It seems to me that you're right; in principle there could be a few sequences with many streaks and many sequences with no streaks, yielding a high expected value and yet a low probability for a streak.

On the other hand, fix $k$ and let $f(n)$ be the number of sequences of length $n$ that contain a streak of heads having length $k$. Then clearly $f(1)=f(2)=...=f(k-1)=0$ and $f(k)=1$. And (unless I screwed up) it's not hard to get the following recursion:
$$f(n)=2^{n-k}+(n-k)2^{n-k-1}-\sum_{j=k}^{n-k-1}f(j)2^{n-k-1-j}$$

The probability $g(n)$ that a randomly chosen sequence of length $n$ contains a string of heads of length $k$ is $f(n)/2^n$, which gives us the following recursion:

$$g(1)=g(2)=...=g(k-1)=0$$
$$g(k)=1/2^k$$
$$g(n)= {1\over 2^{k+1}}\left(2+n-k-\sum_{j=k}^{n-k-1}g(j)\right)$$

The generating function for $g$ is then

$${x^k\over(1-x)(2^k-2^{k-1}x-2^{k-2}x^2-...-x^k)}$$

The question, then, is what the $n$th power series coefficient looks like when $n$ is approximately $e^{2k}$. I don't have a theorem for you, but numerical tests suggest that this is very close to 1. In particular, even for $k=3$, we're looking at roughly the 403'd coefficient, which is approximately .9999999999999975 --- and this increases monotonically with $k$.

In other words, yes, at least one such streak is very likely to occur.