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Dear math stack exchange community,

It's my first post, so hello and please let me know if I haven't posted this question in a proper way- happy to edit it.

So here's an exercise which I'm trying to solve:

We are tossing a fair coin infinitely many times. Let $A_n$ be an event that in first $n$ tosses we had same number of heads and tails. Show that with Probability 1 events $A_n$ will occur for infinitely many values of $n$.

This is an exercise from Borel-Cantelli Lemma's chapter, so I assume I should use it somehow to prove it.

What I already know:

For each $n \in \mathbb{N}:$

$P(A_{2n+1}) = 0$

$P(A_{2n}) = \binom{2n}{n}p^n(1-p)^n$ (in our case $p = 1/2$)

But I cannot use those events in order to show divergence of series of those probabilities, because they are not independent and assumption for B-C Lemma will not apply.

I believe I should construct independent events somehow, but here's where I am stuck- I have no clue how to do it.

Any hints on that? Just in case it's rather expected to do it without prior knowledge about random walks, but any hints/answers are very welcome.

Thank you a lot!

user782750
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  • If we toss a coin infinitely many times, the first $n$ tosses occur exactly once for any given $n.$ So no matter what value of $n$ we choose, $A_n$ can occur at most once. Did you try to paraphrase the problem from the wording in which it was originally presented? – David K May 02 '20 at 22:26
  • In the meantime I realised bad wording and tried to paraphrase already- now it states "infinitely many of $A_n$ events will occur". Is that clear now? Unfortunately I don't have this question in English so I'm trying to translate it. – user782750 May 02 '20 at 22:39
  • Do you mean to say "for infinitely many values of $n$"? – David K May 02 '20 at 22:40
  • Yes, thank you. – user782750 May 02 '20 at 22:42
  • Now I understand what you mean in "what I already know". I hope someone has a better idea what to do next. – David K May 02 '20 at 22:47
  • I edited it again using your wording so it should be easier to understand now. – user782750 May 02 '20 at 22:51
  • Seems like it might be enough to show that the #H - #T returns to 0 (just once) with probability $1$, since then you start over from there, so to speak. – Ned May 02 '20 at 23:28

1 Answers1

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Let $p_k$ denote the probability that the number of heads and tails will ever equalize when there is an excess of $k$ in either direction. By first-step analysis, we have

$$ p_k=\frac12(p_{k-1}+p_{k+1})\;. $$

This is the discretized Laplace equation in one dimension, and the solutions are exactly the linear functions. Since a non-constant linear function eventually leaves the admissible range $[0,1]$ of probabilities, it follows that $p_k$ is constant. Then the boundary condition $p_0=1$ implies that $p_k=1$ for all $k$.

Since the probability for at least one equalization is $1$, the probability for at least $m$ equalizations is $1$ for all $m\in\mathbb N$. Thus the probability for exactly $m$ equalizations is $0$ for all $m\in\mathbb N_0$. By countable additivity, the probability of the union of these events for all $m\in\mathbb N_0$ is also $0$. Thus the probability of its complement, the event that there are infinitely many equalizations, is $1$.

joriki
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