There is at least $2/3$ probability that this question is rather silly, but being an almost absolute beginner in Probability, I will ask it anyway.
Consider the following problem, proposed at AIME 2014:
Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $2/3$ and each of the other five sides has probability $1/15$. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, find the probability that the third roll will also be a six.
Now, the answer is $65/102$, but consider a slight variation to the problem: suppose that the fair die is very rough and the other one very smooth, so that Charles knows which one he has chosen as soon as he touches it, but he does not tell me. So, if I calculate the probability that on his third roll it will be a six, my answer will be $65/102$. Charles on the other hand knows that the probability is $2/3$ (or $1/6$ if he's picked the fair one).
Therefore it seems that we will come up with different probabilities for the same event depending on our knowledge of that event. How is this possible?
Depends on WHEN you ask the question.
If you ask the question before Charles chooses a die, then the answer is 65/102 . It is simply P(launch = 6 | first 2 were 6).
If you ask the question once Charles has picked a die (and he knows what die he has), then the answer is now for P(launch = 6 | first 2 were 6 and I'm using the fair die) or P(launch = 6 | first 2 were 6 and I'm using the cheat die).
It's not that knowledge directly changes the outcome, but rather with knowledge you ask a different question