You throw a fair coin one million times. what is the expected number of strings of 6 heads followed by 6 tails. I tried to solve this problem, and I used something similar to the probability of the  Bernoulli distribution to find how many tosses you need to obtain a probability of almost one for one occurrence. Then I just divide the number of tosses in the problem statement by the number of tosses I get. 
Is there a better way to do this?
Edit: What I tried is
m = $12$
N = $10^6$
Probability of having X = "m heads followed by m tails" is $\frac{1}{2^m}$
Probability of X at exactly at position i is $1/2^m*(1-1/2^m)^i$
Probability of X after p tries is $\sum_{i=0}^{p}1/2^m*(1-1/2^m)^i =1-(1-1/2^m)^{p+1}$
Now when p is big enough the probability is 1. And then I can just divide . But it depends what you define as big enough and lead to speculative results. There must be a solution based on statistics but It's been a while and I forgot quite a bit.

Let's take a simpler example to get the gist: 
flip m times (m = 5), how many times can we get a 2H followed by 2T (n = 2 + 2 = 4). If you think about it, this can only happen in two circumstances, which are: HHTT_ or _HHTT (where _ could either be a H or a T and it would not matter). This can be deduced by m - n + 1. The Expected number of times that this could happen is the E(x) = m-n+1 * prob(HHTT) = 2/(2^4).
Similarly for m = 1000000 and n = 12 we have
E(x) = (1000000-11)/(2^12)  
