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Suppose one independently flips an infinite sequence of fair coins. Let $E_n$ be the event that the $n$-th coin is Heads.

Let $A_n$ be the event that starting from the $n$-th flip one gets $k$ consecutive Heads for some fixed number $k$, i.e., $A_n=\cap_{m=n}^{n+k-1} E_m$. Find $\mathbb{P}\left(A_n\right.$ i.o. $)$, where $\left\{A_n\right.$ i.o. $\}=\cap_{m \geq 1} \cup_{n \geq m} A_n$.

My thought is as follows: find a subsequence $\{A_{n(k)}\}$ of $\{A_n\}$ then prove that the infinite series $\sum\mathbb{P}(A_{n(k)})$ is divergent. By Borel-Cantelli Lemma, we can conclude that $\mathbb{P}\left(A_n\right.$ i.o. $)=1$. However, it does not seem to work after trials. Could anyone tell me that is this idea feasible? I would be thankful if someone could give me your answer.

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    $\begingroup$ An infinite sequence of "heads" and "tails" (even if the coin is biased , but both results have a non-zero probability!) has arbitary long chains of "heads" and also arbitary long chains of "tails" with probability $1$. This implies that we get every such chain infinite many often with probability $1$. And this implies your result. I do not know whether your argument is valid as well. $\endgroup$
    – Peter
    Dec 14, 2022 at 9:05

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Consider $Y_1=1$ if the first $k$ throws are all heads and $0$ otherwise, $Y_{2}=1$ if the throws from $k+1$ to $2k$ are all heads and $0$ otherwise, and so on. Note that now $(Y_n)_{n \in \mathbb{N}}$ are independent identically distributed random variables, s.t. $P(Y_1=1)=1/2^k$, $P(Y_1=0)=1-1/2^k$. So we can say by BC-II: $$\sum_{n \in \mathbb{N}}P(Y_n=1)=\infty\implies P(Y_n=1\textrm{ i.o.})=1$$ Now note $\{Y_n=1\}=\cap_{m=(n-1)k+1}^{nk}E_m=A_{(n-1)k+1}$.

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