Probability that successive $8$ heads in a trial of $10$ tosses. Question is :

A fair coin tossed ten times. What is the probability that we can observe a string of $8$ heads in succession at some time.

I do not even understand this Question properly..
Probability of getting first head would be $\frac{1}{2}$ as we want just $\{H\}$ from $\{H,T\}$
Probability of getting second head would be $\frac{1}{4}$ as we want just $\{H H\}$ from the collection $\{H H, HT ,TH, TT\}$ 
Probability of getting third head would be $\frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{8}$
So, generalizing this further i would like to say that :
Probability of getting $8$ heads consecutively is $(\frac{1}{2})^8$.
But then, I am sure i am missing something as there are not just $8$ coins but there are actually $10$ coins.
So, I should even consider $THHHHHHHHT$ and $TTHHHHHHHH$
I guess in deciding that probability is $(\frac{1}{2})^8$ I have considered case in which only possibility is $HHHHHHHH**$
So, I have realized that i should mix this with something else but i have no clue how to mix that and what to mix...
I would be thankful if some one can help me to clear this issue.
Thank You
 A: We observe a string of 8 consecutive heads if :


*

*We obtain 10 heads HHHHHHHHHH

*We obtain 9 heads HHHHHHHHHT or THHHHHHHHH

*We obtain 8 heads HHHHHHHHTH HHHHHHHHTT THHHHHHHHT HTHHHHHHHH TTHHHHHHHH


Hence $8$ positives cases among $2^{10}$. 
$$P=\frac{8}{2^{10}}=2^{-7}$$
A: There are more cases in which $H$ is flipped at least $8$ consecutive times. You counted: $TTHHHHHHHH$, $THHHHHHHHT$, $HHHHHHHHTT$. There are five additional cases than those you considered cases:
$HHHHHHHHHH$, $THHHHHHHHH$, , $HHHHHHHHHT$, $HTHHHHHHHH$, $HHHHHHHHTH$
Every possible sequence of outcomes in $10$ tosses has a probability of $\left(\dfrac 12\right)^{10}$. $$P = \text{Number of outcomes with 8 consecutive heads}\;\times \left(\frac 12\right)^{10} = 8\times \left(\frac12\right)^{10} = \frac{2^3}{2^{10}} = \dfrac 1{2^7}$$
A: You’re looking for the probability of getting any one of the following outcomes: 
$$\begin{align*}
&HHHHHHHHTT\\
&THHHHHHHHT\\
&TTHHHHHHHH\\
&HHHHHHHHTH\\
&HTHHHHHHHH\\
&\\
&HHHHHHHHHT\\
&THHHHHHHHH\\
&\\
&HHHHHHHHHH
\end{align*}$$
The first five have a string of exactly $8$ heads, the next two have a string of exactly $9$ heads, and the last has a string of exactly $10$ heads. You can easily compute the probability of each of these outcomes, and then you need only add them to get the final result.
A: There are $2^{10}$ possible outcomes to this experiment of flipping $10$ coins. Count how many qualify and divide by $2^{10}$. The number that qualify is fairly small (I found it to be less than $10$), and while you could use formulas, it's more straightforward in this case to just manually count.
It may help to manually count how many outcomes have exactly $10$ heads in a row, then exactly $9$ in a row, then exactly $8$ in a row.
