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

[ed]

This is literally the monte hall problem

[chazus]

This is one of those puzzles that banks on reading comprehension, not math. It gets people to argue over the answer because some people overlook some of the word play.

Basically, it's clickbait, except nobody is getting paid since there's nothing to click.

YOU WONT BELIEVE.

[jackofwind]

It's 50/50.

We've already eliminated the S/S box. So the only possible boxes we could have pulled from are G/G or G/S. Since we have removed one G from one of these two boxes, our remaining options are 0/G or /S, essentially G or S. Since there are only two options and you can only pull one more ball, the chances are 50/50, or 50% probability.

This is dumb.

The point I believe is that you have to take into account what the chance was of you pulling a gold ball from a box in the first place? I dunno, the whole thing is deliberately ambiguous.

[Eldar45]

This is one of those puzzles that banks on reading comprehension, not math. It gets people to argue over the answer because some people overlook some of the word play.

Basically, it's clickbait, except nobody is getting paid since there's nothing to click.

YOU WONT BELIEVE.

Pretty much this.

[Ishimada]

Read "from the same box", 50% right? Either you picked from the gold paired box or the box with the silver one with the gold, and that box has one of those.

Edit: So yeah didn't count in that when you already have picked the specific ball, there is 2 gold balls and 1 silver ball left in the 2 boxes which contain gold balls. Little mind tricker.

[zoefschildpad]

It's 2/3rds. You picked a gold ball, which is can be one of three balls with equal probability. The left gold ball in the left box, the right gold ball in the left box, or the gold ball in the middle box.

In two of those three cases the other ball in the box is also gold.

See also the Monty Hall problem which has a very similar premise.

[Kuntantee]

Damn my iq is 114

This is a fine joke =).

[SinR]

THERE
ARE
FOUR
LIGHTS!

really now why does this thing have to de-cap everything?

[Deldavala]

Except you've eliminated the possibility of the box you're pulling from being the S/S box. So the box you're pulling from is either a G/G box or a G/S box. So its a 50/50 chance.



The wording seems to indicate that you pull a different ball.

Oh you are attacking it wrong. Its similiar to the Monty Hall problem, I think its originally called Betrand box(been some years since I took a class on probability).

Since the ball we pulled is a Gold one, that means the S/S one is gone. The Gold ball we pulled has to either be from the G/S or the G/G box. Since each of these outcomes is equally possible, in 2 out of 3 cases, the Ball will come from the G/G box, meaning in 2 out of 3 cases, the other ball will also be gold.

[Rafoel]

If you REALLY picked box at random at the beginning, then it has 33% chance of being the box with 2 golden balls regardless of what ball you found inside it later.

The riddle is very weirdly worded though (which I guess might be the point... but english is not my mother tongue).

[Wries]

Except you've eliminated the possibility of the box you're pulling from being the S/S box. So the box you're pulling from is either a G/G box or a G/S box. So its a 50/50 chance.



The wording seems to indicate that you pull a different ball.

You also have to account for the fact that you managed to pull the golden ball (in a box that possibly also contains a silver one) on your first random fumbling.

[Ozyorkbourne]

If you REALLY picked box at random at the beginning, then it has 33% chance of being the box with 2 golden balls regardless of what ball you found inside it later.

The riddle is very weirdly worded though (which I guess might be the point... but english is not my mother tongue).

No. You aren't calculating the chance of the box having 1 gold ball in, ergo you needn't worry about the probability of picking a box with a golden ball in. That's already done and pre-dates the question. The box choice has happened meaning there are two boxes you could have chosen.

[Ryme]

Casino owners HATE him!

[Polyxo]

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

[Ozyorkbourne]

Oh you are attacking it wrong. Its similiar to the Monty Hall problem, I think its originally called Betrand box(been some years since I took a class on probability).

Since the ball we pulled is a Gold one, that means the S/S one is gone. The Gold ball we pulled has to either be from the G/S or the G/G box. Since each of these outcomes is equally possible, in 2 out of 3 cases, the Ball will come from the G/G box, meaning in 2 out of 3 cases, the other ball will also be gold.

Same as this: https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

[JackOnPar]

50%. There are only two color types. The two other boxes don't matter.

[Zantos]

Just no. This isn't facebook. Can we not start posting this crap?

[Deldavala]

Same as this: https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

Ye, lots of students in my class had some issues wrapping their head around that one, along with the more popular Monty hall problem and 3 prisoner problem.

[Skulltaker]

The point I believe is that you have to take into account what the chance was of you pulling a gold ball from a box in the first place? I dunno, the whole thing is deliberately ambiguous.

No, you don't but that is a common misconception. The boxes are also almost irrelevant. This is a magicians trick disguised as a math problem.

You have already picked. The Gold ball in your hand is irreleavant, except for the setup, of course. So are the other two boxes. If, at the instant you stuck your hand into one box, the other two would be disintegrated, nothing would change the outcome. You have one golden ball, and the other one that you will have to pick, no way around it, is either gold, or silver.

Choice has already been removed from you at this point. You already picker either the GG or the GS box, and there is only one ball remaining that matters to you - it is either silver, or golden, and the chances are the same.

If you REALLY picked box at random at the beginning, then it has 33% chance of being the box with 2 golden balls regardless of what ball you found inside it later.

The riddle is very weirdly worded though (which I guess might be the point... but english is not my mother tongue).

You already picked one of the two boxes with at least one golden ball in it. The other two boxes are already irreleavant. The remaining ball is either silver, or gold, and therefor, the chance is 50/50.

It changes if you have to put the ball back, though.

[Eldar45]

You also have to account for the fact that you managed to pull the golden ball (in a box that possibly also contains a silver one) on your first random fumbling.

The probability it is asking for doesn't have anything to do with your first random pick. It has to do with the chances of the next ball, from the same box, being gold. And because it was either a gold/gold box that you picked or a silver/gold box that you picked then it is a 50/50 chance.

It changes if you have to put the ball back, though.

But without the problem specifically saying that you put the ball back you have to assume that you don't.