# Random walk or binomial?

1. May 7, 2013

### marina87

Random walk or binomial??

Statement:
A drunk person wonders aimlessly along a path by going forward 1 step and backward 1 step with equal probabilities of ½. After 10 steps,
a) what is the probability that he has moved 2 steps forward?
b) What is the probability that he will make it to his front door within 20 steps before he collapses with the door being 6 steps in front of him.

My approach was to use Binomial in both cases:
a)10C2 (0.5)^10
b)20C6 (0.5)^20

Is that correct? I have been reading about random walk and they sometimes give another equation.
(10+2)/2=6 and thn the result is like this 10C6.
The result is not the same and then I start to have my doubts.

Can some one please tell me if my approach using binomial distribution is right?

2. May 7, 2013

### chiro

Hey marina87.

I think you should retry your problem by defining a sum of random variables with P(X = 1) = P(X = -1) = 1/2.

A binomial random variable only includes 0 and 1 where as it should include -1 and 1.

You can then use a probability generating function to get the probability of having a final sum of +2.

3. May 8, 2013

### NegativeDept

Your method works fine, except I think part b) is answering the wrong question. If the drunk ever gets to +6, then he goes inside and stops walking. I think your answer to b) is the number of paths which end at +6. This neglects any paths which get to +6 at some time $t<20$, but then end up somewhere else at $t=20$.

If I understood the question right, this is a famous hitting time problem. I had to solve a related problem in a stats class, and I still have to think about it a while to do it right. Spoilers: click here and scroll down to "The Maximum Position."