39

I came up with the following question.

Four uniformly random points on a circle are chosen, and line segments are drawn between each pair of points. What is the probability that the longest line segment is between neighboring points?

Exerimental data using Excel suggest that the answer is $\frac23$, and I'm trying to figure out why. I believe the solution may involve order statistics. Or, given that the presumable answer is so simple, maybe there is an elegant argument from symmetry.

I made a Desmos simulator where you can choose four random points on a circle, and the line segments are shown.

My student's wrong method gives the right answer?

I asked my student this question (without mentioning the experimental data), and she said, "There are six line segments, of which four are between neighboring points, so the probability that the longest line segment is between neighboring points is $\frac46=\frac23$."

Huh? Surely that method is wrong: line segments between opposite points should have larger expected length than line segments between neighboring points, so the six line segments should not be treated the same, right?

Weird.

Dan
  • 35,053
  • 13
    The simple method does give the wrong answer to a related slightly different problem. The probability that the shortest line segment is between neighboring points is not $\frac46=\frac23$; the probability is $1$. (A quick way to see this: let the points be $A,B,C,D$ in cyclic order. If $AC$ is the shortest side of $\triangle ABC$, then $\angle ABC$ is the smallest angle, so $\angle ABC \le 60^\circ$. If $AC$ is the shortest side of $\triangle ACD$, then $\angle ADC$ is the smallest angle, so $\angle ADC \le 60^\circ$. But $\angle ABC+\angle ADC=180^\circ$, so one is bigger than $60^\circ$.) – Misha Lavrov Sep 20 '24 at 00:05
  • 3
    An interesting question is for what $n$ this works! It obviously works for $n=2$ and $n=3$, and you've shown that it's also (not so obviously) correct for $n=4$. What are the decoy answer and the right answer for $n=5$? – mjqxxxx Sep 20 '24 at 00:08
  • 2
    @mjqxxxx For $n=5$, the decoy answer is $\frac{5}{\binom{5}{2}}=\frac12$. A quick simulation suggests that the right answer is $\frac{5}{12}$. – Dan Sep 20 '24 at 01:09
  • 4
    Agreed... looks like $f(3)=1=3/3$, $f(4)=2/3=4/6$, $f(5)=5/12$, $f(6)=1/4=6/24$, and $f(7)=7/48$... suggesting $f(n)=n/(3\cdot 2^{n-3})$ in general. – mjqxxxx Sep 20 '24 at 02:15
  • 1
    @mjqxxxx Yes. $\dfrac{n}{3 \cdot 2^{n-3}}$ can be shown to be correct by extending my analysis. – Henry Sep 20 '24 at 09:04
  • 1
    What definition/construction are you using for "uniformly random points in a circle"? IIRC there are at least 3 ways that such points can be constructed and justifiably called 'uniform', each of which produces points with different properties. Which one are you using/expected to use. – Brondahl Sep 20 '24 at 09:50
  • (Off the top of my head ... "generate uniform points in a square and discard points that are generated out of bounds". "Pick a random radius and a random point on that radius". "Pick 2 random points on the circumference, and a random point on the chord that forms". – Brondahl Sep 20 '24 at 09:52
  • 3
    @Brondahl "uniformly random points on a circle" not "in". So on the circumference and a natural uniform distribution either on that length or on the angle at the centre. – Henry Sep 20 '24 at 09:54
  • @Brondahl What Henry just said :) – Dan Sep 20 '24 at 09:55
  • 1
    Oh, yes. Of course. "circle" != "disc". My mistake, sorry :) – Brondahl Sep 20 '24 at 10:40
  • @Dan $\frac{1}{2}$ doesn’t seem so far off from $\frac{5}{12}$. I should read the comments more carefully but the first thing that comes to mind is asking “how far off are the decoy answers” – Sidharth Ghoshal Sep 20 '24 at 13:07
  • 4
    @SidharthGhoshal Henry shows in their answer that the right answer is $\dfrac{n}{3 \cdot 2^{n-3}}$, where $n$ is the number of random points. The decoy answer is $\frac{n}{\binom{n}{2}}=\frac{2}{n-1}$. So the ratio of decoy answer to right answer approaches infinity. – Dan Sep 20 '24 at 13:14
  • "line segments between opposite points should have larger expected length than line segments between neighboring points [...] right?" ... are you sure about that? I can think of cases where they are larger, or smaller; I don't have a sense of "expected". – Don Hatch Sep 21 '24 at 00:52
  • 2
    @DonHatch Indeed it is not obviously true (to me). I used the phrasing "... should ... right?" to express my lack of certainty. But afterward I worked out that the expected length of line segments between opposite points is $12\int_0^1x(1-x)\sin(\pi x)\mathrm dx=\frac{48}{\pi^3}\approx 1.548$, and the expected length of line segments beween neighboring points is $6\int_0^1(1-x)^2\sin(\pi x)\mathrm dx=\frac{6\pi^2-24}{\pi^3}\approx 1.136$. – Dan Sep 21 '24 at 09:42
  • @MishaLavrov actually the two shortest segments are between neighboring points. Why: consider the shortest of the two diagonals of the quadrilateral; it divides the disk into two parts; look at the smaller of those two parts. That part of the disk contains two sides of the quadrilateral, each shorter than that shortest diagonal. – Don Hatch Sep 21 '24 at 21:06

1 Answers1

32

I could not find a justification for a very simple answer, but I can get to $\frac 23$ with this approach.

If the points are $A,B,C,D$ (ignoring order) are on a circle centred at $O$ and if $\angle AOB$ is $\theta \in (0, \pi)$ then $A$ and $B$ are neighbouring points and $AB$ is the longest line segment

  • if $C$ and $D$ are both inside $\angle AOB$ (the red arc below), which happens with probability $\left(\frac{\theta}{2 \pi}\right)^2$
  • or if $\frac23 \pi < \theta < \pi$ and $C$ and $D$ are in an angle size $3\theta-2\pi$ opposite $\angle AOB$ (the pink arc below if it exists), which happens with probability $\left(\frac{3\theta-2\pi}{2 \pi}\right)^2\mathbb I_{\left[\frac23 \pi < \theta < \pi\right]}$

so a combined probability $\left(\frac{\theta}{2 \pi}\right)^2 + \left(\frac{3\theta-2\pi}{2 \pi}\right)^2\mathbb I_{\left[\frac23 \pi < \theta < \pi\right]}$.

circle arcs

Since $\theta$ is uniformly distributed on $(0, \pi)$, the probability of this happening is $$\int_0^{\pi} \frac1{\pi}\left(\frac{\theta}{2 \pi}\right)^2\, d\theta + \int_{\frac23\pi}^\pi \frac1{\pi}\left(\frac{3\theta-2\pi}{2 \pi}\right)^2\, d\theta = \frac1{12} + \frac1{36}=\frac19.$$

But we could have started with any of the ${4 \choose 2}=6$ pairs of points instead of just $A,B$, so the overall probability that some line segment between neighbouring points is the longest line segment is $\dfrac69=\dfrac23$.

If, instead of $4$ points, we have $n$ points then the same analysis applies with the other $n-2$ points all having to appear in the same arc(s), so raising to the power $n-2$ instead of squaring.

This gives the probability $A$ and $B$ are neighbouring points and $AB$ is the longest line segment of $\int_0^{\pi} \frac1{\pi}\left(\frac{\theta}{2 \pi}\right)^{n-2}\, d\theta + \int_{\frac23\pi}^\pi \frac1{\pi}\left(\frac{3\theta-2\pi}{2 \pi}\right)^{n-2}\, d\theta $ $= \frac1{(n-1)2^{n-2}} + \frac1{3(n-1)2^{n-2}}$ $=\frac1{3(n-1)2^{n-4}}.$ But there are ${n \choose 2}$ possible pairs of points, so the overall probability is $\dfrac{n \choose 2}{3(n-1)2^{n-4}}=\dfrac{n}{3 \cdot 2^{n-3}}$, confirming the pattern spotted by mjqxxxx.

Henry
  • 169,616
  • 1
    I found it a bit hard to see what's going on in the picture due to the right angle. May I suggest using 130 degrees instead of 135 degrees for ∠AOB – Don Hatch Oct 05 '24 at 02:50
  • I see a way to replace the first integral with a simple argument that it's 1/12. It's the probability that both C,D are in the "red arc" with respect to A,B. This is the probability that B,C,D are all on the same side of the line through AO (that is, 1/8 + 1/8 = 1/4) times the probability that AB is the longest of AB,AC,AD (that is, 1/3, since all three are equally likely to be longest). So it's 1/4 * 1/3 = 1/12. – Don Hatch Oct 05 '24 at 06:53
  • (And that easily extends to the answer 1/((n-1)*2^(n-2)) for n points, as well.) – Don Hatch Oct 05 '24 at 21:12
  • As for the second integral, it's not hard to see that (for any n) it's always just 1/3 times the first integral: that's because the integrand is taking on the exact same values, but spread linearly over an integration range that's 1/3 as big. More formally: when we rewrite the second integral using change of variable t=3θ-2π, it turns into exactly the first integral (integrand and limits) except that the dθ turns into 1/3 dt. – Don Hatch Oct 05 '24 at 21:25
  • So that gives a proof that avoids evaluating any integrals, although we did still mention the integrals in order to argue that the second term is 1/3 times the first, which seems rather unsatisfying. At this point it's evident that we can turn the whole thing into a purely combinatorial argument that doesn't use any calculus, but that seems to make it more awkard to describe (I don't have a clear way of stating it that way, yet.) – Don Hatch Oct 05 '24 at 21:33
  • The way I'm thinking of seeing the second integral is 1/36 is as follows: it's the probability that C,D are in the "pink arc"; this is the probability that the pink arc exists (i.e. 1/3) times the prob. that C,D are both at angles of < π/2 away from the center of the pink arc (i.e. 1/4), times the probability that b is the largest of b,c,d, where b = 3/2 (θ - 2/3 π) and c,d are the angular distances between the center of the pink arc and C,D respectively. b,c,d are all independent and identitically distributed random variables (uniform on (0,π/2)) so that third probability is 1/3... – Don Hatch Oct 06 '24 at 07:51
  • ... so the second integral is 1/3 * 1/4 * 1/3 = 1/36. Or in general, 1/3 * 1/2^(n-2) * 1/(n-1) = 1/(3(n-1)2^(n-2)), agreeing with your answer. – Don Hatch Oct 06 '24 at 07:52
  • @DonHatch Perhaps you should put that together as an answer – Henry Oct 06 '24 at 07:58
  • 1
    Maybe. I'm not really happy with it, though... it takes too much thinking to convince myself that b,c,d are independent random variables and satisfy all the properties claimed-- in the end, your answer may still be clearer (for the second integral, at least). It's more of a sketch-of-proof, or an existence proof (that there exists a proof without using calculus). I'm optimistic that there's a much simpler calculus-free proof. – Don Hatch Oct 06 '24 at 08:03
  • 1
    @DonHatch I have change the picture (a) to give a different angle as you suggest and (b) to illustrate the $3\theta-2\pi$ caclulation – Henry Oct 06 '24 at 10:54