2

There are four points $A$, $B$, $C$, $D$ which randomly selected on a sphere. Find probability of intersection of shortest arcs (not circles) $AB$ and $CD$.

Shortest arc is a intersection of plane (based on $A$, $B$ and $O$) and sphere. But intersection condition is very hard to formulate.

There is same question: Intersection of two arcs on sphere, but answer is numerical.

Community
  • 1
Michael Galuza
  • 4,730

1 Answers1

1

It would be nice if it was $\frac18$.

Any two distinct great circles intersect at two antipodal points. Take one of the great circles and one of the points of intersection: the probability that the shortest arc between two randomly chosen points includes the point of intersection might be $\frac14$ (consider the two randomly chosen points and their antipodal points splitting the circle into four).

But you also want the probability for the other two randomly chosen points, and there are two points of intersection so the answer might be $\frac14 \times \frac14 + \frac14 \times \frac14$

Henry
  • 157,058
  • It' so simple? I thought that we must find distribution for $f(\theta_1, \varphi_1, \theta_2, \varphi_2)$ and integrate it. Use symmetry, Luke) – Michael Galuza Jun 07 '15 at 12:06
  • I'm not yet convinced that this is correct. If you let $A$ and $B$ be randomly chosen, and then call $\Gamma$ the great circle passing through them, it's not clear at all that $A$ and $B$ are distributed uniformly on $\Gamma$. The reason for this is that the determination of $\Gamma$ depends on $A$ and $B$; it's not as if you were first selecting $\Gamma$ randomly, and then $A$ and $B$ randomly on it. For example, $A$ and $B$ have a smaller chance of being within one degree of each other when being randomly chosen on a sphere than when randomly chosen on a circle. – Keith Jun 08 '15 at 04:53
  • Okay, the answer works out to be correct anyway. The distribution of $A$ and $B$ is not uniform on $\Gamma$. However, the probability of the point of intersection being contained within the arc from $A$ to $B$ is equal to (angle of arc from A to B) / (360 degrees). And the arc from $A$ to $B$ has length $90^{\circ}$ on average, so this probability is $1/4$. The answer is the same in the end. – Keith Jun 08 '15 at 05:06
  • @Keith: Good. All you need is for the angle between $A$ and $B$ to be as likely to be $\theta$ as to be $180^\circ-\theta$. – Henry Jun 08 '15 at 06:14