Conditional probability regarding drawing a marble

The outcomes in the sample space $\{GG,GB, BG\}$ are not equally likely. The probability that we observe $GB$ and $BG$ are each $1/4$ and the probability that we observe $GG$ is $1/2$.

I feel that this problem can be solved intuitively, without calculating through Bayes' Rule. Consider the following possible scenarios:

1. The first marble taken out is green (given). This marble is the initial marble in the bag (assumed for the sake of this scenario). Then, the final draw will reveal a green marble.
2. The first marble taken out it is green (given). This marble is the second marble put into the bag (assumed for the sake of this scenario). The initial marble is green (assumed for the sake of this scenario). Then, the final draw will reveal a green marble.
3. The first marble taken out is green (given). This marble is the second marble put into the bag (assumed for the sake of this scenario). The initial marble is blue (assumed for the sake of this scenario). Then, the final draw will reveal a blue marble.

It seems that these are the only possible scenarios, and 2 out of 3 have the remaining marble as green. So P(remaining marble is green) = 2/3.

Verify with LOTP and Bayes'. Let: A be the event that initial marble is green, B be the event that the first marble taken out is green, and C be the event that the remaining marble is green.

LOTP : $P(C|B) = P(C|B,A)P(A|B) +P(C|B,A^{c})P(A^{c}|B)$ = $P(A|B)$

Bayes': $$P(A|B) =\frac{P(B|A)P(A)}{P(B)}$$ $$P(A|B) =\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}$$ $$P(A|B) = \frac{1*\frac{1}{2}}{1*\frac{1}{2}+\frac{1}{2}*\frac{1}{2}}=\frac{2}{3}$$