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

[DigDog]

However, we are ignoring the occurrences of when a S is picked first but it is still a part of the probability system.

.

The experimental evidence is wrong because the code was badly done. The occurrences where S is drawn are being ignored thus turning your 2/4 chances into 2/3 chances, while we have in truth 2/4 GG, 1/4 SG and 1/4 GS. Thus 50%.

Again, the postulate in OP is a 50% postulate. The language implies it. If you want a 2/3 answer you must state IF you have drawn a gold ball, not "you have drawn a gold ball."

This is because of the 4 possible outcomes 2 are GG and you are ignoring 1/4 of the outcomes, yet you say those outcomes should be factored as possibilities that did not occour.

 

[DigDog]

even though gold was picked, the box with 2 golds in it was twice as likely to be picked as the box with 1 gold and 1 silver in it

Is it?
You have already picked a box BEFORE you draw. Here you are again willfully ignoring the positive results from drawing S but not from owning GS.



In the GGGGGG GSSSSSSS
example you MUST draw G first from whatever box you own. Or alternatively you must pick a box and draw ALL the balls.

You think that drawing gold is deterministic but it isnt. You are specifically told that no SG result can exist, therefore you are not drawing from xS GG or xG SG or xG GS but ONLY from xS xG or xG xS.
The silver ball can never be the first ball in box 2, thus there are only TWO places the ball can be, not three.

 

[DigDog]

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.

If anything, thats bertrands mistake, not mine. Factoring in the likelyhood that you box is gold by artificially removing those occurrences when it isnt, is his mistake, im following the logical mechanism as detailed in the OP.

You can see why if you look at the examples with more balls, ok? This is simple.

The two boxes contain one exclusively gold balls, and the other one gold ball and many silver balls, making it very unlikely to draw the gold ball from the silver box, ok?
In real life, if you draw a ball and it is gold, you are very likely to own the gold box, right?
However this also means those who own the silver box will almost always draw silver, thus being able to predict their next draw. Right?

In the postulate, you pick a box and MUST draw gold, either a random gold from the gold box, or the 1 infinitesimally unlikely gold ball from the silver box.

The loops back to the original determination of "you pick a box" which is 50%.

All the instances where you have the silver box are factored by me as "successful prediction" but not by bertrand. He is purposedly reducing the SG and GS occurrences to be only GS, ignoring the fact that GS is THE ONLY POSSIBLE RESULT with the silver box.
So you will not have 50 people drawing silver, 50 owning silver, and 100 owning gold, but 100 people owning silver and 100 owning gold.


Its not that the guy cant count - its that he cant write.

 

[loafbred]

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

Here's a simpler way of looking at the problem:

1. You picked one box.
2. You removed a gold ball from that one box.
3. You are about to remove the one remaining ball, which is when the question of probability is asked.

The remaining ball can be, without question, either gold or silver. The likelihood of either is the same.

 

[William Gaatjes]

There are three boxes, but the next sentence is about picking a gold ball. That increases the chance of you getting a gold ball again because there are now only two possible boxes.
And you have to pick from the same box again. It is either a gold ball or a silver ball. So it is 50% chance because the silver box with two silver balls are excluded from the choice.

 

[DigDog]

Bertrand is obviously sampling the occurences in which a silver is drawn first, and then saying "that doesnt count".

 

[DigDog]

If you have 200 people sampling, the number of draws needs to be 200, NOT 150. Either that, or you need an equal number of samplings. Otherwise the 50 missing samplings need to be counted together with the 50 silver.

Remember that you are forced to draw both balls. This implies that SG must 100% of times draw G1.

 

[William Gaatjes]

Because you chose a gold ball AND you have to pick from the same box again, the silver balls only box is out of the question.
The probability is from that moment on and not from the starting position with 3 boxes.

If the question is formulated differently, other options apply.

 

[DigDog]

Its that samplings do not occur with the same frequency. When we discard a SG draw, we discard all that system, including the associated G2G3 and G3G2 draws.

So on draw1 Alan and Bobby are given the boxes. Alan draws G1 and Bobby draws G2.
At the second draw Alan AGAIN draws G1 and Bobby draws G3.

We will have two more draws where Bobby draws G2 and G3 but both are discarded as Alan draws S.
A S system is non viable.

All instances where S is drawn are not part of the system if the postulate says so.

But wait you say, does the postulate say so?

well, it implies so. You must draw 100% of the balls from the G2G3 box. You must also draw 100% of the balls from the SG1 box, however you must ignore 50% of the SG1 draws as these start with S.
At this point you must ignore all the samples in that system. You also ignore the G2G3 and G3G2 samples associated.

This means "you have drawn a gold ball; your chances now are calculated in a system where SG2 and SG3 do not exist".

If you have drawn G2, you can only draw G3. If you have drawn G3, you can only draw G2. These occur half as often as G1 because half of the G2 and G3 results happen in the S system, where silver is drawn first and the system tree is discarded.

 

[William Gaatjes]

Some people forget that as soon as the box is mentioned in the question, the box becomes relevant and has priority. And that is why the probability chance is changed in this particular situation because the box comes into play.

 

[William Gaatjes]

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

I am not into java code,
Is this correct :

var secondBallNumberChosen = (firstBallNumberChosen === 0) ? 1 : 0;

3 equal = characters ?

edit:
nevermind.

 

[Flapdrol1337]

You can just write out how you got here.

picked a box at random, each box p=1/3

pick a gold coin from the box.
box1 p=1 so 1/3 * 1=1/3
box 2 p=0.5 , 1/3 * .5= 1/6
box3 p=0, 1/3 * 0 = 0

so the chance of ending up in the first box and drawing a gold coin is twice as high as doing the same in the second. and the chance of doing that in the third is 0

 

[interchange]

I think this really is an IQ test. No matter how much it is explained to the 50% answerers, they don't seem to get it.

I'll rephrase the question where the answer is actually 50% so that maybe people will get it.

3 boxes are arranged, one with two gold balls, one with two silver balls, and the last with the first ball being gold and the second ball being silver. You pick the first ball from a box at random and it is gold. What is the probability that the second ball in that box is also gold?

In that stem, the balls are at a known order. In the original question, they aren't. So you could have chosen either gold ball from the first box or the gold ball from the last box. That's three cases, not two.

 

[William Gaatjes]

Probability can reset you know...

 

[Flapdrol1337]

It doesn't "reset" though. The chance of surviving the second round is much higher (100%) in box1 than in box2 (50%). So you're more likely to be in box1 if you drew gold.

I think this really is an IQ test. No matter how much it is explained to the 50% answerers, they don't seem to get it.

Dunno, it's also that people have to learn to disable their "gut feeling" when doing probability math problems and just analyse it with pen and paper even if it looks simple.​

 

[snoopy7548]

The fact that you've all been arguing over this for days and that I instinctively know it's 50% means that I'm a super genius and you're all plebs.

 

[interchange]

Another way of looking at things.... Consider the overall cases.

GG = 1/3 (pick the box with both gold balls)
SS = 1/3 (pick the box with both silver balls)
GS = 1/3 * 1/2 = 1/6 (pick the box with the mix and get the gold ball first)
SG = 1/3 * 1/2 = 1/6 (pick the box with the mix and get the silver ball first)

So we want to know how often we get GG but only if we picked G first. So that's GG / (GG + GS). WHich is (1/3) / (1/3 + 1/6) = (1/3) / (3/6) = (1/3) / (1/2) = 2/3.

 

[Hitman928]

The experimental evidence is wrong because the code was badly done. The occurrences where S is drawn are being ignored thus turning your 2/4 chances into 2/3 chances, while we have in truth 2/4 GG, 1/4 SG and 1/4 GS. Thus 50%.

Again, the postulate in OP is a 50% postulate. The language implies it. If you want a 2/3 answer you must state IF you have drawn a gold ball, not "you have drawn a gold ball."

This is because of the 4 possible outcomes 2 are GG and you are ignoring 1/4 of the outcomes, yet you say those outcomes should be factored as possibilities that did not occour.

The fact that both the box chosen and the first ball chosen were specifically done at random means you no longer have to state if you have drawn a gold ball. That is implicit in the setup of the problem, by saying the choices were done at random, this is clear language to set up the probability of the system. You would have to say that the box and ball chosen are somehow not random for it to be otherwise.

Edit: If you are compelled to choose a gold ball first, then yes, the probability is 50%, however, that's not what happens in the OP since the choices are random. If you want to say that you are compelled to choose a gold ball, you could just as easily read that the first random choice of a box was not actually random and that you are compelled to choose the first (or second) box and thus the probability is 100% (or 0%) depending on what you want the outcome to be.

 

[interchange]

Dunno, it's also that people have to learn to disable their "gut feeling" when doing probability math problems and just analyse it with pen and paper even if it looks simple.

The fact that you've all been arguing over this for days and that I instinctively know it's 50% means that I'm a super genius and you're all plebs.

https://en.wikipedia.org/wiki/Dunning–Kruger_effect

 

[Fanatical Meat]

Got to be 50% because the two boxes are a possible outcome, not 3 gold balls in a big pool. The next pick is from the same box.
Simplify the problem, there are two boxes on box you has a note saying you are right the other has a note saying you are wrong. Pick one.

 

[loafbred]

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

I think this really is an IQ test. No matter how much it is explained to the 50% answerers, they don't seem to get it.

I'll rephrase the question where the answer is actually 50% so that maybe people will get it.

3 boxes are arranged, one with two gold balls, one with two silver balls, and the last with the first ball being gold and the second ball being silver. You pick the first ball from a box at random and it is gold. What is the probability that the second ball in that box is also gold? In that stem, the balls are at a known order. In the original question, they aren't. So you could have chosen either gold ball from the first box or the gold ball from the last box. That's three cases, not two.

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.

I think this really is an IQ test.

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.

 

[interchange]

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.

I'm not confused. This is a named paradox in every basic probability textbook out there (Bertrand's box paradox). Your criticism assumes that if you chose a gold ball you had equal odds of choosing it from the GG box as the GS box. That's not true. You are twice as likely to have chosen it from the GG box since there are two of them to choose from.

 

[Slew Foot]

even if you ignore occurrences where the first ball picked was silver, you were still 2x as likely to have picked the box with 2 gold balls as your first choice than the box with 1 gold + 1 silver

which doesnt matter, because we are told to ignore everything that happened before the gold ball was pulled out

 

[William Gaatjes]

It doesn't "reset" though. The chance of surviving the second round is much higher (100%) in box1 than in box2 (50%). So you're more likely to be in box1 if you drew gold.

Dunno, it's also that people have to learn to disable their "gut feeling" when doing probability math problems and just analyse it with pen and paper even if it looks simple.​

Well, this to me is the opposite of Occam's razor.
And those cases, people simplify a situation too much.
But now , people make it far too complicated.

If i would ask a question : What is the probability of an asteroid hitting the earth when taking into account every asteroid in the universe ?
If i would ask the same question again but formulated as in the op question: What is the probability of an asteroid hitting the earth when taking into account every asteroid in the universe but only looking at the asteroids in the solar system. It would not be the same answer.

 

[Slew Foot]

I'm not confused. This is a named paradox in every basic probability textbook out there (Bertrand's box paradox). Your criticism assumes that if you chose a gold ball you had equal odds of choosing it from the GG box as the GS box. That's not true. You are twice as likely to have chosen it from the GG box since there are two of them to choose from.

Yes you are confused. We are told to calculate the probability of pulling a second gold ball from the point that a gold ball has already been pulled, NOT to calculate the probability of pulling 2 gold balls in a row.