Waiting for tails then heads with a biased coin

Question

A coin lands heads with probability $\tfrac23$ and tails with probability $\tfrac13$ . You toss it until you first see a tail immediately followed by a head. What is the expected number of tosses?

Short answer

$9/2$ . First wait for a tail (geometric, mean $3$ ). After that, any heads finishes and any further tails just keeps you in the same state, so you then wait for a head (mean $\tfrac32$ ). Total $3+\tfrac32=\tfrac92$ .

Adapted from OpenQuant; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Sample Spaces, Events, and the Probability Axioms
Maximum Likelihood Estimation: Score and Log-Likelihood
Bias, Variance, and Mean Squared Error

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