# Probability of second ball being blue

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.

• Your last paragraph is indeed the correct solution. Commented Mar 3, 2021 at 1:30

So when given that you draw a blue ball there is a probability of $$2/3$$ that there was another blue ball left in that jar.