Fiddler: The Helical Ball?

Instead of requiring a vertical line down the cylinder’s surface, now any helix down the surface is allowed. An example of such a helix passing through all three gaps is shown below.

What is the probability that there exists at least one such helix that can pass through each ring’s gap?

Again, let's assume that we have cylinder with radius one, that the broken rings are embedded within the planes $z=2,$ $z=1$ and $z=0$ and finally that the gap on the ring at $z=2$ is from $\theta \in [0, \pi / 4],$ while the gaps at $z=1$ and $z=0$ are from $\theta \in [\alpha, \alpha+\pi/4]$ and $\theta \in [\beta, \beta + \pi / 4]$ for some $\alpha, \beta \sim U(0,2\pi),$ respectively. Here again we will play a little fast and loose with notation and say that $[\psi_1, \psi_2] \subseteq [0,2\pi]$ should mean the set $[0, \psi_2 - 2\pi] \cup [\psi_1, 2\pi]$ in the case that $\psi_1 \lt 2\pi \lt \psi_2.$

We also see that the helical path from any point $(2, \theta_2)$ to $(1, \theta_1)$ for any $\theta_1, \theta_2 \in [0,2\pi)$ is given by $z(t) = 2-t,$ $\theta(t) = \theta_2 + t(\theta_1 - \theta_2).$ All we need to do is understand where this helical path will hit the plane $z = 0,$ in this case at $\theta(2) = 2 \theta_1 - \theta_2.$ Since any helical path that gets through the upper and middle gaps will have $\theta_2 \in [0, \pi/4]$ and $\theta_1 \in [\alpha, \alpha + \pi/4],$ we see that $\theta_0 = \theta(2) = 2 \theta_1 - \theta_2 \in [2 \alpha - \pi/4, 2 \alpha + \pi/2].$ See the figure below, for instance for the choice of $\alpha = 4.613846199712232.$

So in order to fit through all of the gaps, then we would need to have $$[\beta, \beta + \pi/4] \cap [2 \alpha - \pi/4, 2 \alpha + \pi/2 ] \ne \emptyset,$$ that is, we would need to have $\beta \in [ 2 \alpha - \pi/2, 2\alpha + \pi/2].$ Therefore, the conditional probability of there being a helical path conditional on choice of $\alpha$ is $$p(\alpha) = \mathbb{P} \{ \beta \in [ 2 \alpha - \pi/2, 2 \alpha + \pi/2 ] \mid \alpha \} = \frac{\pi}{2\pi} \equiv \frac{1}{2}.$$ Therefore, the probability of there being at least one such helix that can pass through each of the gaps is $p = \int_0^{2\pi} p(\alpha) \,d\alpha = \frac{1}{2}.$

Ring around the cylinder

A long vertical cylinder has three narrow open rings, each of which wraps around seven-eighths of the cylinder (leaving a one-eighth “gap”). The rings are evenly spaced vertically, but are otherwise randomly rotated about the cylinder’s central axis. For some orientations of the rings, there exists at least one vertical line down the cylinder’s surface that passes through each ring’s gap, as illustrated below.

What is the probability that at least one such vertical line exists?

Let's define some terms. Let's assume we have a unit cylinder and that the broken rings are embedded within the planes $z=2,$ $z=1$ and $z=0.$ Let's, without loss of generality, assume that the opening at the topmost ring is one-eighth of the way around and oriented such that it is from $\theta \in [0, \pi/4].$ Let's further assume that for $\alpha, \beta \sim U(0,2\pi)$ that the middle and bottom rings at z=1 and z=0 are given by $\theta \in [\alpha, \alpha+\pi/4]$ and $\theta \in [\beta, \beta+\pi/4],$ respectively. Here we shall abuse notation slightly and assume that if $\psi_1 \lt 2\pi \lt \psi_2$ then by $[\psi_1, \psi_2] \subseteq [0,2\pi]$ we mean $[0, \psi_2 - 2\pi] \cup [ \psi_2, 2\pi].$

Any vertical line on this cylinder can be written in cylindrical coordinates as $\theta = \theta_0$ for some $\theta_0 \in [0,2\pi).$ If a vertical line can pass through the top and middle gaps, then we must have $[0, \pi/4] \cap [\alpha, \alpha+\pi/4] \ne \emptyset,$ or equivalently, $\alpha \leq \pi/4$ and $\alpha + \pi/4 \geq 0,$ that is $\alpha \in [-\pi/4, \pi/4].$ In order for this line to pass through all three gaps, we must have $$[ \beta, \beta + \pi/4 ] \cap [ \alpha, \alpha + \pi/4 ] \cap [0, \pi/4] \ne \emptyset,$$ or equivalently $\beta \in [ \max \{ 0, \alpha \} - \pi/4, \min \{0, \alpha\} + \pi/4 ].$

Therefore, the probability that a vertical line can pass through all gaps is \begin{align*}p &= \int_{-\pi/4}^{\pi/4} \int_{\max \{0, \alpha\} - \pi/4}^{\min \{0, \alpha\} + \pi/4} \frac{d\beta}{2\pi} \, \frac{d\alpha}{2\pi} \\ &= \int_{-\pi/4}^{\pi/4} \left( \frac{\pi}{2} - |\alpha| \right) \,\frac{d\alpha}{4\pi^2}\\ &= \frac{1}{2\pi^2} \int_0^{\pi/4} \left( \frac{\pi}{2} - \alpha \right), d\alpha\\ &= \frac{1}{2\pi^2} \int_{\pi/4}^{\pi/2} u \,du = \frac{1}{4\pi^2} \left[ \left(\frac{\pi}{2}\right)^2 - \left(\frac{\pi}{4}\right)^2 \right] \\ &= \frac{1}{4\pi^2} \frac{3\pi^2}{16} = \frac{3}{64} = 4.6875\%\end{align*}