Fiddler

The July 31, 2026 problems posed on the Fiddler on the Proof blog go as follows.

From Randi Goldman comes a riddle of rings:

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.

Figure 1
Figure 1

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

Solution to Fiddler

Let’s look at a simulation first.

ringSim <- function(N){
  # Create a 3xN matrix of random uniform RVs
  randRings <- matrix(runif(N*3), nrow=N, ncol=3)
  # A function to be used on each row of the matrix
  # Returns TRUE if all are within 1/8 (using wrapping)
  lineFit <- function(U){
    u1 <- U[1]
    u2 <- U[2]
    u3 <- U[3]
    # The 1 - abs(u1-u2) below deals with wrapping
    d1 <- min(abs(u1-u2),1-abs(u1-u2))
    d2 <- min(abs(u1-u3),1-abs(u1-u3))
    d3 <- min(abs(u2-u3),1-abs(u2-u3))
    
    if(max(d1,d2,d3)<1/8){
      return(TRUE)
    }
    else{
      return(FALSE)
    }
  }
  ## What proportion are TRUE? 
  return(sum(apply(randRings, 1, lineFit))/N)
}

set.seed(42)
ringSim(100000)
## [1] 0.04564
ringSim(100000)
## [1] 0.04796
ringSim(100000)
## [1] 0.04702

The simulation suggests a probability near \(0.047\).

Let the first random ring be selected. Orient it and standardize it so that the opening starts at 0 and the opening is on \([0,1/8]\). Let this be \(I_1\).

The start of \(I_2\) must be anywhere on \([-1/8,1/8]\) in order to intersect \(I_1\). Let that start point be \(s\geq 0\), then \(I_1\cap I_2 = [s,1/8]\), which has length \(1/8-s\). If \(s<0\), then \(I_1\cap I_2 = [0, 1/8+s]\), which has length \(1/8+s\). Since \(s<0\) in the second equation, both lengths can be written \(1/8-|s|\).

The starting point for the third interval must be anywhere on \([-1/8,1/8-|s|]\). This has length \(1/4-|s|\), which represents the conditional probability the third interval hits its starting point correctly. This is conditional on \(s\), which we said above must be on \([-1/8, 1/8]\).

Let’s use symmetry, and compute \[ 2\int_0^{\frac18} \left(\frac14 - s\right) ds \]

This is \[ \begin{aligned} 2*\left[\frac{s}{4}-\frac{s^2}{2}\right]_0^{\frac18} &= 2\left[\frac{1}{32}-\frac{1}{128}\right] \\ &= 2*\left(\frac{3}{128}\right) \\ &= \frac{3}{64} \\ &= 0.046875 \end{aligned} \]

This checks out with the simulation.

Fiddler Extra Credit

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.

Figure 2
Figure 2

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

Solution to Fiddler Extra Credit.

Let’s assume a parametrization of the helix as \(H(t) = (\cos(t), \sin(t), bt)\) as given on the Wikipedia page that was supplied by the problem. For simplicity, we’ll assume that the total path of the helix goes from \(t=0\) to \(t=1/b\).

This means that the angles at the three points up the cylinder (or down) are \(\theta_1=\frac{1}{4b}\), \(\theta_2 = \frac{1}{2b}\), and \(\theta_3 = \frac{3}{4b}\) since they are evenly spaced vertically.

Notice the relationship here of the three angles, and how \(3\theta_1 = \theta_3\), \(2\theta_1 = \theta_2\), or \(\theta_1-2\theta_2+\theta_3 = 0 \mod 2\pi\). These equations, and especially the last, will be helpful. The last is using \(\mod 2\pi\).

Suppose the angle defining the center of openings of the three rings is \(\phi_1\), \(\phi_2\), and \(\phi_3\). For a solution to exist, it must be the case that \(\theta_i \in \left[\phi_i - \frac{\pi}{8}, \phi_i + \frac{\pi}{8}\right]\) for each \(i = 1,2,3\). That is, \(\theta_i = \phi_i + \varepsilon_i\) where \(\varepsilon_i \in \left[-\frac{\pi}{8}, \frac{\pi}{8}\right]\).

For any random three angles \(\phi_1, \phi_2, \phi_3\) on \([-\pi, \pi]\), the expression \(\phi_1-2\phi_2+\phi_3\) can also be anything on that interval (when you mod out by \(2\pi\) and convert those angles on \((\pi, 2\pi]\) to \((-\pi, 0]\)).

However, for a solution to exists, note that \[ \begin{aligned} \phi_1-2\phi_2+\phi_3 &= \theta_1-\varepsilon_1-2\theta_2+2\varepsilon_2+\theta_3-\varepsilon_3 \\ &= -\varepsilon_1+2\varepsilon_2-\varepsilon_3 \end{aligned} \]

must be contained in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).

That is half the interval. So, the probability a solution exists is 1/2.

GIF 1
GIF 1