the probability of those n broken parts of sticks to form a closed polygon? Suppose I have a stick of L length . It is divided into uniformly randomly into n parts.
What is the probability of those n parts to form a closed polygon?
It will be helpful if anyone show the steps to proceed with this problem.
 A: To start with, you might look at this youTube video (starting a at 1:00) to see a different way of computing the answer $1/4$ for the triangle problem.  This readily generalizes to higher dimensions.  
Let us set the length of the stick to be $1$.  If we break the stick into $n$ pieces, we get $n$ positive numbers whose sum is $1$, which we can think of as an ordered $n-$tuple, or as a point $(x_1,\dots,x_n)$ in $\mathbb{R}^n$ that satisfies $0\le x_1\le1,\ i=1,\dots,n\text{ and }\sum x_i=1.$  These points form the standard n-1-simplex, the $n-1-$dimensional analogue of a triangle.  For example, the standard $2-$simplex is the triangle in $\mathbb{R}^3$ with vertices $(0,0,1),(0,1,0),(0,0,1).$  The standard $3-$simplex is a regular tetrahedron sitting in $\mathbb{R}^4,$ and so on.
Note that we can form a polygon out of the $n$ pieces if and only if the longest piece has length less than $1/2.$  The way this relates to the simplex is that the sum of the distances from any interior point to the faces of the simplex is equal to the altitude of the simplex, namely $1$.  This is proved just as in the case of an equilateral triangle.  The volume of an $n-$simplex is proportional to the product of the altitude and the ($n-1-$dimensional) volume of the base.  (The constant of proportionality is $1/n,$ but we don't need that.)  Since all the faces are congruent, the result follows.
So the points inside the simplex corresponding to divisions that result in polygons are those within $1/2$ of all the faces.  For any face, a hyperplane parallel to that face and at distance $1/2$ cuts off a simplex of inadmissible divisions at the opposite vertex.  The excluded points form a simplex similar to the standard simplex, but of half the height, so its volume is $2^{-(n-1)}$ times the volume of the standard simplex.  (Recall that we're working with an $(n-1)-$simplex.)  Since there are $n$ faces, the volume remaining, divided by the volume of the entire simplex is $$\boxed{1-\frac{n}{2^{n-1}}}$$ and that is the probability you seek.
I wrote a little python script to test this by simulation, and it confirms my answer:
from random import random, seed
seed()
successes = 0
trials = 100000
pieces = 7
cuts = pieces-1
for _ in range(trials):
    points = [0]+sorted([random() for _ in range(cuts)])+[1]
    stick = [points[i+1]-points[i] for i in range(pieces)]
    if max(stick) < .5:
        successes += 1
print('actual ',successes/trials, 'expected', 1-pieces*.5**(pieces-1)) 

