1

We toss infinitely many times a coin such that head has probability $p$. Let $A_n$ be an event such that in tosses $2^n$, $2^n+1$, ... , $2^{n+1}-1$ there will be $n$ heads in a row. Prove that if $p<\frac{1}{2}$ then almost surely finitely many $A_n$ will occur and if $p\geq \frac{1}{2}$ then almost surely infinitely many $A_n$ will occur.

My solution:

$P(A_n)=(2^n-n+1)p^n$ (because there are $2^n-n+1$ blocks of length $n$ in $2^n$ block, we put there heads and don't care what happen elsewhere). So for $p\geq \frac{1}{2}$ $P(A_n)$ diverges to infinity as a sequence so $\sum_{k=1}^{\infty}P(A_k)$ is divergent. For $p<\frac{1}{2}$ $\sum_{k=1}^{\infty}P(A_k)$ is convergent as a sum of 3 converging series. $A_n$'s are independent so the result follow from Borel Cantelli lemma.

I think something is wrong, most likely $P(A_n)=(2^n-n+1)p^n$ isn't true, because answer to this question is much more complicated.

Is my solution correct? If something is wrong tell me why.

romperextremeabuser
  • 683
  • 2
  • 7

1 Answers1

1

You are right, $P(A_n)$ is not correct, because you are counting more than once the events in which block of consecutive heads are longer than $n$.

However, since you are counting more cases than needed, you have $P(A_n) \le (2^n-n+1)p^n$, which is enough to conclude the summation is convergent for $p<1/2$.

EDIT: the other case was wrong, wince I was overcounting $TH...HTH...HT$ where $H$ repeats $n$ times. Better to look at the link!

Andrea Marino
  • 5,384
  • I don't believe the second inequality; $(1-p)^2 (2^n-n-1)p^n$ is still an expected value, so it still cannot be a lower bound for a probability. In particular, it can be larger than $1$ when $p$ is close to $1$, and $P(A_n)$ cannot. – Misha Lavrov Apr 18 '23 at 19:24
  • For $n = 10$ and $p = \frac{5}{6}$ your expression $P(A_n) = (1 - p)^2(2^n - n - 1)p^n = 9892578125 / 2176782336 > 4.5$. – Gareth Ma Apr 18 '23 at 19:30