The answer is 50% right?
Damn, would have gotten it wrong.
The question is supposed to trick you into saying 50% chance. However, I think the answer is 2/3, since the fact that you picked a Gold ball implies a 2x likelihood that you’re in the GG box than the GS box. This is the same as that famous question with the doors.
The other way to look at this is to realize that the first pick has 6 distinct balls that you can pick. The fact that the balls are in 3 boxes doesn’t affect the 1/6 likelihood that you will pick any one of the 6 balls. Of course, the second pick is dependent on the first pick.
Let’s say the balls are organized as such: [1, 2] [3, 4] [5, 6], or [G G] [G S] [S S]. You have picked one of balls 1, 2, 3. So, the possible second picks are 2, 1, 4 two of which are G and one is S.
^^
I agree with you Ohai, but it also says we cannot see into the boxes, therefore the organization of the colors and the subsequent narrowing down after the first pick would have to be irrelevant. I think we have to assume the probability is based on picking a ball out of a black box, ignoring the organization/ boundaries of 3 different boxes.
OP… is there a solution somewhere?
where my bae theory!
Ohai is correct and essentially posted the solution. As did DOW. This is basically that problem with the 3 doors. It’s un-intuitive at first but mathematically provable.
^^
Yes, I was overthinking it assuming the participant did not have access to the information that was organization of the different boxes.
Even if the participant doesn’t know the organization of the boxes the probabilities themselves of outcome are the same (2/3, 1/3), the only difference is without the information, the participant might calculate the wrong odds.
Yep… I concur 
always switch doors!!!
This is essentially the Monte Hall problem.
In bridge, a similar idea is known as restricted choice.
brah, change the problem like this in case your intuition is misleading you:
you have 3 boxes. each box has 100 balls. one has 100 gold balls, one has 100 silver balls, and the last one has 50 gold and 50 silver balls. you pick a box at random and take out 50 balls from it at random. they all happen to be gold, LoL. what’s the chance that the next ball you take from the same box will also be gold?
of course it’s not 50%.
I think its 50%
You are in box 1 or box 2 that is given based on the fact that you picked one gold ball already.
Now in the box you have either one gold ball remaining or one silver ball remaining (I assume that you don’t put the ball you picked back into the box before you pick next)
So that would make it 50%.
??
Literally the first post by DoW (second post overall) provided the answer to the question. I find it fascinating that posters continue to debate this and post solutions.
But what if after you draw a ball out the boxes get reshuffled and they have to choose another box at random.
edit: OOOOOOOOOOOOOOOOOOOOOH it says from the same box… Okay. Wow. I’m a dumbass.
google is great! google is good! google is merciful!
…
Then you’re wrong.