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

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!

• +1 : Very nicely presented question: good work shown, good reference to pre-existing mathSE posts, nice explicit questions concerning these posts. Commented May 15, 2021 at 8:37
• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Commented May 15, 2021 at 8:38
• 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. And 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 $\frac n{2^{n-1}}$ Commented May 15, 2021 at 9:28
• @Henry - thank you for the interest. I understand the events are mutually exclusive. Also, I agree that once we have picked a leading point then probability that remaining points lie within a semi-circle is $\frac1{2^{n-1}}$. But once a leading point has all remaining points in a semi-circle - no other point can satisfy this condition i.e. prob of remaining points being in a semi-circle from any other leading point is zero (that is why they are mutually excl). So why are we summing the probabilities of other points? I have tried to draw this parallel in a more familiar coin problem in my ques. Commented May 15, 2021 at 9:59
• (cont'd) ... maybe another way to put it is - the probability that any point 𝑖 is the the correct / desired leading point is 1/n. Therefore, my logic is  P(n points in a semi circle) = P(point 2 to n are in a semi-circle | point 1 is correct leading point) - add this up for all other points  So while we are adding these mutually exclusive events (that only one of them can be a correct starting point) but we also need to multiply the probability that point 𝑖 is the correct leading point Commented May 15, 2021 at 10:20

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.