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? 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. 
 A: 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
$$P(U_3\in du, U_1\in dv)=6(v-u)1_{\{0<u<v<1\}}.
$$ Finally, we have
$$\begin{eqnarray}
P(U_1-U_3<\frac{1}{2})&=&\int_{v-u<\frac{1}{2}}P(U_3\in du, U_1\in dv)\\
&=&\int_0^{\frac{1}{2}}\int_u^{u+\frac{1}{2}}6(v-u)\;dvdu +\int_{\frac{1}{2}}^1\int_u^1 6(v-u)\;dvdu\\
&=&\int_0^{\frac{1}{2}}\int_0^{\frac{1}{2}}6v\;dvdu +\int_{\frac{1}{2}}^1\int_0^{1-u} 6v\;dvdu\\
&=&\frac{3}{8} +\int_{\frac{1}{2}}^1 3(1-u)^2\;du =\frac{1}{2}.
\end{eqnarray}$$
Note: It can be generalized to the case of $n$ points with any distance by noticing that
$$
P(U_n\in du, U_1\in dv)=n(n-1)(v-u)^{n-2}1_{\{0<u<v<1\}}.
$$
A: 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 $1/2$.
A: 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
$$
P_{n,AB}=\int_{0}^{1/2}(1-l)l^{n-2}dl=\frac{l^{n-1}}{n-1}-\frac{l^{n}}{n}\bigg\vert_{0}^{1/2}=\left(\frac{1}{2}\right)^{n-1}\left(\frac{1}{n-1}-\frac{1}{2n}\right)=\left(\frac{1}{2}\right)^{n}\frac{n+1}{n(n-1)}.
$$
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 $P_3=\frac{4}{8}=\frac{1}{2}$.
