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

[BurnItDwn]

I told my wife I have golden balls, she said "no you don't."
Life is no fun sometimes.

 

[Braznor]

The crux of the problem is that you already have crossed one probability event, you have selected a box containing atleast one gold ball. The possibility of the second ball being gold just comes out to the possibility of you having picked the box with one gold ball or the one with two gold balls, so basically 50% since there are only two boxes to have atleast one gold ball inside them. So 50%.

 

[Braznor]

Also balls......

 

[brianmanahan]

i just wrote a program to simulate this problem 100000 times. here is the output:

games with first ball picked = silver: 50023
games with first ball picked = gold, second ball picked = silver: 16757
games with first ball picked = gold, second ball picked = gold: 33220
% times second ball picked was gold: 66.47057646517398

 

[brianmanahan]

here is the simulator where you can see the code and change the number of simulations:

http://js.do/code/225139

click the "Run Simulations" button over on the right side of the screen

 

[Hitman928]

The crux of the problem is that you already have crossed one probability event, you have selected a box containing atleast one gold ball. The possibility of the second ball being gold just comes out to the possibility of you having picked the box with one gold ball or the one with two gold balls, so basically 50% since there are only two boxes to have atleast one gold ball inside them. So 50%.

Again, this is a known statistical problem with a proven solution, which is not 50%. An answer to 50% is the intuitive answer but it is the wrong answer.

Let me ask you this, if one box had 1 million gold coins and the other had 999,999 silver coins and one gold coin. You then reach into a random box and pull out a gold coin, which box would you say is the most probable box you reached into? This isn't the exact same situation, obviously, but the principal of it is the same as the 2 coins per box problem and is what can lead to understanding as to why it is 2/3 and not 50%.

 

[DigDog]

The well known maths are wrong.

The observer - the person who acts on a system - is not deterministic; if it is, you misrepresented the system.

In other words, what YOU did (draw a ball) does not affect the system.

The above example removes the extra ball because YOU MUST DRAW FROM THE SAME BOX.

Lets say that you have the two boxes. Ine is SG and the other is GG.

You MUST decide NOW on the system - you must pick a box and draw one ball from it AND you will be FORCED to also draw the second ball. The other box has been rendered irrelevant.

You draw one ball and it is gold. The other box outside the system is taken away. You are never given the chance to interact with it ergo the system result is 50%.

Lets say you have

Sx and GG
Gx and GG

but the second, untouched box is opened and more balls added.

Sx and GGGGGGGGGBBBBB
Gx and GGGGGPPPPPPPPPPP


Blue and Pink balls, why not.
you are allowed to draw from ONLY the system of Sx or Gx.

Sx and Gx are your system. Other systems in the vicinity do not influence it.

This can be represented, say:

Sx GG is in house 1
Gx GG is in house 2
What percentage of houses have a silver ball.

 

[DigDog]

games with first ball picked = silver

You cant have this.

 

[brianmanahan]

You cant have this.

in reality you can, but just need to disregard those cases when calculating the GS and GG percentages that will occur in the cases where someone DOES happen to pick a gold ball first

 

[paperfist]

Moot. You don't know which box you picked. So only the amount of balls left to choose from matters.

But it says you already picked a gold ball, so the last box isn't even in play since it has silver balls and since you can't switch boxes you're either going to get a silver ball or gold ball.

 

[brianmanahan]

maybe thinking about it this way will make people realize it is %66.6666, not %50

since you got a gold ball first, you are TWICE AS LIKELY to have picked box 1 as you are to have picked box 2

these are the 3 scenarios that result when you picked a gold ball first

 

[highland145]

I told my wife I have golden balls, she said "no you don't."
Life is no fun sometimes.

Be thankful. She might have wanted to sell them.

 

[bigboxes]

50%

 

[bigboxes]

It's good to know who has a low IQ on this board

 

[DigDog]

Again, this is a known statistical problem with a proven solution, which is not 50%. An answer to 50% is the intuitive answer but it is the wrong answer.

Let me ask you this, if one box had 1 million gold coins and the other had 999,999 silver coins and one gold coin. You then reach into a random box and pull out a gold coin, which box would you say is the most probable box you reached into? This isn't the exact same situation, obviously, but the principal of it is the same as the 2 coins per box problem and is what can lead to understanding as to why it is 2/3 and not 50%.

But this is not the system postulated in the original question. You do not have 2 boxes,you have one box. The other box cannot be actioned as it is not contained in the system.
Again, im happy that this is a known problem because my answer is correct and theone in the statistics book is wrong.

There is *no* GG box. There is no SG box. There is only one box which is either Sx or Gx.

The answer is 50%.

 

[brianmanahan]

The answer is 50%.

nope, because you're twice as likely to be in GG than in GS. this "closed system" originated from another system, and you can't ignore the original system.

 

[DigDog]

Look guy im gona rewrite the postulate; it needs to be logically identical,ok?

Two boxes,SG and GG exist. You must chose one from which a Gold ball is extracted, the other ball is removed.

THIS is the postulate. The likelyhood of extracting a gold ball from SGGG is not factored in the system.

So in truth,while the description mentios two boxes, you have one box. You would have two if you could draw from the box with two balls, which you cannot do.

I guess you are getting confused when you draw from GG as you think this leaves you with Gx and SG. THIS IS INCORRECT.

Once you draw from GG you are left with Gx as THE ONLY box you can draw from. Do you understand this?

You either pick to draw from SG, which leads only to Sx,which leads 100% of the time to drawing silver, or you pick to draw from GG, leading to Gx, drawing 100% gold.

50%



Im going to bed. I will only consider worthwhile posts that answer this question:
If you begin with box SG, according to the original postulate, what is the likelyhood that you draw a gold ball?

Hint: the answer is 100%

 

[Mayne]

I posted this question over at the off topic forums at mmo-champion.com. It's now at 44 pages lol. I guess it angered some mods there and I'm banned.

 

[brianmanahan]

I posted this question over at the off topic forums at mmo-champion.com. It's now at 44 pages lol. I guess it angered some mods there and I'm banned.

lol they banned you over that?

 

[Mayne]

lol they banned you over that?

you'd be surprised what they ban you over there. Just another 2 week vacation for me.

 

[Hitman928]

But this is not the system postulated in the original question. You do not have 2 boxes,you have one box. The other box cannot be actioned as it is not contained in the system.

True.

Again, im happy that this is a known problem because my answer is correct and theone in the statistics book is wrong.

False.

There is *no* GG box. There is no SG box. There is only one box which is either Sx or Gx.

After the initial pick of gold, this is true.

The answer is 50%.

False. You're still ignoring that at the beginning of the situation, all options are possible, but we are only considering the occurrences when a gold ball is chosen first.

 

[Thebobo]

Boxers are irrelevant

 

[Hitman928]

Look guy im gona rewrite the postulate; it needs to be logically identical,ok?

Two boxes,SG and GG exist. You must chose one from which a Gold ball is extracted, the other ball is removed.

THIS is the postulate. The likelyhood of extracting a gold ball from SGGG is not factored in the system.

So in truth,while the description mentios two boxes, you have one box. You would have two if you could draw from the box with two balls, which you cannot do.

I guess you are getting confused when you draw from GG as you think this leaves you with Gx and SG. THIS IS INCORRECT.

Once you draw from GG you are left with Gx as THE ONLY box you can draw from. Do you understand this?

You either pick to draw from SG, which leads only to Sx,which leads 100% of the time to drawing silver, or you pick to draw from GG, leading to Gx, drawing 100% gold.

50%



Im going to bed. I will only consider worthwhile posts that answer this question:
If you begin with box SG, according to the original postulate, what is the likelyhood that you draw a gold ball?

Hint: the answer is 100%

I think this post shows where the misunderstanding is. To first answer the bolded, if you start with box SG, you have a 50% of picking either S or G. However, we are ignoring the occurrences of when a S is picked first but it is still a part of the probability system.

You even have experimental evidence that is statistically significant showing it is 2/3. If you don't believe it, try it yourself. Put 2 of one object in one box, like the gold ball, then the same object with another object (the silver ball) in a second box. At random, pick 1 box and 1 object within that box. If it is the "silver" object, you can ignore that try, if it is the "gold" object, pull the second object from the same box and record what the second object is. Do this 1000 times. It will be 2 out of 3 times that when you originally picked "gold", you will pick "gold" again.

 

[brianmanahan]

If it is the "silver" object, you can ignore that try

i think that is his hangup. he is operating in some "closed system" where it was never possible to picked the silver object. but this means he is ignoring reality.

 

[SKORPI0]

nope:

As posted in the OP. you're picking from the same box where you got the gold ball from. You don't know what the other ball is, it's either gold or silver.

If you had the choice between the 2 boxes (which in the OP is not the scenario), of course the correct answer would be 2/3.

Related threads/articles.
https://en.wikipedia.org/wiki/Bertrand's_box_paradox
https://whyevolutionistrue.wordpress.com/2018/02/20/the-answer-is-2-3/
http://datagenetics.com/blog/june72013/index.html