I am having trouble coming up with a solution for this problem:

There is a stick of unit length. We break it into two parts. Now, we pick the bigger one and break it into two parts. I want to calculate the probability that the three pieces form a triangle.

The problem is from "Introduction to Probability, Charles M. Grinstead", Chapter 2.2, Exercise 13

  • $\begingroup$ What have you attempted? $\endgroup$ – ncmathsadist Aug 3 '15 at 20:51
  • $\begingroup$ You will need to add constraints on the likelihood of the pieces being certain sizes... Are the two breaks both uniform on $(0,1)$? Do you pick a break first uniformly then uniformly break the largest of the subsequent pieces...? Not enough information. $\endgroup$ – jameselmore Aug 3 '15 at 20:53

Let $\lambda$ be the length of the bigger piece and we split it into two smaller pieces $\lambda\mu$ and $\lambda(1-\mu)$. It is clear $\lambda$ and $\mu$ are uniform random variables $\sim \mathcal{U}(\frac12,1)$ and $\mathcal{U}(0,1)$ respectively.

In order for the three pieces with lengths $\;1-\lambda, \lambda\mu, \lambda(1-\mu)\;$ to form a triangle, the necessary and sufficient conditions are the fulfillment of following three triangular inequalities: $$\begin{cases} \lambda \mu + \lambda (1-\mu) &\ge 1-\lambda\\ \lambda \mu + (1 - \lambda) &\ge \lambda (1-\mu)\\ \lambda (1-\mu) + (1-\lambda) &\ge \lambda \mu \end{cases} \quad\iff\quad \begin{cases} \lambda \ge \frac12\\ \mu \ge 1 - \frac{1}{2\lambda}\\ \frac{1}{2\lambda} \ge \mu \end{cases}$$ The first inequality is trivially satisfied because we are told to break the bigger piece.
The probability we seek is given by:

$$2\int_{1/2}^1 \int_{1-\frac{1}{2\lambda}}^{\frac{1}{2\lambda}} d\mu d\lambda = 2\int_{1/2}^1 \left(\frac{1}{\lambda} - 1 \right) d\lambda = 2\log 2 - 1 \approx 38.6294\%$$


Let $A$, $B$, and $C$ be the lengths of the three resulting pieces. Let $A$ be the length of the shorter of the two segments from the (presumably uniformly-distributed) initial cut. Clearly $A$ can be written as $A = u_1/2$, where $u_1 \sim \mathcal{U}(0,1)$. Now let $u_2 \sim \mathcal{U}(0,1)$ be independent of $u_1$. We can write $B = (1-A)u_2 = (1-u_1/2)u_2$ and $C = (1 - A)(1 - u_2) = (1-u_1/2)(1 - u_2)$.

In order for these to form a triangle, we need $A\le B+C$, $B\le C+A$, and $C\le A+B$. Substituting the above expressions for $A$,$B$, and $C$ in terms of $u_1$ and $u_2$, and using the fact that $A+B+C = 1$, these three conditions reduce to $$ u_2 \le \frac{1}{2 - u_1} $$ and $$ u_2 \ge \frac{1-u_1}{2 - u_1}\, . $$

The probability that this occurs is given by the area between the two curves defined by making these into equalities: \begin{align} \mathrm{P(triangle)} &= \int_0^1 du_1\, \left(\frac{1}{2 - u_1} - \frac{1 - u_1}{2 - u_1}\right) \\ &= \log(4) - 1\\ &= 0.386294... \end{align}

For what it's worth, I've checked this numerically with 100,000 trials of random "line-segment making", and got 0.38589.

  • $\begingroup$ Ah, achille hui beat me to it with an essentially-identical answer while I was typing. $\endgroup$ – John Barber Aug 3 '15 at 21:33

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