Express probability using its estimation Let a source $S$ generating a binary (and independent) sequence with the following probabilities: $p$ for the symbol '0' and $1-p$ for the symbol '1'.
The estimation of the ratio R (which is a probability) is
$R=\frac{\text{total number of consecutive $n$ zeros}}{\text{total number of zeros}}$.
Example:
Let $S=0010 1010 0001 0000 0011$ and $n=4$. There is 2 subsequences of $n$ consecutive zeros, so $R=\frac{2\times4}{14}=\frac{8}{14}$.
The question is: what would be the expression of $R$ in terms of $p$ and $n$?
 A: The phrase "the estimation of the ratio R (which is a probability)..." is rather confusing (as the title). 
As defined, $R$ is a random variable, with three parameters $M,n,p$ where $M$ is the total number of symbols. I guess you mean that, for $M\to \infty$, we expect $R$ to converge (in probability) to a constant, which depends on  $n,p$.
A slightly informal approach: let divide the sequence in runs (subsequences of equal values), with lengths $a_1,a_2 ... a_T$. Then  $\sum_{k=1}^T a_k =M$ and each $a_k\ge 1$ follows a geometric distribution (with means $1/p$ and $1/(1-p)$ for all-zeroes and all-ones runs respectively). Then, for large $M$, we have 
$$ \frac{T}{2} \frac{1}{p} + \frac{T}{2}\frac{1}{1-p}\approx M \implies T \approx 2M p ( 1-p) \tag1$$
Now, the probability that a given all-zeroes run has length $n$ is $p(1-p)^{n-1}$. Then, on average, we'll have 
$$p(1-p)^{n-1} \frac{T}{2} \approx M p^2 (1-p)^{n} \tag2$$
such runs, which amount to 
$$n M p^2 (1-p)^{n} \tag3$$
total number of $n-$consecutive zeroes. Hence 
$$ R\to n p^2 (1-p)^{n}  \tag4$$
Edited: This answer assumes that a "subsequence of $n$ consecutive zeros" corresponds to a run of exactly $n$ zeros. If (according to your example) you accept runs of larger size (but count them as one if it contains one non-overlapping run of size $n$), then the approach is still valid but it must be modified.
