# Infinite coin flips - sequence of $n$ heads

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.

• What is the probability for a particular $n$? – Henry Feb 7 '18 at 21:31
• @Henry , Well, idk.. a row of n heads can be in the beginning or in the end, so it depends. – Eran Feb 7 '18 at 21:43
• What's the probability for one toss? Here probably less than $p=\frac{1}{3}$ for heads will give a zero probability. Argh, I see the coin is fair. – dEmigOd Feb 7 '18 at 21:45
• What precisely do you mean by "$n$ heads in a row"? At least one run of at least $n$ heads? Or exactly $n$ heads? Can we also have runs of more than $n$ heads? – Robert Israel Feb 7 '18 at 21:51

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.