- #1

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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)?

- Thread starter Fermi-on
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- #1

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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

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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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