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2026-03-17

Problem — 2026-03-17

Medium

Mediumprobability

What is the expected number of fair coin flips needed to get two consecutive heads?

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Let EE be the expected flips from the start, and EHE_H be the expected additional flips after one head. From the start: E=12(1+E)+12(1+EH)E = \frac{1}{2}(1 + E) + \frac{1}{2}(1 + E_H). After one head: EH=12(1)+12(1+E)E_H = \frac{1}{2}(1) + \frac{1}{2}(1 + E). Solving: EH=1+E2E_H = 1 + \frac{E}{2}, and substituting gives E4=32\frac{E}{4} = \frac{3}{2}, so E=6E = 6.