Dtyjlui

This question is normally asked with two points, which is very simple (0.75). But when it extends to 3 points, the 3-dimensional volume is really hard to work out. By simulation, I have already known that the answer is 0.5. But I would like to get a mathematical approach.

Let $X_1,X_2,X_3$ be identically and independently distributed random variables with the uniform distribution on $[0,1]$, and $$U_1>U_2>U_3$$ be its order statistics. Then the problem comes down to finding
$$
P(U_1-U_3<frac{1}{2}).
$$ Note that
$$
P(u<U_3<U_1<v)=P(u<X_i<v,: i=1,2,3)=(v-u)^3
$$ for $0<u<v<1$. By differentiating with respect to $u$ and $v$, we get the joint distribution
$$P(U_3in du, U_1in dv)=6(v-u)1_{{0<u<v<1}}.
$$ Finally, we have
$$begin{eqnarray}
P(U_1-U_3<frac{1}{2})&=&int_{v-u<frac{1}{2}}P(U_3in du, U_1in dv)\
&=&int_0^{frac{1}{2}}int_u^{u+frac{1}{2}}6(v-u);dvdu +int_{frac{1}{2}}^1int_u^1 6(v-u);dvdu\
&=&int_0^{frac{1}{2}}int_0^{frac{1}{2}}6v;dvdu +int_{frac{1}{2}}^1int_0^{1-u} 6v;dvdu\
&=&frac{3}{8} +int_{frac{1}{2}}^1 3(1-u)^2;du =frac{1}{2}.
end{eqnarray}$$

Note: It can be generalized to the case of $n$ points with any distance by noticing that
$$
P(U_nin du, U_1in dv)=n(n-1)(v-u)^{n-2}1_{{0<u<v<1}}.
$$

There must be a problem with the wording of this problem:

"Three points are randomly placed on a line of length 1. What is the probability that the distance of any pair of them is less than 1/2?"

Logically this means: "What is the probability that the distance between $P_1$ and $P_2$ is less than $1/2$ or that the distance between $P_1$ and $P_3$ is less than $1/2$ or that the distance between $P_1$ and $P_3$ is less than $1/2$.

As written, the probability is always $1.0$. No matter where you place the three points, the probability that there exists a pair closer than $1/2$ is $1.0$.

Consider the case where $P_1$ is at $0$, $P_2$ is at $1/2$ and $P_3$ is at $1$. In this set of measure zero, no point is closer than $1/2$ to any other. In every other case, there are two points closer than $1/2$.

I think the OP really wants to ask a different question...

If it is the probability that every pair is closer than $1/2$, then the region is:

and the volume of this within the unit cube is indeed $1/2$.

Let's do the general case of $n$ points, since it's not much harder. Choose a point $A$ to be leftmost and a point $B$ to be rightmost. The probability that their (signed) separation is between $l$ and $l+ dl$ is $(1-l)dl$. At that separation, the remaining $n-2$ points are between the two with probability $l^{n-2}$. Integrating from $l=0$ to $l=1/2$ gives
$$
P_{n,AB}=int_{0}^{1/2}(1-l)l^{n-2}dl=frac{l^{n-1}}{n-1}-frac{l^{n}}{n}biggvert_{0}^{1/2}=left(frac{1}{2}right)^{n-1}left(frac{1}{n-1}-frac{1}{2n}right)=left(frac{1}{2}right)^{n}frac{n+1}{n(n-1)}.
$$
Since we could have chosen $A$ and $B$ in $n(n-1)$ different ways, the total probability is just
$$
P_n=frac{n+1}{2^n}.
$$
This generalizes the known result that $P_2=frac{3}{4}$ and OP's result that $P_3=frac{4}{8}=frac{1}{2}$.