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

[dasherHampton]

If you ask the problem in a way that the actual question is apparent everyone would immediately get it right. The box with 2 silvers is there to distract you and make you feel stupid.

You have a set with three boxes of two balls each in it. As soon as a gold ball is pulled the set becomes one of two boxes, since the box with two silvers can't be in that particular set, per the condition of the original question.

The appropriate question then is "You have TWO boxes, one with two golds and one with a silver and a gold. You draw a gold; what are the odds your next pull from the same box will be a gold?"

That's easy enough for a 1st grader to answer correctly. Thanks for making it convoluted with a red herring, stats nerds.

 

[Schfifty Five]

I've seen this stated several times by two or three different people, and it is patently irrational. It makes no difference which ball you might pick from the G+G box, because they are identical. You can number them, name them, and imagine any number of hidden differences between them, but that cannot alter their influence on the outcome individually. Their similarity is a factor, but neither their position in the box, nor their order of being picked from the box, are relevant. It is only important that they are the same.

What about in craps though? Are you going to tell me rolling a 5 and a 2 vs a 2 and a 5 are the same outcome? Results are the same yes (7 wins or craps out), but they are individually different outcomes/possibilities.

Same with this. Grabbing gold ball #1 first in the 2 goldball box is different than grabbing gold ball #2 first in that same box.

 

[Schfifty Five]

This is like the Monty Hall problem. Should you switch? yes, b/c by switching doors, you will win 2/3 of the time.

 

[loafbred]

Grabbing gold ball #1 first in the 2 goldball box is different than grabbing gold ball #2 first in that same box.

I suspect you're trolling, but I'll listen while you explain the difference.

 

[loafbred]

Thats exactly what happened. If you state "you pick either box, and draw gold" that is the state you are describing.

Maybe it's me, but I don't understand anything that you just said.

 

[DigDog]

Nope, sorry I don't. At what stage are you talking about?

At the point of the question I see the gold box existing in a single state which is twice as likely as the mixed box. I don't imagine the gold box ever existing in multiple states, but there are two possibilities of picking a gold from the gold box and each probability is the same as any other option... .

The gold box can exist as G2G3 or as G3G2. But because the problem wants those that pick the silver box to draw the gold ball first, if you create multiple sets of boxes in each possible state, you would get these possibilities

G1S G2G3
or
G1S G3G2

If the problem simply stated IF you draw a gold ball then bertrand's answer would be correct; and also everyone would get it right.
Because there are not 3 outcomes in the full system, but rather 12, where 6 are double gold, 3 are gold silver, and 3 are silver gold. That 2/3 that bertrand claims is actually 2/4 yes, 1/4 no and 1/4 lol-not-included-here.

 

[Skyclad1uhm1]

I suspect you're trolling, but I'll listen while you explain the difference.

Maybe it is easier to understand if you change the two gold balls to one gold and one bronze, and change the question to the chance of the second draw being gold or bronze after the first draw is either gold or bronze.

Then you have B+G and S+G, the chance of drawing S first from the second box is removed, so you are left with B first + G second as option, with G first + B second, and in the second box G first and S second.

So after you draw either a bronze or a gold ball from a chest, how high is the chance of drawing a gold or bronze one again?
This might make it easier to differentiate between the 2 balls in the 'gold box'.

 

[Tweak155]

I suspect you're trolling, but I'll listen while you explain the difference.

It's different in the fact that it is 2 different events. Notice the problem states that you draw a ball from the box at random... since there are 2 balls to draw from every box, there is a 50% chance you draw gold ball 1 and a 50% chance you draw gold ball 2 (if you randomly picked the box with 2 gold balls). Just as if you picked the box with 1 silver ball and 1 gold ball, it was a 50% chance you drew the gold ball from that box as well, and did not randomly draw the silver ball.

If you picked the box with 2 gold balls logic would tell you there is 100% chance to draw a gold ball the 2nd time. But you can only know that if you know which box you picked, which you don't.

It's mainly important because in order to correctly analyze the probability, you need to consider all the possible outcomes.

 

[Hitman928]

If you ask the problem in a way that the actual question is apparent everyone would immediately get it right. The box with 2 silvers is there to distract you and make you feel stupid.

You have a set with three boxes of two balls each in it. As soon as a gold ball is pulled the set becomes one of two boxes, since the box with two silvers can't be in that particular set, per the condition of the original question.

The appropriate question then is "You have TWO boxes, one with two golds and one with a silver and a gold. You draw a gold; what are the odds your next pull from the same box will be a gold?"

That's easy enough for a 1st grader to answer correctly. Thanks for making it convoluted with a red herring, stats nerds.

Eliminating the box with the 2 silvers is the easy part, pretty much everyone gets that part right

 

[Hitman928]

The gold box can exist as G2G3 or as G3G2. But because the problem wants those that pick the silver box to draw the gold ball first, if you create multiple sets of boxes in each possible state, you would get these possibilities

G1S G2G3
or
G1S G3G2

If the problem simply stated IF you draw a gold ball then bertrand's answer would be correct; and also everyone would get it right.
Because there are not 3 outcomes in the full system, but rather 12, where 6 are double gold, 3 are gold silver, and 3 are silver gold. That 2/3 that bertrand claims is actually 2/4 yes, 1/4 no and 1/4 lol-not-included-here.

You know, in the original formation of the question, "if you draw a gold ball" is exactly how it is stated, so I'm glad you agree with Bertrand that the answer is 2/3.

The only difference in the OP is that the word "if" is excluded, but as it was already explained in this thread, because both picks were explicitly stated as random events, including the word "if" is unimportant, it is inherit in the fact that they were random events. If picking the first gold ball was a forced choice, then it wasn't actually random as the problem stated. If picking the first gold isn't random, but is forced, then picking the first box isn't random either and therefore the whole question is invalid without the possibility of solution as we have no indication how the first choice is be coerced.

Additionally, if you write up the full probability equation, the possibility of picking silver first (from either the SS box or from the SG box) is included. However, picking the SS box becomes essentially an addition of an empty set once you want to calculate the probability of picking gold a second time when the first pick was gold.

 

[Humpy]

including the word "if" is unimportant, it is inherit in the fact that they were random events.

Link?

Because that sounds completely made up.

 

[Hitman928]

Link?

Because that sounds completely made up.

It's easy, explain to me how something can be a random choice of more than one possibility and a forced desired outcome at the same time.

 

[dyna]

*bold added.

I've tried to explain it, but maybe it will work if I just say "nope". You are wrong. There is no ambiguity. You have gotten confused because you don't understand the question or you have gotten confused because you have listened to people who try to people who are trying to create ambiguities.

The question in the OP is "...What is the probability that the next ball you take from the same box will also be gold?"

The question is about "the next ball". As has been stated ad nauseam there are 3 possible "next ball"s out of the 4 total balls in the two boxes which contain at least one gold ball. If, as you state, "the 50% crowd sees the probability start when you have a gold ball" then they only need to read my previous sentence an learn where they went wrong. Two boxes does not equal 50% when there are 3 possibilities. The possibilities are only narrowed down to 3 "when you have a gold ball".

In no way, shape, or form is the question ambiguous. The question however can be interpreted the wrong way...

I'm not confused because I know the answer is 2/3. However, the problem is introduced and projected as a trick question. I'm trying to explain how somebody could see it as 1/2. Trick questions often throw you in different directions with mis-information intended to fool you.

If I told you...You are holding a gold ball.

There are two boxes one with a gold ball and one with a silver. What would you say is the probability you get a gold ball?

You could read the OP question and start the probability at that spot and the answer would be 1/2. It depends on where you start the probability to whether it is 2/3 or 1/2.

 

[Schfifty Five]

I suspect you're trolling, but I'll listen while you explain the difference.

I need to know first if you're on the 1/2 side or the 2/3 side.

 

[Humpy]

It's easy, explain to me how something can be a random choice of more than one possibility and a forced desired outcome at the same time.

I didn't make the claim.

Nearly every "proof" of the answer to the OP has included some sort of unsubstantiated statement in order to arrive at the result.

 

[Humpy]

I'm not confused because I know the answer is 2/3. However, the problem is introduced and projected as a trick question. I'm trying to explain how somebody could see it as 1/2. Trick questions often throw you in different directions with mis-information intended to fool you.

If I told you...You are holding a gold ball.

There are two boxes one with a gold ball and one with a silver. What would you say is the probability you get a gold ball?

You could read the OP question and start the probability at that spot and the answer would be 1/2. It depends on where you start the probability to whether it is 2/3 or 1/2.

Correct. Recognizing that it's a "trick" question leads to more than one valid answer.

Everyone that argues for 2/3 can simply decide to argue for 1/2. Really, I think they should have to make both arguments if they are being honest.

 

[Hitman928]

I'm not confused because I know the answer is 2/3. However, the problem is introduced and projected as a trick question. I'm trying to explain how somebody could see it as 1/2. Trick questions often throw you in different directions with mis-information intended to fool you.

If I told you...You are holding a gold ball.

There are two boxes one with a gold ball and one with a silver. What would you say is the probability you get a gold ball?

You could read the OP question and start the probability at that spot and the answer would be 1/2. It depends on where you start the probability to whether it is 2/3 or 1/2.

This isn't the question that is asked though. It doesn't matter where you start the probability as long as you get the actual setup of the question correct.

 

[Hitman928]

Correct. Recognizing that it's a "trick" question leads to more than one valid answer.

Everyone that argues for 2/3 can simply decide to argue for 1/2. Really, I think they should have to make both arguments if they are being honest.

This is not a trick question, the answer is just counter-intuitive, there's a difference. There is no way to follow the setup of the problem and get to 1/2 and be correct. This is such an easy thing to prove with real life experiments but people just don't want to be wrong. I think the title of the thread makes it even worse because no one wants to think that they have a "low" IQ so they fight tooth and nail that they are right. The requirement of having a 115 IQ to get this answer is totally bogus and was just put in as click bait.

I didn't make the claim.

Nearly every "proof" of the answer to the OP has included some sort of unsubstantiated statement in order to arrive at the result.

The only thing happening at this point in the thread is people who think it is 1/2 create straw men arguments as to why the 2/3 proof isn't right (which requires them to ignore real world experimentational evidence).

As far as "evidence" for my if versus random statements, there's nothing to prove. Either you understand what the words random and probability mean or you don't. As you haven't actually put forth any argument as to why what I said is wrong or how to properly interpret the problem, your comments are vapid and of no worth towards any kind of solution.

 

[Humpy]

There is no way to follow the setup of the problem and get to 1/2 and be correct.

Sure there is.

At the point the question of probability is posed the circumstances have been reduced to one box and two balls.

The reader has to decide whether to include the earlier information or not. Unlike a previous claim, the definition of probability does not require all information about possible outcomes be used. Probability requires a defined sample space, meaning some possibilities can be excluded.

 

[Hitman928]

Sure there is.

At the point the question of probability is posed the circumstances have been reduced to one box and two balls.

The reader has to decide whether to include the earlier information or not. Unlike a previous claim, the definition of probability does not require all information about possible outcomes be used. Probability requires a defined sample space, meaning some possibilities can be excluded.

So you're saying, if you ignore the setup of the problem and parameters outlined therein, you can solve the problem any way you like? OK, I guess that's correct. If that's the case, then my answer is 100% because I pick the GG box every time, I can just ignore all the other details. I win every time in this game and I like that better.

 

[Humpy]

So you're saying, if you ignore the setup of the problem and parameters outlined therein, you can solve the problem any way you like? OK, I guess that's correct. If that's the case, then my answer is 100% because I pick the GG box every time, I can just ignore all the other details. I win every time in this game and I like that better.

That's crazy talk.

I think you would still want to give the odds of the second ball pulled from the box matching the first. After all, that's what the question asks.

 

[Cozarkian]

If I told you...You are holding a gold ball.

There are two boxes one with a gold ball and one with a silver. What would you say is the probability you get a gold ball?

That isn't factually accurate based on the posed question. Yes, there are two boxes, but one of them has two balls and one of them has only one ball. The question is to determine the probability that the ball in the single-ball-remaining box is gold. You've reduced the total number of relevant balls remaining from 4 to 3.

At the point the question of probability is posed the circumstances have been reduced to one box and two balls.

The reader has to decide whether to include the earlier information or not.

Okay, given you have one gold ball in your hand and one box with one ball of unknown color/material. Without further information the question is impossible to answer. The only way to answer the question is to include the earlier information.

 

[Hitman928]

That's crazy talk.

I think you would still want to give the odds of the second ball pulled from the box matching the first. After all, that's what the question asks.

Which is what I'm doing, I just pick the GG box every time so my second pick is always gold. If you can choose to ignore any information from before the first pick, like you suggest, to get to 50%, why can't I choose to ignore that same information to get to 100%?

This is why I said the specific inclusion in the setup that the choices are random have meaning. If they don't and can be ignored, the question is unsolvable. The answer is either 2/3 or is unsolvable, either way you don't get to 50%.

 

[Humpy]

This is why I said the specific inclusion in the setup that the choices are random have meaning. If they don't and can be ignored, the question is unsolvable. The answer is either 2/3 or is unsolvable, either way you don't get to 50%.

"If it's a gold ball" implies randomness still has meaning. "It's a gold ball" implies randomness no longer has meaning to the future question of probability. Ignoring randomness in one instance doesn't require we ignore the result, or the existence of the other balls and boxes, etc. It simply limits what we consider when calculating the probability of the next result.

I don't think I've seen a version of the paradox that left out the word "if" here.

 

[Hitman928]

"If it's a gold ball" implies randomness still has meaning. "It's a gold ball" implies randomness no longer has meaning to the future question of probability. Ignoring randomness in one instance doesn't require we ignore the result, or the existence of the other balls and boxes, etc. It simply limits what we consider when calculating the probability of the next result.

I don't think I've seen a version of the paradox that left out the word "if" here.

You're missing the words before it.

"You pick box at random. You put your hand in and take a ball from that box at random." Including the word "if" certainly helps for clarification, but it is not required. The randomness is explicitly stated in the question, it doesn't need to be implied.

Once again, if the first ball chosen from the box is not random, there is no way to solve this problem without further information. It is either 2/3 or it is impossible to solve, you can't get 50%.