Probability of order statistics Let's say you generate 3 uniformly distributed, independent random numbers on the interval $[0,1]$. Now consider the lengths of the 4 segments made.
What is the probability that the sum of the two medium-length segments is greater than $ 0.5 $?
Example
Let the random numbers be $ 0.5 $, $ 0.3 $, $ 0.1 $. This cuts the interval like so:
|*|**|**|*****|

The sum of the medium-length segments is then $ 0.2 + 0.2 = 0.4 $.
Answer (numerical, no proof)
I ran a computer simulation of this and got the answer: spoiler.
However I can't seem to come up with a derivation of this.
 A: Let us try to compute this probability without actually evaluating any integral.
We begin with the remark that one can realize the three random numbers and the number $1$ itself as the four first points of a homogenous Poisson process. In other words, the lengthes of the four intervals are proportional to $(X_i)_{1\le i\le 4}$ where the random variables $X_i$ are i.i.d. and exponential of parameter $1$. 
Introducing the order statistics $X_{(i)}$ of the sample $(X_i)$, one asks for $p=P(A)$ with 
$$
A=[X_{(2)}+X_{(3)}\ge X_{(1)}+X_{(4)}].
$$
Using the notation $X_{(0)}=0$, the waiting time paradox shows that the increments 
$$
Y_i=X_{(i)}-X_{(i-1)}
$$ 
are independent for $1\le i\le 4$ and that each $Y_i$ is exponential of parameter $5-i$. Since the event of interest is
also
$$
A=[Y_2\ge Y_4],
$$
the probability $p$ is also
$$
p=P(Z\ge 3Z'),
$$ 
where $Z$ and $Z'$ are i.i.d. and exponential of parameter $1$. In other words, $p=P(U\le1/4)$ with 
$$
U=Z'/(Z+Z').
$$
Since $U$ is uniformly distributed in the interval $(0,1)$, $p=1/4.$
Added later on More generally, throwing $n$ points in $(0,1)$ yields $n+1$ intervals. The ordered lengthes of these intervals are proportional to the order statistics $(X_{(i)})_{1\le i\le n+1}$ of an i.i.d. sample $(X_i)_{1\le i\le n+1}$ of exponential random variables. For each $i$, $X_{(i)}=Y_1+\cdots+Y_i$, where the random variables $(Y_i)_{1\le i\le n+1}$ are independent and the distribution of $Y_i$ is exponential with parameter $n+2-i$.
A: The lengths of the segments are uniformly distributed over the unit $3$-simplex (see Simulating uniformly on $S^1=\{x \in \mathbb{R}^n \mid \|x\|_1=1\}$). Thus, the desired probability is the fraction of the part of the unit $3$-simplex with $x_1\le x_2\le x_3\le x_4$ for which $x_2+x_3\ge\frac12$.
The unit simplex has volume $1/6$, so the restriction to one particular permutation of the coordinates leads to a volume of $1/6\cdot1/4!=1/144$. We can check this to make sure the integration bounds are set up correctly:
$$\int_0^{1/4}\int_{x_1}^{(1-x_1)/3}\int_{x_2}^{(1-x_1-x_2)/2}\mathrm dx_3\mathrm dx_2\mathrm dx_1=\frac1{144},$$
as computed here.
The integration bounds under the condition $x_2+x_3\ge\frac12$ are a bit trickier, since there are two possibilities, depending on $x_2$. For $x_2\ge\frac14$, all values of $x_3$ with $x_3>x_2$ fulfill the condition, whereas for $x_2<\frac14$ there's a new lower bound $\frac12-x_2$ for $x_3$. Thus we split the integral into two parts at $x_2=\frac14$, with different lower bounds for $x_3$:
$$\int_0^{1/4}\int_{x_1}^{1/4}\int_{1/2-x_2}^{(1-x_1-x_2)/2}\mathrm dx_3\mathrm dx_2\mathrm dx_1=\frac1{768},$$
$$\int_0^{1/4}\int_{1/4}^{(1-x_1)/3}\int_{x_2}^{(1-x_1-x_2)/2}\mathrm dx_3\mathrm dx_2\mathrm dx_1=\frac1{2304},$$
as computed here and here, respectively. So the desired probability is indeed
$$\frac{\frac1{768}+\frac1{2304}}{\frac1{144}}=144\cdot\frac{4}{2304}=\frac{144}{576}=\frac14\;.$$
