waiting for TT and H? We toss a fair coin until we see one consecutive TT pattern and one H, their order is irrelevant. 
For instance, HHTHTHTT, or TTH.
Let X be the number of tossed required. What is E(X)?
Here is my solution: Let 
$X_H$= number of tosses required to get H.
$X_{TT}$= number of tosses required to get TT.
First, note that $E(X_{TT})=6$. Then
$E(X)=\frac{1}{2}(2+6)+\frac{1}{4}(2+6)+\frac{1}{4}(2+2)=6.5$
Is it correct?

Look at this flow chart (each arrow has probability $\frac{1}{2}$ except the last absorbing state has probility $1$ in its self-loop):

Your answer $6.5$ is confirmed by calculating the expected number of steps for this absorbing Markov chain starting from the empty state and reaching the target state $(\text{has seen } H, TT)$.
It has $Q$-matrix
$$Q = \left[\begin{array}*
0 & \frac{1}{2} & 0 &\frac{1}{2} & 0 \\
0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\
0 & 0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
0 & 0 & 0 & \frac{1}{2} & 0 \\
\end{array}
\right]$$
Sum the first row of $(I-Q)^{-1}$ to get the answer:
$$1+\frac{1}{2} +\frac{1}{2}+3+\frac{3}{2} = \frac{13}{2}.$$
PS. I calculated the inverse with Wolfram Alpha with this input:
inverse ( 1/2*[[2,-1,0,-1,0], [0,2,-1,-1,0], [0,0,1,0,0], [0,0,0,1,-1], [0,0,0,-1,2]])

