Probability of Reaching 1

Question

Consider a particle that performs a random walk on the integers starting at position 0. At each step, the particle moves from position i to position i+1 with probability p, while the probability it moves from i to i−1 is 1−p. If p = $\frac{1}{3}$, find the probability the particle ever reaches position 1.

I thought this was a simple Markov chain problem that I can solve with a recursive expression. My reasoning was as follows, either the particle moves directly from 0 to 1 with probability $\frac{1}{3}$, or it could move back to -1 and then back to 0 with probability $\frac{2}{3}*\frac{1}{3}$. Then it has the same probability to go to 1 as it's back in the initial position. Formulating this:

$x = \frac{1}{3} + (\frac{2}{3}*\frac{1}{3})x$

$x = \frac{3}{7}$

However, the answer is $\frac{1}{2}$. What is wrong with my logic?

where does $i$ come from? I wanna try if I can run a simple simulation for this — Derf, Nov 18 '23 at 03:32
i is just the position on the number line. So we start at I = 0, and want to get to I = 1 — Anon, Nov 18 '23 at 04:31
Duplicate of https://math.stackexchange.com/questions/153123/hitting-probability-of-biased-random-walk-on-the-integer-line — quasi, Nov 20 '23 at 07:47

score 4 · Answer 1 · answered Nov 18 '23 at 05:21

Your recursion should be $x = \frac{1}{3} + \frac{2}{3}\color{blue}{x^{2}}$. This is because once the particle is at $-1$, it needs to go one step right, then one step right again (giving $\color{blue}{x^{2}}$ probability) in order to get to $1$.

In your logic, you assumed that the probability to go from $-1$ to $0$ was $1/3$, but this is not right, because the particle can go from $-1$ to $0$ in more ways than just an immediate step (which would be the $1/3$ probability); for example, it could go left a few more times and then go keep going right to go back to $1$.

Probability of Reaching 1

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