# Calculating probabilities on a spherical map

A black and white colored sphere is given.

We are looking at a random starting point on the sphere below us, which has a certain color. A random rotation can change the color of the spot below us.

What is the probability P that a random rotation of the sphere will change the color of the spot below us from black to white?

Is the following calculation correct:

P = black_surface_fraction * white_surface_fraction

where the "surface fraction" gives the area percentage of the sphere that is black or white?

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You posed two related questions (math.stackexchange.com/questions/38455/…); it would make sense to link them to each other. – joriki May 11 '11 at 14:13

## 2 Answers

This answer assumes that by "random starting point" you mean a starting point drawn from a uniform distribution on the sphere with respect to surface area and by "random rotation" you mean a rotation drawn from a uniform distribution on $SO(3)$ with respect to the Haar measure. In this case the probability you give is correct.

If the points resulting from the random rotations are distributed uniformly, independent of the starting point, then the probability of starting at a black point is given by the black surface fraction, and the probability of ending up at a white point is independently given by the white surface fraction, so the probability of changing colour from black to white is just the product of the two. So we have to show that a uniform distribution with respect to the Haar measure leads to a uniform distribution of the resulting points on the sphere with respect to the surface area, independent of the starting point.

This follows from the invariance of the Haar measure under group multiplication. Fix some point $Q$, and then, given some resulting point $P$, pick some rotation $R$ that rotates $P$ into $Q$. Since rotations are isometries of the sphere, a disk around $P$ will be rotated to a disk around $Q$ with the same surface area. Since the Haar measure is invariant under group multiplication, the measure of the set of rotations that rotate the starting point into the disk around $P$ is the same as the measure of the set of rotations that rotate the starting point into the disk around $Q$. Since $Q$ is fixed, this is in fact independent of $P$. By taking the radius of the disks to $0$, we can conclude that the distribution of the resulting points is uniform.

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You speak of a "random starting point" and a "random rotation" of the sphere, but you must define exactly both terms. The first one seems to admit a rather obvious interpretation, but the second (the rotation) is more delicate. What is "random rotation" ? Different definitions (thought seemingly resasonable) might lead to different results. The Bertrand paradox is a related example.

In general, this is expressed with conditional probabilities:

$P(change) = P(x = B \wedge y = W) + P(x = W \wedge y = B)=$ $= P(y=W | x=B) \; P(x=B) + P(y=B | x=W)\; P(x=W)$

where $x$ is the first point color, and $y$ the second.

If your "random rotation" amounts to selecting a new point with uniform probability over the sphere surphace (independent of the starting point) (but this is not an obvious assumption!), and if the starting point is also selected with uniform probability, this results in:

$P(change) = P(y=W) P(x=B) + P(y=B) P(x=W) = 2 p(B) p(W) = 2 p(B) (1-P(B))$

where $p(B)$ is the white surface fraction .

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The most natural interpretation of a "random rotation" is given by the Haar measure on $SO(3)$. I interpreted the question to in effect ask whether a rotation drawn from a uniform distribution with respect to the Haar measure leads to a uniform distribution of endpoints on the sphere with respect to surface area. (Numerical experiments indicate that it does, but I'm trying to prove it.) – joriki May 11 '11 at 13:53
Right. Obligatory wikipedia link: en.wikipedia.org/wiki/… – leonbloy May 11 '11 at 14:02
You calculated twice the desired probability, since the question only asks for the probability of changing from black to white and not vice versa. – joriki May 11 '11 at 14:53
Thank you for all this! Where can I read more about the Haar measure? – Marc May 11 '11 at 15:05
– joriki May 11 '11 at 15:14