Probability of recurrence for a random walk in $\mathbb Z^3$

Question

Let $\mathbf X(n)$ be a random walk in $\mathbb Z^3$ in the following sense: We start at the point $\mathbf 0=(0,0,0)$ and for each step, we randomly decide in which of the three directions we move by $\pm 1$ step (i.e., there are $6$ possibilities for each step, each with probability $1/6$).

It is well-known that this random walk is transient, i.e. $\mathbf P(\mathbf X(n)=\mathbf 0\text{ for some }n\geq 1) \neq 1$. My question is if this probability can actually be calculated; or if there is some result on what this value is. All sources I found only mention the $\neq 1$ part, but do not comment on the actual value.

Answer 1

Theorem 13.9 of https://www.math.ucdavis.edu/~gravner/MAT135B/materials/ch13.pdf gives the same expression as Eric, and states that it implies the probability of return on $\mathbb Z^3$ of $$0.3405$$ rounded to four decimal places. Meanwhile https://www.abelprize.no/c76018/binfil/download.php?tid=76125 gives the same figure without as much justification

Answer 2

Applying Lemma 2.21 and the formula in Theorem 2.22 of https://www.math.ucla.edu/~biskup/PDFs/PCMI/PCMI-notes-1, we get that the average number of visits to the origin is:

$$1/(2\pi)^3 \int_{[-\pi,\pi]^3} 3/(3-\cos(x)-\cos(y)-\cos(z))\,dx\, dy\, dz$$

That’s around $1.52$.

The probability of going off to infinity is the inverse of that, so around $66\%$.