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I find this easier to visualize with an equivalent way to pick points: drop four (n) lines through the center of the circle. For each line, randomly pick one of the two points that intersect the circle.

Independent of the lines dropped, there are eight (2n) ways to pick adjacent points and sixteen (n^2) different combinations of points.

There are four (n) adjacent points iif the points lie on the same half of the circle (proof by interactive visualization).

So the answer is eight sixteenths (2n/2^n).

Great argument. I found this generalization to higher dimension: https://www.mathpages.com/home/kmath327/kmath327.htm

Probably best to avoid the word ring in a mathematics discussion unless you're talking about the algebraic structure. It's very much a mathematical `keyword`.

> The same decomposition works in higher dimensions.

I don't think the same argument works in higher dimensions. On a circle, we can canonically pick a semicircle corresponding to each point (we have two choices, let's say we pick the clockwise one).

In higher dimensions there's no canonical choice of half-sphere. In odd dimensions one could pick a canonical half-sphere per point but it might turn out that some other non-chosen half-sphere for that point contains all the other points. In even dimensions there isn't even a way to canonically pick a half-sphere for each point (this is a consequence of the Hairy Ball Theorem).

(For all I know the actual numbers might turn out to be the same, I don't know. I'm just saying that the argument doesn't work.)

Well, it's easy to pick a half-n-sphere corresponding to a point on the surface. You just take the half-sphere centered at that point. For our two-dimensional circle, it would be the arc defined by the diameter that is perpendicular to the radius running between "the point" and the center of the circle.

At that point you've lost the ability to say that only one such half-sphere defined by a dropped point can be a valid solution, and you've also lost the ability to say that if a valid solution exists then there must be a valid solution defined by one of the points you want to include in the half-sphere, but you can define a canonical half-sphere for any point.

I was uncomfortable with the idea of picking "random points on a circle" to begin with, because of https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) , but the article doesn't even address whether the concept is well-defined. We can always choose a point on the perimeter deterministically from any chord (...that isn't a diameter), so the ill-definedness of the problem of choosing a random chord seems like it would infect the problem of choosing a random point on the perimeter.

Right, I was taking it as given that the problem of choosing a hemisphere canonically for a point meant "such that the argument works in the same way as for the circle".

Bertrand paradox just doesn't apply here, there's a natural measure on the circle and all higher dimensional spheres. I wouldn't expect an article on this subject to need to make that clarification unless it's dealing with chords or some other situation without a natural measure.

If I choose my four points as the endpoints of two chords chosen by the "random radial point" method described on the wikipedia page, is it still true that the odds of all four being covered by a semicircle are 50%?

I think about this by unwrapping the circle to form a straight line. Then you draw an imaginary point in the middle of the line. Then what are the chances they will all fall on one side of the line or the other? 1/2 because it's divided into two equal lengths.

This approach implies the probability doesn’t depend on N. It only happens to be 1/2 for N=4 (the article goes into this). The trick is that you don’t know beforehand which semicircle all the points can land in, but your unwrapping step assumes you do.

There are two major problems with your idea:

1. Your answer can be "no" when the true answer is "yes". Consider this process with a circle of perimeter "21":

    ---------------------      (unwrap the circle)
    ----------+----------      (bisect the line)
    -**-------+--------**      (drop four points)

2. Your answer of "1/2, because it's divided into two equal lengths" is completely wrong for the scenario that you specify.

Consider the case where we drop a single point. We can do the same procedure:

A. Unwrap the circle;

B. Bisect the line;

C. Drop one point.

But even though the line is still divided into two equal lengths, our one point has a 100% chance of falling either on one side of the bisection point, or on the other side.

For the case where we drop four points, the article already gives the correct answer for your method, which is 1/2^3 (because there are 3+1 points).

Interesting and fun