Choose n points randomly from a circle, how to calculate the probability that all the points are in one semicircle? Any hint is appreciated.

2Have I taken too much of a simplistic view on the problem by thinking the probability is $\left\(\dfrac{1}{2}\right\)^n$? – Noble. Mar 09 '13 at 01:10

3@Noble: Yes, you have  that's the probability that the points are all in one particular semicircle. – joriki Mar 09 '13 at 01:10

1@joriki I thought as much, it's a much more interesting problem then! – Noble. Mar 09 '13 at 01:11

1Hint: Start with a point randomly on the circle and draw a diameter from that point. All you got to do now is ensure that rest of the $n1$ points lie on the same side of the diameter (i.e., on a semicircle). You can place the $n1$ points using a coin toss. – jaysun Mar 09 '13 at 01:12

2@jaysun I dont think this is the correct way, three points could still be in one semicircle even if the last two are on two different sides of the diameter joining the first point and the center. – NECing Mar 09 '13 at 01:14

This is a special case of a theorem of Wendel. This probability will be the same if we only assume the distribution to be centrally symmetric. see e.g. http://mathoverflow.net/questions/33112/estimateprobability0isintheconvexhullofnrandompoints/33132#33132 – Gilles Bonnet May 10 '14 at 12:23

fyi this is a classic quant interview question – JobHunter69 Sep 12 '20 at 18:25
5 Answers
A variation on @joriki's answer (and edited with help from @joriki):
Suppose that point $i$ has angle $0$ (angle is arbitrary in this problem)  essentially this is the event that point $i$ is the "first" or "leading" point in the semicircle. Then we want the event that all of the points are in the same semicircle  i.e., that the remaining points end up all in the upper halfplane.
That's a coinflip for each remaining point, so you end up with $1/2^{n1}$. There's $n$ points, and the event that any point $i$ is the "leading" point is disjoint from the event that any other point $j$ is, so the final probability is $n/2^{n1}$ (i.e. we can just add them up).
A sanity check for this answer is to notice that if you have either one or two points, then the probability must be 1, which is true in both cases.
 2,483
 1
 17
 19

2I don't understand this answer. (Strange, since you think it's a variation on mine. :) I presume the "if" in "if the remaining points end up all in the upper halfplane" is intended to mean "if and only if"? If so, why is that? And why is the final probability simply the number of points times this one probability you calculated? – joriki Mar 09 '13 at 01:46

1@joriki Basically I'm breaking this down into conditional probabilities. The angle around the circle is just an arbitrary assignment, so conditionally pick $i$. All of these conditional probabilities are going to be identical, and there's $n$ of them, so whatever that probability is, multiply it by $n$. That's the easy part. (cont'd...) – Josephine Moeller Mar 09 '13 at 01:48

@joriki Since you've conditionally picked $i$, then you can arbitrarily choose the upper or lower halfplane as your "in the same semicircle" event. That's a coinflip for each point, and they all have to be true, so it's $1/2^{n1}$. (cont'd...) – Josephine Moeller Mar 09 '13 at 01:51

@joriki I was tempted to think you could arbitrarily pick either the lower or upper halfplane, and that these had to be summed up, but then I realized that the "lower" probability would already be covered by the "upper" probability of some other point, so you have to pick either upper or lower, but not both, or else you doublecount. – Josephine Moeller Mar 09 '13 at 01:53

@joriki And it's a variation because my argument uses angles as well, just in a different way. – Josephine Moeller Mar 09 '13 at 01:53

15I think I see now  I find it rather confusingly formulated, but if I understand correctly, you mean something like this: The probability of the remaining $n1$ points being in the semicircle clockwise of a given point is $1/2^{n1}$. These $n$ events (one for each given point) are disjoint, and exactly one of them has to occur for the points to lie in a semicircle; thus the desired probability is their sum. That's a nice argument :) – joriki Mar 09 '13 at 02:02

Yes, that's a good way of putting it. I may incorporate that for clarity in a while, once I can get back to it. – Josephine Moeller Mar 09 '13 at 02:05

4There's a slight variation of this answer that uses the inherent symmetry: instead of picking $n$ points at random, pick $n$ random diameters of the circle and pick the $n$ points by randomly picking one of the 2 poles of each diameter. By essentially the same argument, you have the probability given by $(2n)/2^n=n/2^{n1}$. – sai Mar 09 '13 at 02:16



@joriki In your second comment you say "These $n$ events (one for each given point) are disjoint, and exactly one of them has to occur for the point to lie in a semicircle; thus the desired probability is their sum." What are the events? – Arrow Apr 05 '16 at 14:29

1@Arrow: The events mentioned in the sentence before that: "the remaining $n−1$ points being in the semicircle clockwise of a given point". That defines one event per point, so $n$ events in all, and they're disjoint. – joriki Apr 05 '16 at 15:08

1@Arrow to be clearer, let's call these points $x_{1,\cdots,n}$. If all $x_{2,\cdots,n}$ lie in the clockwise semicircle starting from $x_1$, then it is *not* possible for any other point than $x_1$ to have its clockwise semicircle contain all the other points. (Very simple geometry) That's why they are disjoint. – Vim Feb 13 '19 at 05:12

1Another way to frame this which was more clear to me: we have that the $n$ points all lie on some semicircle, if and only if, for exactly one of the points $\theta_i$ (call it the _leading_ point) all $n$ points lie clockwise from $\theta_i$ in the semicircle formed by the diameter joining $\theta_i$ and the center. If we condition on a particular leading point, we are now working with a _particular_ semicircle, so it's easy to see that each remaining point is either in that semicircle or not. – TheProofIsTrivium Dec 28 '20 at 19:52
Here's another way to do this:
Divide the circle into $2k$ equal sectors. There are $2k$ contiguous stretches of $k$ sectors each that form a semicircle, and $2k$ slightly shorter contiguous stretches of $k1$ sectors that almost form a semicircle. The number of the semicircles containing all the points minus the number of slightly shorter stretches containing all the points is $1$ if the points are contained in at least one of the semicircles and $0$ otherwise; that is, it's the indicator variable for the points all being contained in at least one of the semicircles. The probability of an event is the expected value of its indicator variable, which in this case is
$$2k\left(\frac k{2k}\right)^n2k\left(\frac{k1}{2k}\right)^n=\frac k{2^{n1}}\left(1\left(1\frac1k\right)^n\right)\;.$$
The limit $k\to\infty$ yields the desired probability:
$$ \lim_{k\to\infty}\frac k{2^{n1}}\left(1\left(1\frac1k\right)^n\right)=\lim_{k\to\infty}\frac k{2^{n1}}\cdot\frac nk=\frac n{2^{n1}}\;. $$
 215,929
 14
 263
 474

Why is $ \lim_{k\to\infty}k2^{n}\left(1\left(1\frac2k\right)^n\right)=\lim_{k\to\infty}k2^{n}\left(\frac{2n}k\right)$ true? – sai Mar 09 '13 at 05:09

2@sai: Apply the binomial theorem to the $n$th power  the first term cancels the $1$, the second term yields $2n/k$ and the remaining terms have more than one inverse power of $k$ and thus go to zero as $k\to\infty$. Note that $n$ is fixed; it's not a question of taking $n$ and $k$ to infinity simultaneously; we're just adding a finite number of terms, so the standard rules for adding convergent sequences apply. – joriki Mar 09 '13 at 08:23

sorry for pinging you, but can we use this method for a quadrant (1/4 circle)? – user2838619 Jun 30 '16 at 16:54

1In the first equation mentioned in the answer, shouldn't we get 1  1/k, instead of 1  2/k on the RHS? – user10 Aug 08 '16 at 18:32

It seems to me that there are two errors (that happen to given the correct limit) in the first equation. I think should be $$2k\left(\frac k{2k}\right)^n2k\left(\frac{k1}{2k}\right)^n= \frac{k}{2^{n1}}\left(1\left(1\frac1k\right)^n\right) \to \frac{n}{2^{n1}}$$ – leonbloy Jan 06 '17 at 01:47



@user2838619: Indeed we can; the argument carries through, and the result is $n\left(\frac14\right)^{n1}$, or more generally $nq^{n1}$ for any fraction $q\le\frac12$ of the circle. The arguments in the other answers (as far as they're sound) also carry through. – joriki Jul 11 '18 at 10:00
See
for the general problem (when the points have any distribution that is invariant w.r.t. rotation about the origin) and
for a nice application.
As a curiosity, this answer can be expressed as a product of sines:
Prove that $\prod_{k=1}^{n1}\sin\frac{k \pi}{n} = \frac{n}{2^{n1}}$

"when the points have any distribution that is invariant w.r.t. rotation about the origin", actually the theorem proved by J. G. Wendel pointed out in your first link is more general. We only need a **distribution symmetric with respect to $0$** and the result apply to **any dimension**. – Gilles Bonnet May 09 '14 at 19:06

Also, I think it would be nice if you quote directly in your answer the theorem of Wendel (just in order people looking at the present question don't miss the fact this is a specific case of a much more general "classical" result) – Gilles Bonnet May 09 '14 at 19:31

@sai one one reason it can be expressed as a product of sines is that if you ask for the number of spanning trees on a cycle, which is just $n$, (for the same reason that the numerator is $n$ in this problem because there are $n$ points) then the matrixtree theorem gives the determinant in terms of eigenvalues of the Adjacency matrix on a cycle which are directly related to those sines. I guess you are suggesting there is an iterative way to do it different than [Josephine Moeller](https://math.stackexchange.com/users/950/josephinemoeller) where the product arises naturally ... interesting. – Shannon Starr Feb 01 '22 at 12:07
Bull, 1948, Mathematical Gazette, Vol 32 No 299 (Dec), pp8788 solves this problem in the context of the broken stick problem (he uses polytopes and relative volumes in his argument). Rushton, 1949, Mathematical Gazette, Vol 33 No 306 (May), pp286288 points out that the problem can be restated in terms of placing points at random on the circumference of a circle. Ruston's answer is the clearest I have seen. Place $n$ points randomly on the circumference. Label them $X_1, X_2, ..., X_n$. Open up the circle at $X_n$ and produce a straight line. Label the line $OX_n$ (where $O$ is the part of the circle previously immediately adjacent to $X_n$). There are $n$ line segments: $OX_1, X_1X_2, ..., X_{n1}X_n$. Each segment is equally likely to be longer than half the length of $OX_n$ (and thus correspond to greater than a semicircle of the orginal circle). The probability that the first segment fulfils this condition is the probability that the remaining $n1$ points lie upon the second half of the line $OX_n$. That is $(\frac{1}{2})^{(n1)}$. The probability that there is one segment (note there can be at most one) greater than half the length of the circumference is the sum of the probabilities that each particular segment could be so (because these are mutually exclusive): $n(\frac{1}{2})^{(n1)}$. So, the favorable probability is $1 n(\frac{1}{2})^{(n1)}$.
Another simpler approach,
1) Randomly pick $1$ out of $n$ points and call it $A$ : $\binom n1$ ways
2) Starting from $A$, mark another point $B$ on circumference, such that $length(AB) = \frac12(Cirumference)$ [so that $AB$ and $BA$ are two semicircles]
3) Now out of remaining $(n1)$ points, each point can lie on either $AB$ or $BA$ with probability $\frac12$
4) For ALL the remaining $(n1)$ points to lie on EITHER $AB$ OR $BA$ (i.e., all $(n1)$ lie on same semicircle), the joint probability is $\frac12*\frac12 ...(n1) times$ $=$ $(\frac12)^{(n1)}$
Since #1 above (randomly picking $A$) is an independent event, $\therefore$ $(\frac12)^{(n1)}$ (expression in #4) will add $\binom n1$ times
$\implies$ Required probability is $\binom n1(\frac12)^{(n1)}$ $=$ $n(\frac12)^{(n1)}$
 465
 3
 13

@joriki, Perhaps this would be simpler to understand. I have tried to put the same idea as John Moeller in a different way. – Rahul May 25 '16 at 16:05

1

1