There are $3$ boxes. One box has $2$ blue balls. One box has $2$ red. The last has one blue and one red.
You randomly pick a box and take $1$ ball out. You see that it's blue. What is the probability that the other ball in the box is blue.
The answer is $\frac{2}{3}$ and I confirmed this with Bayes rule. However, I am confused why the answer is not $0.5$. I know that if we aren't given the information that the first ball taken out is blue, then for sure, the second ball being blue has a probability of $0.5$.
However, without referring to Bayes rule, if I account for the fact that the first ball is blue, I can essentially eliminate the box with $2$ reds from my consideration. Because it is surely not possible that we chose that box. That leaves us with $2$ boxes. Then my thought process was that well, we could remove a blue ball from each box (because we just drew one and no longer need to consider it), so now the $2$ boxes each has $1$ ball. Blue in $1$ box and red in another, with each having the same probability of being the second ball, which would make the answer $0.5$, but apparently that's not correct. What is wrong with this thought process?
One thing that I am pondering is, if we know the first ball is blue, then we can remove the box with $2$ reds, but I think knowing this information might also make it so that there's a higher probability that we chose the box with $2$ blues, than the box with $1$ blue and $1$ red. In these $2$ boxes, there are $3$ blue balls total, but because $2$ of these belong in a single box, then that box should have a $2/3$ chance of being chosen conditioned on us knowing that we observed $1$ blue already. Is this right? If so, then I think that answers my previous question.