A coin is tossed 7 times. The probability that at least 4 consecutive heads appear is? A coin is tossed 7 times. The probability that at least 4
  consecutive heads appear is?


I have checked and double checked but I can't figure out which cases I am missing. 

Here are the cases: 


4 consecutive heads appear:


$n_1 = \dfrac{4!}{3!}= 4$


5 heads appear


Subcase 1: 4 consecutive heads appear and the other head is separated.

For example: HTTHHHH or THTHHHH 

Subcase 1 has $4$ possibilities. 

Subcase 2: 5 consecutive heads: 

$\dfrac{3!}{2!} = 3$ 

$n_2 = 4+ 3 = 7$


6 heads appear


There are 6 gaps (including the ends) where we can place the $T$

Let me show those places by G: 

GHGHGHHGHGHG

$n_3 = 6$


All heads appear


$n_4 = 1$ 

$P(\text{4 consecutive heads}) = \dfrac{n_1+ n_2+ n_3 + n_4}{2^7}  = \dfrac{18}{2^7} = \dfrac{9}{64}$

Please let me know my mistake.
The component of $n_2$ counting the ways that five heads appear, one of which is separated from the other four, should be 6 rather than 4:
$$\text{HTTHHHH THTHHHH}\\
\text{HTHHHHT THHHHTH}\\
\text{HHHHTTH HHHHTHT}$$
Thus the probability is $\frac{20}{2^7}=\frac5{32}$.
I would split cases in the following way:

1) Exactly 4 consecutive heads: HHHHTxy, THHHHTx, xyTHHHH, => ten possibilites

2) Exactly 5 consecutive heads: HHHHHTx, THHHHT, xTHHHH => five possibilities

3) Exactly 6 consecutive heads: two possibilities, as the 7th flip must be T

4) Exactly 7 consecutive heads: one possibility

In total 18 possibilities.