re: Solve this OT

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Once you pick the first gold ball, your second pick is either 100% a gold ball or 100% a silver ball. The balls in the other box are inconsequential at this point.

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I agree.

If we labeled the gold balls G1, G2, and G3, and the silver ball S1, what is the set of the possible outcomes for that second ball, given the fact we've picked a gold ball?

You are missing the point of labeling them G1 G2 and S1

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Which side are you on in this whole Pan/Pam dilemma?

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The entire game is decided by which ball you pick first. The rules of the game dictate how the game will play out after you make your first choice and there is ZERO input for the player other than selecting one of 6 balls.

If you pick ball 1, you must also pick ball 2 and the game is over.
If you pick ball 2, you must also pick ball 1 and the game is over.
If you pick ball 3, you must also pick ball 4 and the game is over.
If you pick ball 4, you must also pick ball 3 and the game is over.
If you pick ball 5, you must also pick ball 6 and the game is over.
If you pick ball 6, you must also pick ball 5 and the game is over.

That's it.

Balls 1,2,3 are gold. Balls 4,5,6 are silver.

There are only three ways to pick a gold ball first, and that's to pick 1,2, or 3. This is what we are told. A gold ball HAS been chosen. Of the three ways that can happen, there are two ways that will result in a second gold ball being drawn and 1 way that will result in a silver ball.

Therefore, if you're playing the game and you're holding a gold ball after your first pick, there is a 2/3 probability your second ball will be gold.

That's it. Anything else is wrong. It's a simple matter of counting possible outcomes given the constraints of the problem. The OT can't count.

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The entire game is decided by which ball you pick first.

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Bingo

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There is only one outcome after you've made your first selection according to that logic

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There is but one actual outcome, but it is one of 2 possibilities.

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You can't eliminate G1 or G2 

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You have already eliminated at least 1 and possibly both. Both aren't in play once a gold ball has been selected.

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Christ, you two have been arguing all day over something you both agree is 66.67%.

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10+ pages of the exact same comments back and forth. No budging, yet no insults. It's been amusing.

The problem with you solution is that it doesn't apply to the problem as stated. Your solution would only apply if all the remaining balls were in the same box and you were selecting. Then you would have 3 possibilities -2 gold and 1 silver.

Once a gold ball is eliminated and you have chosen a box, there no longer remains 3 possibilities.

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Your solution would only apply if all the remaining balls were in the same box and you were selecting.

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They may as well be in the same box at that point. You don't know which box you took the gold from so all of the balls in the GG box and the GS box are still in play.

There are six balls. You reach in and pick a gold ball.

Here is what you KNOW.

The box you pulled the ball from is not the box with two silver balls.

You know you have a gold ball, but you don't know if it came from the box with two gold balls or from the box with a gold and a silver ball.

Here's what you know about the ball left in the box you chose.

It can be a silver ball from the Gold ball/silver ball box
It can be a gold ball from the box with two gold balls

It can not be anything else. It can not be the gold ball from the gold ball silver ball box.

So here is the accounting

The remaining ball in the box you pulled the gold ball from
can not be either of the two silver balls that were in the same box.
It can not be the gold ball that was in the gold box, and it can not be the gold ball in your hand.

It can only be a silver ball or the gold ball that was in the box with two gold balls.

50/50 chance it's the gold one