Probability of being the sides of quadrilateral Divide a given line segment into two other line segment. Then cut each of these new line segment into two more line segment. What is the the probability that the resulting line segment are the sides of quadrilateral?
 A: Assuming that the breakpoints are selected according to a uniform distribution, the probability is given by:
$$ 2\int_{0}^{1/2}\frac{1/2-u}{1/2+u}\, du = -1+\log 4 = \color{red}{38,6\%}. \tag{1}$$
Assuming that the initial line segment has unit length, the four segments are the sides of a quadrilateral iff none of them exceeds $\frac{1}{2}$. Proof: let $A$ be the longest segment, $B$ the second longest segment, $C$ and $D$ the remaining ones. If $A> B+C+D$ (that is equivalent to $A>\frac{1}{2}$, since $A+B+C+D=1$) there is clearly no hope. Otherwise, since $B\leq A$ but $B+C+D\geq A$, there is a triangle having sides $A,B,C+D$ and we are done.
To compute the probability, we are free to assume that the first break point is in the second half, by symmetry. To fullfill the previous constraint, we just need that the second break point is chosen in a proper way: if the first break point occurs in $\frac{1}{2}+u$, the second break point must be chosen from $u$ and $\frac{1}{2}$, and we easily get $(1)$.

Notice that $(1)$ strongly depends on the probability ditribution we use to choose the break points. For example, assuming that a human always breaks a unit length stick in a point in the range $30\%$-$70\%$, then the probability to get four sides of a quadrilateral after the process is just $1$.
If the break points are chosen accordingly to a random variable supported on $[0,1]$ having pdf
$$ f(x) = 4\min(x,1-x) = 2-|4x-2| $$
(it is just the convolution of two uniform distributions) then the probability of having four sides of a quadrilateral is given by:
$$2\int_{0}^{1/2}\frac{f(2u)}{u+1/2}\int_{u}^{1/2}f\left(\frac{t}{u+1/2}\right)\,dt\,du=-3+16\log\frac{8192}{6561}=55,2\%.$$
