TITLE: Hitting probability of biased random walk on the integer line
QUESTION [11 upvotes]: Lets say we start at point 1.  Each successive point you have a, say, 2/3 chance of increasing your position by 1 and a 1/3 chance of decreasing your position by 1.  The walk ends when you reach 0.
The question, what is the probability that you will eventually reach 0?
Also, is there any generalization for different probabilities or different starting positions or different rules (say you increase 2 and decrease 1)?
NOTE: I have never taken a course that considered random walks.  So, if possible, could no prior knowledge of random walks be assumed?

REPLY [8 votes]: With fancier tools one can extract a bit more information from the problem. Let $p>\frac12$ be the probability of stepping to the right, and let $q=1-p$. For $n\in\Bbb N$ let $P_n$ be the probability of first hitting $0$ in exactly $n$ steps. Clearly $P_n=0$ when $n$ is even, and $P_1=q$. In order to hit $0$ for the first time on the third step you must go RLL, so $P_3=pq^2$. To hit $0$ for the first time in exactly $2k+1$ steps, you must go right $k$ times and left $k+1$ times, your last step must be to the left, and through the first $2k$ steps you must always have made at least as many right steps as left steps. It’s well known that the number of such paths is $C_k$, the $k$-th Catalan number. Thus,
$$P_{2k+1}=C_kp^kq^{k+1}=C_kq(pq)^k=\frac{q(pq)^k}{k+1}\binom{2k}k\;,$$
since $C_k=\dfrac1{k+1}\dbinom{2k}k$. It’s also well known that the generating function for the Catalan numbers is $$c(x)=\sum_{k\ge 0}C_kx^k=\frac{1-\sqrt{1-4x}}{2x}\;,$$ so the probability that the random walk will hit $0$ is 
$$\begin{align*}
\sum_{k\ge 0}P_{2k+1}&=q\sum_{k\ge 0}C_k(pq)^k\\\\
&=qc(pq)\\\\
&=q\left(\frac{1-\sqrt{1-4pq}}{2pq}\right)\\\\
&=\frac{1-\sqrt{1-4pq}}{2p}\\\\
&=\frac{1-\sqrt{1-4q(1-q)}}{2p}\\\\
&=\frac{1-\sqrt{1-4q+4q^2}}{2p}\\\\
&=\frac{1-(1-2q)}{2p}\\\\
&=\frac{q}p\\\\
&=\frac{1-p}p\;.
\end{align*}$$
For the present case, $p=\dfrac23$, this yields the probability $\dfrac12$. Rounded to four decimal places, the probabilities of first hitting $0$ in $1,3,5,7,9,11$, and $13$ steps are:
$$\begin{array}{rcc}
\text{Steps}:&1&3&5&7&9&11&13\\
\text{Probability}:&0.3333&0.0747&0.0329&0.0183&0.0114&0.0076&0.0053
\end{array}$$
These already account for $0.4829$ (total calculated before rounding) out of the total probability of $0.5$.