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I have attempted this using first step analysis. However, my answer here disagrees with the answer here: Expected number of toss to get a single Head followed by k Consecutive Tails

Can someone please show me how my method is incorrect?

Let $k = $ the number of tosses needed to achieve HTT

Then, $k = P(H) \{P(H)(2+k) + P(T)[P(T)(3) + P(H)(3+k)]\} + P(T)(1+k)$

Substituting $P(H) = P(T) = \frac{1}{2}$ and subsequently solving gives $k=14$.

However, this disagrees with the answer in the above link.

Can someone please point out the mistake in my working? Thanks!

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The binomials $(2+k)$ and $(3+k)$ are wrong. Note that after an $H$ you are in a better position than at the start.

Why don't you set up equations for the expected number $E_0$, $E_H$, and $E_{HT}$ of additional throws at the start, with $H$, and with $HT$ on the stack?

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