Breaking a stick into 3 pieces

Question

Assume that you are breaking a stick into 3 pieces by uniformly and independently selecting two break points 𝑋 and 𝑌. If we denote the event that these pieces form a triangle by T and lengths of the three pieces by 𝐴, 𝐵 and 𝐶, calculate:

a. 𝑃(𝑇 ∩ (𝐴 < 𝐵 < 𝐶)),

b. The probability that the pieces form an equilateral triangle. 

Answer 1

This is a fun problem that highlights the meaning of probabilities of intersections of sets. What is the probability of a given cut forming a triangle AND making an ordered triplet of lengths A,B, and C (from left to right let's suppose). First, let's recall that if you have any 2 events G and H, the probability of both happening is:

P(G∩H) = P(G) * P(H|G) = P(H) * P(G|H)

P(H|G) is the probability of H given G has happened. That means, that when calculating H, we ASSUME that G has already occurred, with probability P(G). Likewise for P(G|H). This formalism doesn't matter if your two events are independent; you just multiply the two probabilities.

Let's say the stick has length = 1, and therefor x and y are uniform random variables ∈ [0,1]

0|----------x-------------y-------------|1

Now picture what would have to be true for a triangle to not occur. If any length is larger than the sum of the other two lengths, then a triangle cannot form. For example if C > A + B you cannot maneuver A and B, joined together, such that C fits to form a triangle. C is too large. Thus, for a triangle to occur, you can show that neither A,B, or C can be greater than 0.5.

So, let's imagine we have already placed x, where x is less than a half. Where can we place y such that no region is larger than 0.5? Well y cannot be less than x, or else the region to the right of x will be > 0.5. Y cannot be greater than 0.5 + x, or else a region to the left of y will be > 0.5. Also, y cannot be between x and 0.5, otherwise the region to the right of y will be too large. No matter what value we pick for x, the stick is always divided into 4 regions regions, 3 of which y cannot be. Since we said that x < 0.5, the average length of the valid region is 1/4. And since y is also a uniform variable, P(T) = 1/4

0|*********x*******0.5------------********|1

y y y

Now let's assume we were able to create a triangle with the two cuts. The lengths of the triangle formed can be ordered with ≤, so let's just label A≤B≤C for the triangle formed. For the given triangle, chose one of the 3 vertices to cut the triangle, such that the lengths are folded back into the stick. You'll find that there are 3 ways to do that, times 2 ways to arrange the order of the regions (forwards or backwards). Out of 6 ways to get back to the stick, only 1 way gets the correct ascending order. This means that if you have a working triangle, there is a 1/6 chance of it happening to have the lengths in increasing order.

Therefor P(A<B<C|T) = 1/6

One can run a Monti Carlo simulation (as I have) to check that the following is correct.

P(T∩A<B<C) = (1/4)*(1/6)=1/24

Answer 2

a. The problem is not clearly stated. One might infer that A is the distance from the "zero" end to x, B is the distance between x and y, and C is the distance from y to the other end. However, that is an assumption. I choose to assign "A" to the shortest piece and "C" to the longest piece. With this approach P(A<B<C) = 1, so the question reduces to finding P(T).

The triangle inequality states that the sum of any 2 sides of a triangle is greater than the remaining side. In breaking the stick, the order that the breaks happen in is not important, so we can just say that we have two random locations and we assign x to the lower one and y to the higher one. One can easily see that for a triangle to be made from segments made from the 2 breaks, the two marks must be on different halves of the stick. Let's call that event "D." The location of the first mark (which could end up being x OR y) is arbitrary, so the probability of the second mark being on the different half is 1/2. That is, P(D) = 1/2.

For simplicity, let us say that the stick has length 1. Another condition must be satisfied: the distance BETWEEN the two marks must be less than 1/2, so that it is not longer than the sum of the two end segments. GIVEN the condition stated in the previous paragraph, the two marks have a uniform probability distribution in their respective halves. Also, y-x has a uniform distribution between 0 and 1, so

P[(y-x) < 1/2] | D) = 1/2

P(T) = P[D ∩ (y-x) < 1/2]

Using Bayes' Theorem, P(T) = P[(y-x) < 1/2 | D] [P(D)] = (1/2)(1/2) = 1/4

b. For any continuous distribution, the probability of an exact value occurring is zero. By extension, the probability that any two or three sides are exactly equal is also zero.