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Statistical Mechanics - Random Walk

  1. Jul 12, 2012 #1
    I'm reading through Reif's "Statistical Mechanics" to prepare for the upcoming semester. Basically, a drunk guy takes N total steps, n1 to the right and n2 to the left. The probability that the current step will be to the right is "p," while the probability that the current step will be to the left is "q=1-p." The probability that one sequence of N steps will have a total of "n1" steps to the right and "n2" steps to the left is "(p^n1)(q^n2)." The total number of possible ways to get that specific number of steps to the right and left after N total steps is "N!/(n1!(N-n1)!) = N!/(n1!n2!)."

    I understand that just fine. What I'm stuck on is why you multiply N!/(n1!n2!) and (p^n1)(q^n2) to get the probability of the drunk ending up with n1 steps to the right and n2 steps to the left. Why isn't it just (p^n1)(q^n2)?
  2. jcsd
  3. Jul 12, 2012 #2
    Picture the guy walking on the axis of integers (starting in 0). [itex]p^{n_1}q^{n_2}[/itex] would be the probability of the guy getting to position [itex]n_1 - n_2[/itex] after [itex]N[/itex] steps, via any one allowed path. For example, if we keep track of the sequence in which the steps were taken, [itex]q \underbrace{p p ... p}_{n_1 \mbox{times}}\underbrace{q q ... q}_{n_2 - 1 \mbox{times}}[/itex] would represent (the probability of) the path that ends in position [itex]n_1 - n_2[/itex] by first taking 1 step to the left, then [itex]n_1[/itex] steps to the right and finally [itex]n_2 - 1[/itex] more steps to the left. So by writing [itex]p^{n_1}q^{n_2}[/itex] you are only looking at one precise path. But since it doesn't matter which path the guy took, we have to add the probabilities of getting there by all possible paths (if there are 2 ways to get to a point, the probability of getting there would be twice as large as the probability of getting there if there was only 1 way). In this case there are [itex]\binom{N}{n_1}[/itex] paths, each with the same probability [itex]p^{n_1}q^{n_2}[/itex], so adding this probability up [itex]\binom{N}{n_1}[/itex] times gives the answer.
    It might be insightful to check this reasoning in the most simple example: [itex]p = q = \frac{1}{2}[/itex] and [itex]n_1 = n_2 = 1[/itex]. What's the probability that the guy ends up in the origin?
    You can also check that this formula defines a probability law, and that just [itex]p^{n_1}q^{n_2}[/itex] alone would not. Given [itex]N[/itex], the probability to take [itex]N[/itex] steps in total must be equal to 1. But the probability of taking [itex]N[/itex] steps in total is the sum of the probabilities of taking [itex]n_1 = 0[/itex] steps to the right, [itex]n_1 = 1[/itex] step to the right, ..., [itex]n_1 = N[/itex] steps to the right. Use the binomial expansion to check that this sums to 1 for the correct formula, but not for just [itex]p^{n_1}q^{n_2}[/itex] alone.
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