Random walks - What is the probability that a path, for most of the time, is close to the origin?

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Let the grid of our possible positions be $\mathbb Z ^2$. Let $(0,0)$ be our starting point.
In each turn, we move with equal probability either one up, down, to the left or to the right.

Let $r\in\mathbb{N}$ be the maximal distance regarding $||\cdot ||_1$, which we see as "close to the origin" .
We define $A_n$ to be the event that a random walk of length $n$ is for more in half of its steps in the square $S:=\{i\in\mathbb N: ||i||_1\le r\}$

I'm looking for the probability that an infinitely long random walk is for more of half its steps in the square $S$, i.e. for the value of: $$ \lim_{n\to\infty} \mathbb P (A_n) $$

Even finding a useful formal characterization of $A_n$ strikes me as difficult.
One not that's probably not that useful is the following.: $$ A_n =\{\pmatrix{\binom {i_{1,1}}{i_{1,2}}\\...\\ \binom{i_{n,1}}{i_{n,2}}}\in(\mathbb Z^2)^n: \binom {i_{1,1}}{i_{1,2}}=\binom 00 \quad\land\quad \forall k\in \{1,...,n-1\}: \left|\left|\binom{i_{k,1}}{i_{k,2}}-\binom{i_{k+1,1}}{i_{k+1,2}}\right|\right|_1 =1 \quad\land \quad \sum_{k=1}^n \mathbf{1}_{ S}\binom{i_{k,1}}{i_{k,2}}\ge \frac n2\} $$

For which we'd have to define for each $n$ the probability space $(\Omega_n,P(\Omega_n),\mathbb P_n)$ with $$\Omega_n:= \{\pmatrix{\binom {i_{1,1}}{i_{1,2}}\\...\\ \binom{i_{n,1}}{i_{n,2}}}\in(\mathbb Z^2)^n: \binom {i_{1,1}}{i_{1,2}}=\binom 00 \quad\land\quad \forall k\in \{1,...,n-1\}: \left|\left|\binom{i_{k,1}}{i_{k,2}}-\binom{i_{k+1,1}}{i_{k+1,2}}\right|\right|_1 =1 \}$$

For the classic formulation with the result space for $A_n$ being $\{\{0,1\}^n\}$ the characterizations for $A_n$ look, as far as I can tell, even more difficult.

What is the value of $ \lim_{n\to\infty} \mathbb P (A_n) $?

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$\mathbb P (A_n)$ is at most twice the average probability of the random walk to be in $S$ (averaged over the time steps up to $n$). The probability for the walk to be in $S$ goes to $0$ for $n\to\infty$, and thus so does its average over the time steps up to $n$, and thus so does $\mathbb P (A_n)$.