I don't have an exact value for you, I'm sorry. However I did spend some time trying to figure out how large it should be. Let's go.

I started my research by simplifying the problem. A natural choice would be to consider the unit interval and the unit disk first. I'll skip those and stick to the sphere. I decided to first consider the boundary of the sphere: the 2-Sphere:

$x^2+y^2+z^2=1$

As you might know the shortest distance between two points without leaving this surface , measured as arc-length, can be found by considering the length of the great circle through both points. This great circle is a one-dimensional object. What's more, if we choose random points uniformly on this object, we might be inclined to think that this distance is also uniformly distributed.

In order to test this, I wrote a short routine, to randomly pick $N$ points on this 2-Sphere

The mean arc-length equals 1.57146 and the boundaries of the 95% confidence interval, based on a normal distribution of the mean arc-length, of this mean arc-length are $1.57146\pm 0.00424$. Which does not suggest that $\pi /2\approx 1.57080$ might be true.

But it did surprise me. Hardly uniformly, is it?

Addendum

In the same manner as shown in my second link (at the bottom) it is possible to find the pdf of this distribution of distance $s$ exactly, it is:

$\frac{\sin \left(s\right)}{2}$

The mean distance is equal to $\pi /2$.

What's more, I also tested the Euclidean distance, i.e. straight lines connecting these points, in a similar fashion:

How about that for a distribution! Did you expect this to happen?
You might be able to calculate the mean by just looking at the graph above.

The pdf of the distribution of the Euclidean distance $s$ equals $s/2$ and its expected value can be calculated as:

${\int }_0^2{s}^2/2ds={\left[s^3/6\right]}_0^2=4/3$

The mean distance equals $1.33174$ and the boundaries for the 95% confidence interval of this mean distance are $1.33174\pm .002930$.

If you do the math and compute $1.96\ast \sqrt{\frac{\text{var}\left(s\right)}{N}}$, using the given distribution, you'll find $.002922$. The numbers add up.

Note: This part (Euclidean distance) is known

Now your question. Here I used a Monte Carlo method to find a uniform distribution of points. I chose $N$ points having an uniform distribution in the unit-cube $[-1,1{]}^3$ and discarded the points outside of the ball:

$x^2+y^2+z^2\le 1$

The distribution of the Euclidean distance:

This one does seem to show a Gaussian distribution. It is not (the tails are too heavy). The bounds for the confidence interval for the mean distance are $1.02833\pm .00234$.

Which tells us that the mean distance is probably slightly larger than $1$.

Addendum. I could find a source describing the distribution

I realise that this is only circumstantial evidence. It might be ‘a little more’ than just an exercise to derive the true distributions. The second 2 distributions are probably known already. I couldn't find/derive the first distribution yet.

Distributions tend to be difficult to derive, let alone calculate its mean exactly, certainly if the degrees of freedom increase. But I agree that I'm curious.

Feel free to explain the first two distributions. Or extend these results.

Footnotes

It depends on how the sphere is represented, and what do you mean by distance.

Assume the sphere has radius 1 (otherwise just scale the formulas).

If the points are given by coordinates $(x,y,z)$ in three-dimensional space (and the center is (0,0,0)), then the distance through the sphere can be computed from the Pythagorean Theorem: $\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2+(z_1-z_2{)}^2}$, and the distance on the sphere can be computed using $\text{arc}\cos (x_1{x}_2+y_1{y}_2+z{1}_1{z}_2)$ (the formula inside parentheses is the dot product, which equals the product of both the vector lengths and cosine of the angle between them).

This representation is likely to be the most convenient when you are doing your computations on a computer (and also in many math computations on paper). There are other representations, for example to specify locations on the surface of Earth we traditionally use longitude $\lambda$ and latitude $\varphi$. You can change these coordinates to $(x,y,z)$ in the following way: $z=\sin \varphi$, x = $\cos \varphi \sin \lambda$, y = $\cos \varphi \cos \lambda$.

(Fun fact: very similar formulas can be used when working with hyperbolic geometry — change some + to - and some trigonometric function to hyperbolic ones.)

I am adding to my initial answer here which very observantly was identified as answering a different question.

The key to solving this problem again is symmetry. A sphere has lots of symmetry.

Pick any point P1 inside the sphere. Now imagine an equatorial plane E1 that cuts through the centre of the sphere. There are acutually infinitely many such planes slicing the sphere. We choose an E1 such that it also contains P1. Note that again there are infinitely many of those planes.

Why? There is a radial projection of P1 onto the surface of the sphere that identifies a unique point P1‘ on the surface of the sphere. E1 also contains P1‘. If we keep E1 fixed, we can rotate the entire sphere along the axis C-P1‘ and E1 is still an equatorial plane.

Now choose P2. It may lie in E1, above it or below. Keep rotating the sphere until P2 lies in E1. Now P1 and P2 both lie in the same plane - and the problem becomes equivalent to finding the average distance of two points in a circle of equal radius.

< Please bear with me. To be continued. >

———- Previous answer ———-

Nice question.

Sounds hard at first … but maybe isn‘t actually.

I would argue that the average distance between two arbitrary points on a sphere is a quarter of the sphere‘s circumference, i.e., pi/2 on a sphere of unit radius.

Here‘s why:

Pick the first point P1. For convenience of argument and without loss of generality spin the sphere so P1 becomes your northpole. Now pick a second point P2 on the sphere. Anywhere.

Note that the sphere is symmetrical about its equator. So for any point P2 that you pick in the upper hemisphere, there is a corresponding point P2‘ in a symmetrical position below the equator. An analogous argument applies if P2 lies below the equator. For any pair P2, P2’ the centre between the two points lies exactly on the equator.

So far, so good.

Now, since all points P2 in the upper hemisphere have such a corresponding point P2‘, for each distance P1-P2 there is a corresponding distance P1-P2‘. The average of those two distances is the distance between the northpole and the centre point between P2 and P2’ - which sits on the equator, as we just saw. Since this applies to all points P2 in the upper hemisphere, we have covered the entire sphere with all P2s from the upper half and their corresponding P2’ from the lower half. The overall average is hence the distance from the northpole to the equator …. which is a quarter of the sphere‘s circumference.

q. e. d.

I assume that you mean circle as the set of points equidistant from a fixed center rather than the set of points contained by a circle? Anyways:

Let’s define a circle of radius r centered at (0, 0) in the Cartesian plane.

Without loss of generality, we fix a point at (r, 0). We wish to integrate the distances between the points with respect to the measure of the minor arc defined by (r, 0) and the secondary point, which we will call $\theta$, from 0 to pi radians and divide by pi radians. Note that the x-coordinate of a point on a circle given $\theta$ is given by $x=r\cos \theta$ and the y-coordinate is $y=r\sin \theta$. Also, note that the distance between two points is given by$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$. Plugging things in, our expression is thus:

$\frac{{\int }_0^{\pi }\sqrt{(r\cos \theta -r{)}^2+(r\sin \theta {)}^2}d\theta }{\pi }$.

WolframAlpha returns $\frac{4r}{\pi }$ for the value of this expression.

Edit: by symmetry, there is not a need to integrate the “bottom half” of the circle.

As Barry Carter’s [now deleted] comment notes, this result is certainly well-known (I’m surprised it has not been asked on Quora before — or at least it is not listed among the “related questions” for this question).

I’ve given a partial development below, but you can find a much more detailed (and general) answer in a paper by Burgstaller and Pillichshammer:

• Link 1
• Link 2

The short answer is: For a circle of radius 1, the average distance is $\frac{128}{45\pi }$.

Assuming that the “random points in a circle” are chosen uniformly throughout the circle — and assuming the circle has radius 1 — the coordinates $(X,Y)$ of such a point has density $f(x,y)=\frac{1}{\pi }$ for $x^2+y^2<1$. Alternatively, we can identify the point by its polar coordinates $(R,\Theta )$, in which case $R$ and $\Theta$ are independent, with respective densities $f\left(r\right)=2r$ for $0<r<1$ and $g\left(\theta \right)=\frac{1}{2\pi }$ for $0<\theta <2\pi$. The distance between two randomly chosen points $A$ and $B$ in the circle is given by

$\begin{array}{}\sqrt{(X_a-X_b{)}^2+(Y_a-Y_b{)}^2}=\sqrt{R_a^2+R_b^2–2R_a{R}_b\cos ({\Theta }_a-{\Theta }_b)}& \end{array}$

so the average distance is found by integrating this expression.

If we use rectangular coordinates, we need a quadruple integral (because there are four variables); if we use polar coordinates, we can note that $m\angle AOB=\Phi =|{\Theta }_a-{\Theta }_b|$ can be treated as if it were uniformly distributed on $[0,\pi ]$. (${\Theta }_a-{\Theta }_b$ has a triangular density on $[-2\pi ,2\pi ]$, but we only need to worry about the cosine of this angle.) So the average value of the distance is $\frac{4}{\pi }{\int }_{r_a=0}^1{\int }_{r_b=0}^1{\int }_{\varphi =0}^{\pi }{r}_a{r}_b\sqrt{r_a^2+r_b^2–2r_a{r}_b\cos \varphi }d\varphi d{r}_{b}d{r}_a$

[edited to correct the final integral; I had forgotten to multiply by $f\left(r_a\right)f\left(r_b\right)h\left(\varphi \right)$.]

What is the average distance between two random point in a circle?

What is the expected distance between two randomly and uniformly selected points on the surface of the unit sphere?

It would seem that there would be 2 possible answers, depending on whether you mean great circle distance or straight line distance. I am not sure what you mean by two points being randomly selected AND uniformly selected at the same time.

For the great circle distance i agree with Joe Rovito’s answer and here is the logic behind it. de is the distance to the equator determined by calling the first point a pole of the sphere. The probability of landing on parallel at distance d1 is the same as the probability of landing on the circle at d2 since the length of the circumference is the same. The average distance of (d1 + d2)/2 = de.

This is true of all the equal size parallel slices and they will all be at equal distance from the equator in order to be equal in size, so the probability is that the great circle distance will be π/2 ≈ 1.57 for a unit circle, or πr/2 for any circle. Half the time the great circle distance will be greater than π/2 and half the time the great circle distance will be less than π/2

The direct line distance is much different.

Here is my stab at it. The straight line distance to equal probability rings is different for this case. Ay any distance, x, from the equator, d1 will equal the

√(h² + (1 + x)²) and d2 will equal √(h² + (1 - x)²) where h = √(1 - x²)

so h² = (1 - x²) Substituting h² and expanding the rest we will get

½ [√(1 - x² + 1 + 2x + x²) + √(1 - x² + 1 - 2x + x²) ] = ½ [√(2 + 2x) + √(2 - 2x)] = ½ [(√2)√(1 + x) + (√2)√(1 - x)] = ((√2)/2) [√(1 + x) + √(1 - x)]

$\frac{\sqrt{2}}{2}[{\int }_0^1\sqrt{1+x}dx+{\int }_0^1\sqrt{1-x}dx]$

$\frac{\sqrt{2}}{2}[{\left(\frac{2}{3}\right(x+1{)}^{\frac{3}{2}})|}_0^1-{\left(\frac{2}{3}\right(x-1{)}^{\frac{3}{2}})|}_0^1]$

$\frac{\sqrt{2}}{2}[(\frac{4\sqrt{2}}{3}-\frac{2}{3})-(0-\frac{2}{3})]=\frac{\sqrt{2}}{2}\cdot \frac{4\sqrt{2}}{3}=\frac{4}{3}$

The probability is that the direct distance will be 4/3. Half the time the distance will be greater than 4/3 and half the time it will be less than 4/3.

Imagine a set of equally space points around a circle, and from one of them, draw the chord to each of the others. The average of their lengths will approximate the answer to this question, which is actually the limit as the number of points goes to infinity. Here you need calculus as well as trigonometry. (Note also that you only need to go half way around to find the average, the other half will have the same value.)

But let’s take just a small number of points to illustrate: 0, 60, 120 and 180 degrees around the circle. The first chord’s length is 0. It’s easy to see that the length of the 60 degree chord is 1 cm, since it makes an equilateral triangle with the center point and each side will be as long as the radius. It’s just as easy to see that the length of the 180 degree chord is 2 cm.

For the length of the 120 degree chord, you need a little trigonometry to find the formula

len = 2 r sin (theta / 2)

where r is the radius (which is 1) and theta is of course the angle. The answer for 120 is about 1.732 cm and the average for all three chords is approximately 1.183. If you did this for every degree from 0 to 180, the average would be about 1.27172. If every 0.1 degrees it would be about 1.27309 (I have a tool that is just right for this sort of thing).

I leave the the final step, the integral over 0 to 180 for 2 r sin theta/2 d theta “to the reader,” but will say that the answer is 4 / pi, approximately 1.27324.

What is the average distance between two random points on a circle?

I was going to answer this question when I noted that it had been merged with a similar looking question:

What is the average distance between two random points in a circle?

Technically, in geometry, a circle is the ‘line’ that forms the perimeter of a disc! Thus, there are no points in a circle, only on a circle.

The ‘on’, the answer for a circle of radius $r$ is simply:

$\frac{4r}{\pi }$

See the answer submitted by Alan Ding on 10 August 2016.

However, if (as most people do) you use circle to mean a disc, then ‘in’ makes sense.

This answer is much more complicated - see that submitted by Darryl Nester.

Given that the answers are so different, I don’t think the questions should have been merged.

I have a different take on this question — at least for the circle portion of the question. It is based upon on arc lengths.

Suppose you have a circle of radius r, centered at the origin (0,0) point in the space. The circumference of the circle is 2(pi)r. Let t be an angle in radians, where t is measured relative a line running between (0,0) and (r,0) and running counter clockwise.

Then, the arc length of angle t = (t/2pi)(2(pi)r) = tr. This is the distance between the (r,0) point and the point (r cos(t), r sin(t), measuring distance along the circle.

Without loss of generality, assume that the first point selected is (r,0). (If not true, you can always redefine the coordinate system to make it true.)

Select a point (r cos(t), r sin(t)) where t is uniformly distributed in the interval [0,pi].

Then, the expected arc length distance between the two points is

In working on this problem, I have implicitly assumed that sin(t) is non-negative (i.e., the second point is above the x axis). If instead, you assume that the selected point is below the x axis, you would get the same answer (due to the symmetry of the problem). Now, it you assume that the point is equally likely to be above the x axis as below the x axis, you would then just average the two answers to get the final answer. If you follow this procedure, you still get the answer above.

Obviously, you will get a different answer if, instead, you define distance between points as the Euclidian distance.

COMMENT

I have not yet tackled the spherical problem. However, I strongly suspect that the answer is the same as above. Here is my reasoning.

You would select one point on the sphere. This is your reference. You select a second point at random.

Here is an interesting observation. You can always draw a Great Circle through two points on a sphere. A Great Circle goes through both points and includes the center of the sphere as the origin. All points lie in the same plane.

Now, if you lift out the plane of the Great Circle and lay it down on a sheet of paper, you would get the same situation as the circle problem above.

This approach is equivalent to first selecting a Great Circle at random from the sphere, and then selecting two points at random along the Great Circle. Thus, if this logic is correct, then you would get the same answer. Again, note that I am defining distances along the surface of the sphere (arc length distances).

UPDATE: Euclidean Distance

If you are interested in Euclidian distance, note that the length of the chord corresponding to an arc of degree t is (2r)sin(t/2). (You get this formula by realizing that the chord and the two radii of an arc form an isosceles triangle.) If you then assume that t has an uniform distribution over [0, pi], you compute the average Euclidian distance as

This result agrees with another Quora answer based upon Euclidian distance. Since this works for a circle, it follows (from the logic above) that this result is also the average Euclidian distance between two random points on a sphere.

UPDATE: Random Sampling on a Sphere

If you read the comments to this post, you will see that the word “random” is ambiguous in the context of a sphere. Terry Moore correctly points out that my approach amounts to the following scenario. You pick a point on the sphere using some random process. You then adjust the coordinate system to label this point the North Pole. Then, to pick the second point, you select its latitude and longitude independently, each from a uniform distribution. Under this process, the conclusion I reached above — that average distances on a circle and a sphere are identical — is correct.

However …

Terry also points out — again correctly — that the curvature of the sphere needs to be taken into account in selecting the second point. For example, consider what might be called a “spherical square”: an area of the sphere whose sides are one degree latitude and one degree longitude in length. Find yourself a globe and look at it carefully. You will quickly see that the area of a spherical square is larger near the Equator than the North Pole. The random sampling of latitude and longitude, in effect, assumes that the areas of all spherical squares are the same.

So, let’s try a new sampling scheme and see what happens. As before, the first point is the North Pole. However, to select the second point, you make a simple observation. The circles of equal latitude are of different circumference at different points of the sphere. Near the North Pole and South Pole, the circles have a circumference that approaches zero in length. At the Equator, the circle has a circumference of 2(pi)r where r is the radius of the sphere. If you index latitude on a radian scale from pi/2 (North Pole) to -pi/2 (South Pole), the circumference of a circle of equal latitude is equal to (2(pi)r)cos(t) where t is the latitude angle.

Suppose now that we choose the latitude of the second point using a probability density f(t) that is proportional to the circumference of the latitude circle. By integrating (2(pi)r)cos(t) from pi/2 to -pi/2, we can figure out the proportionality constant. This leads to f(t) = cos(t)/2. It is symmetrical with a mean equal to zero:

Now, let’s redo the average distance calculations. Observe that each point on a circle of latitude is equidistant from the North Pole for both arc length and for the Euclidean metric. So, we don’t need to worry about the longitude. We just sample from the latitude angles using f(t) and then compute average distance.

Here is the arc length calculation:

Here is the Euclidean distance calculation:

Interestingly, the average arc length remains the same as before. However, the average Euclidean distance has changed slightly from the previous value of (4/pi)r.

This means that the answer to the sphere problem depends on both the method of sampling and the distance metric chosen.

So, I think these are integrals that lend themselves to the ever-friendly Monte Carlo.

Thank you Archimedes, your idea was really, really amazing.

Initialize: Define a circle, of radius 1 to be convenient.

Let reps=some huge number ; current replication=0

While currentreplication<reps:

1. Sample two random points in the circle
1. Note: Nontrivial! Sampling two random numbers within uniform constraints on x and y doesn’t do the task well, you’ll center around the radius.
1. Instead, sample one for the distance between (O,R) and another for the amount of rotation [for one point at a time].
2. Calculate the distance between the two points (I assume Euclidean)
3. Store the distance; then currentreplication=currentreplication+1

Calculate the mean of the stored vector.

note: More reps=better results.

Theoretically, assume you have both vectors of random points. If you randomly shuffle one [or both], and calculate the mean, and do both steps many times, you should actually get a reliable distribution for this distance as well as the point estimate [this of course leads to a standard deviation or a notion of variance]. Make sure the number of replications is sufficient! Permutation Monte Carlo can be fun too.

What happens when the radius changes? [besides needing more points] Well, just vary R and use regression to figure out the relationship between the mean of your samples and the radius.

This tends to be controversial, so… I’m sorry you feel that way.

Great question!

Two points are randomly selected on the circle. What is the expected distance between them? And what will be the expected distance between the two points on a sphere?

Henk Brozius has solved the circle problem.

The sphere version is reminiscent of Borel’s paradox where, if you think of the sphere as the Earth, the probability density given that the point lies on the equator is different from the probability density given that the point lies on a great circle containing a given line of longitude.

For the sphere let’s take the unit sphere centred on the origin. We can take the first point to be the North pole. Use spherical polar coordinates so that $\theta$ is the longitude and $\varphi$ is the latitude.

Now $\theta$ is clearly uniformly distributed, but, given $\theta$, $\varphi$ is not. The reason is that if you know that $\theta$ is between two values (think of a segment of an orange, $\varphi$ is more likely to be near to zero than a right angle because the segment is wider near the equator than near the pole. This is the problem with Gary Russell’s idea that the problem reduces to the circle problem.

The density for $\varphi$ given $\theta$ is $\frac{1}{2}\cos \left(\varphi \right)$. But for this problem it is better to measure $\varphi$ from the North pole so that the density is $\frac{1}{2}\sin \left(\varphi \right)$. The expected distance from the North pole is therefore

${\int }_0^{\pi }2\sin \left(\varphi /2\right)\frac{1}{2}\sin \left(\varphi \right)d\varphi$

$={\int }_0^{\pi }2{\sin }^2\left(\varphi /2\right)\cos \left(\varphi /2\right)d\varphi$

$={\left[\frac{4}{3}{\sin }^3\right(\varphi /2\left)\right]}_0^{\pi }$.

So the expected distance between points is $1\frac{1}{3}$ times the radius.

Most likely you want to find the Great Circle distance between two points whose positions on the sphere are given. A Great Circle is the circle formed by the intersection of a plane passing through the centre of the sphere, and the sphere’s surface. On the Earth the Equator, and all the Meridians of Longitude, are great circles but the lines of equal Latitude other than the Equator are not Great Circles. You may not believe this, but they are called.. wait for it.. Small Circles.

A Great Circle is the closest thing you can get to a straight line on the surface of a sphere, and the shortest distance between two points on a sphere is along the Great Circle joining them. Diistance along a Great Circle is usually given in terms of the angle between the two points, as seen from the centre of the sphere.

Spherical Trigonometry is the study of triangles drawn on the surface of a sphere. It was considered too hard for us at school, but I worked out some of the basic formulae from Plane Trigonometry and some carefully drawn diagrams after I left school. A spherical Triangle ABC will have three sides AB, BC and CA, all of which are parts of Great Circles. The three angles A, B and C add up to more than π radians or 180°. If it’s a very small triangle compared to the sphere then it’s only a tiny bit more than 180°, but a large triangle coverinig nearly half the sphere can have angles that add up to nearly 3π radians, or 540°.

The Earth can be closely approximated by a sphere of radius 6371 km. If you are given the Latitude and Longitude of points A and B you can calculate their distance apart by the following method:

Let the Latitude and Longitude of A be θ₁ and φ₁ respectively
Let the Latitude and Longitude of B be θ₂ and φ₂

You are asking: What is the length of AB along the Great Circle joining them?

We can easily find this by constructing a Spherical Triangle with AB as one of its sides.

Join A to the North Pole N to make the line AN
Join B to the North Pole N to make the line BN
We now have the Spherical Triangle NAB
The angular distance NA is 90° − θ₁ if θ₁ is North Latitude, or 90° + θ₁ if it is South.
The angular distance NB is 90° − θ₂ if θ₂ is North Latitude, or 90° + θ₂ if it is South.
The angle ANB, or simply N, is the difference in Longitude, |φ₁ − φ₂| if φ₁ and φ₂ are in the same hemisphere (E or W), or φ₁ + φ₂ if they are in opposite hemispheres. The | | signs mean ignore the sign of the difference: if it is negative then change it to positive.
If the difference in Longitude is more than 180° then subtract it from 360° to get the shorter path rather than the long path.

You now have a Spherical Triangle of which you know two sides (NA and NB), and the included angle N. You can easily calculate the side AB opposite N from the cosine formula:

cos AB = cos NA cos NB + sin NA sin NB cos N

Remember that a distance in Spherical Trigonometry is measured by the angle it makes at the centre of the sphere. Calculate cos AB and from this obtain the angle AB. It will be between 0° and 180°.
To convert this angle into distance across the surface of the sphere you must first convert the angle to radians. If your answer is in degrees simply multiply by π/180. Then multiply by the sphere’s radius to get distance across the surface. Computer calculations will often give the result of arc cos calculations in radians so no conversion is necessary.

Sea and Air Navigation have another trick for distances: the unit of distance is the Nautical MIle. This is a bit longer than a Statute Mile but it is precisely the distance you have to travel across the surface of the Earth to change your angular position at the centre by exactly 1 minute of arc. So if your calculation of the distance AB on the Earth is, say, 2°35′ of arc, then it is 2 × 60 + 35 = 155 minutes of arc, or 155 Nautical Miles. 1 knot is 1 Nautical Mile per hour. If your ship cruises at 15 knots then it will take 10 hrs 20 mins to travel from A to B.

Picturing the sphere as an Earth-globe: if both points are uniformly random, you can rotate the sphere to make the first point coincide with the North Pole, and the second point fall along 0° longitude;

then the only free variable is the second point’s latitude in $\left[\mathrm{0..}\pi \right]$ radians from the North Pole, which corresponds to $\left[\mathrm{0..}\pi \right]$ units of surface distance from the first point.

The probability of the second point being at a given latitude is proportional to the sphere’s circumference at that latitude, so the cumulative probability is equal to the proportion of the sphere’s area above the given latitude.

Clearly the sphere is mirror-symmetric above and below the equator; so the equator is the point of 50% cumulative probability, and the expected distance is $\pi /2$ units.

Since the circle is on a plane I interpreted the distance between two points A and B “ON” the circumference of the circle with centre in the point O, as the length of the segment joining A and B , i.e. AB =2 r sin (α/2) if r is the radius of the circle and 0<=α<=π the angle between OA and OB. The average value is by integration 4r/π.

As to the sphere, the length of the path between the points A and B on it depends on the arc joining them. The shortest path is the arc A-B belonging to the geodetic joining A and B. This geodetic lies on the intersection (grand circle) between the sphere and the plane identified by A, B and the center of the sphere. The length of this arc is rα and the average value is πr/2.

Please excuse my long winded answer, as I do not recall studying probability involving samples from infinity compared to another sample of infinity. I recall studying samples that are discrete quantities that are finite, such as six sides to a die or numbers of blue balls in a bag. You are going to read my thinking as I think it, stream of conscious style.

First, the question is posed with a sphere. Let me simplify to a circle; the theory wiil be the same but the equations will be different: πr² for the area of a circle vs 4/3 πr³ for the volume of a sphere.

Let me assign C as the constant and D as the diameter of the circle. Starting with C >= D then the probability is 1 (100%). Also if C=0 then the probability is 0. What about values of C in between those values?

Common sense tells us that the probability should be related to the area (Ca) of the circle (C1) defined by C as the numerator and the area (Da) of the circle (D1) defined by D as the denominator. But is common sense borne out mathematically?

It sure feels like it should. If I place a paper target (C1), circle shaped, with smaller diameter than the dart board (D1) over the dart board, then start throwing darts at the thing, it feels as though the probability of hitting the C1 paper target should be Ca/Da. That would satisfy the common sense that the probability would shrink, linearly, from 1 when C=D to 0 when C=0.

But I need to answer the question about the uncountable infinity of point pairs inside C1 versus the uncountable infinity of point pairs inside D1.

It will be easiest to discuss the matter if we assume C1 is centered on the center of D1 (for having C1 off center, but still fully inside D1, only makes the mathematics more complicated but with the same result). Is there a one-to-one mapping of each point pair inside C1 to a point pair inside D1?

The answer is yes, there is. If the centers of our circles are at cartesian 0,0, then the mapping of each point (x,y) in C1 to a point on D1 will be had with (D/C x, D/C y). Similarly there is a mapping, one-to-one, in the other direction.

So the infinities are of the same size.

This is where my thinking gets fuzzy. Common sense tells me that there must be fewer point pairs inside the smaller C1 target than the bigger D1 dartboard. The one-to-one mapping tells me that there are not fewer point pairs. Which is right? How do I come to the result I feel certain, founded on common sense, is right (Probability = Rc²/Rd², where Rc = 1/2 C= radius of C1 and Rd = 1/2 D= radius of D1)?

I'm going to have to leave it to a better mathematician than I am (I haven't done this stuff for 50 years) to solve my fuzzy thinking, but for now I conclude that the probability in the instance of a two dimensional question (circles) gives P = Rc²/Rd² and in the instance of three dimensional question, spheres, as given, would be P= Rc³/Rd³.

As shown in the figure, my problem is that on a sphere, from point A to point B, we can walk along a great circle ACB, which is in a fixed direction. We can also walk along a latitude circle ADB, which is also a fixed direction. That's East or West. So I said that you can reach another point in different directions, and there are infinitely many compromise directions. There are different distances. And there are infinitely many such directions.

Put the origin O on one of the 4 points, so that it’s coordinates are $P_1=$ (0, 0, 0). Draw the x-axis to through O and the second point (assumed it’s distinct from the first point) and define unity as the distance between these two points. Coordinates of the second point are $P_2=$ (1, 0, 0). Choose the y-axis so that the third point belongs to the plane (Ox, Oy). Coordinates of $P_3=$ are $(a,b,0)$ and we need to have $P_1{P}_3=P_2{P}_3=1$ which implies $a^2+b^2=(a-1{)}^2+b^2=1$ hence $a=1/2,b=\sqrt{3}/2$ (choosing the orientation of the y-axis so that $b>0$).

These 3 points form an equilateral triangle with side length 1. The 4th point has coordinates $(x,y,z)$ and we must have $P_1{P}_4=P_2{P}_4=P_3{P}_4=1$ which implies $x^2+y^2+z^2=(x-1{)}^2+y^2+z^2=(x-1/2{)}^2+(y-\sqrt{3}/2{)}^2+z^2=1$. This gives $x=1/2,y=\sqrt{3}/6,z=\sqrt{6}/3$ where the sign of $z$ is made positive by choosing a suitable orientation of the z-axis.
Since we obtained only one set of 4 points, it’s impossible to find a 5th point at distance 1 from the first four points, unless we add one more dimension to the space (one more axis Ot).

Doing so, we can take the coordinates of the first 4 points to be (0,0,0,0), (1,0,0,0), $(1/2,\sqrt{3}/2,0,0)$, and $(1/2,\sqrt{3}/6,\sqrt{6}/3,0)$. The coordinates of the 5th point are $(x,y,z,t)$ and such that $x^2+y^2+z^2+t^2=(x-1{)}^2+y^2+z^2+t^2=1$ and $(x-\frac{1}{2}{)}^2+(y-\frac{\sqrt{3}}{2}{)}^2+z^2+t^2=(x-\frac{1}{2}{)}^2+(y-\frac{\sqrt{3}}{6}{)}^2+(z-\frac{\sqrt{6}}{3}{)}^2+t^2=1$. This gives $(x,y,z,t)=(1/2,\sqrt{3}/6,\sqrt{6}/12,\sqrt{10}/4)$ where the sign of the 4th coordinate can be taken positive by proper choice of the orientation of the t-axis.

This process can be continued for any number of dimensions, and it can be showed that in the Euclidian space of $n$ dimensions you can have exactly $n+1$ points (and not more) such that all $n(n+1)/2$ distances between pairs of points are identical (and positive). This is called an equidistant set.

Maybe, depending on the definition of “points”. There are two general methods to express this: the “Sailing” method, used by ships (Motor or sail, but sail came first by a long shot) and the more abstract geometric method shown by the URL under this question, so I won’t cover that method again.

Using the Sailing method You have to know two things: 1) Where you are, and 2) where you want to go. Where you are is typically expressed as longitude and latitude if we are talking about the Earth (An oblate spheroid, but close enough to a sphere for this discussion) Latitude expresses position on a north/south basis and longitude expresses position on an east west basis.

Lat and Long are really circles around the earth expressed as hours, minutes and seconds. “Helmsman: Steer 41 25 01N and 120 58 57W.” (Note there are several other ways to express Lat and Long, you can see them all here: Latitude and longitude formats.)

So you need to know those two points first. Now, you need to have a location where you wish to go. That determines distance and direction across the surface of the sphere. It is typically expressed the same way. The intersection of the appropriate Latitude and longitude bands.

So, is this 4 points or 2 points or 3 points? Where the infinite set of points on a Latitude circle intersects with a different set of infinite points on a Longitude circle is the conjunction of these two circles’ set of points at the intersection one point or two overlapping points?

Dammed if I know. You decide.