I want to prove that in a sequence W of length n, consisting of 1s and 0s, $P$( in $W$ there is at most $\frac{\log\_2n}2$ consecutive zeroes ) $\leq \frac{K}{n} $ for some constant K. Can anyone help me start on the problem or refer me to some literature that could be of help.


Thank you very much. 



Probability of at most $K$ consecutive zeroes in a sequence of 0s and 1s
Let $a<1<b$. I prove that for large $n$: 


1) the probability that a random sequence of length $n$ has at least $B:=b\log\_2 n$ consecutive zeroes is at most $n^{1-b}$;


2) the probability that a random sequence of length $n$ does not contain $A:=\lfloor a\log\_2 n\rfloor$ consecutive zeroes is at most $e^{-n^{1-a+o(1)}}=O(n^{-M})$ for any $M>0$.


Proofs. 1) For each possible place of $\lceil B\rceil $ consecutive positions consider the event: all positions are 0. Denote these events $E\_1,E\_2,\dots$. The sum of their probabilities does not exceed
$$\frac{n}{2^B}\leqslant n^{1-b}.
$$


2) Choose $m:=\lfloor n/A\rfloor$ disjoint segments of $A$ consecutive places. The probability that none of them contains only zeroes equals
$$
(1-2^{-A})^m\leqslant e^{-m\cdot 2^{-A}}=e^{-n^{1-a+o(1)}}
$$


