# Random Walk Probability number line

I am stuck on this question,

"In a symmetric random walk of a particle on integer number line, starting from origin, what is the probability that it hits $-9$ before $+6$"

Here symmetric implies: P(going left at each step) = 1/2 and ditto for going right.

I could not think of a method for this, I just know that the expected number of steps to reach an integer $k$ can be defined recursively as:

Let $f(k)$ denote the expected number of steps starting from $k$,

$f(0)$ = $\frac{1}{2}$$(1+f(-1)) + \frac{1}{2}$$(1+f(1))$

In general, $f(k)$ = $\frac{1}{2}$$(1+f(k-1)) + \frac{1}{2}$$(1+f(k+1))$

PS: Please excuse me for the poor LaTeX syntax, I'm new to it.

You are asking yourself "how many steps?" when the question is "is position $-9$ reached before $+6$?".
Instead, consider $g(k)$ as this probability starting at $k$. You now have $$g(k) = \tfrac12 g(k-1)+\tfrac12 g(k+1)$$ and the boundary values $$g(-9)=1,g(+6)=0$$ which you should be able to solve
If you were really interested in the number of steps until you stop, the boundary values are $f(-9)=0,f(+6)=0$ and the solution is a little more complicated (you get a quadratic function of $k$, with $f(0)=54$ and it not a coincidence that $54=6\times 9$)
• I defined $g(k)$ as the probability of reaching $-9$ before reaching $+6$ if you start from $k$. So $g(0)$ is precisely the result you want – Henry Aug 8 '18 at 17:27
• Solving this I get $f(0)$ as $\frac{6}{15}$. Just one more question, is there a way to intuitively solve this? since the final results indicate the the probability is ratio of distance from the other end mod$(-6)$ / total distance $(15)$. Many thanks for clarifying this much already! – Mohit Aneja Aug 8 '18 at 17:36
• Let's call $K$ the position at some stage, whether or not you have stopped. Since the steps are equally likely to be left or right, starting from $0$ you have $E[K]=0$ at all times. So if you eventually (with probability $1$) end up at $-9$ or $+6$ then the respective probabilities must be in the ratio $6:9$ to keep the expected position zero, making the probabilities $\frac{6}{6+9}$ and $\frac{9}{6+9}$. See Wikipedia for similar ideas – Henry Aug 8 '18 at 17:58