# After the $n$-th flip one gets $k$ consecutive Heads for a fair coin

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.

• 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. Dec 14, 2022 at 9:05

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 $$\{Y_n=1\}=\cap_{m=(n-1)k+1}^{nk}E_m=A_{(n-1)k+1}$$.