Only people with an IQ of 115 or higher can get this right

[Socialhealer]

The whole point is that you choose a box at random first because you are sticking your hand in a random box. In effect you are choosing the box. There is a 50/50 chance of pulling a ball of any color. Pulling 1 ball out changes nothing on the chances at all its the second ball that effects the chances only after its been pulled. There is still a 50/50 chance of pulling a ball of either color still even after you eliminate 1 box because the first probability is still in effect until after you have taken out the second ball. Only after that is done can you figure out its 2/3. Until then its still 50/50 on what it will be.

no because you discount any tries where you pull a silver ball and remember we're picking a ball first any ball at random.

say you do this 20 times and get exactly even results.

5 picks of the 1st ball gold
5 picks of the 2nd ball gold
5 picks of the 3rd ball gold
5 picks of the 4th ball silver

the 5 silver picks are deleted, the question asked for first ball is gold those don't count.

so we're left with 15 successful picks

10 times you'll get ball 1 or 2, 5 times ball 3 you don't need to actually pick the 2nd ball, you can see here i'll "win" on the 10 picks and lose on the 5 picks = 66%.

[Palinn]

This is a common problem given when explaining/illustrating Baye's theorem for conditional probabilities. Look that up if you're one of the folks trying to better understand why the answer is 2/3.

[Firefall]

This is so much fun.

The real question is, which interpretation of the problem makes one smart with an IQ higher than 115? The people that paid attention is to their stats professors and expertly asserted 66%, or the people that focus more on the boxes and a specific state/point in time and think 50%?

Lol I'm assuming heavy sarcasm but it is kinda interesting how different people approach problem solving. Probably has nothing to do with intelligence but perhaps there is a difference between people who like to discard "extra" information when solving a problem to focus on "what matters," and people who keep the "big picture" in mind the entire time when working through a problem. The latter strategy seems more error prone to me, but is "right" here, while the prior could make you miss something important if you don't also make a habit of stepping back and looking at the big picture from time to time.

Although I think I've heard babies learn to recognize and group things by colors before shapes, so maybe there's more sarcastic dialogue to be made from that haha :P

[Fotmwarlock]

Guys if you still think its 50% imagine it this way.

Imagine there are 1000 balls in each box

Box A is 1000 gold balls

Box B is 1 gold ball, 1 silver ball and 998 non silver or gold balls.

Box C is 1000 silver balls

If you pull a gold ball from a box, what is the probability that it came from Box B vs Box A?

1/3000 vs 1/3

[Thelxi]

I really can't tell if you're actively trolling or just unable to read properly.
OP question is, verbatim: "You pick box at random. You put your hand in and take a ball from that box at random. It's a gold ball. What is the probability that the next ball you take from the same box will also be gold?"
The box is picked randomly. The ball is picked randomly. No mention of a third party chosing a box with a gold ball, nor making sure that your first pick is a gold ball.

Again, IF there was a third party that would ensure that the first ball you pick out of your box was a gold ball, then yes, the probability would be 50%.
But the problem explicitely state that you pick a ball AT RANDOM from your box. Not knowingly.

- - - Updated - - -



Please read my answer on your previous post.
You're wrong, but I do still have hope that you can understand why you're wrong.

It says you pick a box at random, and then a ball at random, but then it gives you that it is a gold ball, invalidating the randomness of the previous two choices. You have a gold ball from either box A or B, that is what is given.

[For_The_Horde]

Which means that if you label the sides of the boxes, and take the gold coin out of the left hand side, the probability is now 1/2.

Again, no it doesn't, you're missing the point. The s/s box is already out, and the g/s box doesn't count if you draw silver first (s/g), that just leaves g/s and g/g. However once you're narrowed down to this point, you have twice the chance to get a double gold draw.



The first box can drawn to get g1/g2 or g2/g1, and the second box can be g/s or s/g. The third box is irrelevant.

[Celista]

well i came in this thread defending the 50% all the way, until i read enough posts and realised the G/G has twice the chance to appear then yeah i'm obviously wrong.

It's still 50% because you are drawing from one box. I understand the rationale behind the other answers but 2/3 is still incorrect.

[Enthusiastic Steward]

It's still 50% because you are drawing from one box. I understand the rationale behind the other answers but 2/3 is still incorrect.

It's 2/3 if you put the first ball back and pick again, which the original question doesn't state one way or the other, hence the argument. At least I think.

[Fotmwarlock]

It's still 50% because you are drawing from one box. I understand the rationale behind the other answers but 2/3 is still incorrect.

If there were only 2 boxes and 1 had gg and the other gs what are your chances of picking a gold ball

[Enthusiastic Steward]

If there were only 2 boxes and 1 had gg and the other gs what are your chances of picking a gold ball

3/4, except you have removed a gold ball from the equation and then committed to that box, so you effectively have one box with one ball which is either gold or silver.

[Senen]

It says you pick a box at random, and then a ball at random, but then it gives you that it is a gold ball, invalidating the randomness of the previous two choices. You have a gold ball from either box A or B, that is what is given.

It doesn't "invalidate the randomness of the previous two choices".

One simple way to see it is to consider what happens if you were to get a silver ball.
If you get a silver ball, what are the chances that the second ball you draw from the same box is also a silver ball? 1/2 or 2/3?
As it's a completely symmetrical problem, your answer should be the same, so 50% for you (and obviously, I argue it's 66%).
So, for you, no matter what color is the first ball we've picked, there's 50% chance that we've picked the box B. This is obviously wrong.

Another simple way to see it is to consider the same experiment but instead of saying "the first ball you picked is a gold ball", the problem states "if the first ball you picked isn't a gold ball, then put it back in its box, shuffle the boxes, and start again until you draw a gold ball".
That changes nothing to probability, but you should then understand that the "It's a gold ball" part of the problem doesn't invalidate the whole randomness of the experiment.

[Enthusiastic Steward]

but you should then understand that the "It's a gold ball" part of the problem doesn't invalidate the whole randomness of the experiment.

But the problem isn't random, it's cluttered. It has given you very specific criteria, just most of it happens to be useless.

Yes, but 50% of the time when you pulled from G/S you'd get a silver and start over. Which means that when you DO get a gold ball, there's a much higher chance that you actually pulled from the G/G box.

The issue is that we're already past that part. The scenario begins after we have already selected a gold ball. The probability of getting a gold ball from the outset is irrelevant, we've already got one.

[Themius]

It's a 50% shot. Unless the balls can somehow merge back and forth between boxes, it's 50/50.

You have to take into account everything.

You don't have a fifty fifty chance of pulling out a gold ball or not out of a single box because there is a chance you have a box with 2 gold balls or a chance you have a box with 1 gold ball. You have to bring that into account. After you have taken out the ball you are left with 3 balls two of which are gold leaving you with a 2/3 chance to pull a gold ball out.

Think about it this way.

when you first start and you are pulling a ball out of a single box you have a 3/6 chance of getting a gold ball. That is a 50 50 chance to get a gold ball. If you eliminate the two silver box, you have a 3/4 chance to pull a gold ball. Taking into account everything once you pull out a gold ball you have a 2/3 chance to pull another gold ball, which is more than a 50% chance.

[Thelxi]

It doesn't "invalidate the randomness of the previous two choices".

One simple way to see it is to consider what happens if you were to get a silver ball.
If you get a silver ball, what are the chances that the second ball you draw from the same box is also a silver ball? 1/2 or 2/3?
As it's a completely symmetrical problem, your answer should be the same, so 50% for you (and obviously, I argue it's 66%).
So, for you, no matter what color is the first ball we've picked, there's 50% chance that we've picked the box B. This is obviously wrong.

Another simple way to see it is to consider the same experiment but instead of saying "the first ball you picked is a gold ball", the problem states "if the first ball you picked isn't a gold ball, then put it back in its box, shuffle the boxes, and start again until you draw a gold ball".
That changes nothing to probability, but you should then understand that the "It's a gold ball" part of the problem doesn't invalidate the whole randomness of the experiment.

I agree I regretted writing it like that the moment I pushed post. It doesn't invalidate the randomness of both box and ball choices. What I meant to say is it invalidates some of the randomness of the problem, and changes what is given. The way I and some others see the problem is this: you have either box A or B in front of you, and have removed one gold ball from the box already. What is the probability you draw another gold ball from the box in front of you? in other words, what is the probability of having picked box A, instead of box B? it is 50%.

Some people here see a clear cut betrand's box paradox and vehemently parrot the 66% or try to explain the math to people, while others say no box C is irrelevant, and it doesn't matter which gold ball you potentially drew from box A, you have either a silver or a gold ball in the box in front of you at this point in time, because you picked either box A or B.

Both perspectives and interpretations are correct, and understood by a growing minority

[Enthusiastic Steward]

You have to take into account everything.

You don't, the field is cluttered with distractions. You mention eliminating the silver box, giving you a 3/4 chance of pulling out a gold ball. The problem is you ALSO have to eliminate 2 of those gold balls, because when the scenario starts they have already been effectively removed (one from the box you pulled it from, the other by the fact that you have to draw from the same box, so whichever box it happened to be only 1 ball remains for you to pick. This ball is either gold or silver.

[Themius]

I agree I regretted writing it like that the moment I pushed post. It doesn't invalidate the randomness of both box and ball choices. What I meant to say is it invalidates some of the randomness of the problem, and changes what is given. The way I and some others see the problem is this: you have either box A or B in front of you, and have removed one gold ball from the box already. What is the probability you draw another gold ball from the box in front of you? in other words, what is the probability of having picked box A, instead of box B? it is 50%.

Some people here see a clear cut betrand's box paradox and vehemently parrot the 66% or try to explain the math to people, while others say no box C is irrelevant, and it doesn't matter which gold ball you potentially drew from box A, you have either a silver or a gold ball in the box in front of you at this point in time, because you picked either box A or B.

Both perspectives and interpretations are correct, and understood by a growing minority

I didn't know about betrand's box and instantly thought 66% because it just makes the most sense when you take everything into account. You have to take into account the higher probability that if you pull a gold ball that the other ball is likely gold, and then the other probability when it comes to the gold/silver box. You have to factor in both the boxes and the balls.

- - - Updated - - -

You don't, the field is cluttered with distractions. You mention eliminating the silver box, giving you a 3/4 chance of pulling out a gold ball. The problem is you ALSO have to eliminate 2 of those gold balls, because when the scenario starts they have already been effectively removed (one from the box you pulled it from, the other by the fact that you have to draw from the same box, so whichever box it happened to be only 1 ball remains for you to pick. This ball is either gold or silver.

And what is the probability of picking another gold ball when there are 2 gold balls of 3 total balls?

[Thelxi]

The chance being 66% actually has nothing to do with the S/S box. That box is irrelevant no matter which camp you're in.

I know it's considering the two gold balls in box A as two different balls that is the crux of the problem. To me a gold ball is a gold ball.

[Senen]

I agree I regretted writing it like that the moment I pushed post. It doesn't invalidate the randomness of both box and ball choices. What I meant to say is it invalidates some of the randomness of the problem, and changes what is given. The way I and some others see the problem is this: you have either box A or B in front of you, and have removed one gold ball from the box already. What is the probability you draw another gold ball from the box in front of you? in other words, what is the probability of having picked box A, instead of box B? it is 50%.

This is correct.
With how you interpret the problem, the answer is, indeed, 50%.

Both perspectives and interpretations are correct, and understood by a growing minority

This is incorrect.
Your interpretation of the problem is false. The conclusion you draw from your interpretation are correct, but your interpretation is false.

The problem stated is:
"There are three boxes. You pick one box at random. Then, you put your hand in and take a ball from that box at random. It happens to be a gold ball."

How you interpret it is:
"There are three boxes. A third party, knowing the content of each box, discard box C and make you chose between box A and box B. Then, a third party look inside that box and knowingly remove a gold ball from it."

Again, in the second problem, there's definitely 50% chance you picked box A and 50% chance you picked box B, you're correct on that.
But no, the first problem and the second problem are vastly different and op's problem is the first one, not the second one.

[Enthusiastic Steward]

I didn't know about betrand's box and instantly thought 66% because it just makes the most sense when you take everything into account. You have to take into account the higher probability that if you pull a gold ball that the other ball is likely gold, and then the other probability when it comes to the gold/silver box. You have to factor in both the boxes and the balls.

- - - Updated - - -



And what is the probability of picking another gold ball when there are 2 gold balls of 3 total balls?

But there aren't three total balls. There is one, because you are required to pick from the same box. In order to pull gold you have 4 balls to work with, 3 are gold 1 is silver. You have removed one gold, so there are two gold and one silver. But you are not picking from three balls because you can't pick from both boxes. You eliminated one gold by picking it up but you also removed two other balls from your pool by picking a box. So one more gold and a mystery ball are no longer possible options and so removed from the field.

[Lewlies]

If you grab 1 ball and keep it....

It's 50%. End of story. Any other answer is asinine.

You discard one box from the start. Leaves you with 2 boxes. You grab one yellow ball and are expected to grab another. You either grab a) silver ball or b) gold ball. No other option. Case closed.


If you grab one ball and put it back...

It's 66%. Simple math.