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Consider a random work on $L=\mathbb Z^2$ endowed with obstacles (i.e each cell $(x,y)$ of $L$ may contain a obstacle, i.e the random walk halts whenever it hits such a cell). Let $P(x,y) = 1$ if cell $(x,y) \in L$ contains an obstacle, and $0$ otherwise. Thus $P$ is a parameter of the problem. Finally, let $\theta(P) \in [0, 1]$ be the probability that the walk started at the origin $(0,0)$ eventually returns there.

Question. Given $\alpha \in [0,1)$, what are general conditions on $P$ which ensure $\theta(P)>\alpha$?

For example, it is known since Polya (see page 9 of this article) that, if $P(x,y) = 0$ for all $(x,y) \in L$ (i.e there are no obstacles), then $\theta(P) =1$.

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    $\begingroup$ A random word, a random work, or a random walk? :-) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 30, 2021 at 15:02
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    $\begingroup$ More to the point: if there is at least one obstacle, there is non-zero probability the random walk will hit it before returning to zero, and so $\theta(P) < 1$ — if I understand correctly the question. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 30, 2021 at 15:05
  • $\begingroup$ Thanks for the input. I meant to ask $\theta(P) > 0$ (and not $\theta(P)=1$). Fixed. $\endgroup$
    – dohmatob
    Commented Jul 30, 2021 at 15:09
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    $\begingroup$ I do not think I have anything interesting to say with regard to Q2, which seems to be a very interesting question, but Q1 again seems to be flawed: if there is a positive probability that one of the neighboring states of the origin is not a trap, then clearly $\theta(P) > 0$, and otherwise $\theta(P) = 0$.. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 30, 2021 at 15:38
  • $\begingroup$ Indeed Q1 was still flawed. Thanks for spotting this. Now removed. Thus my only question is Q2. $\endgroup$
    – dohmatob
    Commented Jul 30, 2021 at 17:44

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