Making one-third and two-thirds events from a coin

Question

Using nothing but a fair coin, describe a way to generate an event that occurs with probability $\tfrac13$ , and one that occurs with probability $\tfrac23$ .

Short answer

Flip pairs and throw away one outcome. Flip twice: call HH a success, HT or TH a failure, and repeat on TT. Success has probability $\tfrac{1/4}{3/4}=\tfrac13$ and failure $\tfrac23$ . (Equivalently, compare coin flips with the binary expansion $0.0101\ldots_2$ of $\tfrac13$ .)

Adapted from Crack, Heard on the Street, Q4.10, p.38; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Sample Spaces, Events, and the Probability Axioms
Applications — Monte Carlo Estimation, Error, and Confidence Intervals
Conditional Probability, Total Probability, and Bayes' Rule

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