TITLE: many expected streaks imply high probability for a streak
QUESTION [5 upvotes]: In CLRS, "Intro to Algorithms" section 5.4.3 the following is shown.  If a fair coin is flipped n times, the expected number of streaks of consecutive heads of length (1/2)log(n) is $ \Theta (\sqrt n)$.  So, for large n, "we expect there are a large number of streaks of length (1/2)log(n)."  
Then the book continues, "Therefore, one streak of such a length is likely to occur."
Is that a correct conclusion?  To conclude that there is high probability for such a streak isn't it neccessary to bound the variance besides for showing a large expectation?
Thanks.

REPLY [2 votes]: The broken logic in the book can bite in many places.  Here is a simple example. If we make a random graph with $n$ vertices and edge probability $3/n$, the expected number of hamiltonian cycles goes to infinity exponentially fast, yet the probability of having any hamiltonian cycle at all goes to 0 exponentially fast.