Can you or “Cantor” you find the distance?

Question: I start with a number line from 0 to 1, and then I remove the middle third. Then I take each of my remaining pieces (the segment from 0 to 1/3 and the segment from 2/3 to 1) and remove their middle thirds. Now I have four segments (from 0 to 1/9, 2/9 to 1/3, 2/3 to 7/9, and 8/9 to 1) and remove their middle thirds. I do this over and over again, infinitely many times.

The points I’m left with are collectively known as the Cantor set. The Cantor set is not empty; in fact, it contains infinitely many points on the number line, such as 0, 1, 1/3, and even 1/4.

It’s possible to pick a random point in the Cantor set in the following way: Start with the entire number line from 0 to 1. Then, every time you remove a middle third, you give yourself a 50 percent chance of being on the left remaining third and a 50 percent chance of being on the right remaining third. Then, when you remove the middle third of that segment, you again give yourself a 50 percent chance of being on the left vs. the right, and so on.

Suppose I independently pick two random points in the Cantor set. On average, how far apart can I expect them to be?

Extra credit: Suppose I independently pick three random points in the Cantor set. Each point has some value between 0 and 1.

What is the probability that these three values can be the side lengths of a triangle?

(Fiddler on the Proof)

Solution

Let the two points start at the middle of the intact unit line and find their positions by repeatedly picking, at random, the midpoint of the left or right third of the line they’re currently on. The sum of all these displacements for each point is the selected Cantor point.

The points are together to start, and can either stay together or move apart. If they stay together then they gain no relative displacement and restart the process at the center of a line one third the length of the original. If they move apart, then they pick up relative distance $\frac23$ and start the next step separated from each other on the two sublines of length $\frac13.$

Once apart, they never come back together. Also, starting a next step apart gives the points new options. The step can produce no relative movement (they move in the same direction) with probability $\frac12,$ move towards each other with probability $\frac14,$ or move apart with probability $\frac14.$ Because the movements are symmetric, the expected additional separation is zero!

So, we just have the equation describing the first step

We can use the scaling property $d_\text{together}(\gamma d) = \gamma \cdot d_\text{together}(d)$ and the fact that $d_\text{apart} = 0$ to close the recursion and solve for $d_\text{together}(1)$:

Extra credit

Three lengths form a valid triangle if each pair of sides satisfy the triangle inequality with the third side, i.e.

Conversely, a triangle will not obtain if one of these is broken like $a > b+c.$ At the very most, one of a trio of numbers can be bigger than the sum of the other two so, if we find the probability that one of the numbers is too big, $P_\text{too big},$ the probability that a triangle forms will be $1-3P_\text{too big}.$

When we pick Cantor numbers, the first division decides whether the number will be on the left or the right side of the first cut. $\text{L}$ numbers will be between $0$ and $\frac13$ and $R$ numbers will be between $\frac23$ and $1.$ Because each strip is a scaled down copy of the cantor set, we can relate the original problem to its cases.

When we make the first decision for the three numbers, $\{a,b,c\},$ we can get one of eight equally likely different outcomes,

• If we get $\text{RRR}$ then all three numbers will be greater than $\frac23$ and less then $1$ which means a triangle is guaranteed, and there is no probability of failure.
• If we get $\text{LLL}$ then we have the same problem again, playing out at $\frac13$ scale. The probability that number $a$ is too big is once again $P(\text{too big}).$
• If we get any of the other scenarios where $a$ is an $\text{L}$ number, the probability $a$ is too long is zero since either of $b$ or $c$ is an $\text{R}$ number and therefore greater than an $\text{L}$ number.
• If we get $\text{RRL}$ or $\text{RLR}$ then $a$ is at least $\frac23$ as is one of the other numbers, and the third number is between $0$ and $\frac13.$ Writing $a$ as $\left(\frac23 + x\right),$ $b$ as $\left(\frac23 + y\right),$ and $c$ as $z,$ the condition for $a$ being too big becomes $\frac23 + x > \frac23 + y+z,$ or $x > y+z.$ Because $x,$ $y,$ and $z$ are scaled down Cantor numbers, this is just the original event and also has probability $P_\text{too big}.$
• The case $\text{RLL}$ surely fails since $b$ and $c$ are at most $\frac13$ and $a \geq \frac23.$

Putting it all together, the probability that number $a$ is too big is $\frac18\left[1 + 3P_\text{too big} \right].$

Solving this, we get $P_\text{too big}=\frac15$ so the probability to form a triangle is ${1-3P_\text{too big} = \frac25.}$

Sadly, I can’t see a clear reason why these two answers ought to coincide.