Expected number of pattern in n coin-tossing without overlapping $N$ is the total number of tossing a fair coin. What is the expected number of occurrence HHH in this $N$ tosses? When overlapping is allowed, e.g., HHHH counts as 2 occurrence, then it is easy to calculate that $$E[X] = \frac{N - 2}{8}$$ 
What if we don't allow overlapping? e.g., HHHH and HHHHH are both counted as only 1 unoverlapped occurrence, while HHHHHH can count as two unoverlapped occurence. How to solve this problem? Thanks!
 A: Let $X_k$ be the number of sequences of exactly $k$ sequential heads in the $N$ tosses. For $k<N$, the expected number of these is $$E[X_k]=2\frac{1}{2^{k+1}}+(N-k-1)\frac{1}{2^{k+2}}=\frac{N-k+3}{2^{k+2}}$$ For $N=k$, the value is $E[X_N]=\frac{1}{2^{N}}$.
Then the expected number of non-overlapping triples is $$E[X]=\sum_{k=1}^N \left\lfloor \frac k 3\right\rfloor E[X_k]$$
That's gonna be messy. Let's say, for simplicity, that $N=3M$. Then you can write it as:
$$ME[X_{3M}]+\sum_{j=1}^{M-1} j\left(E[X_{3j}]+E[X_{3j+1}]+E[X_{3j+2}]\right) =\\\frac{M}{2^{3M}}+\sum_{j=1}^{M-1} j\left(\frac{4(N-3j+3)+2(N-3j+2)+(N-3j+1)}{2^{3j+4}}\right)=\\\frac{M}{2^{3M}}+\sum_{j=1}^{M-1}j\frac{21(M-j)+17}{2^{3j+4}}$$
I'd hate to come up with a closed form for that, but it can be done. WolframAlpha gives:
$$E[X]=\frac{5}{49}\left(\frac{1}{2^{3M}}-1\right)+\frac{3}{14} M$$
Which means $$\lim_{M\to\infty} \frac{E[X]}{3M} = \frac{1}{14}$$
When $N$ is divisible by $3$, then, $$E(X)=\frac{N}{14}-\frac{5}{49}+O\left(\frac{1}{2^N}\right)$$ more generally.
It's probably accurate for all $N$, but I haven't worked the other cases out. (It does seem to work for the values of $N$ that I've tried that are quite small, like $N=13$.)
