# Drunken walk

1. Jan 27, 2005

### jessawells

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. Jan 27, 2005