We flip a fair coin infinite number of times independently.
Does the event "For all $n\in \mathbb{N}$, between flip number $3^n$ and flip number $3^{n+1}$ we get $n$ heads in a row" has positive probability?
I thought about using Borel-Cantelli lemma but cannot see how to organize the sequence of events in such a way that I can use the lemma.
Define a good event $A_n$ - to be $n$ heads in a row somewhere between toss $3^n$ and $3^{n+1}$.
Define event $E_{n, i}$ to be the event that those $n$ heads in a row happen right after toss $3^{n} + i$, so $i$ could be $0, \ldots, 2\cdot 3^{n}-n$.
Probability of each $E_{n, i}$ is $2^{-n}$.
Let's look at $F_{n, i}$, which will be the event $E_{n, jn}$ did not happen for all $j < i$ and $E_{n, in}$ did happen. $F_{n,i}$ are disjoint. Moreover, $F_{n,i} \subseteq A_n$.
$$\mathbb{P}(F_{n,i}) = (1 - 2^{-n})^{i}\cdot 2^{-n}$$
and
$$\mathbb{P}\left(\bigcup\limits_{i= 0}^{\frac{2\cdot 3^n - n}{n}} F_{n, i}\right) = 2^{-n} + (1-2^{-n})2^{-n} + \ldots + (1-2^{-n})^{\frac{2\cdot 3^n - n}{n}}2^{-n} $$
For sufficiently large $n$, we can bound it by $\frac{1}{2}\cdot2^{-n}\cdot\frac{1}{1-(1-2^{-n})} = \frac{1}{2}$.
So, $\mathbb{P}(A_n) \geq \frac{1}{2}$, and all of them independent, so you can apply the second lemma.