1
$\begingroup$

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.

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

2 Answers 2

1
$\begingroup$

This is a variant of the infamous Monty Hall Problem.

The problem you are having is identifying what events are equally likely. If you have six balls in all, three of them blue, and you pull out a blue ball, then there are three possibilities, not two. You could have taken the one ball from the mixed, or you could have taken the first blue ball from the box with two, OR you could have taken the SECOND blue ball from the box with two blue. That's three possibilities, not two.

$\endgroup$
0
$\begingroup$

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?

Yes.

You had an equal probability (no bias) for obtaining each individual blue ball; and two among the three were in the jar that held two blue balls.

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .