Expected Area Problem

We can find the area of the quadrilateral by considering the quadrilateral as two triangles stuck together. If the angle between two toothpicks is $\theta$ , the triangles will have base 1, and heights $a$ and $b$ where $a+b=\sin(\theta)$ . The area will be $\frac a2 + \frac b2 =\frac{\sin(\theta)}{2}$ . We need to notice that if two randomly dropped toothpicks happen to intersect, it is more likely that the angle between them will be near perpendicular, than near parallel. Taking this into account, the expected area will be $\int_0^{\frac{\pi}{2}} \frac{\sin(\theta)}{2} p(\theta)$ where $p(\theta)$ is the proabability that two randomly dropped intersecting toothpicks, intersect at $\theta$ . WLOG, we can say that the first toothpick dropped is vertical. The probability that a second randomly dropped toothpick will intersect the first, will be proportional to the horizontal span of the second toothpick (which equals $\sin(\theta)$ ). Therefore, the probability two randomly dropped intersecting toothpicks, intersect at $\theta$ is $\frac{\sin(\theta) d\theta}{\int_0^{\frac{\pi}{2}} \sin(\theta) d\theta} = \sin(\theta) d\theta$ Therefore the expected area of the quadrilateral will be $\frac 12 \int_0^{\frac{\pi}{2}} \sin^2(\theta) d\theta = \frac{\pi}{8} \approx 0.3927$ The answer that Brilliant wants is not correct. That answer comes from assuming that $p(\theta)$ is constant, which is not true. If we randomly drop two toothpicks onto the plane and they happen to intersect, it will be more likely that they are near perpendicular than near parallel. We can't assume that all values of $\theta$ are equally likely. <edit: This problem has been updated to accept the correct answer>

Moderator note : The correct answer has since been updated.

I have reposted Pedro Cardoso's solution here, copied from his comment to Albert Lau's solution

Given that $\theta$ is the angle between the sticks, and that they intersect, the area of the quadrilateral formed is $\frac{sin(\theta)}{2}$ .

Now, to find the probability that, given that the toothpicks intersect, the intersection angle is theta, consider the toothpicks as two line segments, $AB$ and $CD$ . say $AB$ is fixed on the plane at, say, $A=(0,0)$ and $B=(0,1)$ , and call $M$ the intersection between $AB$ and $CD$ , and make $\theta=AMD$ . Then, imagine moving the point $C$ arround, moving $D$ in the same direction fix $\theta$ constant: What is the region $C$ can be at so that $\theta$ is fixed and $CD$ still intersects $AB$ ? That region is a rhombus, of its angles is $\theta$ , and its sides all measure $1$ . Thus, it's area is $sin(\theta)$ .

Finally, the probability that, given that the toothpicks intersect, the angle between them is $\theta$ is directly proportional to the area of that region, $sin(\theta)$ (To see this intuitively, consider that there are a lot more ways to make the toothpick intersect when the angle between them is, say $46º$ compared to when its $5º$ , because, in the former, there are a lot more places $C$ can be in that keep the toothpicks intersecting). So the probability is the weighted average of $\frac{sin(\theta)}{2}$ , where each area is weighted by $sin(\theta)$ , namely, the probability that that area occurs (or at least, something directly proportional to this probability).

This gives us $\large \frac{\int_0^{\pi}\left(\frac{\sin x}{2}\right)\sin xdx}{\int_0^{\pi}\sin xdx}=0.3926...$

Mendrin: (Some) Monte Carlo simulations support this figure of 0.3926.

If the toothpicks intersect, their position dont matter, only their length and orientation. Proof :

$\begin{aligned} \text{let } \theta \text{ be the acute angle between } AC, BD \\ S_{ABCD} &= S_{\Delta AEB} + S_{\Delta BEC} + S_{\Delta CED} + S_{\Delta DEA} \\ &= \frac{1}{2} \ AE \cdot BE \ \sin \theta + \frac{1}{2} \ BE \cdot CE \ \sin (\pi - \theta) + \frac{1}{2} \ CE \cdot DE \ \sin \theta + \frac{1}{2} \ DE \cdot AE \ \sin (\pi - \theta) \\ &= \frac{1}{2} \ AE \cdot BE \ \sin \theta + \frac{1}{2} \ BE \cdot CE \ \sin \theta + \frac{1}{2} \ CE \cdot DE \ \sin \theta + \frac{1}{2} \ DE \cdot AE \ \sin \theta \quad (\sin \theta \equiv \sin (\pi - \theta))\\ &= \frac{1}{2} \ (AE \cdot BE + BE \cdot CE + CE \cdot DE + DE \cdot AE) \ \sin \theta \\ &= \frac{1}{2} \ (AE + CE) \cdot (BE + DE) \ \sin \theta \\ &= \frac{1}{2} \ AC \cdot BD \ \sin \theta \\ \end{aligned}$

Back to the problem, $AC = BD = 1, \ \text{area } = \frac{1}{2} \sin \theta$

<edit: the solution hereafter is incorrect, please check other solution>

Consider the average value of $\frac{1}{2} \sin \theta , \ 0 \leq \theta \leq \frac{\pi}{2}$

$\begin{aligned} \text{expected area } &= \int_{0}^{\pi/2} \frac{1}{2} \ \sin \theta \ d\theta \ \div \frac{\pi}{2} \\ &= \frac{1}{\pi} \\ &= 0.3183 \end{aligned}$