Proving tail event has probability 1 without Borel-Cantelli

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The following is my tutorial problem:

A (not necessarily fair) coin is tossed infinitely often. Show that every prescribed finite sequence of heads and tails appears infinitely often. (Note: Do not use Borel-Cantelli. Instead, use Kolmogorov's 0-1 Law)


Labeling the event that an arbitrary but fixed sequence of heads and tails as $\mathcal{B}$, It is not difficult to see and prove that $\mathcal{B}$ lies in the tail $\sigma$-algebra. By the Kolmogorov's 0-1 Law, we can conclude that $P[\mathcal{B}] \in \{0,1\}$ as coin tosses are independent. My main difficulty lies in showing that the probability is indeed $1$ without using Borel-Cantelli.

Obviously, it suffices to show that $P[\mathcal{B}] > 0$. I attempt to make various attempts by finding simpler sub-events (i.e. events which are subsets of $\mathcal{B}$) and show that they have non-zero probability, but to no avail. I also see that this problem can be reduced to showing that the probability of head occurring infinitely often is non-zero, as if the sequence has $n$ tosses, one such sub-event would be the event which infinitely many of $(1,\dots,n),(n+1,\dots,2n),\dots$ have the same outcome as the fixed sequence. They are all mutually independent and thus can be willed as successes individually.

Did I miss out some really simple way to show that $P[\mathcal{B}] > 0$?

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Let the probability of a head on each toss be $\ h\ $. Since the assertion isn't true if $\ h=0\ $ or $\ h=1\ $, assume that $\ 0<h<1\ $. Let $\ E_{m,n}\ $ be the event that no heads appear on tosses $\ m+1, m+2,\dots, m+n\ $. Then $\ P(E_{m,n})=(1-h)^n\ $. Let $\ E_m\ $ be the event that no heads occur after toss $\ m\ $. Then $\ E_m \subseteq E_{m,n}\ $ for all $\ n\ge 1\ $, and so $\ P(E_m)=0\ $. Therefore $\ P\left(\bigcup_\limits{m=0}^\infty E_m\right)=0\ $. But $\ \bigcup_\limits{m=0}^\infty E_m\ $ is the event that heads occur only a finite number of times. Therefore, its complement, the event that heads occur infinitely often has probability $\ 1\ $.

The extension of the argument to cover any prescribed finite sequence of heads and tails seems reasonably obvious.