Expected number of coin tosses question To build on my understanding from this question, I was wondering what is the expected number of tosses required to obtain the sequence H-T-H-T-T-H, given that the coin is fair and independent?

Edit: The way I was thinking about it, which I am not sure makes sense. Let $e$ be the expected number of coin tosses. Then, if we get a Tail immediately, expected number is $e+1$. If we get a H and another H, then the expected number is $e+2$. For the third toss, if we have a H,T,T then it is $e+3$ and for the fourth, fifth and sixth it is $e+4$, $e+5$ and $e+6$ respectively. 

So, we have the equation

$$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{64}(e+6)+\frac{1}{64}(6).$$

I get that $e = 126$ which appears to be wrong compared to @Did's Answer.
If you start with $H-H$, you don't go all the way back to the start - you have already begun your second try with the second $H$.
I think it is a little messy:
$$e=\frac12(e+1)+\frac12(f+1)\\
f=\frac12(g+1)+\frac12(f+1)\\
g=\frac12(h+1)+\frac12(e+1)\\
...
$$
then solve the collection of equations.
Probability is (1/2)^6. Approximately 64 attempts (of a sequence of 6 tosses in a row) would be required to gain the desired result.