Why is my reasoning incorrect - probability? In a variant of Russian Roulette, where you put 2 bullets in 2 adjacent chambers, like that:
Image credit: Brilliant.org
Now, first person shoots and survives, you are the second.
The question is: In which scenario are you more likely to survive:

*

*You spin the barrel again, assuming a random spin(each chamber has equal probability).

*Shot, without spinning.

I have tried solving it like that:
Even though I have gotten the correct result I was reasoning incorrectly
I have calculated the probability of 1-st scenario and got 1/3(2 possible ways of killing, 6 barrels), this gives the chance of dying if spin. Then the probability of 2nd scenario, here I got the wrong result.
P(dying in second scenario) = P(1st surviving)*P(you being hit) =
$$ \frac{4}{6} * \frac25 = \frac4{15}$$
Which is approximately 0.27. Not to forget I have also thought that this answer gives the P(dying), but I need P(dying given 1st survived). So, I thought that I need to calculate how big is the chance of this specific outcome of 2nd dying if first survived. To do that I divided it by 2/3 - because that was the probability of the first one surviving. Also, I don't really know why we are dividing, probability by probability.
But the correct answer is 0.25. Reasoning like that:

There are 4 ways to get the specific result, and only 1 way this could happen.
 A: For "spin", you are correct on the probability -- pretty straightforward.
For no spinning, we know the first person didn't land on either bullet. So there are 4 slots left.
Of those slots, you will only die if the first person's spin landed one before the start of the two bullets. So there is only one slot that satisfies this; therefore, the probability you will die is 0.25 --> so you should not spin but just pull the trigger again.
More formal derivation of this using conditional probabilities.
Let $S$ be the event person 1 survived and let $D$ be the event you die if you pull the trigger without re-spinning. Then
$$P(D|S) = \frac{P(D\cap S)}{P(S)}=\frac{1/6}{2/3}=\frac{3}{12}=\frac{1}{4}$$
Your reasoning assumes that you select one of the holes randomly from the remaining $5$ that haven't been tested. But the real situation is much more restricted, hence you got too big of a probability.
Second -- you multiplied the probability of both events but they are not independent, so you can't do this as a general rule. Also, you don't care about the probability of the first person surviving, since we are in a situation where we know they survived, so you really just want a conditional probability.
A: 
P(dying in second scenario) = P(1st surviving)*P(you being hit)

(quote from your question)
holds if you compute the probability of you dying on forehand, before the first person shoots. However, from the wording of the question I think they want you to compute the probability after the other person has shot and you already know that they survived.
After all: that is the moment that you need to make the decision between the two scenarios and hence the most natural moment to calculate the probability.
So you are more interested in calculating:
P(you being hit | 1st surviving)
