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

[interchange]

You pick a ball at random. It turns out to be gold. You either picked ball 1, 2, or 3. If it was ball 1, the other is gold. If it was ball 2, the other was gold. If it was ball 3, the other is silver. 2/3 cases the ball is gold.

Note: you could switch the order of the balls in each bucket to get the same result. The possible permutations for balls 1-4 are: [(1,2)(3,4)][(2,1)(3,4)][(1,2)(4,3)][(2,1)(4,3)]. You get the same result regardless of permutation so the fact that I've put ball #3 first means nothing.

 

[SKORPI0]

If you just picked a box at random and you ignore the box with 2 silvers, then I agree, but that's not the question that was asked.

It would be 2/3 if your 2nd pick is between 2 boxes that remain (ignoring the box with 2 silvers),
but the OP specifically asked "What is the probability that the next ball you take from the same box will also be gold?'
2nd pick will come from the same box as the 1st pick, you can't select the other box.

Coins substituted for balls.

Another thread from Jan 15-16, 2017.
https://warosu.org/sci/thread/8603718

 

[William Gaatjes]

You pick a ball at random. It turns out to be gold. You either picked ball 1, 2, or 3. If it was ball 1, the other is gold. If it was ball 2, the other was gold. If it was ball 3, the other is silver. 2/3 cases the ball is gold.

Note: you could switch the order of the balls in each bucket to get the same result. The possible permutations for balls 1-4 are: [(1,2)(3,4)][(2,1)(3,4)][(1,2)(4,3)][(2,1)(4,3)]. You get the same result regardless of permutation so the fact that I've put ball #3 first means nothing.

That i understand, but you can only choose from one box and you have already a gold ball. From then on, the odds are different.
If you pick 1, you can either pick 2 or 4.
If you pick 2, you can either pick 1 or 4.
That is how it is formulated, from the moment you picked a gold ball and you must pick form the same box.

 

[Flapdrol1337]

I guess the hard part is realizing that introducing the information that you picked a gold coin doesn't just eliminate the silver only box, but also changes the likelihood of which of the other 2 boxes you are in.

 

[interchange]

That i understand, but you can only choose from one box and you have already a gold ball. From then on, the odds are different.
If you pick 1, you can either pick 2 or 4.
If you pick 2, you can either pick 1 or 4.
That is how it is formulated, from the moment you picked a gold ball and you must pick form the same box.

I'm not following your numbers. 1&2 are in the same box. 3&4 are in the same box. If you choose ball 1, then you must pick ball 2 next (gold). If you choose ball 2, then you must pick ball 1 next (gold). If you choose ball 3, then you must pick ball 4 next (silver). So long as the question reads where you are equally likely to pick balls 1, 2, or 3, then the answer is 2/3.

 

[William Gaatjes]

I'm not following your numbers. 1&2 are in the same box. 3&4 are in the same box. If you choose ball 1, then you must pick ball 2 next (gold). If you choose ball 2, then you must pick ball 1 next (gold). If you choose ball 3, then you must pick ball 4 next (silver). So long as the question reads where you are equally likely to pick balls 1, 2, or 3, then the answer is 2/3.

You can only pick two gold balls once not two times after you have chosen a gold ball from a box. From that moment on.

I guess it is just me, I am very flexible thinker in some ways, but a very rigid thinker in other ways.

 

[richaron]

You can only pick two gold balls once not two times after you have chosen a gold ball from a box. From that moment on.

I guess it is just me, I am very flexible thinker in some ways, but a very rigid thinker in other ways.

After you've picked one gold ball, depending on which you chose, you have 3 options:

1) If you picked ball #1 from the box with 2 gold balls you will then get gold ball #2.
2) If you picked ball #2 from the box with 2 gold balls you will then get gold ball #1.
3) If you picked the gold from the box with one of each you will then get a silver.

3 options total. Each the same chance. 2 of these options are the box with 2 gold... 2/3 the box with 2 gold.

The fault with your thinking is that you are confusing the concept of combinations and permutations. There are two ways to pick the gold balls in the box with two gold balls. Picking #1 first is the same chance of picking # 2 first. Two different situations with the same possibility (and each situation has the same chance as picking the gold first from the box with one of each, so 2/3 the box is gold-gold)

 

[bigboxes]

You're confused about what the original question is. Statements were made regarding the contents of the three boxes. A box was blindly chosen. A ball was removed from that box (gold by chance). A probability question was posed regarding the one remaining ball. It necessarily came from one of the two boxes which contained at least one gold ball. The other two boxes are no longer available, and you have one ball to choose from, which is the one remaining in your chosen box. If you chose the G+G box, you'll get another gold ball. If you chose the G+S box, you'll get a silver ball. Those are the only two possible outcomes. That's equal odds.



It's typical of questions on IQ tests, but I can't venture a guess as to what IQ level would render it easily solvable.

As originally posted, it was easy. There's three boxes. Your first draw produced a gold ball. Then the probability question was asked. Since it was stated that you already picked a gold ball you can eliminate the Silver/Silver box as a possibility. Of the remaining boxes, boxes A & B, the probability of picking a second gold ball is 50%.

 

[Schmide]

I'll reduce it another way. There are 2 boxes. One with 2 gold balls and one with 1 gold 1 silver. I choose a ball regardless of which box. My chances of getting each type of ball is 1 in 4 and 3 in 4 respectively. Lets say I get a gold ball. I have 3 balls left in 2 boxes. The remaining balls can exist in any position in either box. If I choose any ball in any box there is 1 in 3 and 2 in 3 chances of getting a gold ball.

The case with 3 boxes is just a mirror image of the sliver side of the gold side.

 

[richaron]

As originally posted, it was easy. There's three boxes. Your first draw produced a gold ball. Then the probability question was asked. Since it was stated that you already picked a gold ball you can eliminate the Silver/Silver box as a possibility. Of the remaining boxes, boxes A & B, the probability of picking a second gold ball is 50%.

2 boxes, yes but that doesn't answer the question.

3 possible outcomes after you've picked that first gold. 2/3 of this options are the box with 2 gold.

I kinda feel like you 50-50 guys are just trolling at this point, since the answer has been explained so many times. But I don't want to start accusing people of that shyte. In fact I'd find it fascinating how how people can Ignore the logic

 

[William Gaatjes]

2 boxes, yes but that doesn't answer the question.

3 possible outcomes after you've picked that first gold. 2/3 of this options are the box with 2 gold.

I kinda feel like you 50-50 guys are just trolling at this point, since the answer has been explained so many times. But I don't want to start accusing people of that shyte. In fact I'd find it fascinating how how people can Ignore the logic

I am not trolling.
I am planning to write a program just like Brian did to see what happens when i change variables here and there and the rules.
Because for me, after i made a pick for a ball that part of the 3 possible outcomes is excluded for the next event that will happen. That event will either produce a gold or a silver ball.
That is what i wrote above as is formulated in the first post.

 

[Humpy]

I kinda feel like you 50-50 guys are just trolling at this point, since the answer has been explained so many times. But I don't want to start accusing people of that shyte. In fact I'd find it fascinating how how people can Ignore the logic

No one is ignoring logic or trolling. Calm down.

There is a disagreement over what the question is asking, not the answer.

 

[bigboxes]

2 boxes, yes but that doesn't answer the question.

3 possible outcomes after you've picked that first gold. 2/3 of this options are the box with 2 gold.

I kinda feel like you 50-50 guys are just trolling at this point, since the answer has been explained so many times. But I don't want to start accusing people of that shyte. In fact I'd find it fascinating how how people can Ignore the logic

I think you 2/3 guys are not able to conceptualize this. You've already pulled a gold ball out of the box you have chosen. That's a given. It does not figure into the probability equation. Since you pulled a GOLD ball out of the box then it can't be the silver/silver box. You've got one ball left to pull (since you've already pulled one GOLD one). There's only TWO boxes left that you could have pulled the GOLD ball from. You can either pull a SILVER ball or a GOLD ball. There are no other options because you have already pulled a GOLD ball on your first pull (that doesn't figure into the probability of pulling a 2nd GOLD ball). There is 1 out of 2 chances that you will pull a 2nd GOLD ball. 1/2 = 50%

 

[Schmide]

One more point to help. The buckets/balls are a distraction. No matter which box or ball I go to the chances are the same for the reduced count.

 

[bigboxes]

Let me put it this way. According to the OP, the probability question is asked AFTER pulling a GOLD ball, not before.

 

[loafbred]

Edit: Disregard the following comments, as I'm now in the "2/3" camp.

After you've picked one gold ball, depending on which you chose, you have 3 options:

1) If you picked ball #1 from the box with 2 gold balls you will then get gold ball #2.
2) If you picked ball #2 from the box with 2 gold balls you will then get gold ball #1.

The variable that you are trying to introduce by differentiating the balls in the G+G box (by numbering them) is not in the OP, and is completely irrelevant.

3) If you picked the gold from the box with one of each you will then get a silver.

3 options total. Each the same chance. 2 of these options are the box with 2 gold... 2/3 the box with 2 gold.

Nope. Two options total. You have one ball to choose from. It will be either gold or silver. It doesn't matter which one of the two gold balls might have been pulled from the G+G box before the probability question was posed. The question simply involves the probability that a gold ball will be pulled from the chosen box, which is now missing one gold ball. It doesn't introduce any variables (like your attempt to number the gold balls). It really is that simple.

 

[Hitman928]

For those still saying it's 50/50, how do you explain that if you actual try it experimentally, the results are 2/3 of the time the second ball is gold?

 

[Hitman928]

Let me put it this way. According to the OP, the probability question is asked AFTER pulling a GOLD ball, not before.

Everyone knows that. If the question was what is the probability of pulling 1 gold ball and then a second from the same box, it would be 1/3.

 

[William Gaatjes]

For those still saying it's 50/50, how do you explain that if you actual try it experimentally, the results are 2/3 of the time the second ball is gold?

Well, that is what i hope to find out.

 

[bigboxes]

Everyone knows that. If the question was what is the probability of pulling 1 gold ball and then a second from the same box, it would be 1/3.

You are majorly confused. You've already pulled one GOLD ball before the probability question is asked. You've only got TWO boxes that you could have pulled the GOLD ball from. That leaves you with EXACTLY TWO boxes that the GOLD ball could have come from. So, you will either pull the GOLD ball from BOX A or the SILVER ball from BOX B. That's 1 out of 2 or 1/2 or 50%.

 

[Hitman928]

You are majorly confused. You've already pulled one GOLD ball before the probability question is asked. You've only got TWO boxes that you could have pulled the GOLD ball from. That leaves you with EXACTLY TWO boxes that the GOLD ball could have come from. So, you will either pull the GOLD ball from BOX A or the SILVER ball from BOX B. That's 1 out of 2 or 1/2 or 50%.

What I was saying is that if the probability question was asked before pulling a gold ball (which is what you said we were confused about) then that would make the probability of pulling 2 gold balls 1/3, not 2/3. I was responding to the situation you said we were using to show that this wasn't what we were doing.

Once again, please answer this question, "For those still saying it's 50/50, how do you explain that if you actual try it experimentally, the results are 2/3 of the time the second ball is gold?"

 

[bigboxes]

What I was saying is that if the probability question was asked before pulling a gold ball (which is what you said we were confused about) then that would make the probability of pulling 2 gold balls 1/3, not 2/3. I was responding to the situation you said we were using to show that this wasn't what we were doing.

Once again, please answer this question, "For those still saying it's 50/50, how do you explain that if you actual try it experimentally, the results are 2/3 of the time the second ball is gold?"

The problem with your reasoning is that there is no GOLD ball in BOX C. So, by the time the OP poses the probability question there are only TWO boxes that the first GOLD ball could have been pulled from. So, if you pulled the GOLD ball from BOX A the second ball would be a GOLD one. If you pulled the GOLD ball from BOX B the second ball would be a SILVER one. Those are the only options you have. TWO. If you can only have two outcomes the probability is 50%. It doesn't matter which GOLD ball you pull from BOX A because the probability question was asked after that event. It's a 1 out of 2 chance at pulling a 2nd GOLD ball out of either BOX A or BOX B.

 

[Hitman928]

More reference material:

https://www.youtube.com/watch?v=CGMc8B60ZpU Has an explanation with visuals.
https://github.com/yourfriendaaron/box-paradox simulation that auto runs or that you can play to demonstrate how the probability works out. Note, must have python installed.

 

[Hitman928]

The problem with your reasoning is that there is no GOLD ball in BOX C. So, by the time the OP poses the probability question there are only TWO boxes that the first GOLD ball could have been pulled from. So, if you pulled the GOLD ball from BOX A the second ball would be a GOLD one. If you pulled the GOLD ball from BOX B the second ball would be a SILVER one. Those are the only options you have. TWO. If you can only have two outcomes the probability is 50%. It doesn't matter which GOLD ball you pull from BOX A because the probability question was asked after that event. It's a 1 out of 2 chance at pulling a 2nd GOLD ball out of either BOX A or BOX B.

Everything you just posted has already been addressed multiple times.

If you are not just trolling, then answer this simple question "For those still saying it's 50/50, how do you explain that if you actual try it experimentally, the results are 2/3 of the time the second ball is gold?"

 

[bigboxes]

If you are not just trolling, then answer this simple question "For those still saying it's 50/50, how do you explain that if you actual try it experimentally, the results are 2/3 of the time the second ball is gold?"

I get how you are thinking. I took statistics in grad school. However, based on the OP's wording of the question you are still wrong. It doesn't matter how many times you copy and paste your reply.