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This question already has an answer here:

The problem

Let's consider the interval $[0, 1]$. We break this interval randomly in two parts. Then we choose the bigger part and again we break it randomly into two parts. At the end we have three parts of the interval.
Breaking points are distributed uniformly.
I am to find the probability of getting a triangle from these three pieces.

My attempt

First of all let's notice that a triangle can be build iff:

  1. $a+b>c$,
  2. $|a-b| < c$,

where $a, b, c$ are lengths of our triangle sights.

Let's denote the breaking points by $x$ and $y$.
Suppose $x > \frac{1}{2}$. We do have three intervals after breaking our segment: $[0, y], [y, x], [x, 1]$.

The first condition of triangle's existence is trivially fulfilled. The second one implies: $$y > x - {1}{2} \text{ and } y < \frac{1}{2}.$$ I calculated that: $$f_{Y|X}(y|x) = \frac{f(x, y)}{\int \limits_{\infty}^{\infty}f_X(x) \mathbb{1}_{[0,x]}(y) \mbox{d}y}.$$ Where $f(x) = \mathbb{1}_{[0,1]}(x)$ and $f(x, y) = \mathbb{1}_{[0,1]}(x)\mathbb{1}_{[0,x]}(y)$.

I don't really know how the integration should look like. I would appreciate any tips or hints.

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marked as duplicate by Andrei, Don Thousand, Michael Hoppe, Pierre-Guy Plamondon, Namaste Dec 17 '18 at 21:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ @M.Nestor while I have seen this question posed here before, and it definitely is a duplicate. The link you post is actually to a different formulation of the problem, where someone's simulation is giving a result other than the expected result. $\endgroup$ – Doug M Dec 17 '18 at 19:32
  • $\begingroup$ @DougM the answer of this problem is exactly twice of that in the linked question b/c we have been asked to break the longer piece. $\endgroup$ – achille hui Dec 17 '18 at 19:47
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Let $x$ and $1-x$ be the length of the two pieces after first split.

Relabel the two pieces if necessary, we can assume $x \sim \mathcal{U}(\frac12,1)$ and the longer piece has length $x$.

Let $xy,x(1-y)$ be the length of pieces after we split the longer piece.

Relabel the two new pieces if necessary, we can assume $y \sim \mathcal{U}(\frac12,1)$.

We now have $3$ pieces of length $xy, x(1-y)$ and $1-x$. In order for them to form a triangle, the conditions are

  • $xy + x(1-y) \ge 1-x \iff x \ge 1-x$
  • $xy + 1 - x \ge x(1-y)$
  • $x(1-y) + 1-x \ge xy \iff xy \le \frac12$

The first two conditions are trivially satisfied. In order to satisfy the third condition. The point $(x,y)$ needs to fall into a shape bounded by two lines $x = \frac12$, $y = \frac12$ and hyperbola $xy = \frac12$. The desired probability is $4$ times the area of this.

$$ 4\int_{\frac12}^1 \int_{\frac12}^{\frac{1}{2x}} dy dx = 2\int_{\frac12}^1 \left( \frac{1}{x} - 1 \right) dx = \log(4)-1 \approx 0.3862943611$$

As pointed out by M.Nestor in comment, a very similar question has been asked before. However, our answer is different from there ($\log 2 - \frac12 \approx 0.19314$) because we have been asked to break the longer piece after the first split. Since breaking the shorter piece "almost never" give you a valid triangle, our answer is exactly twice of the accepted answer in above question.

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  • $\begingroup$ Why can we assume that both $x, y \sim \mathcal{U}(\frac12,1)$? $\endgroup$ – Hendrra Jan 3 at 16:05
  • $\begingroup$ Moreover how did you get that $y > \frac{1}{2}$? $\endgroup$ – Hendrra Jan 3 at 16:32
  • $\begingroup$ This is what "Relabel the two pieces if necessary" refers to. Let's say after first split, the length of the pieces are $p$ and $1-p$. If $p \ge \frac12$, we call the piece with length $p$ piece $1$, the other piece piece $2$. If $p < \frac12$, we call the piece with length $1-p$ piece $1$ instead. Under this naming convention, the length of piece $1$ is $x = \max(p,1-p) \in [ \frac12, 1 ]$. For any $[a,b] \subset [\frac12,1]$, we have $$\mathbb{P}(x \in [a,b]) = \mathbb{P}( p \in [1-b,1-a] \cap [a,b] ) = 2(b-a)$$ This is precisely what one need for $x \sim \mathcal{U}(\frac12,1)$. $\endgroup$ – achille hui Jan 3 at 22:27
  • $\begingroup$ Thank you. I understand almost everything. Can you explain why $\mathbb{P}( p \in [1-b,1-a] \cap [a,b] ) = 2(b-a)$? $\endgroup$ – Hendrra Jan 4 at 21:07
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    $\begingroup$ @Hendrra oops, there is a typo, should be $$\mathbb{P}(x \in [a,b]) = \mathbb{P}(p \in [1-b,1-a]\color{red}{\cup} [a,b])$$ Basically, $x = \max(p,1-p)$, so either $x = p \ge \frac12 \implies p \in [a,b]$ or $x = 1-p \implies p < \frac12 \implies p \in [1-b,1-a]$. $\endgroup$ – achille hui Jan 4 at 22:42

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