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Random walk or binomial?

  1. May 7, 2013 #1
    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. jcsd
  3. May 7, 2013 #2

    chiro

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    Science Advisor

    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.
     
  4. May 8, 2013 #3
    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."
     
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