Your walk is always transient when you use the symmetric condition of reflection in both directions, and the argument is essentially given in the previous answer. But when the condition of reflection is *asymmetric*, so that you have reflection only when you are going out, an infinite number of reflectors will give recurrent behavior for arbitrarily small reflection probability in Euclidean space (LATER EDIT: so long as the $R_k$ do not grow exponentially fast--- in this case, you can have a transition--- you also have a good transition on the Cayley graph).

### Electrostatic criterion for recurrence

The recurrence/nonrecurrence of a random walk is a time-independent problem, and can be solved by finding a steady-state solution. This has a simple electrostatic analog, see this answer: http://physics.stackexchange.com/questions/8149/collision-time-of-brownian-particles/14837#14837

The basic principle is as follows: consider a small sphere and a large sphere which absorb Brownian particles. Inject particles at a random position in a sphere of unit radius, and ask what is the probability that they are absorbed by the large sphere as opposed to the small one. Recurrence is when the small sphere always absorbs first, in the limit of an infinite large sphere.

In steady state, the particle density obeys Laplace's equation with the boundary condition that the potential function is zero on the small inner sphere, and on the outer sphere, with a kink on the unit sphere which represents the incoming flux of injected particles.

The flux of particles which are first absorbed at infinity is given by the integrated gradient of the potential (the electric flux) at large distances, and the flux which is first absorbed by the small sphere is given by the gradient of the potential at small distances (the electric flux at small distances). The flux is normalized by the values of the electrostatic potential at the small and large radius (because the potential has to vanish on these two spheres, because they are absorbing).

The upshot of this is that the walk is recurrent if and only if the potential from a small sphere is divergent at infinity (Later edit: this is true for homogenous lattices--- see below). This is true in 1d, where it is linearly divergent, and in 2d, where it is log divergent, but fails in 3d and above, where you approach a constant asymptotic limit at long distances. The constant limit makes the flux at infinity nonzero, because the normalization from the zero potential metal boundary condition is not infinite.

### One Way Reflectors

When you add two-way reflectors, the steady state distribution is unchanged, because if you have an infinitesimally thin reflecting plane, in order for there to be zero flux through it, the density of walkers on the left and on the right of the plane have to be equal. This means that the plane does not change the flux, and there is no difference from the no-reflector case. The walk is still transient with two-way reflectors (this is the content of the previous answer).

When you have a *one way* reflector, however, the condition is different. Now the density on the interior has to be bigger right on the reflector by a ratio of (1/1-p) compared to the exterior to allow the flux through the plane to balance. Call the small ratio $A_p = 1-p$.

In order to have zero flux to infinity in steady state, the distribution of particles with a unit source must have zero r-derivative away from the jumps at the reflectors. This means that all the density falloff must come at the jumps, and the condition that the sphere at infinity is at zero potential gives the restriction:

$$ \prod_k A_{p_k} = \prod (1 - p_k) = 0 $$

When this infinite product is zero the walk is recurrent. When it is not zero, the walk is transient.

EDIT: Or so I thought! The last equation is obviously completely wrong, as pointed out by Peter Shor, and seconded by anonymous downvoters. Thank you for catching the mistake I was blind to.

There is nothing wrong with the method above, it is only that asymptoting to zero potential at large distances does not guarantee that an infinitesimal flux will go to zero at a finite time (this was counterintuitive for me). If you asymptote to zero slowly enough, you can still have a nonzero flux at large radii without ever hitting zero.

The correct condition on the big conductor is that it must have zero flux through it when it is very large. For normal growth rates, to get zero flux, all you need is the potential to go to zero. But for exponentially fast growing $R_k$, to zeroing out the potential on a large sphere, you can still require a nonzero flux even if the sphere starts out at an infinitesimal potential, just because it is so enormously huge (thanks again to Peter Shor for discovering this stupid error--- it does not require a modification to the method, only to the faulty analysis at the very end).

So one asks, given that there is an outward flux $q$, what is the radius of the sphere at which the potential is first zero? If there is no such radius for small enough $q$ (it is allowed to asymptote to zero, so long as it never reaches it--- this was my mistake of before) then the random walk is recurrent.

The outward flux is the electric charge, so the solution going out is of the form

$$ \phi(r) = {q\over r} + A $$

At $R_1$, it is attenuated by $a_1 = 1-p_1$, so that

$$ \phi(R_1) = a_1 ( {q\over r} + A) $$

It continues along for larger r with the same q, but starting at the attenuated height, so that the solution for $r>R_1$ is:

$$ \phi(r) = ({q_1\over r} + A') $$

Where the constant A' is given by

$$ A' = a_1 A + (1-a_1) (- {q\over R_1}) $$

i.e., it is the weighted average of the previous value of A with the negative value in parentheses, with weight given by $a_1$. The q can be factored out if you redefine A multiplicatively to absorb it. The value of A after each of the transitions is given by a similar weighted average

$$ A_{n+1} = a_n A_n + (1-a_n) ( - {1\over R_n} ) $$

This linear difference equation can be solved by standard methods, in particular defining

$$ A_n = B_n \prod_{k=0}^n a_k$$

Then

$$ B_{n+1} - B_n = { (1-a_n)\over \prod_{k=0}^n a_k } ( - {1\over R_n} ) $$

And the condition that A becomes negative after a finite number of steps for arbitrarily small q tells you that the above series is divergent:

$$ \sum_{n=1}^\infty {(1-a_n) \over \prod_{k=0}^n a_k } {1\over R_n} = \infty $$

The divergence of this series is the condition for recurrence. In the special case of constant $a_k = 1-p$, then the series has terms

$$ {p\over (1-p)^n} {1\over R_n} $$

Which is a diverging exponential unless R_n is growing faster than $1\over (1-p)^n$. So for exponentially growing $R_n$, you *do* get a nontrivial phase transition, contrary to what I wrote initially.

### MORE OF LATER EDIT: Cayley Graph

On the Cayley graph (infinite binary tree), instead of 3 space, the problem does have a phase transition in p, because by having every radius be one-way reflecting, and tuning p, you can make the radial walk unbiased after a path-dependent time reparametrization. There is a steady stream outward with probability 2/3, but if p is 1/2, then half of the outgoing 2/3 is brought back on the next step, so that you get a 1/3 probability for going inward, a 1/3 probability for going outward, and a 1/3 probability of coming back after two steps, which is an standard unbiased diffusion after you merge the two steps on the returning path into one.

So for p<1/2 you have transience, and for p>1/2 you have rec