What is the probability of flipping at least 3 heads or 3 tails in a row over the course of 10 flips? I want to find the chance of a streak of 3 or more of the same side occurring within 10 flips.

My idea:

Number of favorable outcomes / Number of possible outcomes
Number of favorable outcomes / 2^10


To find the number of favorable outcomes:

First, I find all possibilities of the 7 non-streaking flips:

2^7


Then I find all the possibilities of the streak of 3 flips:

It could be HEADS or TAILS
It could be in any of the 8 positions between the non-streaking flips
2 sides * 8 positions = 16


My wrong answer:

2^7*16 / 2^10 = 2


What did I do wrong?
It's easier to work out the complementary probability, that all streaks
have length at most $2$. Then there will be $j$ streaks of length $2$
and $10-2j$ streaks of length $1$, $10-j$ streaks altogether. Here $0\le j\le 5$. Out of the $10-j$ streaks you must choose $j$ to have length $2$
and you can choose to start with a head or tail. Therefore there are
$$2\binom{10-j}j$$
sequences with $10-2j$ streaks of length $1$ and $j$ of length $2$.
The probability you seek is then
$$1-\frac1{2^9}\sum_{j=0}^5\binom{10-j}j.$$
Let ${\mathscr{S}}_{n}$ be the set of sequences of $n$
flips that don't contain a streak of $3$. We are going
to compute $\left|{\mathscr{S}}_{n}\right|$ by induction on $n$. First
note that ${\mathscr{S}}_{1} = \left\{H , T\right\}$ and
${\mathscr{S}}_{2} = \left\{H H , H T , T H , T T\right\}$, so that
$\left|{\mathscr{S}}_{1}\right| = 2$ and $\left|{\mathscr{S}}_{2}\right| = 4$.

In order to give an induction formula, we partition

$${\mathscr{S}}_{n} = {\mathscr{V}}_{n} \cup  {\mathscr{W}}_{n}$$

where ${\mathscr{W}}_{n}$ is the subset of sequences that end with a double
streak, such as $H H$, and ${\mathscr{V}}_{n}$ is the set of sequences
that don't end with a double streak. Obviously, $\left|{\mathscr{V}}_{2}\right| = 2$
and $\left|{\mathscr{W}}_{2}\right| = 2$.

We create elements of ${\mathscr{S}}_{n+1}$ by appending a flip to an element
$s \in  {\mathscr{S}}_{n}$. If $s \in  {\mathscr{W}}_{n}$, there is only
one way to complete it without creating a streak of $3$ and
the result an element of ${\mathscr{V}}_{n+1}$. If $s \in  {\mathscr{V}}_{n}$, there
are two ways to complete it, one way creates an element of ${\mathscr{V}}_{n+1}$,
the other way creates an element of ${\mathscr{W}}_{n+1}$.

It follows that we have the following induction relations

$$\left\{\begin{array}{rcl}\left|{\mathscr{V}}_{n+1}\right|&amp;=&amp;\left|{\mathscr{V}}_{n}\right|+\left|{\mathscr{W}}_{n}\right|\\
\left|{\mathscr{W}}_{n+1}\right|&amp;=&amp;\left|{\mathscr{V}}_{n}\right|
\end{array}\right.$$

It is now obvious by induction that $\left|{\mathscr{V}}_{n}\right| = 2 {F}_{n}$ and
$\left|{\mathscr{W}}_{n}\right| = 2 {F}_{n-1}$, where ${F}_{n}$ is the
$n$th Fibonacci number. Indeed the result is true when
$n = 2$ and the relation ${F}_{n+1} = {F}_{n}+{F}_{n-1}$ ensures
the induction.
The conclusion is that

$$\boxed{\left|{\mathscr{S}}_{n}\right| = \left|{\mathscr{V}}_{n}\right|+\left|{\mathscr{W}}_{n}\right| = 2 {F}_{n}+2 {F}_{n-1} = 2 {F}_{n+1}}$$

In particular, $\left|{\mathscr{S}}_{10}\right| = 2\times{F}_{11} = 178$.

It follows that the required probability is 

$$1-\frac{178}{{2}^{10}} = \frac{423}{512}$$

Incidentally, combining this with @LordSharkTheUnknown's reasoning yields the conjectural result (which proof by induction looks easy).
$$
\boxed{\sum_{j\ \le\ n-j}\binom{n-j}{ j} = F_{n+1}}
$$
This formula is also given in the Wikipedia page about binomial coefficients.