Why are we not multiplying 1/n conditional probability of selecting disjoint points in the question - "prob. of N points within a semi-circle"

Question

This is my first question on stackexchange - so apologies in advance if I haven't been able to follow the best practices while asking this question.

I am trying to understand the solution to question "Probability of n points lying within a semi-circle". The question has been asked and answered here - Probability that n points on a circle are in one semicircle.

I am struggling with a statement in this answer https://math.stackexchange.com/a/325168/927405. I understand that the n points are disjoint in a way that only one of the n points can have all the points within a semi-cicrle in particular angular direction. My question is - why are we not multiplying (1/n) for the (conditional) probability of selecting the right point?

I am drawing a parallel that while calculating probability of different outcomes in 2 coin tosses - we say that if I get heads on first toss, prob. of getting tails in 2nd toss is 1/2. Likewise we can get a TH. HT and TH are both disjoint but while calculating final answer - we don't say that if I choose heads first, then the prob of tails in second toss is 1/2 and we can also have a TH and since both are disjoint - let's add them up 1/2 + 1/2 and prob. of different outcomes would be 1 (which we know is incorrect answer because we need to multiply the probability of getting first heads / tails too!)

Similarly, why are we not multiplying 1/n for the probability of selecting the right starting point in this "N point in a semicircle question"? (which will lead to the answer $\frac1{2^{n-1}}$)

Thank you so much!

Answer 1

Compilation of comments:

First see that the events are mutually exclusive (at least when $n >2$). So the probability any one of them occurs is the the sum of the probabilities each of them occurs, not the average.

The probability say that point $2$ is the leading point of a semicircle is $\frac1{2^{n-1}}$ not $\frac1{n2^{n-1}}$, and due to symmetry/exchangeability the same is true for each of the other points, making the final answer $\frac n{2^{n-1}}$.

If you throw a $6$-sided die, the probability of it being even is the probability of it being $2$ plus the probability of it being $4$ plus the probability of it being $6$, i.e. the sum of probabilities of mutually exclusive events.

With the circle points, you have the probability point $1$ leads a semicircle plus the probability point $2$ leads a semicircle plus ... . The probability $\frac1{2^{n-1}}$ that point $2$ leads a semicircle of all the other points already takes into account the probability it is the correct point.