n) \mapsto (x_1, x_2 + m, x_3 + 2m, \dots, x_n + (n-1)m). \]

Now, $f(S)$ has volume $(1-(n-1)m)^n/n!$, and since $f$ is clearly volume preserving (t is linear and the corresponding matrix has determinant $1$), then $S$ has the same volume, and therefore the probability that all the points have distance at least $m$ is $(1-(n-1)m)^n$ if $(n-1)m < 1$, and $0$ otherwise. If we want the probability of any two points having distance less than $m$, then it is \[ 1 - (1-(n-1)m)^n \] if $(n-1)m < 1$, and $1$ otherwise.