On a circle, choose $6n$ $(n\in\mathbb{Z^+})$ uniformly random points and label them $a_0,a_1,a_2,\dots,a_{6n-1}$ going anticlockwise, with $a_0$ chosen randomly.
Draw three chords:
- Chord $a_0 a_{3n}$
- Chord $a_n a_{4n}$
- Chord $a_{2n} a_{5n}$
Here is an example with $n=5$. The rightmost point is $a_0$. The centre of the circle is shown.
Let $P(n)=$ probability that the triangle formed by the three chords contains the centre of the circle.
What is $\lim\limits_{n\to\infty}P(n)$ ?
Intuition
When I came up with this question, I had no intuition as to whether $P(n)$ should increase or decrease, as $n$ increases.
Intuitively, as $n$ increases, the expected area of the triangle should decrease, which tends to make $P(n)$ decrease. But at the same time, the triangle should "get closer" to the circle's centre (e.g. the expectation of the distance between the triangle's centroid and the circle's centre, should decrease), which tends to make $P(n$) increase. It was not clear to me which of these opposing factors should dominate.
Simulations
Simulations, with $5\times10^6$ sets of three chords for each value of $n$, yielded the following estimated probabilities:
$P(1)\approx 0.0625\space$ (I prove that $P(1)=\frac{1}{16}$ below.)
$P(2)\approx 0.0741\overset{?}{=}\frac{2}{27}$
$P(3)\approx 0.0786$
$P(4)\approx 0.0809$
$P(5)\approx 0.0824$
$P(6)\approx 0.0832$
$P(10)\approx 0.0849$
I also got $P(100)\approx 0.0872$ based on a simulation with $10^6$ sets of three chords. I tried to get an approximation for $P(1000)$, but my computer started overheating (I'm using Excel for my simulations).
Here is a plot of estimated $P(n)$ against $n$.
So as $n$ increases, it seems that $P(n)$ is approaching some number strictly between $0$ and $1$.
Due to the naturalness of my question's geometric construction, I suspect that $\lim\limits_{n\to\infty}P(n)$ has a closed form, and it should be an interesting number worth finding. I also suspect that $P(n)$ is a rational sequence.
My attempt
I have only been able to prove that $P(1)=\frac{1}{16}$. Here is my proof.
Suppose that:
- When we choose the six random points on the circle, instead of having a continuous distribution of points to choose from, we have $2k$ evenly spaced points to choose from, where $k$ is a large integer.
- The circumference of the circle is $2k$.
- A point can be chosen more than once.
Let $x=$ distance from $a_0$ to $a_3$ along the circle going anticlockwise.
Let $y=$ distance from $a_2$ to $a_5$ along the circle going anticlockwise.
Let $z=$ distance from $a_1$ to $a_4$ along the circle going anticlockwise.
If the triangle contains the centre of the circle, then either
- $x\le k$ and $y\le k$ and $z\ge k$, or
- $x\ge k$ and $y\ge k$ and $z\le k$.
These two configurations are shown respectively below, where $a_0$ is the rightmost point in each diagram.
By symmetry, each of these happens with equal probability, so we have:
$$P(1)=2\times P(x\le k \land y\le k\land z\ge k)$$
To see these distances more clearly, imagine cutting the circle at $a_0$ and straightening the circle into a line segment, so that the points from left to right are $a_0, a_1, a_2, a_3, a_4, a_5$.
To understand the next part, it may be easier to first work with specific numbers, so let's suppose $k=100$, and suppose $\color{red}{x=10}$. Then $90\le y \le 100$.
- If $y=90$, there is only $\color{blue}{1}$ possible combination of locations for $a_1$ and $a_4$ (because we require $z\ge 100$).
- If $y=91$ then there are $(1)+(1+2)=\color{blue}{4}$ possible combinations of locations for $a_1$ and $a_4$.
- If $y=92$ then there are $(1)+(1+2)+(1+2+3)=\color{blue}{10}$ possible combinations of locations for $a_1$ and $a_4$.
- $\cdots$
- If $y=100$ then there are $(1)+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+11)=\color{blue}{286}$ possible combinations for $a_1$ and $a_4$.
So for $\color{red}{x=10}$, the total number of possible combinations for $a_1$ and $a_4$ is $\color{blue}{1+4+10+\dots+286}=1001$.
The numbers $1,4,10,\dots,286$ are the tetrahedral numbers. The sum of the first $x+1$ tetrahedral numbers is $\binom{x+4}{4}$. So the total number of combinations of $a_1, a_2, a_3, a_4, a_5$ that satisfy $(x\le k \land y\le k\land z\ge k)$ is $\sum\limits_{x=0}^{k}\binom{x+4}{4}$.
So the probability that $(x\le k \land y\le k\land z\ge k)$ is $\sum\limits_{x=0}^{k}\binom{x+4}{4}$ divided by the total number of ways to choose $a_1, a_2, a_3, a_4, a_5$, which is $\binom{2k}{5}$.
To change from a discrete back to a continuous distribution of available points along the circle, we take the limit as $k\to\infty$.
Remembering that $P(1)=2\times P(x\le k \land y\le k\land z\ge k)$, we have:
$\begin{align} P(1)&=2\lim\limits_{k\to\infty}\frac{\sum\limits_{x=0}^k\binom{x+4}{4}}{\binom{2k}{5}}\\ &=2\lim\limits_{k\to\infty}\frac{\sum\limits_{x=0}^k\frac{x^4}{4!}}{\frac{2^5}{5!}k^5}\\ &=2\lim\limits_{k\to\infty}\frac{\frac{1}{4!}\cdot\frac{1}{5}k^5}{\frac{2^5}{5!}k^5}\\ &=\frac{1}{16} \end{align}$