16. New Year's Tradition (from 9th grade. 3 points). On one island tribe, there is a tradition - during a ritual dance, the shaman throws up three thin straight rods of the same length, tied in a shape resembling the letter П. Adjacent rods are connected by a short thread and can freely rotate relative to each other. The rods fall on the sand, forming a random figure. If a self-intersection occurs (the first and third rods cross each other), the tribe will face poor harvests and various troubles in the coming year. If there is no self-intersection, the year will be successful, bountiful, and happy. Find the probability that the rods will predict good fortune for the year 2017.
Solution. Let the broken line formed by the rods be $A B C D$. Let $\alpha=\angle A B C$ be the angle between the first

![](https://cdn.mathpix.com/cropped/2024_05_06_e03b48fad60e6548aec7g-10.jpg?height=683&width=694&top_left_y=1523&top_left_x=1246)
link and the second, and $\beta=\angle B C D$ be the angle between the second and the third (see Fig.1). We can assume that $0 \leq \alpha \leq \pi$ and then $0 \leq \beta \leq 2 \pi$. The elementary outcomes are pairs $(\alpha ; \beta)$. On the coordinate plane $\alpha \mathrm{O} \beta$, they fill a rectangle $G$ (see Fig.2), with the probability of a point $(\alpha ; \beta)$ falling inside a certain figure being proportional to the area of that figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_e03b48fad60e6548aec7g-11.jpg?height=520&width=506&top_left_y=248&top_left_x=501)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_e03b48fad60e6548aec7g-11.jpg?height=517&width=685&top_left_y=250&top_left_x=1028)

Fig. 2

The first and third links $A B$ and $C D$ intersect at some point $K$ only if the ray $C D$ lies inside the angle $B C A$, and the ray $B A$ lies inside the angle $C B D$, that is,

$$
0 \leq \angle B C D < \angle B C A, \quad 0 \leq \angle A B C < \angle C B D
$$

Considering that triangles $A B C$ and $C B D$ are isosceles, we get two symmetric conditions: $0 \leq \beta < \frac{\pi}{2} - \frac{\alpha}{2}$ and $0 \leq \alpha < \frac{\pi}{2} - \frac{\beta}{2}$.

On the coordinate plane $\alpha \mathrm{O} \beta$, these inequalities define a quadrilateral $F$ inside the rectangle $G$. The probability of self-intersection is $\mathrm{P}(F) = \frac{S_{F}}{S_{G}} = \frac{1}{12}$. Therefore, the desired probability is $1 - \frac{1}{12} = \frac{11}{12}$.

Answer: $\frac{11}{12}$.