What kind of distribution would one get by rotating a Galton board with constant angular velocity?
Probability – Rotating Galton Board
pr.probability
Related Solutions
This is a partial answer to Q2, and suggests to me that there is a physical arrangement which would give a yes answer to Q3.
If you use ordinary pins, you can probably get a dyadic approximation with some arrangement. Let me suggest using weighted pins as a partial solution, and then perhaps someone can implement a close enough approximation to a weighted pin with a series of dyadic pins.
So normalize things so that the function f has integral one over the interval [0,1], and is to be approximated by 2^k bins. Suppose p in [0,1] is the fraction of balls needed to represent the function on [0, 1/2], equivalently p is the integral of f from [0,1/2]. Then place a weighted pin very high such that it dumps p of the balls toward the pin over the interval [0, 1/2]. (You may want to put a divider right under this pin so that the ball doesn't jump to the [1/2,1] side.) Now recurse (k-1) more levels. Working backwards from this to get a horizontal arrangement should be clear, and of course one can use the physics of the situation to change the endpoints from dyadic rationals to something more appropriate to the desired function f.
It may be possible to emulate the bias by ever so slight horizontal adjustments of the pins, but you need to place the later pins just so that their bias accomodates the various trajectories of the incoming ball. But of course we have infinite precision pins and balls, so what's to worry?
Gerhard "Likes The Unreality of Mathematics" Paseman, 2011.07.06
James correctly identified percolation theory as the place where something like this is studied seriously. But let's do an elementary calculation.
Each possible path consists of $4n-1$ squares and is uniquely specified by saying which $2n-1$ of the $4n-2$ squares other than the first is vertically above the square before. Thus, there are exactly $$\binom{4n-2}{2n-1}$$ possible paths. Each path appears in a random board with probability $2^{-4n+1}$. Therefore, the expected number of paths is $$2^{-4n+1}\binom{4n-2}{2n-1} \sim \frac{1}{\sqrt{8\pi n}},$$ where the last expression comes from Stirling's formula.
Since the expected number of paths goes to 0, the probability that there is at least one path goes to 0 at least as fast. A quick simulation shows that James is correct that the probability goes to 0 exponentially fast (maybe slightly faster than $2^{-n}$).
Best Answer
In the presence of rotation, the original diffusion process is biased by the centrifugal "force", corresponding to a repulsive harmonic potential. Diffusion in a one-dimensional harmonic potential is solved, for example, here. Although the discussion there implicitly assumes a positive spring constant $f$, corresponding to an attractive harmonic potential, I have checked that the given solution also applies at negative $f$, corresponding to the repulsive case. In short, the distribution continues to be Gaussian, but with an increased width compared to the non-rotating board.