Random Walk ( Drunk Instance)

1. Jan 30, 2009

FourierX

1. The problem statement, all variables and given/known data

Say a drunk starts making his steps of equal distance from a lamppost. Assuming that each of the steps are of equal distance, and N as the total number of steps, what is the probability of him/her ending at the lamppost? Find the probability when N is even and also for odd.

2. Relevant equations

PN(a) = ( N! p (N+m)/2 q(N-m)/2 ) / [{(N+m)/2}! {(N-m)/2}!]

whree,
a = integer
p = probability of drunk being in the right side of the lamppost
q = probability of drunk being in the left side of the lamppost

3. The attempt at a solution

Derivation of the equation is quite straightforward. I am worried about my answer for this particular problem however. Since the drunk starts from the lamppost (x=0), when the N is even, he can land back to the lamppost. However, if N is odd, he can not land back at 0 (as he/she has to land back to an odd number). I do not know if my understanding is correct. Any clue ?

Berkeley

2. Jan 30, 2009

Focus

I have no idea what this PN(a) = ( N! p (N+m)/2 q(N-m)/2 ) / [{(N+m)/2}! {(N-m)/2}!] is. But you are correct about the odd ones, after odd amount of steps he can only be at an odd integer. Thus the lesson to be learned is when you are drunk always walk an odd number of steps.

3. Jan 30, 2009

FourierX

"Thus the lesson to be learned is when you are drunk always walk an odd number of steps. "

Did you miss ODD in the sentence?

Are you saying that when you start from odd point, you end of up at and when you start from even, you end up at even? If so, yeah the formula says so.

4. Jan 30, 2009

Dick

Sure. Odd will never get you back where you started. But you never defined what m is. Why not define m to be the number of steps taken to the right? Then you only get back where you started if the number of steps to the left is also m. So N=2m. Get rid of some of those strange expressions like (N+m)/2.