# When tossing a fair coin, what is the expected number of tosses to get a Head, followed by two Tails?

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 $$k=14$$.

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?