Thanks to Gene and Dan and NeilB, Gary regained his nerve and ran the problem, with, I think, considerable success, even though it completely stumped his regulars. From whom I'm hoping to hear some howls of incredulity. http://wordplay.blogs.nytimes.com/2014/07/07/wcup2 (Click on Enhancer.) On p4/7 of my "solution" I quoted Dan's, below, which became badly garbled when I pasted it in. Hopefully, Gary will be able to fix it. I forgot to mention using this result with a physical ball as an off-the-wall π estimator. --rwg On Sat, Jul 5, 2014 at 6:06 AM, Bill Gosper <billgosper@gmail.com> wrote:

Dan Asimov wrote: <

RWG wrote: <

For a ceremonial match, a soccer ball is colored with a map of the Globe instead of a traditional pattern. At the start of the match, the ball rests on its south pole, with its lat 0, long 0 point aimed due east. At the end of the match, the ball lies forgotten on the pitch, in a truly random orientation. For any orientation, there will always be an axis through the center about which a single rotation will restore the original north up, 0,0 east orientation. What is the (surprising) expected magnitude of this required rotation?

...

RWG SPOILER SPACE

.....

.....

.....

.....

.....

.....

.....

.....

.....

.....

.....

.....

SOLUTION: ---------

We want to find the average distance in the rotation group SO(3) between an arbitrary rotation and the identity, where distance is measured by the angle t of the rotation, 0 <= t <= pi.

(*) Every rotation f in SO(3) is represented by either of exactly two unit quaternions, which are antipodal.

This works by first identifying S^2 with the unit sphere in S^3 that is the equator with respect to the poles +-1. Then for any x in S^2, the rotation f_q(x) is defined by

f_q(x) = q' x q

where q' denotes the reciprocal of q (for a unit quaternion, just its conjugate).

If t is the least angle between q and the real axis (so that 0 <= theta <= pi/2), then the rotation f_q is by an angle of 2t.

We can now find the average rotation angle from the identity among all rotations in SO(3). By (*), SO(3) corresponds almost exactly to the hemisphere Hem^3 of S^3 nearest to the identity . . . except that those points farthest from the identity, forming a copy of a unit 2-sphere, need to be identified by the antipodal map to get just one quaternion for each rotation. But since this 2-sphere is measure 0 inside Hem^3, we can ignore this nicety to find the average rotation angle.

All the quaternions q of Hem^3 at angle t > 0 from the identity form a locus that is a 2-sphere of radius = sin(t). For all these q, the rotation angle of the rotation f_q is 2t., and their 2-volume is 4 pi sin^2(t).

The total 3-volume of the (unit) 3-sphere S^3 is 2 pi^2, so that of Hem^3 is just pi^2. So we get for the probability density of the angle t

d(t) = 4 pi sin^2(t) / pi^2

and so the expected rotation angle is

E(2t) = Integral_{0,pi/2} (2t) 4 pi sin^2(t) dt / pi^2

= (8/pi) Integral_{0,pi/2} t sin^2(t) dt

= pi/2 + 2/pi .

Which is pretty mystical. Makes you wonder if there isn't some good reason it turns out like that.

--<Dan

In perfect agreement with http://people.csail.mit.edu/bkph/other/Stanford_AI_WP_79_Salamin.pdf .

So *two* of you would've gotten it, if you're both Numberplayers!-)

What a neat variation on Buffon's needle--no stripes required. No ball required

either if we Monte Carlo. Of course, it might be hard to write the latter w/o

using π. If so, maybe it would converge if we start with 3 and modify the Monte

Carlo as our knowledge of π improves.

--rwg