```From:           Robin Chapman <rjchapman@mailandnews.co.uk>
To:             sci.math
Subject:        RE: Geometrical probability question.
Date:           Fri, 20 Jul 2001 09:48:00 -0700
```

```Original Message From mathwft@math.canterbury.ac.nz (Bill Taylor):
> There are some good geometrical probabilists around here, so I'm
> betting one of them can answer this, which I don't recall seeing
> before, though it seems quite "natural"...
>
> Three points are chosen independently at random on a homogeneous
> projective plane; (equivalently, on a sphere with opposite points
> identified).  Each pair of points is connected by the shortest great
> circle segment possible.
>
> What is the probability that the loop so formed goes globally round the PP?
> [i.e. that it cannot be contracted to a point.]

Well this is bludgeon, but it works. Wlog take one point to be the
north pole and the other two points chosen at random (uniformly wrt
to surface area) in the northern hemisphere. The probability se
seek is that of these two points having an angular separation of at
least pi/2. Let theta be the angle from the first point to the
north pole. The probability that the 2nd point has a distance pi/2
from the first is proportional to the area of the intersection of
the northern hemisphere with the hemisphere opposite the 1st point
which is  2R^2 theta. The probability is of this thus theta/pi. Hence
the probability we seek is integral theta/pi dS over the northern
hemisphere divided by its area. Using spherical polars this is
(1/pi) int_0^{pi/2} theta sin theta d theta = 1/pi.

------------------------------------------------------------
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen
```