Alternate proof for two random walks in $d$ dimension meeting

Question

I was coming up with an alternate approach to finding the probability of two random walks starting at origin in $d$ dimensions. Let $S_n$ be the Manhattan distance between the points at time $n$. Since both points change the Manhattan distance by $\{ -1,1 \}$ with equal probability, $S_n$ is a random walk with step $X_n$ being $2$ and $-2$ with probability $0.25$ each, and $0$ with probability $0.5$.

Now, since $S_n$ is a one-dimensional random walk, independent of the original dimension $d$, it must return to the origin. If $S_n$ is $0$, it means the random walks intersect with probability $1$.

But doesn't this contradict a result by Polya as given in the Wikipedia link for random walk. Any help is appreciated.

Another variation of this question which was also asked by Pólya is: "if two people leave the same starting point, then will they ever meet again?"[10] It can be shown that the difference between their locations (two independent random walks) is also a simple random walk, so they almost surely meet again in a 2-dimensional walk, but for 3 dimensions and higher the probability decreases with the number of the dimensions.>

Answer 1

If the points are at $(3,8,9)$ and $(7,8,9)$ after $n$ steps then your $S_n=|7-3|+|8-8|+|9-9|=4$ and

• the probability it becomes $S_{n+1}=2$ is $\frac1{36}$ (they both move towards each other),
• the probability it stays at $S_{n+1}=4$ is $\frac{10}{36}$ (one moves towards the other but the other moves away)
• the probability it becomes $S_{n+1}=6$ is $\frac{25}{36}$ (they both move away from each other).

So $S_n$ is not a symmetric random walk. Another issue is that the probabilities depend on the relative positions.