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

[Hitman928]

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.

The probability of picking two golds at the start is 1/3. Only after having drawn the first gold ball does the probability increase to 2/3.

I suggest browsing the thread a little. The reason it is 2/3 is given on pretty much every page in various forms and there are 3 independent computer simulations provided to prove it.

 

[Cerb]

The probability of picking two golds at the start is 1/3. Only after having drawn the first gold ball does the probability increase to 2/3.

I suggest browsing the thread a little. The reason it is 2/3 is given on pretty much every page in various forms and there are 3 independent computer simulations provided to prove it.

I did, which is why I worded my response the way that I did. The chances of picking two gold balls from the start is 1/3. The chances of picking two silver balls from the start is 1/3. The chances of picking two gold balls or two silver balls from the start is 2/3. IE, that reasoning for the classic paradox is valid, and the 2/3 answer correct, for Bertrand's original problem. What make's Bertrand's Box counter-intuitive at first is the psychological manipulation of the reader, using the gold coin in the first selection round as a precondition for the question, but not a given for the calculation. If his problem had been written the way I wrote it, the result would not be counter-intuitive at all.

In the OP's scenario, the question does not make your selection of a gold ball a conditional, but an absolute. Your chances of having picked the first gold ball are 100%, and so your chances of having chosen the box with two silver balls is already 0%. Your box either has a single silver ball in it, or a single gold ball, and even chances as to which of the two boxes it was.

 

[zinfamous]

ok that only took 20 seconds and no math. this is really just a simple word problem.

 

[paperfist]

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

Because it doesn't say which box and there's 2 that are possibilities, GG & GS.

 

[dave_the_nerd]

I did, which is why I worded my response the way that I did. The chances of picking two gold balls from the start is 1/3. The chances of picking two silver balls from the start is 1/3. The chances of picking two gold balls or two silver balls from the start is 2/3. IE, that reasoning for the classic paradox is valid, and the 2/3 answer correct, for Bertrand's original problem. What make's Bertrand's Box counter-intuitive at first is the psychological manipulation of the reader, using the gold coin in the first selection round as a precondition for the question, but not a given for the calculation. If his problem had been written the way I wrote it, the result would not be counter-intuitive at all.

In the OP's scenario, the question does not make your selection of a gold ball a conditional, but an absolute. Your chances of having picked the first gold ball are 100%, and so your chances of having chosen the box with two silver balls is already 0%. Your box either has a single silver ball in it, or a single gold ball, and even chances as to which of the two boxes it was.

What I think you're missing is the initial box selection and the likelihood of pulling that silver ball out of the G/S box. (If you randomly pull a gold ball out of the selected box, it's more likely that you've managed to randomly select the G/G box in the first action.)

 

[Cozarkian]

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.

The answer if you start with only two boxes is still 2/3. You are ignoring that there are 2 different gold balls in the first box you could have picked, but only 1 in the second. Try these on for size:

1. There are two boxes, {Red, Yellow} and {Purple, Silver}. You pick a box and draw a ball. It is not Silver. What are the odds the second ball in that box is not Silver? (The correct answer is 2/3).

2. There are two boxes {100 gold balls}, {1 gold ball, 99 silver balls}. You pick a box and draw a ball. It is gold. What are the odds the next ball you draw from the same box will be gold? (The correct answer is 100/101).

In the OP's scenario, the question does not make your selection of a gold ball a conditional, but an absolute. Your chances of having picked the first gold ball are 100%, . . .

The question says you didn't pick a silver ball, not that it is impossible to do so. You chose randomly, which means it was possible to pick a silver ball, but in hindsight, you did not. The fact that you know you didn't pick a silver ball provides new information that allows you to determine there is a 2/3 chance you have the double gold box.

The idea that the odds are 2/3 if the question states "if you pick a gold ball" and 1/2 if the question states "you pick a random ball. It is gold." is absurd. It means if you were to do a real-life experiment, before you make your selection the odds of getting double gold if you pick gold is 2/3, but after you actually draw a gold ball the odds fall to 1/2, because it is no longer an "if."

Here is a formulation where 1/2 would be correct: There are 3 boxes . . . . You pick a box at random and hand it to me. I look inside, select a ball, and show it to you. If there is a gold ball in the box I must select it. I show you a gold ball. What are the odds the other ball in the box is gold?

 

[Hitman928]

Because it doesn't say which box and there's 2 that are possibilities, GG & GS.

2 Boxes but 3 possible outcomes from your first choice.

I'll say it for like the 5th time, if you really think it is 50% why do all simulations and even real world physical trials say it is 2/3? If you're really convinced it is 50%, why not try the experiment yourself? It literally takes less than 5 minutes to setup with things almost every human being in an even remotely civilized country will have laying around the house.

https://www.youtube.com/watch?v=CGMc8B60ZpU - Has an explanation with visuals.
https://github.com/yourfriendaaron/box-paradox - simulation that auto runs or that you can play to demonstrate how the probability works out. Note, must have python installed.

 

[Cerb]

The answer if you start with only two boxes is still 2/3. You are ignoring that there are 2 different gold balls in the first box you could have picked, but only 1 in the second. Try these on for size:

1. There are two boxes, {Red, Yellow} and {Purple, Silver}. You pick a box and draw a ball. It is not Silver. What are the odds the second ball in that box is not Silver? (The correct answer is 2/3).

2. There are two boxes {100 gold balls}, {1 gold ball, 99 silver balls}. You pick a box and draw a ball. It is gold. What are the odds the next ball you draw from the same box will be gold? (The correct answer is 100/101).



The question says you didn't pick a silver ball, not that it is impossible to do so. You chose randomly, which means it was possible to pick a silver ball, but in hindsight, you did not. The fact that you know you didn't pick a silver ball provides new information that allows you to determine there is a 2/3 chance you have the double gold box.

No, by knowing the gold ball was chosen as part of the absolutely true premise, all attempts in which a silver ball was chosen are irrelevant.

The idea that the odds are 2/3 if the question states "if you pick a gold ball" and 1/2 if the question states "you pick a random ball. It is gold." is absurd. It means if you were to do a real-life experiment, before you make your selection the odds of getting double gold if you pick gold is 2/3, but after you actually draw a gold ball the odds fall to 1/2, because it is no longer an "if."

It's not absurd, it's what is proposed. But, the way it is proposed, is that, "you picked a gold ball." So, the start is knowing that you chose Gx from either {G1,G2} or {G3,S1}, and the question is basically whether or not you chose G3.

Here is a formulation where 1/2 would be correct: There are 3 boxes . . . . You pick a box at random and hand it to me. I look inside, select a ball, and show it to you. If there is a gold ball in the box I must select it. I show you a gold ball. What are the odds the other ball in the box is gold?

No, it works even without that oracular ability. It's 1/2 because the given information precludes having the chance of ever getting to select the {S,S} box, nor the silver ball from the {G,S} box. As asked/proposed, the language provides an assumption that reset the probabilities between the initial description of the boxes, and the second ball removal.

Here's my simulation of the problem in the OP, which as written is not Bertrand's paradox, as I read it, in python, with the code made to be as descriptive as possible:

import random;

box = [ ['gold', 'gold'], ['gold', 'silver'], ['silver', 'silver'] ];

gold_second = 0;
silver_second = 0;
rounds=100000;

for i in range(rounds):
    box_selected = random.randint(0,2);
    if box_selected < 2: # you didn't choose the box with only silver, ignoring...
        ball_first = random.randint(0,1);
        ball_second = -1;
   
        if box_selected == 0: # if you chose the gold/gold box, order doesn't matter
            if ball_first == 0:
                ball_second = 1;
            else:
                ball_second = 0;
        else: # if you chose the gold/silver box, you already chose the gold ball
            ball_first = 0;
            ball_second = 1;
   
        if box[box_selected][ball_second] == "gold":
            gold_second += 1;
        else:
            silver_second += 1;

print ("Rounds attempted: " + str(rounds) );
print("Gold: " + str(gold_second) );
print("Silver: " + str(silver_second) + "\n\n");

I didn't even bother converting to percent, because my results remained within +2% of equal, each time.

 

[interchange]

No, by knowing the gold ball was chosen as part of the absolutely true premise, all attempts in which a silver ball was chosen are irrelevant.

It's not absurd, it's what is proposed. But, the way it is proposed, is that, "you picked a gold ball." So, the start is knowing that you chose Gx from either {G1,G2} or {G3,S1}, and the question is basically whether or not you chose G3.

No, it works even without that oracular ability. It's 1/2 because the given information precludes having the chance of ever getting to select the {S,S} box, nor the silver ball from the {G,S} box. As asked/proposed, the language provides an assumption that reset the probabilities between the initial description of the boxes, and the second ball removal.

Here's my simulation of the problem in the OP, which as written is not Bertrand's paradox, as I read it, in python, with the code made to be as descriptive as possible:

import random;

box = [ ['gold', 'gold'], ['gold', 'silver'], ['silver', 'silver'] ];

gold_second = 0;
silver_second = 0;
rounds=100000;

for i in range(rounds):
    box_selected = random.randint(0,2);
    if box_selected < 2: # you didn't choose the box with only silver, ignoring...
        ball_first = random.randint(0,1);
        ball_second = -1;
  
        if box_selected == 0: # if you chose the gold/gold box, order doesn't matter
            if ball_first == 0:
                ball_second = 1;
            else:
                ball_second = 0;
        else: # if you chose the gold/silver box, you already chose the gold ball
            ball_first = 0;
            ball_second = 1;
  
        if box[box_selected][ball_second] == "gold":
            gold_second += 1;
        else:
            silver_second += 1;

print ("Rounds attempted: " + str(rounds) );
print("Gold: " + str(gold_second) );
print("Silver: " + str(silver_second) + "\n\n");

I didn't even bother converting to percent, because my results remained within +2% of equal, each time.

No the language doesn't say that. It says the ball you chose was gold. It doesn't say the box you chose contained a gold ball. Therefore, the probability is reset when the silver balls are eliminated, not when the box with both silver balls is eliminated. There are 3 gold balls to choose from. 2/3 of them have another gold ball in the box with them. The answer is 2/3.

 

[Cerb]

What I think you're missing is the initial box selection and the likelihood of pulling that silver ball out of the G/S box. (If you randomly pull a gold ball out of the selected box, it's more likely that you've managed to randomly select the G/G box in the first action.)

That assumes that the first action is being considered as more than a way to set up the second's premise. The first action is part of the probability for the classic problem. The way I am reading this problem, you chose a gold ball, and the problem to be solved is the chance of it having been from the gold and silver box, or the gold and gold box.

 

[Cerb]

No the language doesn't say that. It says the ball you chose was gold. It doesn't say the box you chose contained a gold ball. Therefore, the probability is reset when the silver balls are eliminated, not when the box with both silver balls is eliminated. There are 3 gold balls to choose from. 2/3 of them have another gold ball in the box with them. The answer is 2/3.

Since you cannot select one of the other boxes, and you know the contents of each possible box, the box available is directly tied to the ball chosen. The probability appears to be reset after the gold ball is chosen, because (A) that is when you know the silver/silver box was not selected (which is also what makes the original problem 2/3, by not allowing such a reset), and (B) the wording of the problem describes the question beginning with all of that knowledge, not including the prior chances. I read the problem in the OP as making the silver/silver box a red herring, and precluding having selected the silver ball from the gold/silver box, as 100% true assumptions.

IMO, the ultimate solution to the problem is the lulz being had by whomever was clever enough to create and spread the original image.

 

[Hitman928]

Since you cannot select one of the other boxes, and you know the contents of each possible box, the box available is directly tied to the ball chosen. The probability appears to be reset after the gold ball is chosen, because (A) that is when you know the silver/silver box was not selected (which is also what makes the original problem 2/3, by not allowing such a reset), and (B) the wording of the problem describes the question beginning with all of that knowledge, not including the prior chances. I read the problem in the OP as making the silver/silver box a red herring, and precluding having selected the silver ball from the gold/silver box, as 100% true assumptions.

This is where your trouble is. Yes, the silver silver box is a red herring, to use your term. However, the silver ball from the second box is not excluded from the probability. We know this because the second choice, the choice of which ball you get from the box, is described as random. If it is described as a random choice between two objects, they each have a 50% probability of being picked.

If the gold ball from the second box was a forced choice, then the choice was not random. If the choice of ball was not random, then the word random has no meaning (or unknown meaning) in context of the problem. If the word random has no meaning, then the choice of box is also not random. If the choice of box is not random, then there's no way to calculate probability because all you know is that you must choose gold, but you don't know what box you're going to pick. So you could just as easily argue that it is 100% as it is 50%, there's no way to know conclusively.

Edit:

This is also what you are putting in your code.

       else: # if you chose the gold/silver box, you already chose the gold ball
           ball_first = 0;
           ball_second = 1;

You are forcing the choice of a gold ball when picking the second box, but this isn't what the problem says. It says the choice is random.

Setup the code so that it strictly follows the question. Make all 3 boxes available (not necessary, but just for completeness), and make the box choice random. Make the choice of ball random. Compile all the results so that when a silver ball is chosen first, the result is ignored. However, when a gold is chosen first, calculate how many times the ball came from the first (or GG) box. It will be 2/3.

 

[SKORPI0]

Setup the code so that it strictly follows the question. Make all 3 boxes available (not necessary, but just for completeness), and make the box choice random. Make the choice of ball random. Compile all the results so that when a silver ball is chosen first, the result is ignored. However, when a gold is chosen first, calculate how many times the ball came from the first (or GG) box. It will be 2/3.

The color of the 1st pick can be gold or silver (black/white), etc. Ask the probably of getting the same color as the 1st pick, for the 2nd pick.
Should be the same, 2/3 or 66.667%.

 

[Cerb]

You are forcing the choice of a gold ball when picking the second box, but this isn't what the problem says. It says the choice is random.

"It's a gold ball." That invalidates randomness that happened prior. That bit of code is not an error, but an important part of how I am reading the OP's problem (and again, note that I consider the OP's problem to be a meme, intentionally modifying the well-known classic problem). You got a gold ball from one of two boxes. One that has another gold ball in it, and one has a silver ball. What are your chances of the next ball being gold? Everything before that is more or less fluff.

 

[Cozarkian]

It's 1/2 because the given information precludes having the chance of ever getting to select the {S,S} box, nor the silver ball from the {G,S} box.

There is nothing that states you can't pick a silver ball, it just asks you to consider the scenario where the ball you actually picked is gold. In fact, the question states you pick randomly and can't see into the box, which is incompatible with a mechanism to prevent a silver ball from being drawn.

You can try the question as presented in real life. If you draw a silver ball, put it back, mix up the boxes, and try again. When you get a gold (because you absolutely will - it isn't an if), the odds that you have the double gold will be 2/3.

As asked/proposed, the language provides an assumption that reset the probabilities between the initial description of the boxes, and the second ball removal.

As asked/proposed, the language provides an assumption that asks you to consider only the 50% of trials where a gold ball is selected, it does not in anyway state that 100% of trials will result in a gold ball - that would be incompatible with the condition that the selection is made randomly.

The question presented is not comparable to this: You come to an intersection. You go left. (Here the odds of going left are 100%).

The question is comparable to this: You come to an intersection. You decide to flip a fair coin. On heads you will go left. On tails you will go right. You flip the fair coin. It comes up heads. You go left. (Here the odds of going left are not 100%, because even though going left is actually what happened, a fair coin was used to make the decision, which means the odds of going left were only 50%).

As for your code:

for i in range(rounds):
box_selected = random.randint(0,2);
if box_selected < 2: # you didn't choose the box with only silver, ignoring...

Here you ignore trials where double silver is picked.

ball_first = random.randint(0,1);
ball_second = -1;

if box_selected == 0: # if you chose the gold/gold box, order doesn't matter
if ball_first == 0:
ball_second = 1;
else:
ball_second = 0;
else: # if you chose the gold/silver box, you already chose the gold ball
ball_first = 0;
ball_second = 1;

But here instead of ignoring the trials where silver is picked, you eliminate their possibility (I realize this was deliberate - but why treat it differently than the selection of box 3?). The odds for picking any single ball should be 1/6, but you've doubled the odds of picking box 2, gold to 1/3.

 

[Cerb]

There is nothing that states you can't pick a silver ball, it just asks you to consider the scenario where the ball you actually picked is gold. In fact, the question states you pick randomly and can't see into the box, which is incompatible with a mechanism to prevent a silver ball from being drawn.

The possibility of that has been precluded, though.

As asked/proposed, the language provides an assumption that asks you to consider only the 50% of trials where a gold ball is selected, it does not in anyway state that 100% of trials will result in a gold ball - that would be incompatible with the condition that the selection is made randomly.

It relies on ignoring those trials, though, because you have a gold ball already.

Here you ignore trials where double silver is picked.

Exactly.

But here instead of ignoring the trials where silver is picked, you eliminate their possibility (I realize this was deliberate - but why treat it differently than the selection of box 3?). The odds for picking any single ball should be 1/6, but you've doubled the odds of picking box 2, gold to 1/3.

Because I'm reading it a lot like your intersection problem, right down to having two choices: you either got it from box GG, or box GS, but don't have enough information to determine how you got there. So, let me put it this way: you got a gold ball from the box, and know it came from one of two boxes. Pick that box. The randomness from the possibility of the GG vs. GS selection is past, at this point.

 

[Hitman928]

"It's a gold ball." That invalidates randomness that happened prior. That bit of code is not an error, but an important part of how I am reading the OP's problem (and again, note that I consider the OP's problem to be a meme, intentionally modifying the well-known classic problem). You got a gold ball from one of two boxes. One that has another gold ball in it, and one has a silver ball. What are your chances of the next ball being gold? Everything before that is more or less fluff.

So, let me put it this way: you got a gold ball from the box, and know it came from one of two boxes. Pick that box. The randomness from the possibility of the GG vs. GS selection is past, at this point.

This makes no sense whatsoever. If randomness is invalidated, then the probability is 100% because I pick the GG box every time, prove me wrong.

 

[JujuFish]

Here's another way to look at it:

Instead of picking gold balls, pick non-silver balls. Let's recolor the gold balls.

[Red, Blue] [Yellow, Silver] [Silver, Silver]

You pick a ball at random. It's not silver. What are the odds the remaining ball is also not silver?

Here are the only possible ways to have pulled not-silver:

Red (remainder is Blue)
Blue (remainder is Red)
Yellow (remainder is Silver)

As you can see, the odds are 2/3.

 

[Cozarkian]

After a random selection, you are holding a gold ball != It is impossible to pick a silver ball. (the outcome is the same, you will be holding a gold ball, but the journey is different.)

I agree, if it is impossible to pick a silver ball, then the answer is 50%. Here, the question does not say impossible. It just asks you to assume you didn't.

You are retroactively eliminating the condition that a choice was made randomly, just because you know the outcome of that random choice. By doing so, you are changing the question.

 

[Cozarkian]

The possibility of that has been precluded, though.

It relies on ignoring those trials, though, because you have a gold ball already.

Exactly.

Because I'm reading it a lot like your intersection problem, right down to having two choices: you either got it from box GG, or box GS, but don't have enough information to determine how you got there. So, let me put it this way: you got a gold ball from the box, and know it came from one of two boxes. Pick that box. The randomness from the possibility of the GG vs. GS selection is past, at this point.

To help you see that the fact a gold ball was chosen does not preclude the possibility of drawing a silver one, consider the following variation:

there are two boxes. Each box has two compartments. In one box, both compartments hold a gold ball. In the other box, one compartment holds a gold ball, and the other compartment holds a silver.

You flip a coin twice. On heads, you select the box or compartment on the left; on tails, you select the box or compartment on the right. After both flips, you are holding a gold ball. What are the odds that the ball in the other compartment in the same box you picked is also gold?

 

[Lanyap]

Whoever has the highest IQ in ATOT, please give us the correct answer.

 

[brianmanahan]

Whoever has the highest IQ in ATOT, please give us the correct answer.

paging @spaceman

 

[aigomorla]

You guys its a known paradox equation.
If you still insist on saying its 1/2, that is the paradox and you flat out lost.

https://en.wikipedia.org/wiki/Bertrand's_box_paradox

However the original OP's picture was WORDED INCORRECTLY.

The 'paradox' is in the probability, after choosing a box at random and withdrawing one coin at random, IF that happens to be a gold coin, of the next coin drawn from the same box also being a gold coin.

The original paradox says IF THAT... and not IT.
So as i said, the worded portion is very misleading.

Anyhow that box is screaming MEOW, just don't open it, because there is a hidden camera somewhere and its obviously someone who put in a bluetooth speaker inside that box messing with you.

 

[Carson Dyle]

However the original OP's picture was WORDED INCORRECTLY.

The original paradox says IF THAT... and not IT.
So as i said, the worded portion is very misleading.

Makes no difference. It's worded just fine. It's not misleading and it's not ambiguous.

"You take a gold ball from a box. What is the probability that the next ball will be gold?"

is exactly the same as

"If the ball that you've taken is gold, what is the probability that the next ball will be gold?"

 

[aigomorla]

If the ball that you taken -> then goto B.
You took a ball -> already at B.

Theres a big difference in wording.

In the OP, it was stated the ball you took was gold.... so your already at B.
The original paradox eq, says if the ball you took was gold, then... which means you were at A and if you finish req, then goto B.

That is the confusion.