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I am working with random walks and http://mathworld.wolfram.com/RandomWalk2-Dimensional.html says that

"Amazingly, it has been proven that on a two-dimensional lattice, a random walk has unity probability of reaching any point (including the starting point) as the number of steps approaches infinity."

I have been trying to prove this result with an expression for the probability distribution of a 2D lattice random walk,

$P(N)$=$\frac{N!}{(n_r)!(n_l)!(2^N)}$, where $n_r+n_l=N$

$N$ is the total number of steps

$n_r$ is the number of steps to the right and

$n_l$ is the number of steps to the left

and I substituted the value of N as 5000 and approximated the value of the factorial using Sterling Approximation. Then, I looked at different combinations of $n_r$ and $n_l$, such as (1000,4000); (2000,3000) etc and then calcualted the value of the probability function. The probability didn't even come close to 1 for any of the values. Is this the correct way to prove this statement?

Also can this process be applied to 3D random walks to find the probability that the walk will pass through the origin?

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    You're looking at the wrong probability. The probability that the walk is at your chosen point after exactly $N$ steps goes to $0$. What goes to $1$ is the probability that the walk is at the point at some time from $0$ to $N$. – 2017-02-24
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    so what would the probability that the walk passes through a point (for example the origin) be as the number of steps increases? – 2017-02-24
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    I don't think there's a closed form expression for it. – 2017-02-24

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