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

[interchange]

It's a question of whether you view the two selections as a unit or as separate events. Similar to the old "100 coin flips" model.

The odds of flipping heads 100 times in a row is billions to one (or higher). If you somehow manage to flip 99 heads in a row you could beat impossible odds with the final flip - IF you view the 100 flips as a set of 100 events.

But taken as an individual event the odds of the 100th flip being heads is 50/50, even after 99 straight "heads" flips. All preceding flips are irrelevant.

This problem embraces the same philosophy. If you see the two selections as a single event the odds are 2/3. If you view the second selection as it's own event the odds are 50/50, the box with two silvers having become irrelevant.

That's not accurate. In the heads/tails problem there is only one coin for which to calculate probability on. At the start of the probability question here, there are 3 balls for which to calculate from. So long as either ball from the first box could have been selected, the answer is 2/3.

An analogy would be saying you flip 2 coins simultaneously 49 times and they each come out heads. What is the probability that the next time you flip both coins they both also become heads? The answer is 1/4 for that one.

 

[dasherHampton]

No, you're making it too complicated. As soon as a gold ball is pulled the box with two silvers becomes irrelevant and alters the probability of the second event.

Seen as a separate event the odds of the second pull being gold are 50/50. Seen as the sum of two events it's 2/3.

 

[Vindris]

50% is the correct answer.

jk 2/3

 

[loafbred]

No, you're making it too complicated. As soon as a gold ball is pulled the box with two silvers becomes irrelevant and alters the probability of the second event.

Seen as a separate event the odds of the second pull being gold are 50/50. Seen as the sum of two events it's 2/3.

The problem with focusing only on the second event is that you can't. The fact that there were two twin boxes and one mixed, in the first event, will unavoidably influence the odds of pulling a ball from a twin box, every time. That influence doesn't vanish after you've pulled the first ball. I was determined to discard any influence from the first event, when the thread first started, but finally looked further in.

 

[dasherHampton]

The second event consists of a random pull of either one silver ball or one gold ball. Unless magic is somehow involved that must be classified as a 50/50 proposition.

If you combine the two events then the chance pulling a gold ball obviously increases.

I'm agreeing with everybody, only trying to show where the disconnect lies.

 

[Humpy]

Probability means the ratio between a given outcome versus all possible outcomes.

Defining all possible outcomes (i.e. sample space) is often the first step in these problems.

When the language is sloppy you don't have a statistics problem, you have an argument of opinion.

As a silly example, from the OP I have no reason to believe the balls are of equal weight and texture. It has to be assumed that the second ball isn't so heavy and slippery that it can actually be taken from the box.

Some are assuming that the question is really asking "What is the probability, if a gold ball is picked first, that the next ball will also be gold". Others see it as it literally says "What is the probability that the next ball will also be gold."

 

[SKORPI0]

Now I get it..... .
Best way to solve the puzzle is differentiate the G balls, one could be of different size.
But this was not specified in the original post, that the balls are different. If they are, it changes
the result.

1st pick

Gg, GS, Ss -
result you get G or g from one of the 3 boxes.
========================================
2nd pick. - at this point you have to specify which gold coin was picked. G or g.

X/x is picked G/g.

3 possible results from the 1st pick

Xg, GS, Ss - 1
Gx, GS, Ss - 1
Gg, XS, Ss - 0

2 out of 3 chances of getting G/g

. .

 

[Hitman928]

The second event consists of a random pull of either one silver ball or one gold ball. Unless magic is somehow involved that must be classified as a 50/50 proposition.

If you combine the two events then the chance pulling a gold ball obviously increases.

I'm agreeing with everybody, only trying to show where the disconnect lies.

That's the problem though, that's not the second event. Misunderstanding this is why people think it is 50/50.

 

[dasherHampton]

OK, let's frame this another way.

Divide the task between two people. One person does the first pull and one does the second. The second person is sequestered in their own room and has no knowledge of previous events.

The first person pulls a gold with the balls set up as described. The box is then moved to another room where the second person does their pull.

The second person knows nothing about the parameters of the puzzle or what the first person pulled. They are only told that the ball they are pulling is either silver or gold.

They must guess before their pull whether the ball will be silver or gold. Since the remaining ball could be either gold or silver the second puller has a 50% chance of being correct. Nothing that happened previously can affect those odds.

 

[Hitman928]

OK, let's frame this another way.

Divide the task between two people. One person does the first pull and one does the second. The second person is sequestered in their own room and has no knowledge of previous events.

The first person pulls a gold with the balls set up as described. The box is then moved to another room where the second person does their pull.

The second person knows nothing about the parameters of the puzzle or what the first person pulled. They are only told that the ball they are pulling is either silver or gold.

They must guess before their pull whether the ball will be silver or gold. Since the remaining ball could be either gold or silver the second puller has a 50% chance of being correct. Nothing that happened previously can affect those odds.

Is the second person confined to picking from the same box that the first person pulled from as outlined in the OP?

*It wouldn't actually matter in the end for this particular case, but it may change how you get there.

 

[WelshBloke]

PowerEngineer is correct, maybe try to reread the OP.

Read it.

It asks you what the probability of making that last choice from the system is.

The system at that point is two boxes one with a gold ball and one with a silver ball.

 

[IronWing]

Read it.

It asks you what the probability of making that last choice from the system is.

The system at that point is two boxes one with a gold ball and one with a silver ball.

Actually the system is just one box with either a gold ball or a silver ball. How the system reached that state is superfluous to the question asked.

 

[Hitman928]

Read it.

It asks you what the probability of making that last choice from the system is.

The system at that point is two boxes one with a gold ball and one with a silver ball.

At the end of the first step, you don't know what is in the boxes. Initially, there are 2 balls, in each box.

Box 1 : Gold Ball (a) + Gold Ball (b)
Box 2 : Gold Ball (c) + Silver Ball

One gold ball is pulled. So you could have:

Box 1 : Gold Ball (a) + Gold Ball (b)
Box 2 : Silver Ball

or

Box 1: Gold Ball (a)
Box 2: Gold Ball (c) + Silver Ball

or

Box 1: Gold Ball (b)
Box 2: Gold Ball (c) + Silver Ball

3 Possible outcomes from the first step, none of which are 2 boxes, one with a Gold ball and one with a Silver ball.

 

[WelshBloke]

At the end of the first step, you don't know what is in the boxes. Initially, there are 2 balls, in each box. One gold ball is pulled. So you could have:

Box 1 : Gold Ball (a) + Gold Ball (b)
Box 2: Silver Ball

or

Box 1: Gold Ball (a)
Box 2: Gold Ball (c) + Silver Ball

or

Box 1: Gold Ball (b)
Box 2: Gold Ball (c) + Silver Ball

3 Possible outcomes from the first step, none of which are 2 boxes, one with a Gold ball and one with a Silver ball.

You have two boxes (because the one with two silver in is discounted because it's not relevant to the question)

The two boxes are [G,G] and [G,S]

A gold ball is removed from a box and the question is what's the probability of the next ball being gold. The two options, at that point, are [x,G] or [x,S] (the x being the removed gold ball.

 

[IronWing]

At the end of the first step, you don't know what is in the boxes. Initially, there are 2 balls, in each box.

Box 1 : Gold Ball (a) + Gold Ball (b)
Box 2 : Gold Ball (c) + Silver Ball

One gold ball is pulled. So you could have:

Box 1 : Gold Ball (a) + Gold Ball (b)
Box 2 : Silver Ball

or

Box 1: Gold Ball (a)
Box 2: Gold Ball (c) + Silver Ball

or

Box 1: Gold Ball (b)
Box 2: Gold Ball (c) + Silver Ball

3 Possible outcomes from the first step, none of which are 2 boxes, one with a Gold ball and one with a Silver ball.

Nope, you only have one box and you already pulled a gold ball out of it.

So your choices are either you have:
Box 1: 1 gold ball

or

Box 2: 1 silver ball

Forget combinations of boxes, you only have one box, you just don't know which box is the box you have.

 

[Hitman928]

You have two boxes (because the one with two silver in is discounted because it's not relevant to the question)

The two boxes are [G,G] and [G,S]

A gold ball is removed from a box and the question is what's the probability of the next ball being gold. The two options, at that point, are [x,G] or [x,S] (the x being the removed gold ball.

Let me ask it to you this way, if I pull the gold ball from the box that is [G,S], how then do I have a box with 1 Gold ball and another box with 1 Silver ball when there are 2 remaining gold balls in the first box?

 

[Roger Wilco]

After the initial step, you determine two things that affect the known probability of the final step:

1. The box you pulled the gold ball from cannot be the SS box.
2. You were 2x more likely to pick the gold ball from the GG box when you completed the initial step.

Because you were 2x more likely to pick the gold ball from the GG box, you are now more likely to pull a second gold ball from that box in the final step. This means the probability cannot be 50% for the final step, but rather a second gold ball is the favored outcome.

 

[Hitman928]

xG, GS, SS - 1
Gx, GS, SS - 1
GG, xS, SS - 0
GG, Sx, SS - 0

2 out of 4 success

OP question clearly specifies 2nd pick is the same box (picked G marked x) you picked the gold from, you cannot pick from another box. .

In your setup it should be,

xG, GS, SS - 1
Gx, GS, SS - 1
GG, xS, SS - 0

2 out of 3 success

This final possibility doesn't exist. You are counting the gold ball in the second box twice by adding it.
GG, Sx, SS - 0

 

[WelshBloke]

Let me ask it to you this way, if I pull the gold ball from the box that is [G,S], how then do I have a box with 1 Gold ball and another box with 1 Silver ball when there are 2 remaining gold balls in the first box?

It doesn't matter. The question is asked at the point where there is only the choice of a gold ball in a box or a silver ball in a box.

How you get to that bit doesn't matter because of the way the question is phrased.

The question in the OP is an excellent troll, it's much better than the airplane on the treadmill one.

 

[SKORPI0]

In your setup it should be,

xG, GS, SS - 1
Gx, GS, SS - 1
GG, xS, SS - 0

2 out of 3 success

This final possibility doesn't exist. You are counting the gold ball in the second box twice by adding it.
GG, Sx, SS - 0

changed my post above. .
The initial confusion is about the Gold balls, you have to differentiate which Gold ball was picked from the box #1, left or right.
This changes the result to 3 possibilities, not 2 as I initially thought. I rest my case.. it's 2/3.

 

[Hitman928]

It doesn't matter. The question is asked at the point where there is only the choice of a gold ball in a box or a silver ball in a box.

How you get to that bit doesn't matter because of the way the question is phrased.

The question in the OP is an excellent troll, it's much better than the airplane on the treadmill one.

Nope, the question in the OP is Bertrand's Box Paradox with some grammatical errors. It is a known problem with a known solution, which is 2/3.

It is mathematically proven. It is experimentally proven. If you do not believe it, try the experiment yourself. After a sufficient number of trials, the second ball drawn will be gold roughly 2/3 of the time.

 

[Hitman928]

changed my post above. .
The initial confusion is about the Gold balls, you have to differentiate which Gold ball was picked from the box #1, left or right. This changes the result to 3 possibilities, not 2 as I initially thought. I rest my case.. it's 2/3.

Correct. I noticed you distinguished them as different sizes which is fine to help visualize it, but they don't have to be. They can be identical balls, as long as they are 2 distinct balls which is what is stated in the OP.

 

[WelshBloke]

Nope, the question in the OP is Bertrand's Box Paradox with some grammatical errors. It is a known problem with a known solution, which is 2/3.

Those 'grammatical errors' change the question. That's why they are put in there, they arent there by accident.

 

[sandorski]

Even though I got the right answer, I now realize I came to the conclusion incorrectly. The 50%ers make some compelling arguments, but it's called a "Paradox" for a reason.

 

[IronWing]

I read Betrand's Box article. I'm going to admit error but I'm not going to be graceful about it. Stupid balls.