Probability of special configuration of ones in a binary string

Question

Consider the sequence $(X_i)_{1 \leq i \leq L}$ of i.i.d. random variables, where $X_1 \in \{0,1\}$ and $P(X_1 =1) = p$. For a $k \in \mathbb{N}$ define the event $A_{k,L}$ as "all ones in the sequence (if any) appear in groups of $k$ adjacent ones". This means that with $k=3$ and $L=5$, the sequences $00000$ and $01110$ (for example) are in $A_{k,L}$, but $01000$, $10110$ and $11110$ are not in $A_{k,L}$. The string $11110$ is not in $A_{k,L}$, as the ones appear in a group of $4 \neq k$ adjacent ones.

Question: For which values of $k$ (if any), $P(A_{k,L})$ remains positive in the limit of $L \rightarrow \infty$?

Hint: Using combinatorial arguments one could show that, $$P(A_{L,k})= \sum\limits_{n=0}^{\left\lfloor{L \over k+1}\atop{}\right\rfloor} {L - nk \choose n}p^{nk}\left(1 - p\right)^{L - nk}. $$

Answer 1

To elobrate on an already very good answer for $k \geq 2$.

Split the string of Length $L$ into $\lfloor L/k \rfloor$ blocks of three denoted $A_{1},A_{2},...A_{\lfloor L/k \rfloor}$. Let $B$ be the event that the string is not in the set $A_{k,L}$ and let $A$ be the event that one of these three blocks contains the string 010. Clearly $A \subseteq B$ and so $P(A) \leq P(B)$. Now the probability of $A^{\mathsf{C}}$ is given by $P(A^{\mathsf{C}})=(1-1/8)^{\lfloor L/k \rfloor} \rightarrow 0$ as $L \rightarrow \infty$ and so $P(A) \rightarrow 1$ as $L \rightarrow \infty$. Consequently $P(B) \rightarrow 1$ as $L \rightarrow \infty$.

Answer 2

None. Divide the string into groups of length 3, possibly ignoring a few digits at the end. For $k \ge 2$, each group has a $1/8$ probability of being $010$, causing $A_{k,L}$ to not happen. For $k < 2$, instead consider the possibility where the group is $111$. ($A_{k,L}$ can also fail to occur in other ways, but we're setting a bound.) Each $1/8$ event has to not happen for $A_{k,L}$ to not happen, and the probability of them all not happening goes to $0$ as $L$ goes to $\infty$.