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

[mindless1]

Or, your reasoning is wrong so your disproof is invalid.

If you actually try the situation yourself, just as described in the problem, 2/3 of the time the second ball will be gold and 1/3 of the time the second ball will be silver.

lol, no it won't, because you've already chosen one gold ball.

If it were a 50% chance, the second ball you pull should be silver half the time and gold half the time.

It is. Once you have already pulled out one gold ball. You might fail at understanding this for a 2nd reason I didn't realize at first, that you are still considering G3? If you have already chosen a gold ball, there is no G3 remaining.

If the real life (perfectly repeatable) result of experimentation doesn't match your logical conclusion but matches someone else's, yours is the one that is incorrect.

Show this. I'll look at it for maybe 30 seconds even though however that conclusion was reached, is impossible, just to see where their test went wrong. It's probably what I already stated, that they're treating the same choice as two different choices. ie there is no G1 and G2, only G.

Hint: That is also what is shown in the video, only "G". If G1 and G2 are to be differentiated then they have to be applied to specific coins, and then you didn't pull out a "G" coin, rather you pulled out a "G1" or "G2" coin.

 

[Hitman928]

lol, no it won't, because you've already chosen one gold ball.



It is. Once you have already pulled out one gold ball. You might fail at understanding this for a 2nd reason I didn't realize at first, that you are still considering G3? If you have already chosen a gold ball, there is no G3 remaining.



Show this. I'll look at it for maybe 30 seconds even though however that conclusion was reached, is impossible, just to see where their test went wrong. It's probably what I already stated, that they're treating the same choice as two different choices. ie there is no G1 and G2, only G.

There are already 2 simulations provided in this thread you can look for and try yourself, both of them come out to be 2/3. However, you don't need anyone else, you can do it yourself, it's not that hard to recreate. By saying I'll only look for 30 seconds suggests you don't want to find the answer, you're convinced you're right and that's that, any other answer is impossible and therefore shouldn't even be considered, even if reality says different, you're right and reality is wrong.

 

[Hitman928]

Hint: That is also what is shown in the video, only "G". If G1 and G2 are to be differentiated then they have to be applied to specific coins, and then you didn't pull out a "G" coin, rather you pulled out a "G1" or "G2" coin.

You are being extremely pedantic with this reasoning. There are two boxes. Within each of those two boxes exists two distinct objects. Those objects are of either classification "A" or classification "B". Within box one, there are two objects of classification "A". Within box 2, there are two objects, one of classification "A" and one of classification "B". The use of gold coins (or balls) is just to have something relatable for the reader to visualize. Additionally, G1 and G2 are just labels to help visualize the scenario, it doesn't matter what you label them or if you label them, they are two distinct objects, both of the same classification. When the object is pulled (at random), it is of classification "A". That object could have been from box 2 or it could have been either one of the 2 objects in box 1. Since only 1 object was pulled, that means there is left in the boxes, 1 object of classification "B" and 2 objects of classification "A".

When you describe that G1=G2. . . therefore 50%, this is exactly the flawed reasoning why this was labeled as a paradox, just that most people do it intuitively without realizing it whereas you explicitly state the reasoning but still fail to realize you fell into the trap of the paradox.

 

[SKORPI0]

you have never dealt with a 52 year old woman that is scottish and has giant fake books. ..and she is 6'2"

I think you mean boobs.

 

[Hitman928]

I think you mean boobs.

I'm pretty sure he means books. Have you ever had to deal with giant fake books before? Where do you put them? They don't fit on a book shelf. They take up tons of space in a closet, they're a tripping hazard if left on the floor and you can't even read them! I'd hate to have a 6'2" sister with giant fake books to deal with when she wants to move in with me .

 

[Darwin333]

the best is her son will be sleeping in my bedroom. it never ends at my place.

Dang, if I was him I'd burn the mattress and if I was you I'd burn the mattress when I got it back. Then again, considering your past living arrangements no wonder how much family spunk you have been sleeping in

 

[destrekor]

I think you mean boobs.

I'm pretty sure he means books. Have you ever had to deal with giant fake books before? Where do you put them? They don't fit on a book shelf. They take up tons of space in a closet, they're a tripping hazard if left on the floor and you can't even read them! I'd hate to have a 6'2" sister with giant fake books to deal with when she wants to move in with me .

I mean, where else do you think Fake News comes from? Fake Books, of course!

 

[paperfist]

You are being extremely pedantic with this reasoning. There are two boxes. Within each of those two boxes exists two distinct objects. Those objects are of either classification "A" or classification "B". Within box one, there are two objects of classification "A". Within box 2, there are two objects, one of classification "A" and one of classification "B". The use of gold coins (or balls) is just to have something relatable for the reader to visualize. Additionally, G1 and G2 are just labels to help visualize the scenario, it doesn't matter what you label them or if you label them, they are two distinct objects, both of the same classification. When the object is pulled (at random), it is of classification "A". That object could have been from box 2 or it could have been either one of the 2 objects in box 1. Since only 1 object was pulled, that means there is left in the boxes, 1 object of classification "B" and 2 objects of classification "A".

When you describe that G1=G2. . . therefore 50%, this is exactly the flawed reasoning why this was labeled as a paradox, just that most people do it intuitively without realizing it whereas you explicitly state the reasoning but still fail to realize you fell into the trap of the paradox.

You're only picking from one box though...........

Just so everyone understands the Mayneism, being terrible at reading comprehension and statistics is a good thing.

Oh I thought it was code for he has no bloody idea what the answer is.

 

[Hitman928]

You're only picking from one box though...........

This is true, however, how it works with the boxes has already been explained multiple times over. My post you quoted was a direct response to mindless1's reasoning of how many balls are in play as part of the system. it wasn't meant to explain the entire problem, only to address one issue of it.

 

[local]

You're only picking from one box though...........

That is correct. As a convert to the 2/3 religion I will tell you what changed my mind.

Since it is already determined that a gold ball was pulled it is twice as likely that you pulled one from box 1 than box 2 and impossible to pull one from box 3. The part I missed was that half the time you pulled a random ball from box 2 it would be silver and thus discarded from the equation but every time one was pulled from box 1 it would stay in play. Meaning that over a significant number of random pulls you would have box 1 in front of you twice as often as box 2 leaving you with a 2/3 chance that box 1 is the one that you are picking from on any given chance.

 

[mindless1]

You are being extremely pedantic with this reasoning.

Nope, just doing what the task was.

There are two boxes. Within each of those two boxes exists two distinct objects. Those objects are of either classification "A" or classification "B". Within box one, there are two objects of classification "A". Within box 2, there are two objects, one of classification "A" and one of classification "B". The use of gold coins (or balls) is just to have something relatable for the reader to visualize. Additionally, G1 and G2 are just labels to help visualize the scenario, it doesn't matter what you label them or if you label them, they are two distinct objects, both of the same classification.

You were fine up to this point, except that we have to label them something. Since there is no stated difference between G1 and G2, it is invalid to treat them as separate choices when either choice has the same outcome.

Look at it this way: You have two chicken legs and a slice of pizza in the fridge. You're either going to eat a chicken leg or pizza. Do you have three choices of what to eat or only two? What you do not understand is that the problem with this *puzzle* was the misapplication of probability at a low level, invalid within the context of the question, or rather, correctly applied would result in 50%.

When the object is pulled (at random), it is of classification "A". That object could have been from box 2 or it could have been either one of the 2 objects in box 1. Since only 1 object was pulled, that means there is left in the boxes, 1 object of classification "B" and 2 objects of classification "A".

Yes, this aligns with your claim and mine.

When you describe that G1=G2. . . therefore 50%, this is exactly the flawed reasoning why this was labeled as a paradox,

Wrong. The only description of the balls is gold vs silver, The only necessary outcome is a gold ball, not WHICH gold ball.

just that most people do it intuitively without realizing it whereas you explicitly state the reasoning but still fail to realize you fell into the trap of the paradox.

This is a case where people who have an IQ of 115, shouldn't argue with those who have one much higher, or to put it another way, can't see the forest for the trees.

It is very sad that you can't accept when you're wrong. Perhaps you should start over and read the paradox more carefully so that you eventually realize why. "What is the probability that the next ball you take from the same box will be gold?" Whichever ball you might have chosen from the box that has two gold balls, there is only one ball remaining, you can't go back and count the other ball again and say maybe it was G1 or maybe it was G2 to calculate, as if that doubles the gold ball probability in that box. If you use both G1 and G2, to label two gold balls then the ball already picked is one of those two designations, and only the other remains.

It had to be one or the other if you label them differently, even if you take this multiple G ball labeling stance which I already indicated was invalid within the puzzle context when there is no way to differentiate between the two gold balls, when even the puzzle itself, labels them only as gold or silver.

It's a little bit crazy that I have to explain this.

 

[brianmanahan]

It is very sad that you can't accept when you're wrong.

you are "simplifying" that which cannot be simplified. the actual performance of this scenario and measurement of the outcomes disagrees with your answer. when you pick a gold ball, real life odds that you pick another gold ball approaches 2/3 the more times you do it.

 

[mindless1]

^ It's called algebra and it's slightly more difficult than the invalid probability equation, but still pretty simple.

There is no disagreement with 50% except by those who can't think logically or follow directions, who were duped by the implied validity to count the same ball you already picked, again after it is out of the equation.

Measurements of outcome will always be 50% if you follow the stated mission instead of the misapplied logic and misapplied probability calculation. There are only two states remaining possible. gold or silver, not which gold.

The real life odds do not change the slightest bit the longer you do it, unless there is some other variable involved, something crazy added like only men are tested and one box was pink, or the boxes are identical but one is closer to the participants' left hand after they were handed a beverage into their right hand.

 

[Hitman928]

^ It's called algebra and it's slightly more difficult than the invalid probability equation, but still pretty simple.

There is no disagreement with 50% except by those who can't think logically or follow directions, who were duped by the implied validity to count the same ball you already picked, again after it is out of the equation.

Measurements of outcome will always be 50% if you follow the stated mission instead of the misapplied logic and misapplied probability calculation. There are only two states remaining possible. gold or silver, not which gold.

If you have such a superior intellect than anyone in this thread as well as the entire mathematical world, then this should be very easy to prove.

I'm waiting. . .

 

[paperfist]

That is correct. As a convert to the 2/3 religion I will tell you what changed my mind.

Since it is already determined that a gold ball was pulled it is twice as likely that you pulled one from box 1 than box 2 and impossible to pull one from box 3. The part I missed was that half the time you pulled a random ball from box 2 it would be silver and thus discarded from the equation but every time one was pulled from box 1 it would stay in play. Meaning that over a significant number of random pulls you would have box 1 in front of you twice as often as box 2 leaving you with a 2/3 chance that box 1 is the one that you are picking from on any given chance.

I'm gonna take my 113 IQ and use it to hire a mathematician from India off of upwork.com to come in here and solve this problem before Infinity Wars II: The Wrath of Thanos comes out.

 

[mindless1]

^ I already proved it. You just didn't have an open enough mind to learn anything.

Last time: Even if you designate each ball to have a different label such as G1 or G2, whichever ball you picked, is one of those two, and only the other of those two remain.

You can't then claim that both picking G1 and then G2, -AND- picking G2 and then G1, are possible at that point. It has to be one or the other depending on which ball you already picked.

Superior intellect, meh it's just one puzzle and an extreme easy one. Am I superior to some at this type of puzzle? Apparently.

 

[brianmanahan]

Measurements of outcome will always be 50% if you follow the stated mission instead of the misapplied logic and misapplied probability calculation.

lol, just give it a try 100 or 200 times in real time. you'll see it's roughly 2/3.

performed ~50k times with a JS script doing randomization (http://js.do/code/225139):

games with first ball picked = gold, second ball picked = silver: 16882
games with first ball picked = gold, second ball picked = gold: 33485
% times second ball was gold when first ball was gold: 66.48202195882224

 

[brianmanahan]

If you have such a superior intellect than anyone in this thread as well as the entire mathematical world, then this should be very easy to prove.

superior intellect and yet a username of mindless1

 

[mindless1]

lol, just give it a try 100 or 200 times in real time. you'll see it's roughly 2/3.

performed ~50k times with a JS script doing randomization (http://js.do/code/225139):

games with first ball picked = gold, second ball picked = silver: 16882
games with first ball picked = gold, second ball picked = gold: 33485
% times second ball was gold when first ball was gold: 66.48202195882224

If you write an equation to do it with javascript using the same faulty logic that counts G1 and G2 both separately and still possible after one of the two is removed, sure you wrote an equation invalid to this puzzle.

 

[Hitman928]

^ I already proved it. You just didn't have an open enough mind to learn anything.

Last time: Even if you designate each ball to have a different label such as G1 or G2, whichever ball you picked, is one of those two, and only the other of those two remain.

You can't then claim that both picking G1 and then G2, -AND- picking G2 and then G1, are possible at that point. It has to be one or the other depending on which ball you already picked.

Superior intellect, meh it's just one puzzle and an extreme easy one. Am I superior to some at this type of puzzle? Apparently.

You proved nothing, you just said what you thought was the reason and injected, "trust me, i'm really smart". Prove it, show it to be true through actual experimentation, it's really easy to construct the experiment either through simulation or the real world.

If you write an equation to do it with javascript using the same faulty logic that counts G1 and G2 both separately and still possible after one of the two is removed, sure you wrote an equation invalid to this puzzle.

You can't just say, the answer is not what I like so the program must be wrong. Say why it's wrong and propose the proper solution. . . again, proof is consistently lacking in your posts.

 

[mindless1]

superior intellect and yet a username of mindless1

mindless1 can mean different things to different people. Some may see it as an insult, and I am fine (oblivious) to that. Others may see mushin, discarding the clutter in your mind to reach a state where you see more clearly.

 

[Hitman928]

^ I already proved it. You just didn't have an open enough mind to learn anything.

Last time: Even if you designate each ball to have a different label such as G1 or G2, whichever ball you picked, is one of those two, and only the other of those two remain.

You can't then claim that both picking G1 and then G2, -AND- picking G2 and then G1, are possible at that point. It has to be one or the other depending on which ball you already picked.

Superior intellect, meh it's just one puzzle and an extreme easy one. Am I superior to some at this type of puzzle? Apparently.

No one is claiming what you say they are claiming.

What's really funny is that you think your logic is in any way a novel solution to the problem, it's not. Your same logic is EXACTLY the reason people get this wrong all the time and have been proven wrong over and over again.

 

[Hitman928]

I'm gonna take my 113 IQ and use it to hire a mathematician from India off of upwork.com to come in here and solve this problem before Infinity Wars II: The Wrath of Thanos comes out.

There is zero controversy in the mathematical world over the solution to this problem, none. The solution to this problem has been long understood and accepted by mathematicians the world over. The only controversy comes from arm chair mathematicians who think they've outsmarted the world and see an answer where no one else could. Unsurprisingly these people never actually try the experiment, otherwise they would quickly realize they were wrong.

 

[mindless1]

You proved nothing, you just said what you thought was the reason and injected, "trust me, i'm really smart". Prove it, show it to be true through actual experimentation, it's really easy to construct the experiment either through simulation or the real world.

Trust, me, if I meant exactly "trust me, I'm really smart", I would have written that. You seem to be good at making errors, slight deviations then going in one direction with them, which may account for why when you were duped by the invalid probability equation, your mind closed to the issue.

You want what exactly? I should set up a video camera and make a youtube video, perhaps with a notary or a judge present to validate the authenticity, then post the video, then have you just make up nonsense to refute it?

Thanks for that laugh.

You can't just say, the answer is not what I like so the program must be wrong. Say why it's wrong and propose the proper solution. . . again, proof is consistently lacking in your posts.

It is sad that your mind is this far disconnected from reality. I have told you multiple times why the result was invalid. You can't pick ONE of something unique (uniquely named gold coin) and then claim it is still there to pick again for the purposes of the calculation. This is such an incredibly basic simple thing, yet you still argue against it because your mind is closed to reexamining where you went wrong.

Further, what do I care if you disagree? I don't really, because you don't want to learn anything, so it was a bit wasteful to reply, but I did because many people do still have an open enough mind to learn things, especially why an error was made, can be more important to understand than the error itself.

 

[Hitman928]

Trust, me, if I meant exactly "trust me, I'm really smart", I would have written that. You seem to be good at making errors, slight deviations then going in one direction with them, which may account for why when you were duped by the invalid probability equation, your mind closed to the issue.

You want what exactly? I should set up a video camera and make a youtube video, perhaps with a notary or a judge present to validate the authenticity, then post the video, then have you just make up nonsense to refute it?

Thanks for that laugh.

I'm not saying you have to prove it to me, I've already done the experiment and proved it out, so has the rest of the math world. I'm saying do it for yourself, it's a very simple experiment to do, it can be setup in under 5 minutes.

It is sad that your mind is this far disconnected from reality. I have told you multiple times why the result was invalid. You can't pick ONE of something and then claim it is still there to pick again for the purposes of the calculation. This is such an incredibly basic simple think that it is ludicrous that you still argue against it.

The code doesn't pick one of something and then claim it is still there. That's the point. If you don't believe it is proper code, show us where the code is flawed because this purported flaw isn't in the code.

Further, what do I care if you disagree? I don't really, because you don't want to learn anything, so it was a bit wasteful to reply, but I did because many people do still have an open enough mind to learn things, especially why an error was made, can be more important to understand than the error itself.

You are literally the only one arguing against it being 2/3 in this thread anymore, you're the only one disagreeing. I don't care if you disagree or not, I just thought initially you might want to know the actual solution to the problem but it's clear that isn't the case and now it's just fun seeing you build up straw men as to why our solution is flawed.