A probability involving side lengths of a random triangle on a disk: Is it really $\frac37$?

Question

Choose three uniformly random points on a disk, and let them be the vertices of a triangle. Call the side lengths, in random order, $a,b,c$.

What is $P(a^2<bc)$ ?

A simulation with $10^7$ such random triangles yielded a proportion of $0.4285833\approx1.00003\times\frac{3}{7}$ satisfying $a^2<bc$, leading me to believe that the probability is $\frac37$.

I find this alleged probability interesting because it seems that the number $7$ rarely appears in the answers to natural geometry or probability questions.

Context

Recently I have learned that some probabilities related to circles, of the form $P(x^2<yz)$, have simple rational values (for example, a probability about a random triangle inscribed in a circle, and a probability about a random pentagram inscribed in a circle). So I wondered if there is a probability like this related to a disk. I may have found one.

My attempt

Let the boundary of the disk be $x^2+y^2=1$.

Using disk point picking, let the three points be

• $C\left(\sqrt{r_1}\cos\theta_1,\sqrt{r_1}\sin\theta_1\right)$
• $B\left(\sqrt{r_2}\cos\theta_2,\sqrt{r_2}\sin\theta_2\right)$
• $A\left(\sqrt{r_3},0\right)$

where each $r$ is a uniformly random real number from $0$ to $1$, and each $\theta$ is a uniformly random real number from $0$ to $2\pi$.

Let $a=BC$, $b=AC$, $c=AB$.

This gives:

$$P(a^2<bc)=P\left(r_1+r_2-2\sqrt{r_1r_2}\cos(\theta_1-\theta_2)<\sqrt{\left(r_1-2\sqrt{r_1r_3}\cos\theta_1+r_3\right)\left(r_2-2\sqrt{r_2r_3}\cos\theta_2+r_3\right)}\right)$$

I do not know how to set up an integral, nor any other way to calculate the probability. Apparently, after the dust settles, we should be left with $\frac37$.

Answer 1

Simulation with your sample-size of $n=10^7$ gave you an observed proportion $\hat p=0.4285833,$ so an approximate confidence interval for $p:=P(a^2<bc)$, with, say a $95\%$ "confidence level", is $$\hat p\pm 1.96\,\sqrt{\hat p(1-\hat p)\over n}=(0.4283, 0.4289)$$

Although $3/7=0.42857...$ does lie in this interval, the interval only esimates $p$ to about three digits of precision. Simulations using a larger sample-size suggest, with a very high degree of confidence/credibility, that $p\ne {3\over 7},$ the difference showing up in the fourth decimal place. Here's a picture of what I find with sample-size $10^{11}$:

The posterior distribution is $\text{Beta}(a,b)$, with $(a,b)=(n\hat p+{1\over 2}, n(1-\hat p)+{1\over 2}).$

The $100(1-\alpha)\%$ credibility interval is therefore $$(B_{\alpha\over 2},B_{1-{\alpha\over 2}})$$ where $B_q$ is the $q$-th quantile of the $\text{Beta}(a,b)$ distribution.

The $100(1-\alpha)\%$ confidence interval is $$\hat p\pm z_{\alpha\over 2}\,\sqrt{\hat p(1-\hat p)\over n}$$ where $z_q$ is the $q$-th quantile of the standard normal distribution.

With the sample-size this large the various approaches to confidence/credibility intervals for $p$ should all give practically the same results (assuming a non-informative prior distribution in the Bayesian methods); e.g., they agree to about six decimal places for a $99.9\%$ confidence/credibility interval, namely
$$0.428790 \pm 0.000005 = (0.428785, 0.428795).$$

(I found no OEIS entries for decimal expansions in that interval, so I can't guess at the exact value.)