re: Solve this OT

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This is pretty straight forward with no gotcha or ambiguity

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Not really. It's an intentional paradox with mathmatical arguments for both. It's intentional presented that way.

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How can you say this? You cannot look into the boxes.

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Because the question tells tou one box had 2 gold, one 2 silver and one a gold and silver.

If you pick gold, you know 100% that the box does not ahve 2 silver,so that box can be eliminated.

You are left with 3 balls and you KNOW 2 are gold and one is silver. The question tells you that

66.7% .It’s a variation of the Monty Hall problem.

The trick is that they gave you part of the answer already, which is that you selected a gold ball. Knowing that, you know you selected either box #1 or box #2. But given that box #2 has only one gold ball, the gold ball you selected more likely came from box #1 which is why you now have a greater than 50% chance of selecting another gold ball.

In other words, knowing that you have a gold ball already, there are three possible outcomes:

1) Pick up gold ball #2 (1st box, you’re already holding ball #1)
2) Pick up gold ball #1 (1st box, you’re already holding ball #2)
3) Pick up silver ball (2nd box, you’re holding the gold one).


2 of those 3 possibilities mean you’re picking up a gold ball. Hence, 2/3 = 66.%

Nah Cold Drink explains it perfectly.

Whichever box you originally selected from is the box you're selecting from the second time.

So you're stuck with that one box. Correct?

You know it isn't the 2 silver box. That box is gone.

So of the 2 remaining, after having selected a gold ball, the second ball would either be a gold from box 1 or a silver from box 2.

Right?

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NOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!

You have a gold ball in your hand, so the box either has 1 gold ball or 1 silver ball left in it. 50% chance.

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The question isn't asking what collor the remaning ball in the box is. If that were the question, yes its 50% gold 50% silver. BUT THAT'S NOT THE QUESTION.

The question is what are the odds the the next ball you pick is gold. You know of the 3 balls, 2 are gold and one is silver so without knowing what box you have, its 2/3 or 67% odds that the remaining ball is gold.

It's not the box paradox or the Monty hall because this question specifically states your second draw HAS to come from the same box as the first draw ...

So you are stuck with the odds that you had at the start which is 1/3.

3 balls left. 2 are gold. 2/3. It’s fairly simple, guys.

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66.7% .It’s a variation of the Monty Hall problem.

The trick is that they gave you part of the answer already, which is that you selected a gold ball. Knowing that, you know you selected either box #1 or box #2. But given that box #2 has two gold balls, you were more likely to have selected one of those in box #1 which is why you have a greater than 50% chance of selecting another gold ball.

In other words, knowing that you have a gold ball already, there are three possible outcomes:

1) Pick up gold ball #2 (1st box, you’re already holding ball #1)
2) Pick up gold ball #1 (1st box, you’re already holding ball #2)
3) Pick up silver ball (2nd box, you’re holding the gold one).


2 of those 3 possibilities mean you’re picking up a gold ball. Hence, 2/3 = 66.%

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Ok this makes sense. I'm wrong.

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It's not the box paradox

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It’s the exact same wording.

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There are three boxes:

a box containing two gold coins,
a box containing two silver coins,
a box containing one gold coin and a silver coin.
The 'paradox' is in the probability, after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, of the next coin drawn from the same box also being a gold coin.

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