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Fermi-on
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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)?
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)?