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

[Wermys]

After thinking about this, could it be 1/3 instead 2/3. The logic I am using is that there are 6 balls 3 gold 3 silver, once you select a box, that eliminates the silver box, but that also eliminates a gold ball. So the original chance was 50/50 of drawing a gold ball or silver ball. Now if you still include the silver balls because you have to pick up from the same box you have effectively eliminated 2 of the 3 boxes. So the original probably changes because of this. You don't redraw the box, you go from the original box you selected in the first place. So now 2/3 boxes contain no gold balls so that leaves a 1/3 a chance of it being a gold ball on the next draw, and 2/3 of it being silver now based on the original random chance draw. Eh going back to my original its a 50/50 shot. You have to go from current probability what what it would have been from the start.

[Notshauna]

Nah it's just incorrectly settled question.
The way it is described it is ofc. 50%.

When you choose the box and don't know then 1st ball is golden yet - it's 2/3.
After the step of picking the ball is done the whole question is rebuilt to
"There are 2 possible variants - the chosen box is GG or GS, what is the chance it is GG"

If you use The Bayes' theorem for solving - note that by the definition of this question Probability of 'We have drawn a golden ball first" is 100%.

The same goes to flipping the coin in a row:

What is the chance a coin will show heads 2 times in a row?
It's 0.5*0.5 = 0.25

3 times in a row?
It's 0.5*0.5*0.5 = 0.125

But after you flip a coin once and it shows heads - what is the chance it will show heads AGAIN?
It's 0.5 as usual, because flipping the coin each time is an independant experiment.
If it shows heads again - the next heads is expected at 50% rate again

This is an explicitly dependent experiment though, as the result of the first draw is what the second draw's probability is exclusively dependent on the first because it's based on you drawing a gold ball then drawing another from the same box.

[schwarzkopf]

Correction OP - only those with an education in statistics, who speak English will get the above correct.

[dlhak]

It's poorly written and confusing probably deliberately since it's a really lame puzzle. If you have already picked out a gold ball, then the probability of finding another in the same box is either 100% or 0%

But since we don't know which box we're supposed to have picked one out of...

2 of the three boxes have at least one golden ball.

If you have already picked one, that means either the second ball is the second golden ball in box a, or the silver ball in box b ... so it should be 50% chance.

[cubby]

Especially when a bunch of folks, including myself, linked to the exact mathematical proof.

The probability is 2/3.

You can blame me for part of that.

I have to admit that I'm still a bit baffled at the probability theory vs the practical percentages. I'm sure you're going to dice me up for saying that, but I'm still not seeing it - even with the mathematical proof.

[Jayburner]

and you would be correct.

[cubby]

Like I mentioned earlier, I think it's particularly hard for people to break away from the idea of "Previous results don't affect the next result" seeing as how that concept is beaten into your head at every turn whenever any probabilities are discussed.

In that sense, it might actually trip up "I know a little math" people more than "I don't know any math" people.

What really, really bothers me is that I'm fully aware of previous results not affecting future ones - I just had a chat with my son last night about basic coin flipping probability. And me, looking at it through the past doesn't affect the future, still gives me the 50% answer. And I'm reading the damn theorem.

I think people have a blockage when it comes to probability theory - because it's not as intuitive as regular math.

[BreathTaker]

This is an explicitly dependent experiment though, as the result of the first draw is what the second draw's probability is exclusively dependent on the first because it's based on you drawing a gold ball then drawing another from the same box.

What I'm trying to say is that the question on the image in this topic is not a original Bertrand's box paradox.
Someone messed up retelling the question so that the question implies that you don't draw the first ball at all. It's already drawn by the author by my hand, and thus it is history and had 100% chance, so there are no 3 boxes to choose from.
The question starts at the moment I only have one box to choose from and with only one ball in it

[Deleted]

People with 150+ IQ worked out that solving this problem would be a waste of time. People with 200+ IQ aren't on MMO champion.

[Kasierith]

You can blame me for part of that.

I have to admit that I'm still a bit baffled at the probability theory vs the practical percentages. I'm sure you're going to dice me up for saying that, but I'm still not seeing it - even with the mathematical proof.

If you incorporate the original choice of the box into it, there are six options for what comes out with the capitalized being the ball you chose after you pick your box.

Gg gG Gs sG Ss sS

Of these, there are three fail states, sG, Ss, sS.

Of the remainder, you have three options with equal likelihood of having occurred.

Gg gG Gs
1/3 1/3 1/3

Box A (two golds) = gG + Gg, 1/3 + 1/3 = 2/3
Box B (one silver one gold) = Gs, 1/3

So the GG box is disproportionately favored over the GS in the original selection if such a selection is made because 50% of the time that you pick GS you fail. If you open a box at random and the first one you pull out is gold you're probably going to get gold for the second pick as well.

[Darsithis]

What really, really bothers me is that I'm fully aware of previous results not affecting future ones - I just had a chat with my son last night about basic coin flipping probability. And me, looking at it through the past doesn't affect the future, still gives me the 50% answer. And I'm reading the damn theorem.

I think people have a blockage when it comes to probability theory - because it's not as intuitive as regular math.

Yes, and traditionally things that are "atomic", that is, don't affect anything around them are usually the same. You flip a coin, it's either going to be heads or tails, no matter how often you do. Each flip is completely new, separate, and unique. However, in this case your first action has an effect on the subsequent action because your choice locked you into a specific next action. Since you chose a gold ball, you are more likely to get another because you are more likely to have chosen the two-gold box the first time.

It's only because your future actions are affected that you get this probability.

[Endus]

You can blame me for part of that.

I have to admit that I'm still a bit baffled at the probability theory vs the practical percentages. I'm sure you're going to dice me up for saying that, but I'm still not seeing it - even with the mathematical proof.

You're thinking of how many boxes there are.
You should be thinking about how many gold balls you could have grabbed, and what each possibility means.

Like I said; you brute force probability by listing each possible outcome and counting. There's three gold balls you could have grabbed. Only one means the other ball in their box is silver. 2 out of 3 times, it's gold.

If the question were "how likely is it the first ball you grab from any box is gold", the answer would be 50%, but that's not the question.

[Fojos]

In my mind, this is real simple: Let's call the three gold balls A, B and C. You know you've already picked up one of them, which means there's three options:

1: You've picked A, B is in the same box. You win.
2: You've picked B, A is in the same box. You win.
3: You've picked C. The other ball is silver, you lose.

Out of the three options, two result in you winning. 2/3, 66%.

One of few answers in this thread that is correct and didn't add a lot of useless fluff that means nothing.

[Notshauna]

You can blame me for part of that.

I have to admit that I'm still a bit baffled at the probability theory vs the practical percentages. I'm sure you're going to dice me up for saying that, but I'm still not seeing it - even with the mathematical proof.

Pretty much only thing you can do with that is to either accept the 2/3rds and quiet your beliefs or test it yourself. If you still doubt it, I'd highly recommend testing it experimentally and recording the results.

[HumbleDuck]

What I'm trying to say is that the question on the image in this topic is not a original Bertrand's box paradox.
Someone messed up retelling the question so that the question implies that you don't draw the first ball at all. It's already drawn by the author by my hand, and thus it is history and had 100% chance, so there are no 3 boxes to choose from.
The question starts at the moment I only have one box to choose from and with only one ball in it

conditional probability would still be 1/(1+0+0.5) = 2/3

Unless I'm missing some Jlock trickery.

[Jayburner]

We can not go to bed this easily.

[LilSaihah]

2/3rds, because 2/3 of the gold balls have a gold ball companion.

I'm going to bed Jay, I have work tonight.

[Celista]

One of few answers in this thread that is correct and didn't add a lot of useless fluff that means nothing.

It's inaccurate, you couldn't have picked C as you chose a gold ball and C is silver/silver.

- - - Updated - - -

Nah it's just incorrectly settled question.
The way it is described it is ofc. 50%.

When you choose the box and don't know then 1st ball is golden yet - it's 2/3.
After the step of picking the ball is done the whole question is rebuilt to
"There are 2 possible variants - the chosen box is GG or GS, what is the chance it is GG"

If you use The Bayes' theorem for solving - note that by the definition of this question Probability of 'We have drawn a golden ball first" is 100%.

The same goes to flipping the coin in a row:

What is the chance a coin will show heads 2 times in a row?
It's 0.5*0.5 = 0.25

3 times in a row?
It's 0.5*0.5*0.5 = 0.125

But after you flip a coin once and it shows heads - what is the chance it will show heads AGAIN?
It's 0.5 as usual, because flipping the coin each time is an independant experiment.
If it shows heads again - the next heads is expected at 50% rate again

I'm not sure why no one addressed your post as it is the correct answer.

[Notshauna]

What I'm trying to say is that the question on the image in this topic is not a original Bertrand's box paradox.
Someone messed up retelling the question so that the question implies that you don't draw the first ball at all. It's already drawn by the author by my hand, and thus it is history and had 100% chance, so there are no 3 boxes to choose from.
The question starts at the moment I only have one box to choose from and with only one ball in it

Two things first of all the question does say you pick the first ball. Second even if someone else did draw the first ball, you still have a 2/3rds chance of drawing a gold ball from the same box as they did, because they still drew a gold ball from the same situation and impacted the probability for you to draw one.

[m4xc4v413r4]

Since the first ball is gold it has to be the first 2 boxes, so if you're on the GG box it's 1, if it's the GS box it's 0, so the answer is the probability is 0.5 (50%)