Probability of creating a triangular frame from a straight stick If you have a straight stick and you break it in exactly two places what is the probability that the three pieces if put together can create a triangular frame?
This is exactly the question my professor gave on our statistics and probability final.
I analyzed it as best as I could (I wrote a whole page filled with assumptions and limitations) but I would like to know if anyone can come up with a scientific valid answer.
 A: There are many ways to break a stick, but let's assume that the first break is made uniformly at random along the length of the whole stick to create a first piece of size $x$, and that the remaining piece is broken at another point chosen uniformly at random along its length. Call the total length $L$.
First note that each piece must be less than half the length of the stick, so assume $x < \frac{L}{2}$. The remaining two pieces will satisfy the triangle inequality if and only if they are both greater than $\frac{L}{2}-x$ in length. This means that the break must be made within a section of length $x$ in the middle of the remaining piece; the probability of this happening is therefore $\frac{x}{L-x}$. Integrating and dividing by the total length $L$, we have:
$$P = \frac{1}{L}\int_{\frac{L}{2}}^L\frac{x}{L-x}dx=\ldots=\log 2 - \frac{1}{2}$$
The probability under these assumptions is therefore approximately $0.19$.
A: Create a ternary plot $T$ representing the lengths of the three pieces. Let us assume that the stick will be broken in such a way that all points on $T$ are equally likely. The region of $T$ where the three pieces can be reconstituted to form a triangular frame is the equilateral triangle inside $T$ whose vertices are the midpoints of $T$. This triangle is exactly one quarter as big as $T$, so the probability under these assumptions is $0.25$.
A: Without loss of generality we may assume the stick to be of unit length. Let $X_1$ and $X_2$ be two random positions where the stick was broken. Then length of three sticks obtained are $$
 \ell_1 = \min(X_1,X_2), \quad \ell_2 = \max(X_1,X_2) - \min(X_1,X_2), \quad \ell_3 = 1-\max(X_1,X_2)
$$ 
It is possible to form a triangle out of these sticks, provided triangle inequalities hold true:
$$
    \ell_3 < \ell_1+\ell_2 \quad  \ell_1 < \ell_2+\ell_3 \quad \ell_2 < \ell_1+\ell_3
$$
which are satisfied provided
$$
 \Omega =  \left(   0 < X_2 < \frac{1}{2} < X_1 < \frac{1}{2} + X_2 \right)    \lor 
  \left( X_2 - \frac{1}{2} < X_1 < \frac{1}{2} < X_2 < 1 \right)
$$

The probability of this event, given the joint pdf of $(X_1,X_2)$, $f(x_1,x_2)$ equals:
$$
   \mathbb{P}(\Omega) =  \int f(x_1,x_2) \mathbf{1}_\Omega\left(x_1,x_2\right) \mathrm{d} x_1 \mathrm{d} x_2
$$
In the case of uniform distribution, $f(x_1, x_2) = 1$, and the probability is easily seen to equal $\frac{1}{4}$. 
A: I'm not any good at math, nor do I understand much about probability, but I've always approached this problem from graphically. 
Lets say the length of the stick is 1 and assume that the cuts on the sticks are distributed uniformly. Lets call the distance from the left end of the stick to the first cut $x$, and the distance from the stick to the second cut $y$, and lets assume that the stick is of length 1: 

We know that the sum of any two sides of a triangle must be greater than the third side. Thus we have: 
$$x + (y-x) > 1-y$$
$$(y-x)+(1-y)>x$$
$$x+(1-y)>y-x$$
And simplifying:
$$y>.5$$
$$.5>x$$
$$x+.5>y$$
Let us also add the condition that the $y$ is the first plus second piece starting from the left, and x is always the first piece, so:
$$y>x$$
We also know that the most the $x$ piece can be in length is 1, and the least is 0, same with y, so we add conditions:
$$1>x>0$$
$$1>y>0$$
If you graph out the last 3 conditions I gave then that graphically will give you the whole area where it is possible for you to get when you do the two cuts, and then the first 3 conditions I gave will give you the piece of that area that allows you to make a triangle. I don't know how to draw a graph, but you should get a smaller triangle that is 1/4 the bigger triangle, meaning .25 chance of getting a triangle with 2 uniformly distributed cuts. 
