Solving a Probability Problem for Thermodynamics: Stirling's Approximation

[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.

 

[Alkatran]

"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...