Random walk

Paul has once again drunk too much and been driven home by taxi. At home, he still has n = 2 n = 2 steps to his doorstep. Since Paul is completely drunk, he makes a step forward (to home) with probability q = 1 2 q = \frac{1}{2} and a step back with probability 1 q = 1 2 . 1-q = \frac{1}{2}. If he gets back into the taxi (which is only one step away from cell 0 in the diagram), the taxi driver drives him directly to the next police station, so he can spend the night in the sobering-up cell.

What is the probability P P of Paul reaching his front door (in percent) to 3 decimal places?


Bonus problem: Solve the problem for arbitrary n n and q q .


The answer is 33.333.

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1 solution

Markus Michelmann
Oct 17, 2017

All possible outcomes can be represented by a tree diagram The numbers indicates the probabilitiy of Paul reaching the corresponding state. Paul gets to his front door on the first try with a probability p 1 = 1 4 p_1 = \frac{1}{4} , on the second try with p 2 = 1 16 p_2 = \frac{1}{16} , and so on. For the m m th try, the probability reads p m = 1 4 m P = m = 1 p m = 1 4 m = 0 1 4 m = 1 4 1 1 1 / 4 = 1 3 \begin{aligned} p_m &= \frac{1}{4^m} \\ \Rightarrow \quad P &= \sum_{m = 1}^\infty p_m = \frac{1}{4}\sum_{m = 0}^\infty \frac{1}{4^m} = \frac{1}{4} \frac{1}{1 - 1/4} = \frac{1}{3} \end{aligned} with the help of the geometric series m = 0 r m = 1 1 r \sum_{m= 0}^\infty r^m = \frac{1}{1 - r} .

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