Does a simple symmetric random walk in an orthant always exits the orthant with probability 1?

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Let $$S = \{(x_1,x_2,\ldots,x_d) \in \mathbb{Z}^d : x_i > 0, i=1,\ldots,d \}$$ and consider a simple and symmetric random walk starting in a point $x_0 \in S$.

I wish to know more properties about the probability $P(x_0)$ that this random walk always stays in $S$. From Polya's recurrence theorem, I know that $P(x_0) = 0$ for $d=1$ and for $d=2$.

What can be said for $d \ge 3$?

  1. In particular, is there an explicit expression for $P(x_0)$?
  2. If not, it is true that $P(x_0) > 0$ for all $x_0 \in S$ ?
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There is at least one direction in which the walk takes infinitely many steps. (In fact it almost surely takes infinitely many steps in all $d$ directions, but we don’t need that.) The movement in this direction is a simple symmetric random walk in one dimension. The probability that it stays on the same side of the origin forever is $0$. Thus the probability that the $d$-dimensional walk stays in the same orthant forever is also $0$.