>>12936665The same way that it's a conditional given that the first ball you pick is blue. You just do.
(More precisely, you consider only the outcomes where you both pick a blue ball for your first choice, AND in which you pick the second ball from the same box as the first.)
For example, here are all the following sequences for what balls you can pick. The blue blue box is "box 1", the blue red box is "box 2", and the red red box is "box 3". In parenthesis is the total, non conditional probability for that outcome.
box 1 blue, box 1 blue (1/9)
box 1 blue, box 2 blue (1/18)
box 1 blue, box 2 red (1/18)
box 1 blue, box 3 red (1/9)
box 2 blue, box 1 blue (1/18)
box 2 blue, box 2 red (1/18)
box 2 blue, box 3 red (1/18)
box 2 red, box 1 blue (1/18)
box 2 red, box 2 blue (1/18)
box 2 red, box 3 red (1/18)
box 3 red, box 1 blue (1/9)
box 3 red, box 2 blue (1/18)
box 3 red, box 2 red (1/18)
box 3 red, box 3 red (1/9)
As you can see, this all adds up to a total probability of 1.
The condition of "given your first ball picked is blue" narrows these down to the state space as follows. The conditional probabilities have all been divided by 1/2, their total probability, to form the new conditional probabilities.
box 1 blue, box 1 blue (2/9)
box 1 blue, box 2 blue (1/9)
box 1 blue, box 2 red (1/9)
box 1 blue, box 3 red (2/9)
box 2 blue, box 1 blue (1/9)
box 2 blue, box 2 red (1/9)
box 2 blue, box 3 red (1/9)
The second condition of "[given] you pick from the same box", which is on-top of the condition of your first ball being blue, the state space is once more restricted down to the following. This time the probabilities are divided by 1/3, their total probability, to normalize the new conditional probabilities.
box 1 blue, box 1 blue (2/3)
box 2 blue, box 2 red (1/3)
As you can see, the only remaining outcome where you choose blue twice is when you picked from box 1, for a probability of 2/3.
