So, I'm not ready to sit down and try to do the calculus, but surely the probabilities for 3, 4, 5, and 6 are all the results of the sums of a few definite triple integrals. For 3 in particular, there's really just one triple integral, whose value has to be multiplied by 8 to get P(3). That integral gives the probability of a triangle being clipped off by any particular corner. The integral is a little bit hairy because it's a spherical one, and needs to have the "d theta d phi" term adjusted to give constant density over the sphere. This problem would have been fine on an 18.01 problem set, no? On Sun, Oct 15, 2017 at 9:53 PM, James Propp <jamespropp@gmail.com> wrote:
I googled for five minutes and looked at half a dozen articles; I saw various models of random polytopes, but none of them appeared to be of the form "choose a random plane that intersects a polytope nontrivially and then take that intersection".
I also didn't find the resilt about random polygons that Warren mentioned (the one that says that a random polygon has 4 sides on average). So if Warren or anyone else can provide more leads, I'd appreciate it!
Jim Propp
On Sunday, October 15, 2017, Warren D Smith <warren.wds@gmail.com> wrote:
I believe there is a known published model of "random polytopes" in which the exact expected value is 4.
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