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Drunken walk

  1. Jan 27, 2005 #1
    i'm stuck trying to figure out this probabilities problem for my thermodynamics class. the question is:

    consider an idealized drunk, restricted to walk in one dimension (eg. back and forward only). the drunk takes a step every second, and each pace is the same length. let us observe the drunk in discrete timesteps, as they walk randomly - with equal probability - back or forward.

    a) suppose we have 2 non-interacting drunks who start out in the same location. What is the probability that the drunks are a distance A apart after M timesteps? (use stirling's approximation if you need to)

    b)suppose instead that the 2 drunks started a distance L apart. Find the probability that the drunks meet at precisely the Mth timestep.


    i know that the probability of a binary model system is given by:
    P = multiplicity of system / total # of accessible states
    = g (M, s) / 2^M

    where g is the multiplicity and s is the spin excess (# of forward steps - # of backward steps")

    = M! / [(1/2M+s)! (1/2M-s)! 2^M]

    using stirling's approx. for large M, this becomes,

    P (M,s) = sqrt[2/M(pi)] exp[-2s^2/M]


    i'm not sure where to go from here and i'm really confused. the formula i wrote takes care of the M timesteps, but how do i factor in the distance A? how should i go about doing this question? I would appreciate any help. Thanks.
     
  2. jcsd
  3. Jan 27, 2005 #2

    Alkatran

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    "Thinks in Head"

    Let's try this from drunk 1's frame.
    Drunk 2 can either move 0 relative, 2 relative, or -2 relative. There's a 1/2 chance of 0 and a 1/4 chance for each 2.
    So half the time he doesn't move relative to you, a quarter of the time he moves away (assuming he isn't at the same spot) and the other quarter he moves closer.

    Interesting...
     
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