Bernoulli trials - proving that a sequence will occur infinitely many times Let's consider the infinite numbers of Bernoulli trials.
The probability of the success ($S$ - success, $F$ - failure) is equal to $p$.
I am to prove that the possibility of getting the infinite amount of the following occurrence - $SSSFFF$ is equal to $1$.

My attempt:
1)
I tried to solve the problem by redefining the success. Now for me it would be not $S$ but the sequence $SSSFFF$. The probability of such sequence would be equal to $p^3(1-p)^3 > 0$.
$E[X = \text{success at k-th trial}] = \frac{1}{p^3(1-p^3)}$ so I would need $E[X]$ trials to get my "new" success. Of course after infinite numbers of them I would get infinite numbers of successes.
2)
$P(\text{getting at least one redefined success in n trials}) = 1 - P(\text{only F in n trials}) = 1 - (1 - p^3(1-p^3))^n \rightarrow 1$.
Thus because of the "no memory rule" we will get infinite numbers of "new" successes.
Are my solutions correct? If no, why?
 A: Your solution is almost there, but you need something to tie it all together.  One way to do so is as follows: if there are only finitely many redefined successes, let $k$ equal the number of redefined successes.  By the no memory rule (also known as the strong Markov property), the trials after the $k$th success are distributed the same as taking them all independently.  The probability there is at least one redefined success in this remaining sequence, is at least the probability of having at least one in the last $n$ for each $n$; using your calculation $(2)$ completes the proof.
I strongly disagree with using the central limit theorem here, as it says significantly more than you need, and some work is needed to get from there to your answer since the CLT gives convergence in distribution rather than almost sure convergence.  If you're going to use any machinary on this, the second Borel-Cantelli lemma proves it immediately.  If you haven't seen that before, you can use the strong law of large numbers which is perhaps what the commenter really meant.
A: First of all, there is no such thing as an infinite number of Bernoulli trials. Any time "infinity" is used in mathematics, it's being used as a shorthand for some concept. So the first task is to find some interpretation "infinity". One possible interpretation is:
Given any k, $\epsilon$, there exists an N such that after N Bernoulli trials, the probability of getting at least k instances of the sequence is greater than 1-$\epsilon$. (Note that the probability is not equal to 1, it just goes to 1).

$E[X = \text{success at k-th trial}] = \frac{1}{p^3(1-p^3)}$ so I would need $E[X]$ trials to get my "new" success. 

After E[X] trials, the expected number of successes is 1. But that does not mean that after E[x] trials, you will have a success. You will need a lot of work to go from this to proving the desired conclusion.
One strategy is to consider each Bernoulli trial to be a Level1 trial. Then consider every six Bernoulli trials to be a Level2 trial in which there is $p^3(1-p)^3$ probability to get the given sequence. This ignores the possibility of the sequence "straddling" two different groups of six, but since we can make N as large as we want, and we don't have to get the exact probability, just show that it's larger that 1-$\epsilon$, that's not important, we can just concentrate on finding a lower limit of the probability.
You can then define a Level3 trial as being b Level2 trials, for some b, with a Level3 success being at least one success among the Level2 trials within the Level3 trial. You can then make the probability of a Level3 success arbitrarily close to 1 by increasing b.
Next, you can take a Level4 trial to be k Level3 trials, where success is now having success in all Level3 trials. So a Level4 success now constitutes k instances of the sequence SSSFFF. 
You can calculate the probability of a Level4 success in terms of b, p, and k, and then all you have to do is show that given any $\epsilon$, you can find b such that the probability of a Level4 success is greater than 1-$\epsilon$. 
