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Comment by jochi427 | Hacker News Reader

I remember learning about the probability of returning to the origin in a 2D random walk versus a 3D random walk when I took stochastic processes. After we proved with probability 1 you return to the origin in a 2D walk (and with probability 0 you return in 3D) my professor said "that's why you hand a drunk man the keys to a car and not an airplane when he leaves the bar". After checking wikipedia it looks like he riffed off this quote from Shizuo Kakutani: "A drunk man will find his way home, but a drunk bird may get lost forever".

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That's interesting, about the probability being zero in 3D. Is this on an integer lattice? The source that cannot be cited on HN without loss of karma says that the probability of returning to the origin in Z^3 is approximately 0.34.

I don't see how it could possibly be zero, even for reals, unless you're relying on the idea that the probability of any given real emerging from a uniform RNG is zero. That would seem to apply in 2D as well.

I'm sure I am just misremembering -- it was definitely on Z^3 so I guess its actually 34%. Thanks for letting me know

Here's how to formulate the question in continuous space/time:

Random walks can be defined on continuous space and time as a probability distribution on functions R -> R^n (Brownian motion in n dimensions).

We can then ask whether Brownian motion beginning at the origin will ever revisit it i.e.

Given 2D Brownian motion X such that X(0)=(0,0), the probability that there exists t>0 such that X(t)=(0,0) is 1.

Given 3D Brownian motion X such that X(0)=(0,0,0), the probability that there exists t>0 such that X(t)=(0,0,0) is 0. (This is more clearly true when it doesn't begin at the origin, but it's almost certainly not at the origin at t=1, and you can divide the half open interval (0,1] into a countable number of intervals, each of which have 0 probability of passing through the origin.)

Random walks in 2D are space filling curves; random walks in 3D are not.

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