Four coin steps, back where you started?

Question

A marker starts at 0 and takes 4 steps. Each step is +1 or -1, each with probability 1/2, independently. What is the probability the marker ends back at 0?

Accepted Answer

To finish at 0 after 4 steps you need exactly two +1 steps and two -1 steps, so the ups and downs cancel. The number of ways to choose which 2 of the 4 steps go up is \binom{4}{2} = 6, out of 2^4 = 16 equally likely step sequences: \frac{6}{16} = \frac38. Takeaway. Short random-walk endpoints are just coin-flip counting: to land at 0 after 2n steps the probability is \binom{2n}{n}/2^{2n}, here 6/16 = 3/8.

Four coin steps, back where you started?

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