how does the probability change based on past independent events Lets say we have a coin with side 1 and side 2, and these are the only two results we can get flipping that coin.
When flipping the coin, we know that:
$$Pr(result = 1) = \frac{1}{3}$$
$$Pr(result = 2) = \frac{2}{3}$$
And the events are always independent

What that means, as I know, that we will get the result 1 every three flips on average, also that if we flip the coins three times, (*) the probability that none of the three results was 1 is 8/27.

Now we want to flip the coin three times, we flip twice and the results for now are: 2 2
I'm a little bit confused about what we can say about $Pr(result = 1)$ at this point, on the first hand these events are independent so the probability should stay as it is, on the other hand I'm having a feeling that the probability should be more than 1/3, or at least that we can say something else about it or about the three flips, because of the "average" and because of bullet (*)
 A: The problem you have stumbled across has clear links to some of the most famous and counter-intuitive results in probability theory. The short answer to your question is that given that the events are independent,
$$
Pr(\text{result}=1)=\frac{1}{3}
$$
irrespective of what has come up beforehand. Think about what it would mean if this probability suddenly changed after a few flips. The coin would have to somehow 'remember' what happened earlier on and make sure that everything averages out nicely. Unfortunately, we tend to avoid anthropomorphising coins in mathematics!
It is important to be clear about what we mean by the word 'average'. If we say that the average number of $1$'s that come up in $3$ flips is equal to $1$, then what we mean is that if we consider dozens upon dozens of cases where you flip $3$ coins, and each time you compute
$$
m=\frac{\text{Number of times result = 1}}{3}
$$
then the mean value of $m$ should be roughly $1/3$. Not exactly $1/3$, but close enough. If this still feels a little abstract, then perhaps my computer simulation will convince you:

These are genuine results I got after flipping three of your biased coins $10$ times in a row. As you can see, the value of $m$ varies. Sometimes we might even see '$111$' come up, even though this only has a $\frac{1}{27}$ chance of happening. When we calculate the mean value of $m$, then we get that $m_\text{mean}=0.2\overline{3}$. So we didn't get exactly $10$ occurrences of $1$, but this is hardly surprising. So remember, averages tell you about happens in the long run, but can be unreliable within the space of a few flips, and certainly won't tell you what is about to occur. Let me know if you have any questions.
