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Given $N$ points drawn randomly on the circumference of a circle, what is the probability that they are all within a semicircle?

I'm confused. Let's choose an arbitrary point, and consider the semicircle to the right of that point. The probability $N-1$ points are in that area is $\frac{1}{2^{N-1}}$. We need to do this for an infinite number of starting points, since we chose an arbitrary point as the starting point to begin with, so we need to multiply the probability of choosing a random point by $\frac{1}{2^{N-1}}$ and sum over all possible starting points. I'm pretty sure this works but I'd never give this kinda solution in an interview.

My friend suggested I start off with $N$ random points and use each of the $N$ points as starting points. And so the answer is $\frac{N}{2^{N-1}}$. But to me this doesn't make sense, because we're not considering points that are not those $N$ starting points. He said we can "zoom in" on a particular configuration, but I don't get why.

No one is dealing with the content of my post; I obviously read the duplicates and it didn't help, as I make clear above.

beginner
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    Hypothetically, if points 1 and 2 form the diameter of a circle, then point 3 will determine the pertinent semicircle. Then the probability for each of the remaining points being in the semicircle is (1/2), so the overall probability is $~\displaystyle \frac{1}{2^{n-3}}. ~$ The difficulty is that there is no way of determining how close point 2 (or any other point) will approach forming a diameter, with point 1. So, my analysis, although a reasonable first step, does not really effectively attack the problem. – user2661923 Jul 15 '23 at 17:01
  • @Klaus No, it doesn't address why we can just examine $N$ starting locations. – beginner Jul 15 '23 at 18:00
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    The $N$ events of no other points in the semicircle to the left of the selected point are mutually exclusive so their probabilities can be added, and since their probabilities are each $\frac1{2^{N-1}}$ (as you say) then this can be multiplied by $N$. – Henry Jul 15 '23 at 21:07

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