Random Walk And Stochastic-Processes Assume that $P(X_i = 1) =1/2, P(X_i =-1)= 1/4,\text{ and }P(X_i = 0)=1/4$.
Consider the random walk starting at 1 given by 
$$S_n = 1 + X_1 + X_2 + \cdots + X_n$$  where  $X_1,X_2, ...$ are i.i.d.
What is the probability that the random walk ever reaches $0$?
I have tried to solve this using Binomial Theorem, but I do not have an equal probability here!
I would appreciate it so much if you could help me solve this problem. 
 A: Suppose it reaches $0$ in n steps. We can calculate the probability using the multinomial distribution. Out of n steps some k will be +1, k+1 will be -1, and n-2k-1 will be $0$.
Hence, $P(S_n=0) = \sum_{k=0}^{(n-1)/2} (1/2)^k (1/4)^{n-k}\frac{n!}{k! (k+1)! (n-2k-1)!}$
A: Let P1 be the probability it ever reaches 0, starting at 1, and P2 the probability it ever reaches 0, starting at 2.
To reach 0 from 2, it has to reach 1 from 2, then 0 from 1.
Reaching 1 from 2 is the same as reaching 0 from 1.
So the probability $P2=P1^2$
Now, start at 1, and make one move.  The chances are 1/3 you are at 0 (success!) and 2/3 you are at 2.  So $$P1=\frac13(1)+\frac23(P2)\\1-3P1+2P1^2=0\\(1-2P1)(1-P1)=0$$
So P1=1 or 1/2. :(
A: The probability of ever reaching zero does not depend on the probability to not move at all. Thus you can instead consider the walk with a $2/3$ chance to go right and a $1/3$ chance to go left. There is a nice general solution to this problem. Let $n \in \mathbb{N}$, then the probability to hit $0$ before hitting $n$ starting at $0<x<n$ satisfies:
$$u_n(x)=\frac{2}{3} u_n(x+1)+\frac{1}{3} u_n(x-1).$$
Adjoin this to the boundary conditions $u_n(0)=1,u_n(n)=0$. Then the recurrence relation can be solved explicitly. (Hint: the general solution is of the form $c_1 r_1^x + c_2 r_2^x$, where $r_1,r_2$ are the roots of $\frac{2}{3} r^2-r+\frac{1}{3}$.) Then the desired quantity is $u(1)=\lim_{n \to \infty} u_n(1)$.
