The probability that $3$ random points on the circumference form a right-angled triangle?

Question

In my probability theory course, I dealt with a similar problem which asks for the probability that $3$ random points on the circumference of a circle lie on the same semi-circle. But it makes me think of the following:

Given a circle, we randomly mark $3$ distinct points on the circumference and name them $X_1, X_2$ and $X_3$ respectively. What is the probability that $\Delta X_1X_2X_3$ is a right-angled triangle?

Here is my work. Please tell me if there is any problem in my solution. Feel free to leave me your solution so that I can learn from you.

Solution:

Let $E$ be the event that "$\Delta ABC$ is a right-angled triangle" and let $E_i$, $i=1,2,3$, be the event that "Starting from $X_i$, traversing the circle counterclockwise, one can visit all the points by turning exactly $180$ degrees".

I then claim that $$P(E) = P(E_1)+P(E_2)+P(E_3).$$ We then consider $P(E_i)$ by conditioning on the position of $X_i$, $$P(E_i) = \int_0^{2\pi} P(E_i | X_i = x) \frac{1}{2\pi} dx.$$ Furthermore, we consider $P(E_i | X_i = x)$. For a fixed $X_i$, if we can visit all three points in exactly $180$ degrees, one of the remaining points, say $X_j$, should be exactly 'opposite' to $X_i$, which forms a diameter with $X_i$. Thus there is only one possible position of $X_j$. For the last point, we note that $X_iX_j$ divides the whole circle into two equal semi-circles. For $E_i$ to occur, we must have the last point lying in one of the semi-cirlces. Thus, I claim that $$P(E_i | X_i = x)=\frac{1}{2\pi} \frac{1}{2}$$ which quickly implies $$P(E)=\frac3{4\pi}.$$

It seems strange to me. Thanks in advance for leaving your ideas!

Answer 1

The probability is $0$, assuming that each point is chosen with uniform distribution on the circle.

For a right-angled triangle you need two of the points to be the two endpoints of a diameter. The probability for that is obviously $0$, since there isn't any range of possibilities, and the distribution is continuous.

For anyone who is not satisfied so far, below is a more rigorous argument.

Consider the following question, which is pretty much the same, only simpler to answer in this format: When choosing two points on a circle, where the two choices are independent, and each point has a uniform distribution, what is the probability that the chosen points are the two endpoints of a diameter?

Having chosen one of the points, we need the other one to be exactly the first one's antipodal. As we know, uniform distribution yields that the probability that a point is between $\alpha$ and $\beta$ is just $$P(\alpha\leq p\leq\beta)=\frac{1}{2\pi}(\beta-\alpha).$$As $\alpha$ and $\beta$ get closer to one another, the probability gets smaller, and eventually, if $\alpha=\beta$, the probability vanishes.