Random walk?

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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?
 

Answers and Replies

  • #2
Office_Shredder
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Can you explain how you got a and b? a) looks like you tried to calculate the probability that out of the 10 steps he took, 2 of them were forward steps, which would result in being 8 steps backwards from his starting position
 
  • #3
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I used combinations and yes that's what I tried and how I saw it at the beginning. The probability of try 10 times and only success two times but then I had my doubts. I am not sure if I am approaching the problem in the correct form because he can move two steps backyards and then he is going to move 4 steps forward.
 
  • #4
Office_Shredder
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That's not how I read the problem. He has to end up 2 steps forward from his starting position after taking 10 steps, which means he needs 6 forward steps and 4 backwards steps (which will get you a 10C6 in the solution)


As an aside, for your interpretation you do have to divide by an additional 2 to eliminate that problem
 
  • #5
Ray Vickson
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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?
In (a), if he takes f steps forward and b steps backward, you need f-b=2 and f+b=10. With the correct f and b, the binomial will apply.

In (b) you need to account for the fact that if he reaches the front door the walk stops; that is, the walk stops when he reaches the front door or takes 20 steps, whichever comes first. Therefore, a simple binomial will not apply.
 
  • #6
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That's not how I read the problem. He has to end up 2 steps forward from his starting position after taking 10 steps, which means he needs 6 forward steps and 4 backwards steps (which will get you a 10C6 in the solution)


As an aside, for your interpretation you do have to divide by an additional 2 to eliminate that problem
So I don't have the correct combination. Its not 10C2 its 10C6. That is what I obtain using
N=#steps
X=#steps forward

y=(N+X)/2 --> 10Cy I use the Y in the combinatorial.
 
  • #7
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In (a), if he takes f steps forward and b steps backward, you need f-b=2 and f+b=10. With the correct f and b, the binomial will apply.

In (b) you need to account for the fact that if he reaches the front door the walk stops; that is, the walk stops when he reaches the front door or takes 20 steps, whichever comes first. Therefore, a simple binomial will not apply.
@RAY I understood the first part (a) but I have breaking my head with the second part. I have been thinking in how to do this but everutime that I think that after 6 steps forward or reach position 6+ I can not get forward I stop.
 
  • #8
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That's not how I read the problem. He has to end up 2 steps forward from his starting position after taking 10 steps, which means he needs 6 forward steps and 4 backwards steps (which will get you a 10C6 in the solution)


As an aside, for your interpretation you do have to divide by an additional 2 to eliminate that problem
That makes sense and it fix find but I am at zero can I go to position -1? I want to understand/interpret correctly the problem.
 
  • #9
Ray Vickson
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@RAY I understood the first part (a) but I have breaking my head with the second part. I have been thinking in how to do this but everutime that I think that after 6 steps forward or reach position 6+ I can not get forward I stop.
One way to get the probability is through a recursive technique. Let f(i,m) = probability of eventually reaching the door, given the man is in state i and has m possible steps left; here, "state" refers to position, so in state i he is i steps from the door. The answer you want is f(6,20). Now look at what happens in one step: from state i he goes to state i+1 or state i-1, each with probability 1/2. Thus, f(i,m) = (1/2)*f(i+1,m-1) + (1/2)*f(i-1,m-1) for i > 1. For i = 1 we have f(1,m) = (1/2) + (1/2)*f(2,m-1).

Starting from f(1,1) = 1/2, and f(i,1)=0 for i >= 2 we can get f(i,m) for all i and all m <= 20. Note that we do not need to have large values of i, since the total number of steps allowed is <= 20. In fact, the maximum number B of backward steps is given by B + 6 + B = 20, or B = 7. That means we do not need i larger than 6+B = 6+7 = 13. In other words, i <= 13 throughout.

You can fairly easily program the recursion in a spreadsheet, starting from the initial values f(1,1) = 1/2, f(i,1) = 0 for 2 <= i <= 13.
 

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