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

[Socialhealer]

Im sitting with 1 box in front of me. That could be box A or box B. I have 1 gold ball in my hand. I reach into box A or B (Don't know which). The remaining ball is either gold from box A. Or silver from box b. That is 2 possible options.

ok answer this since you want to make it about boxes.

box A is a winner 100% of the time and you have twice the chance to pick it because it has twice the number of gold balls.

box B is a loser 100% of the time, but you only have 1 chance to pick it when you get the gold.

if you play 9 games with this logic, 6 times you'll have box A infront of you, 3 times you'll have box B infront of you.

what is the probability you'll pull a second gold ball?

[Deleted]

Thought I'd skim the first few pages for people to accept that 2/3s was correct answer, on page 39 there's still people arguing its 50% lol. Probably doesn't help that there are a lot of people early on saying its 2/3s but for completely faulty reasoning based on 3 balls left.

[Melchior]

This is why I like my earlier scenario, because making them people instead of balls forces the reader to understand that they are all unique. The two gold balls in the first box are different unique balls, which so many people are ignoring. They are thinking that you can only choose Box 1 or Box 2, when in reality, you can choose Box 1, Box 1, Box 2.

[Deldavala]

There isn't a chance for the first ball in the question you already have it drawn you are moving on to the second draw.

Read the text on the picture.

Step 1: Pick a box at random
Step 2: Pick a ball at random
Step 3: Observe the ball
Step 4: If ball is Golden what is the chance that the other ball in the same box is golden?

Meaning the step where you get the golden ball is crucial to calculating the probability.

If you dont understand this concept read up on https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox and come back to us.

[Delekii]

It doesn't matter which gold I took from box A as I already have already taken it out. I'm not choosing one of the 3 remaining balls, I'm choosing the only remaining ball that is in my chosen box.

It does matter which gold you took? Or rather, it matters that there are two different goods in that box, they are not interchangeable?

[Socialhealer]

There isn't a chance for the first ball in the question you already have it drawn you are moving on to the second draw.

we're not saying that were asking the probability of which box that first gold ball came from.

if you had 2 boxes, one had 10 gold balls, the other 1 gold 9 silver, and you picked a gold ball at random.

you win 10 out of 11 times, because the chance you picked the box with 10 gold balls is far far greater than picking the box with 1 gold out of 10.

i dunno why i'm trying to explain this if you don't get it i'm probably just confusing you lol.

[Delekii]

There isn't a chance for the first ball in the question you already have it drawn you are moving on to the second draw.

That you already have a gold ball gives zero information on WHICH gold ball you have. 66% is not the chance of getting a gold ball; it's the chance that you got a gold ball from the GG box given that you definitely got a gold ball.

[Osmin]

That you already have a gold ball gives zero information on WHICH gold ball you have. 66% is not the chance of getting a gold ball; it's the chance that you got a gold ball from the GG box given that you definitely got a gold ball.

You're almost there. Yes, there's a 66% chance you pulled a ball from the GG box, leaving a 33% chance you picked a gold ball from the GS box, and a 0% chance you pulled the silver ball from the GS box.

[Senen]

Probably doesn't help that there are a lot of people early on saying its 2/3s but for completely faulty reasoning based on 3 balls left.

Even now most 2/3 people arguing that the reason why it's 2/3 is that there's three gold balls, two of them in box A. And that it's about the balls, not the boxes.

You could have only one gold ball in box A, 200 gold balls and 200 silver balls in box B and 1000 silver balls in box C, you would still have 2/3 chances of having picked box A, given that you've picked a gold ball.

[kenoathcarn]

Ok we can ignore box 3, leaving us 2 boxes. Your'e saying there's a 50/50 chance you pick box one or two. That's true before you pick up the first gold ball. Once you do, and are holding a gold ball, then there is a 2/3 chance that you picked from the first box of all gold, because 2 of those balls are in that first box, and one of them is in the second. The fact that you're holding a gold ball already has cancelled one of the possibilities of the 2nd box.

see here:

https://www.mmo-champion.com/threads...1#post49776204

code:
https://pastebin.com/VhrqQuQw

run here:
https://www.tutorialspoint.com/compile_java_online.php

credits to haxartus (original post linked above)

This is the question modeled on a computer.

it simply discards any draws where the first draw was silver, and if the first draw was gold it logs the second draw.

Results from 1000000 simulation runs :
Golden:666568
Silver:333432

Gold is picked twice as often as silver. this isnt a mistake, you are literally arguing against Brute force Computation and the Consensus behind the Bertrand's box paradox

[Delekii]

Even now most 2/3 people arguing that the reason why it's 2/3 is that there's three gold balls, two of them in box A. And that it's about the balls, not the boxes.

You could have only one gold ball in box A, 200 gold balls and 200 silver balls in box B and 1000 silver balls in box C, you would still have 2/3 chances of having picked box A, given that you've picked a gold ball.

It's still about the chance of picking a given colour within each box.

It IS because of the number of balls, this wouldn't change if the ratios of balls changed. There is a 100% chance to take gold from a, 50% from b and 0% from C.

[Senen]

What? No. Simply no.

Yes. Simply yes.
The only relevant information about boxes is that box A have 100% gold balls, box B has 50% gold balls, box C has 0% gold ball. That's enough to say you are twice as likely to have picked box A than box B if you happen to pick a gold ball. Hence 2/3 probability.

[Delekii]

Yes. Simply yes.
The only relevant information about boxes is that box A have 100% gold balls, box B has 50% gold balls, box C has 0% gold ball. That's enough to say you are twice as likely to have picked box A if you happen to pick a gold ball.

I edited my reply, but there is no difference in expressing the result fractionally or as a percentage. Listing out potential results is a tool to help people conceptualise it, nothing more.

You could do the same exact thing with your example. It would justtake much longer to list the ball iterations.

[Senen]

I edited my reply, but there is no difference in expressing the result fractionally or as a percentage. Listing out potential results is a tool to help people conceptualise it, nothing more.

You could do the same exact thing with your example. It would justtake much longer to list the ball iterations.

Yes and no. Let's say box A has 1 gold ball, box B 1 gold ball and 1 silver ball and box C 2 silver balls.
I'm fairly confident that even among the 2/3 people you could find people arguing that in this example it's 1/2 because there's only two gold balls.
It's not. It's still 2/3, because even if there's only two gold balls so two cases to consider, those cases are not equiprobable.

[Sanguiris]

50%. Double gold is 100%, 1 gold and 1 silver is 50%, double silver is 0%. 100+50+0= 150/3=50

[Osmin]

see here:

https://www.mmo-champion.com/threads...1#post49776204

code:
https://pastebin.com/VhrqQuQw

run here:
https://www.tutorialspoint.com/compile_java_online.php

credits to haxartus (original post linked above)

This is the question modeled on a computer.

it simply discards any draws where the first draw was silver, and if the first draw was gold it logs the second draw.



Gold is picked twice as often as silver. this isnt a mistake, you are literally arguing against Brute force Computation and the Consensus behind the Bertrand's box paradox

...I don't understand the distinction you're trying to make here. How are we not saying the same thing? 6 balls, 3 of them golden, we know it's one of those 3 because it was given in the question that it is gold, and only 1 of those 3 resulting in the other ball being silver. Isn't that what I said, seems to be what you're saying, and giving us the same answer? Where's the argument?

[Nutrition]

Again, this is the logic that leads into, "It's 100% to be tails because that's what I got".

You weren't given the gold ball by magic. You weren't given the gold ball by someone with knowledge of the boxes. You DREW the gold ball. You cannot ignore the accompanying probability of drawing said gold ball just as you can't ignore the probability of the coin flip.

In fact, that's ALL the question is really asking. The wording is just approaching it from the other direction to prey on "common sense" ways of thinking about it. The question is really, "When you draw a gold ball, what is the probability that it came from any given box?"

Again, this is proven math. We've posted numerous mathematical proofs, computer simulations and thought experiments on the issue. The reality is that when you draw a gold ball, there's a 66% chance that it came from the G/G box. (And yes, when you draw a silver, there's a 66% chance it came from the S/S box.)

Yeah I thought it over and reread the question a few times. Your right it isn't 50/50 I was working off the two draws being independent.

[Deleted]

lol failed your own math you pull A you win, you pull B you win, you pull C you lose, you pull the ball not the box, read the damn question you pick a BALL not a box.

Nope. You pick the box, THEN you pull a ball. Heed your own advice, and learn to read.

[haxartus]

Let's think about something:

If the probability of picking each box is exactly the same, what is the difference, under a million runs, between having 2 boxes, as described by the OP (GG/GS), and having just one box with 2 golden balls and one silver ?

[Darsithis]

50%. Double gold is 100%, 1 gold and 1 silver is 50%, double silver is 0%. 100+50+0= 150/3=50

Unfortunately, no. I thought that, too, at first, but it's 2/3. The reason why is that you chose a box at random and then pulled a gold ball. It's more likely (since the ball was gold) that you selected the gold/gold box, which means it's more likely that the next ball will be gold. To think about it another way, what are the odds you chose the gold/gold box?

It's known as Bertram's Box Paradox.