Current Events > IQ Test

(+) Yeah! (+)

https://imgur.com/GJOUZWR

There are 3 boxes. Each box contains 2 balls.

One box contains 2 gold balls, another box contains 2 silver balls, and the final box contains one gold ball and one silver ball.

You pick a box at random. You put your hand in and take a ball from that box at random.

It's a gold ball.

What is the probability that the next ball you take from the same box will also be gold?

Note: You can't see into any of the boxes.
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I'll post the correct answer and explanation later.
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50%, aka 1/2 if I understood the problem correctly and there are no shenanigans with the wording.

You need to discard the box with no gold balls inmediately, since it's useless. That leaves only two choices, either you picked the box with two gold balls and will get a gold one, or you picked the other one and it will be a silver ball.

I'm kinda drunk so I may be wrong though :3
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1/2.

You either draw a gold ball or don't.
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I voted 50/50.

After the first draw you know the box you drew from is either of the two with gold balls in, it just come down to 50/50 chance it's the one with 2 gold balls or not.

HagenEx posted...

You need to discard the box with no gold balls inmediately, since it's useless.

RedZaraki posted...

What is the probability that the next ball you take from the same box will also be gold?

https://imgur.com/jkAvZSa

reading comprehension is hard. ayy lmao
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myzz7 posted...

HagenEx posted...

You need to discard the box with no gold balls inmediately, since it's useless.

RedZaraki posted...

What is the probability that the next ball you take from the same box will also be gold?

https://imgur.com/jkAvZSa

reading comprehension is hard. ayy lmao

No not really he means the box without any gold balls is now irrelevant as there's no gold balls in it.

50/50

You could only have drawn a ball from box 1 or 2. Since box 3 does not have a gold ball, you can dismiss it entirely from the equation.

The only question now is if you drew a ball from box 1 or 2.

So, its a 50/50 chance.

Not that hard.
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The balls are a planet and the box is a room outside.
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I had to look up the answer. Interesting. I won't spoil it.

Looking it jogged my memory; I *have* seen it before. It must have been a while back since I got it wrong.

luigi13579 posted...

I had to look up the answer. Interesting. I won't spoil it.

Looking it jogged my memory; I *have* seen it before. It must have been a while back since I got it wrong.

I probably got it wrong as well, in mathematic's they don't see the number of boxes as relevant just the number of balls left IE 2/3 of the time you'll pick a gold ball.

It's a trick question really.

Newhopes posted...

luigi13579 posted...

I had to look up the answer. Interesting. I won't spoil it.

Looking it jogged my memory; I *have* seen it before. It must have been a while back since I got it wrong.

in mathematic's they don't see the number of boxes as relevant just the number of balls left IE 2/3 of the time you'll pick a gold ball.

But why? You're picking from the same box that you already took a ball from, which means there are only 2 outcomes. It can't be 2/3 because the ball you already picked takes out two choices, aka a golden ball from box#1 and golden ball from box#2 out of the equation, which leaves only the golden ball from box#1 and the silver one from box#2.

Once again, I'm drunk, but I think I'm right.
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2/3

There are three situations in which you pick a gold ball first. Two of those three situations leave another gold ball in the same box.

EDIT: I think I know why some people think the answer is half. I'm guessing they're failing to account for the reduced likelihood of choosing a gold ball when initially selecting from the box with both a gold and silver ball, compared to initially selecting from the box with two gold balls.

tag for solution

voted for 2/3 because thats how likely it was that the first ball came from the double gold box

Apparently most people think 1/2 half tho so idk
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masterbarf posted...

There are three situations in which you pick a gold ball first. Two of those three situations leave another gold ball in the same box.

Pretty much. It's more likely that you've picked the first box since it contains two gold balls.

Since the secret's out, here's the Wikipedia article:

https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

They don't factor in the likelihood that someone will not complete the test because they will run off with the gold and silver balls to sell them. Do I have the highest IQ?

g980 posted...

voted for 2/3 because thats how likely it was that the first ball came from the double gold box

That's about the stone skinny of it.

8-bit_Biceps posted...

They don't factor in the likelihood that someone will not complete the test because they will run off with the gold and silver balls to sell them. Do I have the highest IQ?

I have the best IQ in the history of IQs, maybe ever. That I can tell ya.

luigi13579 posted...

I have the best IQ in the history of IQs, maybe ever. That I can tell ya.

Of course Trump posts here.

wanted to vote 2/5 but since that wasn't even a choice...

6 balls

3 balls are gold

take away 1 gold ball, draw again

5 balls

2 balls are gold

2/5

Rika_Furude posted...

50/50

You could only have drawn a ball from box 1 or 2. Since box 3 does not have a gold ball, you can dismiss it entirely from the equation.

The only question now is if you drew a ball from box 1 or 2.

So, its a 50/50 chance.

Not that hard.

please be joking
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Choco posted...

Rika_Furude posted...

50/50

You could only have drawn a ball from box 1 or 2. Since box 3 does not have a gold ball, you can dismiss it entirely from the equation.

The only question now is if you drew a ball from box 1 or 2.

So, its a 50/50 chance.

Not that hard.

please be joking

you have 5 words to explain how i'm wrong, otherwise you're trolling.
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dbf50 posted...

wanted to vote 2/5 but since that wasn't even a choice...

6 balls

3 balls are gold

take away 1 gold ball, draw again

5 balls

2 balls are gold

2/5

Your second ball has to be from the same box as the first ball.
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HINT #1:

You choose a box at random, and then a ball from that box at random. And the second pull must be from the same box. (This isn't new info, just a rephrasing of the rules)

HINT #2:

If your first ball is a gold ball, there is a 2/3 chance that you obtained it from box 1.
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7
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This is literally the "two children" question but with gold and silver balls instead of male and female kids.
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Ive got two balls for you.
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But this question is objectively easy, its 1/2

CirocObama posted...

i was thinking 1/2, but im assuming there's something I'm missing.

You've already picked a gold ball, so you know the next ball is either going be silver, or gold.

But I guess it's 2/3

The secret lies in this question:

Which gold ball did you pick?
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HINT #3:

Don't look at the first pull as you picking a box and that's it. This isn't the case. Look at the first pull as you picking one of the six balls.
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@Rika_Furude posted...

Choco posted...

Rika_Furude posted...

50/50

You could only have drawn a ball from box 1 or 2. Since box 3 does not have a gold ball, you can dismiss it entirely from the equation.

The only question now is if you drew a ball from box 1 or 2.

So, its a 50/50 chance.

Not that hard.

please be joking

you have 5 words to explain how i'm wrong, otherwise you're trolling.

5 words aren't enough because you have a fundamental misunderstanding of stochastics

if you draw a ball from the gold+silver box, the chance of getting the gold ball is only 50%

if you draw a ball from the 2gold box, the chance of getting a gold ball is 100%

the boxes themselves have equal probability when we don't know the result of drawing, but since we know a gold ball was drawn, we have more information and know it was more likely the 2gold box

tl;dr: see the post tc made above this one
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An explicit explanation.

Let's draw numbers on the balls to track them, shall we?

Box 1, Gold 1 = #1
Box 1, Gold 2 = #2
Box 2, Gold = #3
Box 2, Silver = #4
Box 3, Silver 1 = #5
Box 3, Silver 2 = #6

You choose a box at random. This is 1/3 probability. You choose a ball at random. This is 1/2 probability (within that box).

Thus, you have a 1/6 probability of pulling any specific ball here.

HOWEVER. The rules state YOU PULLED A GOLD BALL. Yes? Guaranteed, you are holding either ball #1, #2, or #3 in your hand. AND they all had equal probability of occurring. Thus, 1 out of 3 odds each.

So the odds you are holding each ball right now is:

#1 = 1/3
#2 = 1/3
#3 = 1/3
#4 = 0 (we did not pull a silver ball)
#5 = 0 (we did not pull a silver ball)
#6 = 0 (we did not pull a silver ball)

Everyone follow so far?

Let's look at what happens in each of the three cases. GIVEN that you know you pulled a gold ball:

1/3 times you Pulled #1: Your next pull is guaranteed to be #2, also a gold ball. = 100% gold
1/3 times you Pulled #2: Your next pull is guaranteed to be #1, also a gold ball. = 100% gold
1/3 times you Pulled #3: Your next pull is guaranteed to be #4, a silver ball. = 0% gold

Let's add these up.

(1/3 * 100%) + (1/3 * 100%) + (1/3 * 0%) = (1/3 * 1.0) + (1/3 * 1.0) + (1/3 * 0.0) = 2/3
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Crazy that some said 50/50
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2/3
Silver ball box doesn't factor in.
So it's just a box with 2 gold, and one with a gold and silver, so 4 balls remain with one being silver. You removed one gold ball, leaving 2 gold and a silver left in the boxes. With no other info, the chances of getting another gold ball are 2/3.
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KogaSteelfang posted...

2/3
Silver ball box doesn't factor in.
So it's just a box with 2 gold, and one with a gold and silver, so 4 balls remain with one being silver. You removed one gold ball, leaving 2 gold and a silver left in the boxes. With no other info, the chances of getting another gold ball are 2/3.

Right answer, wrong reason. Youre pulling from the same box the second time
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RedZaraki posted...

An explicit explanation.

Let's draw numbers on the balls to track them, shall we?

Box 1, Gold 1 = #1
Box 1, Gold 2 = #2
Box 2, Gold = #3
Box 2, Silver = #4
Box 3, Silver 1 = #5
Box 3, Silver 2 = #6

You choose a box at random. This is 1/3 probability. You choose a ball at random. This is 1/2 probability (within that box).

Thus, you have a 1/6 probability of pulling any specific ball here.

HOWEVER. The rules state YOU PULLED A GOLD BALL. Yes? Guaranteed, you are holding either ball #1, #2, or #3 in your hand. AND they all had equal probability of occurring. Thus, 1 out of 3 odds each.

So the odds you are holding each ball right now is:

#1 = 1/3
#2 = 1/3
#3 = 1/3
#4 = 0 (we did not pull a silver ball)
#5 = 0 (we did not pull a silver ball)
#6 = 0 (we did not pull a silver ball)

Everyone follow so far?

Let's look at what happens in each of the three cases. GIVEN that you know you pulled a gold ball:

1/3 times you Pulled #1: Your next pull is guaranteed to be #2, also a gold ball. = 100% gold
1/3 times you Pulled #2: Your next pull is guaranteed to be #1, also a gold ball. = 100% gold
1/3 times you Pulled #3: Your next pull is guaranteed to be #4, a silver ball. = 0% gold

Let's add these up.

(1/3 * 100%) + (1/3 * 100%) + (1/3 * 0%) = (1/3 * 1.0) + (1/3 * 1.0) + (1/3 * 0.0) = 2/3

I am so confused.
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All that math matters if you're numbering the balls. However, what if it's the color/material that's important?

The image in OP clearly says probability of pulling another gold ball, which gold ball it is DOESN'T MATTER IMO.

Both 1/2 and 2/3 are correct, depending on how you interpret the wording.
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@Rika_Furude whatsa matter afraid to respond?
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dbf50 posted...

wanted to vote 2/5 but since that wasn't even a choice...

6 balls

3 balls are gold

take away 1 gold ball, draw again

5 balls

2 balls are gold

2/5

Reading comprehension problem or just trolling? I hope it's the latter.
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I voted 1/2 but it might be 2/3 now that I am reading some more comments. I thought it might be a trick like the Monty Hall problem, but I didn't see how it could be like that.
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RedZaraki posted...

An explicit explanation.

Let's draw numbers on the balls to track them, shall we?

Box 1, Gold 1 = #1
Box 1, Gold 2 = #2
Box 2, Gold = #3
Box 2, Silver = #4
Box 3, Silver 1 = #5
Box 3, Silver 2 = #6

You choose a box at random. This is 1/3 probability. You choose a ball at random. This is 1/2 probability (within that box).

Thus, you have a 1/6 probability of pulling any specific ball here.

HOWEVER. The rules state YOU PULLED A GOLD BALL. Yes? Guaranteed, you are holding either ball #1, #2, or #3 in your hand. AND they all had equal probability of occurring. Thus, 1 out of 3 odds each.

So the odds you are holding each ball right now is:

#1 = 1/3
#2 = 1/3
#3 = 1/3
#4 = 0 (we did not pull a silver ball)
#5 = 0 (we did not pull a silver ball)
#6 = 0 (we did not pull a silver ball)

Everyone follow so far?

Let's look at what happens in each of the three cases. GIVEN that you know you pulled a gold ball:

1/3 times you Pulled #1: Your next pull is guaranteed to be #2, also a gold ball. = 100% gold
1/3 times you Pulled #2: Your next pull is guaranteed to be #1, also a gold ball. = 100% gold
1/3 times you Pulled #3: Your next pull is guaranteed to be #4, a silver ball. = 0% gold

Let's add these up.

(1/3 * 100%) + (1/3 * 100%) + (1/3 * 0%) = (1/3 * 1.0) + (1/3 * 1.0) + (1/3 * 0.0) = 2/3

Okay I believe this is correct. Another way to think of the final answer is to say that if one were to run this experiment thousands of times, they would pull out another gold ball 2/3 of the time that the first one they pull out is gold.
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About 1/tree fiddy

g980 posted...

KogaSteelfang posted...

2/3
Silver ball box doesn't factor in.
So it's just a box with 2 gold, and one with a gold and silver, so 4 balls remain with one being silver. You removed one gold ball, leaving 2 gold and a silver left in the boxes. With no other info, the chances of getting another gold ball are 2/3.

Right answer, wrong reason. Youre pulling from the same box the second time

I know it's the same box, the only balls left are 2 gold and 1 silver. The odds you're pulling out another gold ball, even from the same box is still 2/3. There's no way to know which box you're pulling from until after you pull the second ball, so you just have to look at the odds of the remaining balls and not the boxes.
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2/3 this is a variant of the Monty Hall problem
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FL81 posted...

2/3 this is a variant of the Monty Hall problem

No it is not.

It's a variant of the Boy or Girl paradox.

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
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Sativa_Rose posted...

RedZaraki posted...

An explicit explanation.

Let's draw numbers on the balls to track them, shall we?

Box 1, Gold 1 = #1
Box 1, Gold 2 = #2
Box 2, Gold = #3
Box 2, Silver = #4
Box 3, Silver 1 = #5
Box 3, Silver 2 = #6

You choose a box at random. This is 1/3 probability. You choose a ball at random. This is 1/2 probability (within that box).

Thus, you have a 1/6 probability of pulling any specific ball here.

HOWEVER. The rules state YOU PULLED A GOLD BALL. Yes? Guaranteed, you are holding either ball #1, #2, or #3 in your hand. AND they all had equal probability of occurring. Thus, 1 out of 3 odds each.

So the odds you are holding each ball right now is:

#1 = 1/3
#2 = 1/3
#3 = 1/3
#4 = 0 (we did not pull a silver ball)
#5 = 0 (we did not pull a silver ball)
#6 = 0 (we did not pull a silver ball)

Everyone follow so far?

Let's look at what happens in each of the three cases. GIVEN that you know you pulled a gold ball:

1/3 times you Pulled #1: Your next pull is guaranteed to be #2, also a gold ball. = 100% gold
1/3 times you Pulled #2: Your next pull is guaranteed to be #1, also a gold ball. = 100% gold
1/3 times you Pulled #3: Your next pull is guaranteed to be #4, a silver ball. = 0% gold

Let's add these up.

(1/3 * 100%) + (1/3 * 100%) + (1/3 * 0%) = (1/3 * 1.0) + (1/3 * 1.0) + (1/3 * 0.0) = 2/3

Okay I believe this is correct. Another way to think of the final answer is to say that if one were to run this experiment thousands of times, they would pull out another gold ball 2/3 of the time that the first one they pull out is gold.

Probability gets wonky when you add conditions.

Like, for example, if I were to say how many heads would you get on average in a single coin flip. The answer is 1/2 except you can't flip a half of a head so it's not a real result, but that's how the math works out.

People tend to think about these problems as tangible things with a sequence of events within a specific occurrence and not as a more general abstract math problem.

You could repeat this exercise with 1 million extra boxes of 2 silver balls and the result is unchanged.
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Tyranthraxus posted...

Sativa_Rose posted...

RedZaraki posted...

An explicit explanation.

Let's draw numbers on the balls to track them, shall we?

Box 1, Gold 1 = #1
Box 1, Gold 2 = #2
Box 2, Gold = #3
Box 2, Silver = #4
Box 3, Silver 1 = #5
Box 3, Silver 2 = #6

You choose a box at random. This is 1/3 probability. You choose a ball at random. This is 1/2 probability (within that box).

Thus, you have a 1/6 probability of pulling any specific ball here.

HOWEVER. The rules state YOU PULLED A GOLD BALL. Yes? Guaranteed, you are holding either ball #1, #2, or #3 in your hand. AND they all had equal probability of occurring. Thus, 1 out of 3 odds each.

So the odds you are holding each ball right now is:

#1 = 1/3
#2 = 1/3
#3 = 1/3
#4 = 0 (we did not pull a silver ball)
#5 = 0 (we did not pull a silver ball)
#6 = 0 (we did not pull a silver ball)

Everyone follow so far?

Let's look at what happens in each of the three cases. GIVEN that you know you pulled a gold ball:

1/3 times you Pulled #1: Your next pull is guaranteed to be #2, also a gold ball. = 100% gold
1/3 times you Pulled #2: Your next pull is guaranteed to be #1, also a gold ball. = 100% gold
1/3 times you Pulled #3: Your next pull is guaranteed to be #4, a silver ball. = 0% gold

Let's add these up.

(1/3 * 100%) + (1/3 * 100%) + (1/3 * 0%) = (1/3 * 1.0) + (1/3 * 1.0) + (1/3 * 0.0) = 2/3

Okay I believe this is correct. Another way to think of the final answer is to say that if one were to run this experiment thousands of times, they would pull out another gold ball 2/3 of the time that the first one they pull out is gold.

Probability gets wonky when you add conditions.

Like, for example, if I were to say how many heads would you get on average in a single coin flip. The answer is 1/2 except you can't flip a half of a head so it's not a real result, but that's how the math works out.

People tend to think about these problems as tangible things with a sequence of events within a specific occurrence and not as a more general abstract math problem.

You could repeat this exercise with 1 million extra boxes of 2 silver balls and the result is unchanged.

Exactly.

Here's an example:

You flip a coin. The first result is Heads. What is the probability that the second flip is also heads?

The answer is still 1/2. Every single flip is 1/2.

It doesn't matter that your previous flip was heads. Future flips do not tend to be more likely tails because you got heads already. They are all equally heads and tails forever.

All probability means, is that given an INFINITE dataset, the result set TRENDS towards 1/2 heads and 1/2 tails.
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If it was infinite, it would be exactly 0.5 each. No need to say trend.
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