Geometry and PDF around the unit stick broken into 3 pieces and the expected length of the longest piece?

Question

This post is related to previous SE posts, could anyone please explain the final geometry steps i.e. the link between the geometry and the pdf.

• Average length of the longest segment : this part : " By rewriting this probability as area in the unit square, I get " @TCL.

@TCL gave a very clear explanation for almost the same problem with : If a 1 meter rope is cut at two uniformly randomly chosen points, what is the average length of the smallest piece? but could you please explain this part : " which is (1−3a)2(1−3a)2. Note that for these sets to be nonempty, one must have 0≤a≤1/3 0≤a≤1/3 "

• Stick of unit length is broken into three random pieces, what is the expected length of the longest piece? : this part : " A bit of geometry will give you the result " @Canardini

Another very quick reply to a very similar problem : @ Hans Parshall

• If a 1 meter rope is cut at two uniformly randomly chosen points, what is the average length of the smallest piece?" This part : It's not hard to show that they all have probability ... / our joint PDF is given by f(x,y)=6f(x,y)=6"

Answer 1

I will be using this: Stick of unit length is broken into three random pieces, what is the expected length of the longest piece?.

Right before the "A bit of geometry will give you the result", the answer there had the joint probability $$2P(X\leq a,1-Y\leq a,Y-X\leq a,X<Y).$$ In order to solve that joint probability, we need to find out the situations in which all four probabilities are true at the same time. Now, $$X\leq a,1-Y\leq a,Y-X\leq a,X<Y$$ are also a system of inequalities of variables $X$ and $Y$, and thus can be plotted on a plane with lines as boundaries. The area enclosed by the four inequalities as well as the bounding unit square is the joint probability for any particular $a$.

If n cuts are made, instead of 2, then the inequalities will be of n different variables, in dimension n bounded by hyper-planes of dimension n-1 instead, and you are to find the hyper-volume enclosed.