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This is like monty hall, kind of.
The first choice revealed information - that you are more likely to have chosen box 1 over box 2 and that you did not choose box 3. Let's consider if the problem was different, if box 1 had 1000 golden balls, and box 2 was the same - 1 gold, 1 silver. You put your hand in and take a ball from a box at random and it's gold. You were more likely to have put your hand in box 1 anyways at this point. What are the odds that if you draw from the same box it will be gold? Clearly the number is greater than 50% since in this example theres 999 left in box 1.
So I think to the original problem, there's a 66% chance you pull again and get gold, because you were already more likely to have your hand in box 1, and if you have your hand in box 2, there's a 0% chance to pull gold.
