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.
________________________________ From: Fred Lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Tuesday, September 17, 2013 4:47 PM Subject: Re: [math-fun] Probability that all molecules of a gas are in one half of the container
I'm not saying that Veit's method is wrong --- but it's not obvious that it doesn't fail as a result of a Bertrand-style paradox. WFL
On 9/17/13, Veit Elser <ve10@cornell.edu> wrote:
On Sep 17, 2013, at 3:02 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
But what is the probability if any hemisphere is allowed? I'm stuck on this problem.
In two dimensions, the probability that all n molecules lie in some semicircle is 2n/2^n.
Here's a trick that works in any dimension. Sample the points in two stages. First sample a set of n axes that pass through the origin. Next sample one point on each axis. It turns out that the homo-hemisphere property doesn't depend on the choice of axes at all, only on which side of the origin each point is placed in the second stage. Calculating your probability is now an exercise in combinatorics.
-Veit
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