Suppose $3$ points are drawn uniformly at random from $[0,1]$. Call them $x_1,x_2,x_3$ with $x_1\leq x_2\leq x_3$. I am interested in the distances between them.
Fix $0<a<b<1$. I want to compute the probability $P$ that $$x_2-x_1 < a, \ \ \ \ x_3 -x_2 < a,\ \ \ \ x_3-x_1 > b$$ In other words, the two pairs of adjacent points are $a$-close to each other, but the two outside points are $b$-far apart. Clearly this can't happen unless $b<2a$.
Since we have a uniform distribution, we can integrate the constant density $\equiv 1$ over the appropriate region in $[0,1]^3$. Now, $x_2$ must be within $a$ of $x_1$, and $x_3$ has to be at least $b$ greater than $x_1$ but not more than $a$ above $x_2$. This gives the following:
$$P = \int_0^{1-2a}\int_{x_1}^{x_1+a}\int_{x_1+b}^{x_2+a}\ dx_3\ dx_2\ dx_1$$
The problem: $x_{1}$ doesn't have to be less than $1-2a$; it should be able to be as large as $1-b$ (well technically, arbitrarily close to it but still less than). For instance, if $a=.2, b=.3$ then an acceptable set of points would be $.69,.85,1$. Clearly there are $2$ cases to consider, but I'm having a hard time explicitly differentiating the cases in order to write down two integrals. It seems the second case should look like
$$\int_{1-2a}^{1-b}\int_{x_1}^{x_1+a}\int_{x_1+b}^1\ dx_3\ dx_2\ dx_1$$
However I don't feel that this is correct. Specifically, it doesn't look like the above guarantees that $x_3-x_2<a$. Are these the only two cases to consider? What are the appropriate bounds in the second integral? Is there a better way to approach this problem?
How about this order? (Using physics notation, where we place the differential form next to the integral sign)
$$\int_0^{1-b} dx_1 \int_{x_1+b}^{\min(1,x_1+2a)} dx_3 \int_{x_3-a}^{x_1+a} dx_2$$
That seems reasonable to simplify. To get the desired probability, don't forget to normalize by the volume of the simplex $x_1<x_2<x_3$, which is $\frac{1}{3!}$.