Toss a coin repeatedly and independently. Show that heads eventually appears. Show that any fixed finite sequence such as HTHHHT eventually appears? I'm not getting the answer for the second part where I have to prove about the fixed finite such as HTHHHT will eventually appears. Please help me with this part.
 A: Do you know the Borel-Cantelli lemmas?  In particular, the second one?
It says that if $E_1,\ldots,E_n$ are independent events, and
$$
P(E_1)+P(E_2)+\cdots=\infty,
$$
then with probability 1 infinitely many of the events $(E_n)_{n=1}^{\infty}$ occur.
What if $E_n$ is "elements $6n+1,\ldots,6n+6$ are HTHHHT"?
You could, of course, come up with a proof that doesn't use Borel-Cantelli.  For any $n$, $P(E_n)=\frac{1}{2^6}$, and these events are independent. Using this, what's the probability that at least one event $E_1,\ldots,E_n$ occurs?  What does this probability do as $n\to\infty$?
A: Let $X_n$ denote the event that the particular $6$-long pattern occurs starting at position $n$, so $P(X_n)=1/2^6$.
Let $Y_n$ be the event that none of the events $X_1,\ldots,X_n$ occur. It is tempting to claim that
$P(Y_n)=(1-1/2^6)^n$, but this is false! $X_i$ and $X_j$ are dependent if $|i-j|<6$, since the two windows overlap.
However, if $|i-j|\ge6$ then $X_i$ and $X_j$ are independent, as they depend on distinct dice throws. So we have
$$P(Y_n)\le P(\neg X_n) P(\neg X_{n-6}) P(\neg X_{n-12})\cdots = (1-1/2^6)^{\lfloor n/6\rfloor}\to 0$$
as $n\to \infty$. Thus the pattern will eventually appear "almost surely" (i.e. with probability one.)
