For f <= 1/2 the probability is n f^(n-1), i.e. the enhancement factor from allowing any semicircle vs. a fixed semi circle is n/f. I haven't solved the problem for f > 1/2, but my intuition agrees with yours on the location of the breakpoints. -- Gene
________________________________ From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Tuesday, September 17, 2013 7:00 PM Subject: Re: [math-fun] Probability that all molecules of a gas are in one half of the container
Thanks, Veit. I almost understand this.
In any case I'll add an interesting footnote: On the unit circle C one can ask, What is the probability that for n points chosen at random from C, there exists some arc of length 2pi*f (i.e., a fraction f of the circumference) that contains all n points. Denote the answer by P(f,n).
For a fixed n, as f approaches 1 the formula P(f,n) passes through a countable number of fractions f_1 < f_2 < . . . < f_k < . . . < 1 at which the P(f_k,n) is not differentiable, even though it's continuous everywhere in f. (I think the f_k are at 1/2, 2/3, 3/4, 4/5, etc.)
--Dan
On 2013-09-17, at 6:35 PM, Veit Elser wrote:
On Sep 17, 2013, at 8:05 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
The distance of each particle from the center can be individually scaled by positive numbers without affecting the "contained in a hemisphere" property. Therefore, I do not see how Veit's suggestion is helpful. A little more detail, please.
We have n axes through the origin and these define n orthogonal hyperplanes, also passing through the origin. In 2D we have n lines that subdivide the unit circle into 2n arcs, in 3D we have n planes that divide the unit sphere into N(3,n) spherical polygons, etc. Let's put off for now the evaluation of N(3,n).
The significance of the sphere subdivisions is the following. When you make the binary choice for each axis on the placement of the point with respect to the origin you are adding a constraint on the position of the common hemisphere. Specify the position of the common hemisphere by its "north pole". Suppose this north pole lies within some spherical subdivision. This uniquely determines the binary choice on each axis so that the corresponding point lies within the common hemisphere. So how many of the joint binary choices have _some_ common hemisphere? Answer: the number of spherical subdivisions. (There is a bijection between the set of valid axis choices and the subdivisions of the sphere by hyperplanes.)
The probability there exists a common hyperplane is therefore N(d,n)/2^n, which equals 2n/2^n in 2D. Finding the number of spherical subdivisions for arbitrary d and n involves writing recursion formulas and recognizing the solution is a sum of binomial coefficients. In 3D there is a direct approach using Euler's formula. With n intersecting planes we get V=n(n-1) vertices on the sphere and E=2(n-1)n arcs that bound F spherical polygons. From Euler we get F=2-V+E=n^2-n+2=N(3,d). The probability is therefore (n^2-n+2)/2^n.
-Veit
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