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

[Schfifty Five]

Edit: Nvm, misread!

 

[aigomorla]

OK i get where the confusion is between me and other people.

The 2/3rds is taking into account the first gold ball that you already pulled is included in the Probability. <--- which is correct because its the Bertrand Box Paradox.
The 1/2 is taking into account that the first gold ball pulled is given, and not calculated in the Probability. <---- which all the 50% people think, because we assume given is given.

 

[Hitman928]

OK i get where the confusion is between me and other people.

The 2/3rds is taking into account the first gold ball that you already pulled is included in the Probability. <--- which is correct because its the Bertrand Box Paradox.
The 1/2 is taking into account that the first gold ball pulled is given, and not calculated in the Probability. <---- which all the 50% people think, because we assume given is given.

Yep, you got it!

 

[BudAshes]

Obvious answer is pick the boxes up, gold is twice as heavy as silver so the heaviest one has 2 gold balls.

 

[brianmanahan]

If you do a reverse image search you'll see about 20 other forums went ape s$&@ over this problem including the body building forum from which the image is from.

lol body building forum has been a goldmine for so much stuff like this

 

[paperfist]

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%.

Yeah box, not boxes

 

[Hitman928]

Yeah box, not boxes

What's you're point? You have to pull both balls from the same box, that's part of the probability.

 

[Humpy]

I think this is where I'll probably stop and agree with y'all that the answer is 5/7nds.

 

[interchange]

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.

Actually it's not important at all. You can change the word "gold" to "not silver" in the problem and it's obvious that it wouldn't change the problem at all. Therefore, you could make them any color you want if it helps visualize the problem.

But because it's not important at all, they could still be completely indistinguishable. But because you can't tell the difference between 2 objects, it does not make them 1 object. They're still different balls. The only thing that's important is that you could choose either ball from the first box.


Here's another way to approach the problem. Let's ignore the constraint for now that the first choice you made resulted in a gold ball. Therefore, we are left with:
P(GG) + P(GS) + P(SG) + P(SS) = 1

But we're only interested in the states where gold is picked first. Specifically we want P(GG) under the conditions that gold is picked first. So that's:
Answer = P(GG) / (P(GG) + P(GS))

Since the problem is symmetric, it's easy to see that exactly half the time we'll pick a gold ball first. So:
P(GG) + P(GS) = 1/2

So our Answer = P(GG) / (1/2) = 2*P(GG)

So how do we find P(GG)? Well that's really simple. There are 3 boxes. Only 1 of the boxes has 2 gold balls. So P(GG) occurs when you pick the box with 2 gold balls. So that's a simple 1/3.

So our Answer = 2*P(GG) = 2*(1/3) = 2/3

 

[IHateMyJob2004]

2/3 .... OK I just ran through the outcomes(6)

A pull -> then B
G->G
G->G
G->S
S->G
S->S
S->S

Ignore the data set starting with S:
G->G
G->G
G->S

The confusing part is that you will pull the G from the GG box more often than the G from the GS box.

 

[purbeast0]

2/3 .... OK I just ran through the outcomes(6)

A pull -> then B
G->G
G->G
G->S
S->G
S->S
S->S

Ignore the data set starting with S:
G->G
G->G
G->S

The confusing part is that you will pull the G from the GG box more often than the G from the GS box.

LOL nice edit.

I still think it's wrong.

You have 1 ball left to pick, and it is either gold or silver.

Therefore, 50%.

 

[repoman0]

LOL nice edit.

I still think it's wrong.

You have 1 ball left to pick, and it is either gold or silver.

Therefore, 50%.

I see why the thread is so long. People explain why this is not the case in hundreds of different ways

 

[purbeast0]

I see why the thread is so long. People explain why this is not the case in hundreds of different ways

Well they are wrong and solving a problem that wasn't asked.

 

[repoman0]

Well they are wrong and solving a problem that wasn't asked.

Nope. You're solving the wrong problem.

 

[SKORPI0]

GG / GS / SS

you picked G from 1 box.
with 3 possible results of box you picked from.

_G / GS / SS /
G_ / GS / SS /
GG / _S / SS /

out of these 3 possibilities, 2 have a G

_G, G_, and _S
----------------------------------------------------------------
My initial thought was 1/2 or 50%. Since to me, you didn't differentiate between the 2 G
that was in 1 box. Same thing with the two S in the other box. So the results were.

G_/ GS / SS
GG/ _S/ SS

G_ and _S, 2 possibilities of the box you picked the G from. 50% chance.

Now creating this in the real physical world was confusing to me and getting 2/3, or maybe to a lot of other folks too.
I'll do my own experiment with 3 boxes, shuffle them around for random picks. not knowing which ball/coin is in which. Do it for 100 times and count the results.
-----------------------------------------------------------------------

 

[Drako]

Threads like this remind me of this:

"Think of how stupid the average person is, and realize half of them are stupider than that."

- George Carlin

 

[Cozarkian]

Actually it's not important at all. You can change the word "gold" to "not silver" in the problem and it's obvious that it wouldn't change the problem at all.

That might be a great way to illustrate it for the people saying there are only 2 balls left, one gold, one silver (and for those saying 50% is possible because of ambiguity in the question).

There are 3 boxes. Each box contains 2 balls, with colors as follows:

Red/Yellow
Purple/Silver
Silver/Silver

"You pick box at random. You put your hand in and take a ball from that box at random. It's [not] a [silver] ball. What is the probability that the next ball you take from the same box will also [not] be [silver]?"

With the above modification, the ball you are holding is either red, yellow, or purple. If it is red, there is a 100% chance the next ball will not be silver. If it is yellow, there is a 100% chance the next ball will not be silver. If it is purple, there is a 0% chance the next ball will not be silver. Which gives us (1 + 1 + 0) / 3 = 2/3.

 

[Humpy]

Threads like this remind me of this:

"Think of how stupid the average person is, and realize half of them are stupider than that."

- George Carlin

OMG! George was wrong about that.

If he had an IQ above 115 he wouldn't have confused mean with median.

 

[Hitman928]

OMG! George was wrong about that.

If he had an IQ above 115 he wouldn't have confused mean with median.

First thought I had too when I read that quote.

With that said, unfortunately, I think he knew the difference but also knew most audiences (especially when drunk) wouldn't know what he meant if he said median. So rather than have to explain median first and ruin the approach and timing of the joke, he just used average because most people would think it's right anyway and those who know better would probably still laugh anyway knowing what he meant.

 

[aigomorla]

I see why the thread is so long. People explain why this is not the case in hundreds of different ways

lol...

its because of how the problem is worded.

As i said, its worded to sound that you would not take into account the first ball pulled.
But as Humpy pointed out, the solution to the paradox is you need to account the first ball pulled because it was stated you pulled that at random.

Hence that is the paradox.

It is like Schrodinger cat.
If we didn't hear the cat at the start, we would automatically assume the cat is dead.
However there is something going MEOW in the box, so we must assume the cat is alive, or some higher power is just messing with us.

 

[repoman0]

lol...

its because of how the problem is worded.

As i said, its worded to sound that you would not take into account the first ball pulled.
But as Humpy pointed out, the solution to the paradox is you need to account the first ball pulled because it was stated you pulled that at random.

Hence that is the paradox.

I get why it's confusing to people, especially if they haven't been introduced to conditional probability. I got the "intuitive" 50% answer at first too until I wrote out in P(A|B) notation like I outlined a page ago and realized you have to take into account the higher probability of having picked from GG in the first place

 

[dave_the_nerd]

Interesting. I took discrete math in 2014 and I've already forgotten most of it.

Oh well, you can always brute-force it.

#BertrandsBoxTest

from random import randint

Boxes = [["G","G"],["G","S"],["S","S"]]

boxesSelected = 0
firstGold = 0
secondGold = 0

for iteration in range(0,10**4):
    # Select a box.
    ThisBox = Boxes[randint(0,2)]
    boxesSelected += 1
 
    # Pick a coin from the box
    Coin1Index = randint(0,1)
    Coin2Index = int(not Coin1Index)
 
    if ThisBox[Coin1Index] is "G":
        firstGold += 1
        if ThisBox[Coin2Index] is "G":
            secondGold += 1
 
         
print boxesSelected, "boxes selected."
print "First coin was gold", firstGold, "times."
print "Second coin was gold", secondGold, "times."
print "If the first coin in a selected box is gold, then"
print "the second is also gold", float(secondGold)/float(firstGold)*100.00, "% of the time."

Run enough iterations, enough times, and it pretty much zeroes in on 2/3.

 

[Hitman928]

Interesting. I took discrete math in 2014 and I've already forgotten most of it.

Oh well, you can always brute-force it.

#BertrandsBoxTest

from random import randint

Boxes = [["G","G"],["G","S"],["S","S"]]

boxesSelected = 0
firstGold = 0
secondGold = 0

for iteration in range(0,10**4):
    # Select a box.
    ThisBox = Boxes[randint(0,2)]
    boxesSelected += 1
 
    # Pick a coin from the box
    Coin1Index = randint(0,1)
    Coin2Index = int(not Coin1Index)
 
    if ThisBox[Coin1Index] is "G":
        firstGold += 1
        if ThisBox[Coin2Index] is "G":
            secondGold += 1
 
        
print boxesSelected, "boxes selected."
print "First coin was gold", firstGold, "times."
print "Second coin was gold", secondGold, "times."
print "If the first coin in a selected box is gold, then"
print "the second is also gold", float(secondGold)/float(firstGold)*100.00, "% of the time."

Run enough iterations, enough times, and it pretty much zeroes in on 2/3.

Obviously you coded it wrong, otherwise it would have been 50%

Sorry, couldn't help myself, it's been a long thread. This is actually the third coded simulation that has been posted in this thread and all show 2/3, but my first sentence is the response given by the remnant still clinging onto hope that it's not 2/3.

 

[dave_the_nerd]

Obviously you coded it wrong, otherwise it would have been 50%

Sorry, couldn't help myself, it's been a long thread. This is actually the third coded simulation that has been posted in this thread and all show 2/3, but my first sentence is the response given by the remnant still clinging onto hope that it's not 2/3.

Oh, cool. I didn't read the whole thread - shouldn't have been surprised to see somebody else code it, given the crowd around here.

I figured it wrong, myself, but then looked up that Bertrand Paradox thing and got schooled, so I figured I'd test it out. TIL and stuff.

 

[Cerb]

1/2.

2/3 is the chance of picking two balls of the same color, gold or silver, from the start of the scenario. The question in the OP does not occur at the start of the scenario, but after you have selected a gold ball, the way I am reading it. That means that boxes {G,G} and {G,S} are all that matter to the probability of the next ball's selection. IE, as written in the OP, the {S,S} box is a red herring, since it is no longer a valid option. No amount of reasoning that includes the {S,S} box in the calculation, which will lead to 2/3, is valid, IMO, given the way the question is placed and written. The givens to the problem include that the {S,S} box has already been excluded, so it may as well have just started off with two boxes, {G,G} and {G,S}, instead.