Most likely length of random lines inside a sphere.

Question

Take a sphere in $\Bbb R^3$ with diameter $d$.
Now secants are drawn randomly through the sphere.
Consider tangents as secants here. Any secant is equally likely. Also this.
$L$ is the length of such a randomly drawn secant inside the sphere. (For tangents $L=0$)
What is the most likely value for $L$?

I have an explanation here which says that the answer is

$d$, the diameter of the sphere

which is somewhat puzzling, since it is

the longest possible distance.

Could anybody explain this without using too many terms from probability theory?

Answer 1

WLOG, $d=2$. Starting from the premise that all secants are equally likely, the secant direction doesn't matter and WLOG we can assume them to be vertical.

Then we assume that the piercing point of the secant in the equatorial plane is uniformly distributed.

The probability of a piercing point being at distance $r$ from the center is $\dfrac{2\pi r\,dr}\pi$, and this corresponds to a secant length $2\sqrt{1-r^2}$. Conversely, the probability of the length being $l$ is

$$2\sqrt{1-\frac{l^2}4}\frac{\dfrac{2l\,dl}4}{2\sqrt{1-\dfrac{l^2}4}}=\frac l2\,dl.$$

The most probable length is indeed achieved for $l=2$, while the average length is

$$\overline l=\frac{\int_0^2l\frac l2dl}{{\int_0^2\frac l2dl}}=\frac43.$$

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An intuitive explanation of the paradox is that the probability of a given length is proportional to the projection on the equatorial plane of a slice of the surface of the sphere of height $\Delta l$. The area of that projection is maximal at the pole (projection factor close to $1$) and null at the equator where the surface is vertical.

While $l$ grows, the projection factor increases, but the area of the ring decreases. It turns out that the first effect (green) is stronger than the second (blue), so that the projected area just grows linearly (red).

Answer 2

Since you can't possibly determine what distribution the drawing machine follows, take it to be most like to follow Gaussian distribution. (Since most random unclassified things like errors and noise etc.. follow Normal Distribution). Take $-\infty $ to be the centre (since a circle is symmetric about its centre) and $+\infty$ to be the other end of a radius(on the circumference).

It means that the most probable length of the secant is $$2\sqrt{\left\{{d\over 2}\right\}^2-\left\{{d\over 4}\right\}^2}$$

Even if you consider things fair and to follow Uniform distribution, The expectation would be the same i.e. the secant at a distance of half the radius from the centre.

Answer 3

The argument that the diameter has the greatest probability derives from how you define "choose randomly".

If you place a unit circle around the origin of an $(x,y)$ plane and then pick a random point along the x-axis then the most likely point to be chosen on the circle is where it meets the x-axis, at $y=0$ and $x=\pm 1$ because if you imagine projecting the points on the circle vertically downwards onto the x-axis, they are densest at that point.

If however you define "choose randomly" as picking a point on the y-axis then the most likely point to be chosen is at $x=0$ and $y=\pm 1$, because you project horizontally and you get the converse result.

If however you define "choose randomly" as choosing a random angle around then origin then all points become equally likely.

The moral of this story is to make sure you choose a method of "random choice" which is unbiased in respect of the result you are attempting to quantify.

Answer 4

Since the circumference at the widest point of the sphere is the longest circumference, it has the greatest probability space of all secants. However... this is not strictly true. Paradoxically, there is only one diameter while there are two of the point infinitessimally either side of the diameter, so the most likely length is actually in the limit asymptotically approaching the diameter.

Answer 5

Now that the distribution has been clarified by your link to MathWorld, the problem is well-defined; the solution is in fact linked to in the "See also" section of that MathWorld article: Sphere Line Picking. It gives the density of the secant lenght $l$ as $\frac12l$ (for the unit sphere), so indeed the longest possible length is the most likely.

An intuitive way to see this is to fix one point at the north pole and to note that each Cartesian coordinate of a point uniformly randomly picked on a sphere has uniform density. Applying this to the $z$ coordinate and noting that the length changes most slowly with $z$ at the south pole then yields the result.