Knocking on doors until someone agrees

Question

A survey worker walks a street knocking on doors one at a time. Each household independently agrees to answer the survey with probability $1/6$ , and the worker keeps knocking until someone says yes.

On average, how many doors must be knocked on to get the first agreement (counting the door that agrees)?

Accepted Answer

Each door is an independent trial with success probability p = 1/6. We want the expected number of trials until the first success. Use the one-step argument. The first knock always costs 1. With probability p it succeeds; with probability 1-p it fails and the situation resets, leaving an expected E more knocks: E = 1 + (1-p)\,E \quad\Longrightarrow\quad pE = 1 \quad\Longrightarrow\quad E = \frac{1}{p}. With p = 1/6, E = \frac{1}{1/6} = 6 	ext{ doors.} Takeaway. The expected number of independent tries to reach the first success is 1/p; a 1-in-6 chance takes 6 tries on average.

Knocking on doors until someone agrees

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