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In Bernoulli percolation on the square lattice $\mathbb{Z}^2$ every edge is kept with a probability $q$ and erased with probability $1-q$. It is classical that whenever $q > \frac{1}{2}$ there is an infinite cluster of edges that has some positive density.
In particular, there is a probability that $0$ is part of an infinite path. See https://www.ihes.fr/~duminil/publi/2017percolation.pdf for more on Bernoulli percolation.

Now, suppose you know $q$ and a configuration with this $q$ is sampled, but you do not get to see it. Instead, you start at zero and you can start walking along the edges, but you destroy the edges that you walk along. Every time you get to a new vertex you get to see the edges incident to that vertex and choose which one to walk along (and destroy).

Suppose that you start in the infinite cluster, what is the optimal probability $p(q)$ that you can achieve for walking on an infinite path?

I suspect that it is often 0. If not, what are good algorithms for this exploration?

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  • $\begingroup$ How are the steps in your walk chosen? If you chose them at random, then you will get into a trap and the walker will find itself in a finite cluster with probability one. $\endgroup$
    – Viktor B
    Oct 10, 2022 at 11:37
  • $\begingroup$ You get to choose them based on what you have seen so far. $\endgroup$
    – Frederik Ravn Klausen
    Oct 10, 2022 at 11:42
  • $\begingroup$ You may also use your of your knowledge of $q$ and some randomness if you prefer. $\endgroup$
    – Frederik Ravn Klausen
    Oct 10, 2022 at 11:43
  • $\begingroup$ If you only have the information about what you have seen, then you will get stuck with probability one. Each step may be the last step, you have no way of knowing that. $\endgroup$
    – Viktor B
    Oct 10, 2022 at 12:28
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    $\begingroup$ Just a comment. You are picturing the entire configuration as being sampled in advance. An alternative model that is completely equivalent is that you just randomly sample the bonds coming out of the vertex each time you arrive at a new vertex. $\endgroup$
    – Anthony Quas
    Oct 10, 2022 at 22:54

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