Coinflips, get maximum 1 head consecutively If I flip a coin 6 times, what is the probability of getting maximum 1 head consecutively?
My thinking goes like this:
H T T T T T 
T H T H T H
H T H T H T
T T T T T H
T H T H T T
H T T H T H
T T T T H T

Are some of the instances where this would happen. 2^6 = 64, 7/64 = 10% (I figure there is more as well but this is a tedious way of approaching this)
So I figured I could use 6c1, but this equals 6 and in my above example I can list 7 instances where this is true.
Why won't 6c1 suffice? What am I missing?
 A: The  Goulden-Jackson Cluster Method is a convenient method to derive a generating function for problems of this kind.

We consider words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the set $\mathcal{B}=\{HH\}$ of bad words which are not allowed to be part of the words we are looking for.
We derive a function $F(x)$ with the coefficient of $x^n$ being  the number of wanted words of length $n$.
  According to the paper (p.7) the generating function $F(x)$  is
  \begin{align*}
F(x)=\frac{1}{1-dx-\text{weight}(\mathcal{C})}
\end{align*}
  with $d=|\mathcal{V}|=2$, the size of the alphabet and with the weight-numerator $\mathcal{C}$ with
  \begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[HH])
\end{align*}
We calculate according to the paper
  \begin{align*}
\text{weight}(\mathcal{C}[HH])&=-x^2-\text{weight}(\mathcal{C}[HH])x
\end{align*}
  \begin{align*}
\text{weight}(\mathcal{C}[HH])=-\frac{x^2}{1+x}\\
\end{align*}

It follows:

A generating function $F(x)$ for the number of words built from $\{H,T\}$ which do not contain the subword $HH$ is
  \begin{align*}
F(x)&=\frac{1}{1-dx-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-2x+\frac{x^2}{1+x}}\\
&=\frac{1+x}{1-x-x^2}\\
&=1+2x+3x^2+5x^3+8x^4+13x^5\\
&\qquad+\color{blue}{21}x^6+34x^7+55x^8+89x^9+144x^{10}+\cdots\tag{1}
\end{align*}
We conclude out of $2^6=64$ words of length $6$ there are $64-21=43$ valid words containing no subword $HH$. The probability is
  \begin{align*}
\frac{43}{64}\doteq 67.19 \%
\end{align*}

Note: The last line (1) was calculated with the help  of  Wolfram Alpha and we see the coefficients  are the  Fibonacci numbers $F_n$  starting with $F_2=1$.
The generating function $F(x)$ is 
\begin{align*}
F(x)=\frac{1+x}{1-x-x^2}=\sum_{n=0}^\infty F_{n+2}x^n
\end{align*}
and the number of binary strings of length $n=6$ which do not contain the substring $HH$ is the Fibonacci number
$$
F_{8}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{8}-\left(\frac{1-\sqrt{5}}{2}\right)^{8}\right)=21
$$
