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Comment by Xcelerate | Hacker News Reader

One might ask the question: what is the probability that you will return to your starting position over the course of an infinite random walk? On a 1 dimensional or 2 dimensional lattice, that probability is 1. What's crazy though is that for a 3D lattice, the probability is not 1 — it's about 0.3405.

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I really love this proof. It's a great example of using maths to prove a counter-intuitive result. They way to prove it is rather clever, and made me appreciate what mathematicians do a lot more.

Shame I've never seen it shared online. I was actually hoping the submitted article was a proof of this, but you can't have everything in life.

Wow, do you have any proofs for this? I'm especially curious about the generalized n-dimensional case.

There is some more information and references here: http://mathworld.wolfram.com/PolyasRandomWalkConstants.html

If you find it GP, I'd love to see where the integral constructed comes from, since that's the clever part rather than the evaluation.

Here's a reference I found for one way to do it: http://www.math.nus.edu.sg/~matsr/ProbII/Lec6.pdf (Theorem 2.1). You define the Green's function G(x, y) = \sum_n Pr_x(S_n=y), where x and y are 3-vectors and Pr_x(S_n=y) is the probability that an n-step random walk starting at x ends up at y. If you have an infinite random walk starting at 0, then G(0, 0) is the expected number of times that the walk returns to 0. That's what the mathworld link calls u(3). You can use Fourier inversion to compute G(0, 0) -- the link gives the gnarly details. It's pretty cool.

You're a scholar and a gentleman, merci buckets

The probability of returning to the origin follows a very smooth logarithmic curve for dimensions 3 - 8 [copied values from the mathworld link]

image: http://imgur.com/CL8MXej

Thanks a lot. Had to save this for further reference the moment I saw it.

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