Probability of drunks being apart in 1D random walk

[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.

 

[wais]

Hi there,

I can understand your confusion with this problem. It can be quite challenging to incorporate both the time steps and distance A in the probability calculation. However, I will try to break it down for you and hopefully, it will make more sense.

Firstly, let's define the variables in the problem:

M = number of timesteps
A = distance between the two drunks
L = initial distance between the two drunks
s = spin excess (# of forward steps - # of backward steps)

Now, let's focus on part a) of the problem. We want to find the probability that the drunks are a distance A apart after M timesteps. To do this, we need to consider all the possible paths that the drunks can take to reach this distance A.

Let's say that the first drunk takes x number of forward steps and y number of backward steps. This means that the second drunk will take (M-x) forward steps and (M-y) backward steps to reach the distance A. Since each step is of equal length, we can say that:

x + (M-x) = A
and
y + (M-y) = A

Solving these equations, we get x = (M+A)/2 and y = (M-A)/2. Now, we can use these values to calculate the multiplicity of the system using the formula you mentioned above:

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

Substituting this in the formula for probability, we get:

P(A,M) = g(M,s)/2^M = (M!)/[(1/2(M+A))!(1/2(M-A))!2^M * 2^M]

Simplifying this, we get:

P(A,M) = (M!)/[(1/2(M+A))!(1/2(M-A))!]

Now, for part b) of the problem, we want to find the probability that the drunks meet at precisely the Mth timestep. This means that they must be at the same location after M timesteps, which is A = 0. Using the same approach as above, we can calculate the probability as:

P(0,M) = (M!)/[(1/2(M+0))!(1/2(M-0))!]

Simplifying this, we get