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How do I prove that the answer to this question: If N objects are placed randomly in a ring, what is the probability that they all lie together along an arc of length that is a fraction, f < 1/2, of the circumference or shorter?

is: $Nf^{N-1}$.

This problem comes up in astronomy in the guise of what the chances are that these N planets in the sky are above the horizon at some point in a day. It can also come up in coverage problems and length of gaps of coverage from independent Earth-observing satellites.

eshaya
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2 Answers2

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The probability that objects $2$ to $N$ are all in the arc of fraction $f$ of the circumference clockwise from object $1$ is $f^{N-1}$. Similarly for the probability that all objects except number $k$ are in the arc of $f$ clockwise from object $k$. The probability of the union of $N$ mutually exclusive events is the sum of the probabilities of the events.

[EDIT] This is assuming $f < 1/2$, so only one of the objects can be the first in the arc.

Robert Israel
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  • But, how can these be N mutually exclusive events, if each event is based on the positions of exactly the same set of points? That seems correlated to me. – eshaya Aug 10 '23 at 23:03
  • @eshaya you need the additional constraint that $f<\frac{1}{2}$. The formula $Nf^{N-1}$ is also only valid when $f<\frac{1}{2}$ – Alan Abraham Aug 10 '23 at 23:43
  • @AlanAbraham - Good catch. – eshaya Aug 11 '23 at 01:32
  • Here is one thing that may be making the events independent. If you work from low angles to high angles, when you compute the distance (angle) from the 2nd object to the 1st, you add $2 \pi$ to the angle of the 1st, so the distance is different than taking the distance from the 1st to the 2nd. Same for the 3rd to the 2nd, etc. In fact, one only needs to test those pairs to know if the criterion is met. – eshaya Aug 11 '23 at 01:39
  • Say the objects are labelled $A, B, C, \ldots$. If all objects except $A$ are in an arc of fraction $f$ of the circumference clockwise from $A$, then this is not true of $B$, or $C$, or $\ldots$. That makes the events mutually exclusive. – Robert Israel Aug 11 '23 at 01:42
  • I answered the question for $f=1/2$ here: https://math.stackexchange.com/questions/2786367/probability-that-polygon-formed-by-n-points-on-a-circle-contain-the-center-of-th/2786431#2786431 The key point is not that certain events are independent, but that they are in fact disjoint. You cannot have multiple points be the starting point, so the probabilities that a given point is the starting point ($f^{N-1}$) can be added up, because there is no double counting possible. – Ingix Aug 11 '23 at 10:36
  • Should we have "objects from 2 to N" in the first sentence? – Snoop Aug 11 '23 at 16:56
  • @Snoop Yes, thanks for the catch (edited). – Robert Israel Aug 13 '23 at 05:08
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The request of having $N$ points that lie in a fraction of size $f$ of the circumference is satisfied if and only if there is a point $p_1$ such that the others $N-1$ points lie on the fraction of size $f$ that starts at point $p_1$ and goes on clockwise. This is easy to prove because whatever set of $N$ points that lie in a fraction of size $f$ has two points that are distant from each other no more than $f*len(circumference)$ and all the others that lie between those two, so one of the two points has all the others that lie in a fraction of size $f$ that starts at that point and goes on clockwise and the other one has the same property but with the fraction that goes on anticlockwise. On the other hand if there is a point $p_1$ such that the other $N-1$ ones lie on the fraction of size $f$ that starts at point $p_1$ and goes on clockwise the property of having $N$ points that lie on a fraction of size $f$ of the circumference is obviously satisfied. Now we have to calculate the probability that one of the points has all the properties of $p_1$ (we call this ==$p_1$). If the points are $P_1,...,P_n$ this is the probability $p(E)$ of the event $E=(P_1==p_1) \cup(P_2==p_1) \cup... \cup(P_n==p_1)$ which is $p(P_1==p_1)+p(P_2==p_1)+...+p(P_n==p_1)$ given that $p(P_j==p_1 \cap P_i==p_1)=0 $ for any $j \neq i$. In other words if one of the points satisfies the properties of $p_1$ all the others do not. This is because if we take two points there are two fractions of the circumference that link those that sum up to $len(circumference)$, so if one of the two is $p_1$ this has clockwise distance $d$ to the other, that has clockwise distance to $p_1$ $h= len(circumference)-d$. $d\le f*len(circumference) \implies h\ge(1-f)*len(circumference)>$$>f*len(circumference)$. This makes it impossible for the second point to satisfy the properties of $p_1$, in fact $p_1$ lies in a fraction of the circumference that starts at that second point and goes clockwise which size is greater than $f$. Of course $p(P_1==p_1)=p(P_2==p_1)=...=p(P_n==p_1)$, so $p(E)=N*p(P_j==p_1)$. The fraction of the circumference starting at $p_1$ of size $f$ that goes clockwise identifies a specific fraction of size $f$, thus the probability of having the following $N-1$ points that lie on that fraction is $f^{N-1}=p(P_j==p_1)$ hence $p(E)=Nf^{N-1}$. We conclude that the probability of having $N$ points that lie on a generic fraction of size $f$ is the probability of having one point that satisfies the properties of $p_1$, which is $p(E)=Nf^{N-1}$.

SummertimeS4dness
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