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Let's say you have a one dimensional random walk, for example integers from 0 to infinity on the number line, and you start at some value $n$ with a probability $P(0)$ to take a step towards zero and a probability $P(\infty)$ to move towards infinity.

My intuition tells me that no matter how small $P(0)$ is, if you keep taking steps forever you will inevitably at some point hit the bound (zero). I also believe that if this is indeed the case, you will actually hit the bound infinitely many times.

The way I think about it is that even though the number of steps from zero changes at each step you take, you will at some point hit a sequence of consecutive steps towards zero such that you will hit it, no matter how unlikely (as long as $P(0) \neq 0$, of course).

Is my intuition correct? And is there a way to prove what is actually the case?

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    If $P(0) < \frac12$ then there is a positive probability of never reaching $0$, even starting from $n=1$. For intuition, say $P(0) = \frac13$, then at time $T$ you are on average $T/3$ away from the starting point, but the standard deviation is only $O(\sqrt{T})$. – 2017-01-07
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    @ErickWong well if P(0) < 1 in general. – 2017-01-07
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    @ErickWong Really? Wow, that seems crazy to me. Could you explain where my justification for believing otherwise doesn't work? It seems to me that no matter how small the chance of taking n steps towards the bound, you're bound (no pun intended) to hit that sequence of n steps eventually. – 2017-01-07
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    @jeremyradcliff Yes, no matter how far you are there is a positive probability of hitting $0$. But if you are far away enough, then it's plausible to believe that the sum of those probabilities is $<1$, which makes it less than inevitable. – 2017-01-07
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    @SakethMalyala What do you mean? If $P(0) = \frac12$ then you will hit the bound infinitely many times. – 2017-01-07
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    @ErickWong, that makes sense thank you. I erased my previous comment because you already addressed it above. – 2017-01-07
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    This famous problem is very similar, https://en.wikipedia.org/wiki/Gambler%27s_ruin#Unfair_coin_flipping, I think you can probably model your problem by taking the limit as $n_2\to\infty$. – 2017-01-07

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Suppose you start at some location $i>0$. Let $A$ be the event that we ever visit the location one step to the left of this (that is, to location $i-1$). So, $P[A]$ is the probability that we ever visit $0$, given we start at 1. By the repeated structure of the problem, $P[A]$ is also the probability that we ever visit $1$, given we start at 2. By the law of total probability: $$ P[A] = \underbrace{P[A|\mbox{first step left}]}_{1}p_0 + P\underbrace{[A|\mbox{first step right}]}_{P[A]^2}(1-p_0) $$ Define $q=P[A]$. The above equation reduces to: $$q =p_0 + q^2(1-p_0)$$
Solving the quadratic for $q$ gives two possible solutions: Either $q=1$ or $q = \frac{p_0}{1-p_0}$. Which one is the answer? Notice that $q$ is a probability, so it must be in the interval $[0,1]$. So, if $p_0\geq 1/2$, then $q=1$ is only possible choice. As Erick mentions in his comments above, if $p_0<1/2$, it can be shown that $q<1$ (although this is not obvious) and so $q=\frac{p_0}{1-p_0}$ in that case. So in Erick's example, if $p_0=1/3$ then $q = \frac{1/3}{2/3} = 1/2$.