>It took ChatGPT three tries to get this "riddle" for 10 year olds right

>It took ChatGPT three tries to get this "riddle" for 10 year olds right
Two times I had to tell it that it was wrong. There is no need to be scared of AI if you are even slightly more intelligent than the average moron out there.

 ChatGPT Wizard Shirt \$21.68 Beware Cat Shirt \$21.68
 ChatGPT Wizard Shirt \$21.68
1. 9 months ago
Anonymous

>a learning algorithm learned

Good job moron, are you going to show me how to flush my toilet next

• 9 months ago
Anonymous

God I hope he does the shits really pilling up at my place and I really haven't figured out this flushing thing.

2. 9 months ago
Anonymous

Did you use gpt4?

• 9 months ago
Anonymous

That shit can't code and therefore can't solve problems. It can only give a solution if one was already in the beliefs database, otherwise it will spit gaslighting.

3. 9 months ago
Anonymous

We first must determine the probability of having picked each box given that we've drawn a gold coin. This is a classic example of a problem requiring Bayes' Theorem.

Let's denote:

G1: The event of drawing one gold coin.
G2: The event of drawing another gold coin from the same box.

B1: The event of picking the box with two gold coins.
B2: The event of picking the box with two silver coins.
B3: The event of picking the box with one gold coin and one silver coin.

From the problem, we know that P(B1) = P(B2) = P(B3) = 1/3, since the box is picked randomly.

The probability of drawing a gold coin given that we've picked each box is:

P(G1|B1) = 2/2 = 1 (since there are two gold coins in the first box)
P(G1|B2) = 0/2 = 0 (since there are no gold coins in the second box)
P(G1|B3) = 1/2 = 0.5 (since there is one gold coin and one silver coin in the third box)

According to Bayes' Theorem, the probability of having picked each box given that we've drawn a gold coin is:

P(B1|G1) = P(G1|B1) * P(B1) / P(G1)
P(B3|G1) = P(G1|B3) * P(B3) / P(G1)

Where P(G1) is the total probability of drawing a gold coin:

P(G1) = P(G1 and B1) + P(G1 and B2) + P(G1 and B3)
= P(G1|B1) * P(B1) + P(G1|B2) * P(B2) + P(G1|B3) * P(B3)
= 1 * (1/3) + 0 * (1/3) + 0.5 * (1/3) = 1/3 + 0 + 1/6 = 1/2

So we have:

P(B1|G1) = 1 * (1/3) / (1/2) = 2/3
P(B3|G1) = 0.5 * (1/3) / (1/2) = 1/3

Given that the first coin drawn is gold, the probability that the box chosen is B1 is 2/3, and the probability that the box chosen is B3 is 1/3. It's impossible that we have chosen B2 since the first coin drawn is gold.

cont..

• 9 months ago
Anonymous

Now, given that we've picked B1 (two gold coins), the probability of drawing another gold coin is 1 (P(G2|B1) = 1), since there's only gold coins left in the box. And given that we've picked B3 (one gold and one silver coin), the probability of drawing another gold coin is 0 (P(G2|B3) = 0), because after picking a gold coin from B3, there's only a silver coin left in the box.

Therefore, the total probability of drawing another gold coin given that we've already drawn a gold coin is:

P(G2|G1) = P(G2 and B1|G1) + P(G2 and B3|G1) = P(G2|B1 and G1) * P(B1|G1) + P(G2|B3 and G1) * P(B3|G1) = 1 * 2/3 + 0 * 1/3 = 2/3.

-----------------------------------------

Can any mathgay confirm? My understanding is the answer is 50:50. Why does GPT4 shit the bed on this?

• 9 months ago
Anonymous

[...]
The correct answer is indeed 1/2, because you aren't asked what is the probability of picking a random box, then getting gold, then getting silver. First you took gold from a box, then you want to know what is the upcoming probability that it is silver. So either you took a gold from the GG box, and it's 0, or you took it in the GS box, and it's 1. Since you picked the box at random initially, you have 1/2 chance of being left with either box since possibility 3 is fake. It's ((1/3) / (2/3)) for GG and ((1/3) / (2/3)) for GS.

Let me end the thread very quickly with an explanation that even brainlets like you will understand. Imagine that the box in the middle has one gold ball but 100,000,000,000,000,000,000,000,000,000,000,000 silver balls. You pick a box at random, then you pick a gold ball. Now you start to think “huh.. it would be really darn lucky to pick the gold ball out of the box that has 99.99999999999999% chance of giving me silver, so I probably picked the box that only has gold, which will give me a gold ball every single time that I pick it.” And you would be correct. The same reasoning applies in the original problem. Box 1 gives you gold 100% of the time, and Box 2 gives you gold 50% of the time, so you’re twice as likely to pick a gold ball from box 1.

2x + x = 1
x = 1/3

• 9 months ago
Anonymous

Congrats, you are illiterate.

• 9 months ago
Anonymous

>picks gold ball
>”hurrr there’s a 50% chance I got it from the big ol’ box with a million silvers I reckon hurr durrrr
PLEASE bet money, I’m begging you. We’ll arrange the problem in real life. We’ll keep it simple and add 100 small silver beads in the second box, with 1 gold bead. Then we’ll pick the boxes randomly until we pick a gold bead randomly. I will bet \$10 against your \$1 that the next pick will be gold. And we can do this over and over until you would like to stop. If you’re right, then I should pay you \$10 half the time, and you should pay me \$1 half the time. Do you accept this deal? Let’s actually make this a PRACTICAL experiment and discover the truth. Put your money where your mouth is, I wanna get rich

• 9 months ago
Anonymous

You know it's an old problem, right? You could google it. OR you could read the solution that was outlined to you and try to understand it instead of sperging out despite being wronger than a Black person in a victorian play.
Hell, you could even numerically simulate the problem and you'd still get 50%.

• 9 months ago
Anonymous

Let’s run the experiment then. We’ll keep picking boxes at random and balls at random until we get a gold ball. Then we’ll do the bet. If the next ball is gold, you give me \$1. If the next ball is silver, I’ll give you \$1.25. We can repeat to infinity. Are you afraid?

• 9 months ago
Anonymous

I'm setting up the experiment now. I'm curious to see what happens.

• 9 months ago
Anonymous

Well it seems you were right. I concede. I used dice and everything, honestly unbelievable.

• 9 months ago
Anonymous

Also thanks. Awesome trick.

you should now try to understand the logic behind the results. It’s good to be logical for when you aren’t always able to simulate the experiment, but notice how easy it was to find the truth without even having to meaninglessly argue over each other. True math is a science because it can be tested and falsified. You cannot do the same with most “philosophy” or modern mathematical inventions like actual infinities, undefinable numbers, etc.

• 9 months ago
Anonymous

Also thanks. Awesome trick.

• 9 months ago
Anonymous

>We’ll arrange the problem in real life.
When and where? I'll take this bet but you fricking better bring money. Hell, I'll fly out to wherever you say if I have an ironclad contract that you'll pay out.

Or, you could stop being a dishonest israelite and actually try the experiment yourself with the boxes. Record it.

• 9 months ago
Anonymous

Now, given that we've picked B1 (two gold coins), the probability of drawing another gold coin is 1 (P(G2|B1) = 1), since there's only gold coins left in the box. And given that we've picked B3 (one gold and one silver coin), the probability of drawing another gold coin is 0 (P(G2|B3) = 0), because after picking a gold coin from B3, there's only a silver coin left in the box.

Therefore, the total probability of drawing another gold coin given that we've already drawn a gold coin is:

P(G2|G1) = P(G2 and B1|G1) + P(G2 and B3|G1) = P(G2|B1 and G1) * P(B1|G1) + P(G2|B3 and G1) * P(B3|G1) = 1 * 2/3 + 0 * 1/3 = 2/3.

-----------------------------------------

Can any mathgay confirm? My understanding is the answer is 50:50. Why does GPT4 shit the bed on this?

The correct answer is indeed 1/2, because you aren't asked what is the probability of picking a random box, then getting gold, then getting silver. First you took gold from a box, then you want to know what is the upcoming probability that it is silver. So either you took a gold from the GG box, and it's 0, or you took it in the GS box, and it's 1. Since you picked the box at random initially, you have 1/2 chance of being left with either box since possibility 3 is fake. It's ((1/3) / (2/3)) for GG and ((1/3) / (2/3)) for GS.

• 9 months ago
Anonymous

>From the problem, we know that P(B1) = P(B2) = P(B3) = 1/3, since the box is picked randomly.
This part is irrelevant. The problem states that you've already picked a box with a gold ball in it. The problem starts after that event.

• 9 months ago
Anonymous

Yes, just like Monty Hall also starts after the host has opened a door.

4. 9 months ago
Anonymous

The problem I'm running into is the wording, "You pick a box at random." The issue here is that this implies that you choose a box, and THEN pick a ball out of it. If you were selecting from 2 boxes without caring which box you pick, THEN the answer is 1/3, but I am convinced that based on the wording of the problem, that the answer is 1/2. I have provided a visual explanation of the other case.

5. 9 months ago
Anonymous
• 9 months ago
Anonymous

It's very simple, the probability of picking a box that satisfy all the criteria in the first step is [math]p_1[/math]. The probability that the next criteria is satisfied after the first step is then [math]p_2[/math] and the overall probability is [math]p_1cdot p_2[/math]. In this case the relevant numbers are [math]dfrac{2}{3}[/math] and [math]dfrac{1}{2}[/math] so their product is [math]dfrac{1}{3}[/math]

• 9 months ago
Anonymous

this is correct and no one can refute the logic of the answer without sounding like an illogical idiot

• 9 months ago
Anonymous

this is correct and no one can refute the logic of the answer without sounding like an illogical idiot

P1 is not considered, and is irrelevant to the question that has been asked.
>What is the probability that the next ball you take
>Next ball
>Next
>Ball
You are illiterate.

6. 9 months ago
Anonymous

chatgpt is a statistical parrot, it doesn't comprehend, it just puts words next to each other that make statistically sense to be next to each other in any given context.
but since coding languages are so limited compared to human languages, putting words next to each other that statistically make sense to be next to each other can yield often decent results, so if you're a code monkey whos only qualification is that you can write code, chatgpt is probably gonna take your job in the next 10-15 years

• 9 months ago
Anonymous

humans are also statistical parrots. But our training took millions of years.

• 9 months ago
Anonymous

You do realize that no part of your brain is anywhere near as intelligent as your whole brain working together, right? What do you think happens when you connect multiple (potentially specialized) instances of GPT-4 together in just the right way?

Never underestimate nonlinear emergence.

• 9 months ago
Anonymous

>when you connect multiple (potentially specialized) instances of GPT-4
You get garbage.
See: Nothing Forever

7. 9 months ago
Anonymous

Am I moronic or is it all of you? Once you pick out a gold ball, the box MUST BE one of the first two. Since one has silver and one has gold, it has to be 1/2, right? It is already certain that you did not pick from the 3rd box.

• 9 months ago
Anonymous

Statisticians may use pilpul to say otherwise but yes it's 50/50.

• 9 months ago
Anonymous

This is me

[...]
Let me end the thread very quickly with an explanation that even brainlets like you will understand. Imagine that the box in the middle has one gold ball but 100,000,000,000,000,000,000,000,000,000,000,000 silver balls. You pick a box at random, then you pick a gold ball. Now you start to think “huh.. it would be really darn lucky to pick the gold ball out of the box that has 99.99999999999999% chance of giving me silver, so I probably picked the box that only has gold, which will give me a gold ball every single time that I pick it.” And you would be correct. The same reasoning applies in the original problem. Box 1 gives you gold 100% of the time, and Box 2 gives you gold 50% of the time, so you’re twice as likely to pick a gold ball from box 1.

2x + x = 1
x = 1/3

I get it now. Thanks.

8. 9 months ago
Anonymous

It's a third.

We know we picked a gold ball, eliminating the box with only silver balls. The remaining balls (so the three that you didn't pick from the two boxes that contain gold balls) are two gold balls, and a silver ball. So there's three options you could pick from. The one we're looking for is the silver one, which is only one third of the total. The chance you'll get a silver ball is one third.

Now let's see if it works. if we picked a gold ball from the box with two gold balls, then what I'm saying (two gold balls and a silver ball left) is true, but the same is true if you'd picked from the mixed ball.

The chance of the next ball being grabbed form the same box being silver is one third.

• 9 months ago
Anonymous

A box contains 2 balls not 3. Its 50/50. The amount of balls in the non silver one doesnt matter, would have been 50% even if it had 10 golden ones

• 9 months ago
Anonymous

• 9 months ago
Anonymous

There are only two outcomes. Either the remaining ball is silver or it isn't. The nature of the problem's setup eliminates all other possible choices.

• 9 months ago
Anonymous

Weighted coin, gets 'em erry tiem

• 9 months ago
Anonymous

Here's the code, the answer is 1/3. Generalizing is left as an exercise for the reader

• 9 months ago
Anonymous

you dumb fricking nonce, you programmed a different problem

• 9 months ago
Anonymous

Point out the error, it's only ~300 characters with spaces. it should be very easy to audit and verify

• 9 months ago
Anonymous

first of all it's written like a moron. you must peruse BOT a lot. secondly, the code is written fine, there aren't any bugs. as i said, it's a solution to a different problem. the fact you can only assess your code as whether it has bugs or not is part of the issue. you're a blind code monkey who has no idea what it is he's actually programming. you clearly don't understand math.

• 9 months ago
Anonymous

modify it to model the real problem then, it should be easy enough, it's simply a matter of counting

• 9 months ago
Anonymous

i'm not modifying your homosexual ruby code. you need to stop thinking about programming as whether it has bugs or not, and more as a tool to solve a problem. and before you can use it to solve a problem, you need to understand what that problem is. enjoy your career in webdev, dumbass. stick to BOT

• 9 months ago
Anonymous

so what is the correct code then bruv? you should be able to do it right?

• 9 months ago
Anonymous

>he needs code to understand the problem
jesus christ. anon, this problem is easily solved on pen and paper. again, i'm trying to help you understand that programming is a tool. it's overkill for this problem, and is the wrong tool to use. just stop posting, you're embarrassing yourself.

• 9 months ago
Anonymous

so u can't code then huh? just as i thought. it's too late for u then, might as well just stop breathing

• 9 months ago
Anonymous

Don't be like that, his problem is not the reasoning nor the code. His problem is the little twist in the problem description in OP, i.e. an interpretation problem. You have to explain it to him slowly.

• 9 months ago
Anonymous

i can do that, but i can't be nice to him about it. here let me try again.

https://i.imgur.com/xhMOx35.png

Here's the code, the answer is 1/3. Generalizing is left as an exercise for the reader

hey, dumbfrick: if it's known you picked a gold ball, why are you including the box with two silvers as a possibility?

• 9 months ago
Anonymous

It's more than that actually. I have adjusted my code to be wrong in the way his is wrong, but added your condition that the silver-silver box is not considered. The simulation still gives 1/3.
The problem is that the problem statement does not actually include picking a gold ball at random, it forces you to pick the gold ball. If you pick at random, you are really doing the same thing as picking any one random ball from [g g g s] and discarding if it's s. In fact the statement says, effectively, "if the box was [g g] then remove either gold ball of your choice. If the box was [g s], remove the gold ball". The 'you pick a gold ball at random' is meant to confuse you because it says 'you pick at random'. But if the outcome is decided, the outcome is actually not random.

• 9 months ago
Anonymous

Briefly, the problem is that you did not normalize by the probability of getting gold in the situation setup.
This is what the other anons mean when they say it's a different problem that you're solving here. If you do rejection sampling as you do here, you are 2x as likely to randomly select the gold-gold box at random and not reject, than the gold-silver box. To do it right, you can reject the silver box, but then you must delete one gold ball from the remaining box (rather than randomly pick and reject). That's because the question says: you picked it at random AND IT WAS GOLD. This is the trick part of the question that's designed to frick you up/create debate/troll u.
You fell for it hook, line and sinker.
I wrote shit code on purpose (python) that is hopefully clearly readable to show you what the correct simulation looks like.

• 9 months ago
Anonymous

why r u skipping the 1st index?

• 9 months ago
Anonymous

oops, supposed to be 2. Doesn't change the result.

• 9 months ago
Anonymous

Your code is equivalent to the following

• 9 months ago
Anonymous

because you skip box 3 entirely and then randomly sample from the 2 remaining boxes. that's not what the problem is saying. the problem is about making 2 random choices and in your problem statement there is effectively only 1 choice being made. you're solving a different problem

• 9 months ago
Anonymous

>because you skip box 3 entirely
Yes, that is called rejection sampling. You take random samples from the whole state and ignore samples that are invalid (in this case, a sample is valid only if we end up picking a gold ball, which is only possible if we pick one of the other two boxes).
The problem that I tried to explain to you a few times so far but that you don't seem to grasp, and it's why you get the (wrong) 1/3 answer, is that you don't do two pickings, you only do one.
YOU ONLY PICK THE BOX AT RANDOM
THE GOLD BALL IS ALWAYS PICKED AS GOLD NO MATTER WHAT BOX YOU PICK AND THUS NOT RANDOM
To be more accurate you could say that if you have the [gg] box your probs are [0.5 0.5] but if it's the [gs] box it's [1 0]. But in your simulations, you sample [0.5 0.5] every time.

• 9 months ago
Anonymous

i don't think you understand rejection sampling at all because all you're doing is just counting how many times "random.randint(0, 1)" is equal to "1"

• 9 months ago
Anonymous

The one who doesn't understand is you.
The reason why the problem reduces to counting how many times we picked box 1 is because that's exactly what the probability tree maps to.

• 9 months ago
Anonymous

ya i don't think that's right. my code is clearly correct

• 9 months ago
Anonymous

"the anon above me is a homosexual"
Well, my code is bug free. It must be correct that you're a homosexual. Stop being a homosexual.
>Pwn2h4

• 9 months ago
Anonymous

> count(random.randint(0,1) == 1)

that's your code with extra steps, you're a moron if you think that's the actual problem statement

• 9 months ago
Anonymous

No my code is
"the poster who wrote >15457003 is a homosexual"
Sword swallowing, cum guzzling, fudge packing homosexual. The code is undeniable.

• 9 months ago
Anonymous

i think you're just extra moronic, it's better if you just stop breathing, it's better that way for everyone

• 9 months ago
Anonymous

"the user who wrote >15457015 is an incel"
Man, it feels good to be right with infallible code. I'm sure your high paying webdev salary will get you laid some day, pal. You'll just have to pay for the sex.

• 9 months ago
Anonymous

It is correctly solving a problem not asked by the OP.

> count(random.randint(0,1) == 1)

that's your code with extra steps, you're a moron if you think that's the actual problem statement

The whole point of the problem in OP is that it's not actually a probability question, it's a problem statement interpretation question. It's a trick.

• 9 months ago
Anonymous

no it's not, the code i wrote is exactly what is being asked and any moron that thinks it's a trick to count "random.randint(0,1) == 1" is obviously brain damaged

• 9 months ago
Anonymous

I give up, you're too moronic for this. The other anon was right.

• 9 months ago
Anonymous

What is this relevant to?

getting the right answer. it's ok, not everyone can write code to simulate probabilities of events

• 9 months ago
Anonymous

So you're just trolling by pretending to be stupid? You could have said so to begin with.

• 9 months ago
Anonymous

• 9 months ago
Anonymous

• 9 months ago
Anonymous

He's not pretending.

• 9 months ago
Anonymous

>the problem is about making 2 random choices
Actually in the problem the choice is predetermined, not random. Your first pick will always be gold. That's why autistic people get it so wrong, they don't understand the context and attempt to solve it without understanding.

• 9 months ago
Anonymous

"the user who wrote >15457015 is an incel"
Man, it feels good to be right with infallible code. I'm sure your high paying webdev salary will get you laid some day, pal. You'll just have to pay for the sex.

u jelly?

• 9 months ago
Anonymous

Jealous of an incel homosexual who doesn't understand probability? Nah.

• 9 months ago
Anonymous

u sound jelly tho

• 9 months ago
Anonymous

What is this relevant to?

9. 9 months ago
Anonymous

g1 --> g2
g2 --> g1
g3 --> s
2/3

10. 9 months ago
Anonymous

It's either silver or it's gold. 50/50

• 9 months ago
Anonymous

the code does not lie, the answer is 1/3

• 9 months ago
Anonymous

>the code does not lie
so how does chatgpt lie?

• 9 months ago
Anonymous

chatgpt is a israeli psy-op so it couldn't not lie

11. 9 months ago
Anonymous

It's 50% you fricking tards, holy shit.

• 9 months ago
Anonymous

it's 1/3

12. 9 months ago
Anonymous

>It took ChatGPT three tries to get this "riddle" for 10 year olds right
Considering professors failed the Monty Hall problem, I'm not surprised.

13. 9 months ago
Anonymous

[...]

Thats not how it works moron, it doesn't learn as you use it.

14. 9 months ago
Anonymous

This issue between saying it is 1/3 vs 1/2 stems from parsing the problem. Either you parse it philosophically, or you parse it mathematically.

If you follow the steps outlined, you can either discard the results that don't conform to the conditional, or you just invert your result from then on, turning a first ball silver to a gold, thus inverting the next ball chosen so you can save time instead of resetting. This will, ultimately, yield an answer of 1/3. 2/3 of the time it will be a second gold ball.

Now, the fun part that will, undoubtedly, piss off all the autists. It can be argued that the conditional where you MUST draw a gold ball implies that there are only two boxes, and that the boxes are stacked.
>But anon, you need to number the balls, so g1, g2 and g1, s1.
You can't do that because, due to the conditional, s1 functionally does not exist at this time, you cannot draw it in your initial choice, it would violate the conditional. there never was a choice. you were bound by fate (the conditional) to draw a first ball gold. the die was cast, free will is an illusion, and epstein did not kill himself.

• 9 months ago
Anonymous

Even if s1 doesn't exist if you number the gold balls you are still more likely to pick g1 or g2 instead of g3

15. 9 months ago
Anonymous

This is a really dickish way of obscuring the three-doors "riddle"

16. 9 months ago
Anonymous

False, the middle box contains one gold and one silver.

• 9 months ago
Anonymous

Proof?

• 9 months ago
Anonymous

By inspection of OP

17. 9 months ago
Anonymous

It’s 1/3 because you pick the first ball AT RANDOM.

You have a 100% chance of getting a gold ball from the first box. You have a 50% chance of getting a gold ball from the second box. Therefore, if you have randomly selected a gold ball, it is twice as likely that you have selected the first box.

• 9 months ago
Anonymous

This assumes you're allowed to keep the first gold ball you picked instead of putting it back into the box. The correct answer (not making that assumption) is 1/4.

• 9 months ago
Anonymous

If it doesn't say you put it back then why would you assume you put it back? The ball is taken and, without further assumptions, remains taken.

18. 9 months ago
Anonymous

Midwit, your intellectual input on AI is useless. So I will just make a joke:
You are God. You pick a country at random. Without looking you put your hand in and pull out a person. It is a smelly Black person. What are the chances you have picked a country in Africa?

19. 9 months ago
Anonymous

Everyone who says it's 50-50 because of some clever twist in the wording is conveniently unable to actually explain what clever twist that is. It's 1/3. It's a known problem and it was solved ages ago. The only real trick is it usually asks for the inverse, but then the answer would be 2/3 and still not 50-50.

• 9 months ago
Anonymous

>I don't know what a conditional is
It's okay, plenty of other morons ITT. The problem as stated in the op is not Bertrand's box (nor it's inversion). In Bertrand's box the conditional is that IF you picked gold, etc. Here there is no IF—you DID pick gold. Full stop. That changes the probability outcome.

• 9 months ago
Anonymous

See

[...]
Let me end the thread very quickly with an explanation that even brainlets like you will understand. Imagine that the box in the middle has one gold ball but 100,000,000,000,000,000,000,000,000,000,000,000 silver balls. You pick a box at random, then you pick a gold ball. Now you start to think “huh.. it would be really darn lucky to pick the gold ball out of the box that has 99.99999999999999% chance of giving me silver, so I probably picked the box that only has gold, which will give me a gold ball every single time that I pick it.” And you would be correct. The same reasoning applies in the original problem. Box 1 gives you gold 100% of the time, and Box 2 gives you gold 50% of the time, so you’re twice as likely to pick a gold ball from box 1.

2x + x = 1
x = 1/3

>picks gold ball
>”hurrr there’s a 50% chance I got it from the big ol’ box with a million silvers I reckon hurr durrrr
PLEASE bet money, I’m begging you. We’ll arrange the problem in real life. We’ll keep it simple and add 100 small silver beads in the second box, with 1 gold bead. Then we’ll pick the boxes randomly until we pick a gold bead randomly. I will bet \$10 against your \$1 that the next pick will be gold. And we can do this over and over until you would like to stop. If you’re right, then I should pay you \$10 half the time, and you should pay me \$1 half the time. Do you accept this deal? Let’s actually make this a PRACTICAL experiment and discover the truth. Put your money where your mouth is, I wanna get rich

• 9 months ago
Anonymous

see

>We’ll arrange the problem in real life.
When and where? I'll take this bet but you fricking better bring money. Hell, I'll fly out to wherever you say if I have an ironclad contract that you'll pay out.

Or, you could stop being a dishonest israelite and actually try the experiment yourself with the boxes. Record it.

• 9 months ago
Anonymous

You can try the experiment, yourself. I and

Well it seems you were right. I concede. I used dice and everything, honestly unbelievable.

already know what will happen. I recommend making a YouTube video to prove everyone wrong. But it won’t happen lol.

• 9 months ago
Anonymous

Run the experiment that we’re talking about. The middle box has 99.999999% silver balls. In other words, that box will virtually never give you gold balls. Every time you get a gold ball, you will instinctually think “I must have picked the first box.” It doesn’t matter if you say “If I pick a gold ball” or “I have picked a gold ball.” You picked a box, you picked a ball, and it was gold. There’s a 99.99999% change that the next ball will be gold as well. It’s so simple you don’t even have to test it

i ran the experiment in real life using shoe boxes and coins. did it 172 times. 80 times it came up silver. unlikely come from 1/3 chance, but likely from 1/2 chance. don't write code, use real life dummy

• 9 months ago
Anonymous

now take a video and post proof. No verbal arguments necessary

• 9 months ago
Anonymous

Ah, so you've managed to fool yourself into believing that your interpretation of a semantic ambiguity is correct even though it directly conflicts with what is explicitly stated.

The entire scenario is hypothetical. The "if" is implicit. "You have picked a gold ball at random" and "if the ball you picked randomly is gold" are functionally identical. You did pick gold - randomly. You need to understand semantics before you can try to understand conditionals.

• 9 months ago
Anonymous

>The "if" is implicit.
proof?
>"You have picked a gold ball at random" and "if the ball you picked randomly is gold" are functionally identical
wrong.

• 9 months ago
Anonymous

Run the experiment that we’re talking about. The middle box has 99.999999% silver balls. In other words, that box will virtually never give you gold balls. Every time you get a gold ball, you will instinctually think “I must have picked the first box.” It doesn’t matter if you say “If I pick a gold ball” or “I have picked a gold ball.” You picked a box, you picked a ball, and it was gold. There’s a 99.99999% change that the next ball will be gold as well. It’s so simple you don’t even have to test it

• 9 months ago
Anonymous

>proof?
A working understanding of semantics. Me saying "you picked a gold ball" is not a statement of fact. It is asking you to consider a hypothetical scenario in which you picked a gold ball. That is exactly the same as "if you picked a gold ball", which is, after all, also asking you to consider a hypothetical scenario in which you picked a gold ball.

It's also moot because if you ask someone to select a box at random and take a ball from it, and that ball happens to be gold, and you ask them at that point (at which "you have picked a gold ball" actually would a statement of fact) what the probability of getting a silver ball is, the fact that the gold ball was obtained from a random box still renders the answer 1/3. That's the part you're (deliberately?) overlooking. You may think that if something is implicit that leaves it open to interpretation, but the word "random" is not open to interpretation here. You acquired your ball randomly. That's the condition you have to understand to properly apply conditional probability here.

• 9 months ago
Anonymous

>hypothetical

now take a video and post proof. No verbal arguments necessary

or you could just do it yourself. let's be real. you won't accept video proof since you'll just assume it's edited or that the boxes or rigged or some shit.

• 9 months ago
Anonymous

Its logically impossible to perform a selection both randomly and non-randomly simultaneously. The implicit "if" is required, else the entire thing is nonsense.

• 9 months ago
Anonymous

>Its logically impossible to perform a selection both randomly and non-randomly simultaneously
Wrong babe

• 9 months ago
Anonymous

Ah, so you've managed to fool yourself into believing that your interpretation of a semantic ambiguity is correct even though it directly conflicts with what is explicitly stated.

The entire scenario is hypothetical. The "if" is implicit. "You have picked a gold ball at random" and "if the ball you picked randomly is gold" are functionally identical. You did pick gold - randomly. You need to understand semantics before you can try to understand conditionals.

>The "if" is implicit.
proof?
>"You have picked a gold ball at random" and "if the ball you picked randomly is gold" are functionally identical
wrong.

Seems to be a fair critique. If we're starting from a single box that contains either 1 gold ball or 1 silver ball, and there's no logical connection that would weight the gold box more likely than the silver box, then clearly it's 1/2 either way.

• 9 months ago
Anonymous

>and there's no logical connection that would weight the gold box more likely than the silver box, then clearly it's 1/2 either way.
Yes, but the trick is, there is actually such a logical connection, and it is the knowledge that the box was selected randomly, then the ball was obtained, and you're picking from the same box again.

In other words, if you come into the room and you see someone in front of two boxes with a gold ball in his hand, and you pick a box, you'll be right 1/2 the time. But if you ask that person "which box did you take that from" you can confidently be correct 2/3 of the time.

• 9 months ago
Anonymous

Makes sense. Then it's the condition "same" box that makes it impossible to avoid the logical connection.

• 9 months ago
Anonymous

> In other words, if you come into the room and you see someone in front of two boxes with a gold ball in his hand, and you pick a box, you'll be right 1/2 the time.
only if you include the third box. With two boxes, there are 3 gold balls and 1 silver, so the probability would actually be 3/4

• 9 months ago
Anonymous

One ball is in the hands of the person and you would be picking a box to pick from first and then taking a ball from that.

But also for us to reduce it down to two boxes, someone actually has to check the boxes to remove the all-silver one, which is exactly the mistake 50-50 people are making: they are trying to select for only the gold boxes, reasoning that if it happened, it could not have happened otherwise.

>hypothetical
[...]
or you could just do it yourself. let's be real. you won't accept video proof since you'll just assume it's edited or that the boxes or rigged or some shit.

We aren't talking about reality. Show me the box. Show me the ball. It's not here. We are instructed to imagine it.

• 9 months ago
Anonymous

kek. which must be why you're scared to do the experiment for real. you know deep down that reality can reject your schizo math babble

• 9 months ago
Anonymous

"kek" indeed.

It's in the second person. Me reading it means it's about me. The mere fact of having a picture on BOT narrate to me that I picked a gold ball does not actually make that into reality. I did not, in reality, pick a gold ball. I am using my imaginative faculties to put myself in a hypothetical scenario in which that is the case. And, using those same faculties, I can reason that if I were to be in that scenario, under the conditions specified, I can reasonably state the probability asked to be 1/3. Which is what it would work out to in reality. Hypothetically.

• 9 months ago
Anonymous

>Which is what it would work out to in reality. Hypothetically.
dumbass

• 9 months ago
Anonymous

Alas, we can only discuss this hypothetical reality hypothetically, because it is not factual reality. Just as, hypothetically speaking, you could be correct. But in actual reality, you're not.

• 9 months ago
Anonymous

Not him but what if the question were "next ball you take from any box"?

• 9 months ago
Anonymous

5/9

• 9 months ago
Anonymous

Bzzzt. 3/5. You've assumed that the outline of a box makes it more or less likely to choose a ball in that box.

• 9 months ago
Anonymous

Why ask me, specifically, when I've been talking about semantics first and foremost?

The all-gold box will always yield gold and the all-silver box will always yield silver. But in 1/3 of the cases, the mixed box will yield silver and in 2/3 of the cases, it can go either way. It seems to me that

5/9

is correct. You're not choosing a random ball. You're choosing a box to take a ball from.

• 9 months ago
Anonymous

>You're choosing a box to take a ball from.
No, I mean you choose 1 of the 5 remaining boxed balls with equal chance. Not you choose 1 of the 3 boxes with equal chance then you choose 1 of the 1 or 2 balls in that box with equal chance.

• 9 months ago
Anonymous

Yes, that's the problem. You should, in fact, do the latter.

• 9 months ago
Anonymous

Latter is the OP problem, not what we're talking about.

• 9 months ago
Anonymous

I've made a helpful graphic. Adding the probabilities is an exercise left for the reader. I hope it's not too taxing, although judging from some of the responses in this thread, you never know...

• 9 months ago
Anonymous

Oof.. Sorry to hear you're a moron. Better luck next time.

• 9 months ago
Anonymous

The baseless and unsupported opinions of addlepated siimpletons hold no value to me.

• 9 months ago
Anonymous

Nevertheless you evaluate stupidity as a pleasure. Good luck.

• 9 months ago
Anonymous

True, I do delight in your antics, why do you think I'm here?

• 9 months ago
Anonymous

My guess is that you're annoyed by something. Something you just can't let go of

• 9 months ago
Anonymous

Do you prefer to be thought of as amusingly misguided or annoyingly dense? To be fair, you do rather vacillate between them.

• 9 months ago
Anonymous

This is right. Anyone speaking to the contrary is illiterate or intentionally arguing what they think to be (incorrect) semantics, but the wording is unambiguous.

according to this methodology, you cannot test such an answer, because the conditions of picking the box and the ball are in some realm where probability does not apply, and you could pick the second box even if you only had a %0.00000000000000001 chance of doing so. In other words, it is a completely impractical and therefore nonsensical interpretation. In the real world, if you set up the boxes and the balls, and pick a gold ball, you should bet money that the next ball you pick will be gold. Or you can just continue to be a moron and lose your money

• 9 months ago
Anonymous

No, all you'd have to do to "test" the answer by its own assumptions is sum a million random bits and divide by two.

• 9 months ago
Anonymous

This is right. Anyone speaking to the contrary is illiterate or intentionally arguing what they think to be (incorrect) semantics, but the wording is unambiguous.

• 9 months ago
Anonymous

The wording IS unambiguous when it states the box was selected randomly, yes.

• 9 months ago
Anonymous

Ah, so you've managed to fool yourself into believing that your interpretation of a semantic ambiguity is correct even though it directly conflicts with what is explicitly stated.

The entire scenario is hypothetical. The "if" is implicit. "You have picked a gold ball at random" and "if the ball you picked randomly is gold" are functionally identical. You did pick gold - randomly. You need to understand semantics before you can try to understand conditionals.

there's only two boxes. it's 50%. that's clearly the answer the question is demanding. don't think of it in problems more generally or problems you know but what the question wants from you. you seem hung up on something you already know and assuming that's what this is. it's not. you need to put aside ego and obey the test.

• 9 months ago
Anonymous

>there's only two boxes. it's 50%.
Same as Monty Hall, right?

• 9 months ago
Anonymous

Different problem, different solution

• 9 months ago
Anonymous

There's only two doors, why isn't it 50%? If you know the answer, then you should also know why what you said before is a huge crock of shit.

• 9 months ago
Anonymous

Different problem, different solution.

• 9 months ago
Anonymous

You don't know what you're talking about.

Latter is the OP problem, not what we're talking about.

The only condition that was changed was that you are allowed to pick any box. You still have to pick a box. What you're asking is what if the remaining balls were all dumped into the same box and you take one from there.

• 9 months ago
Anonymous

OP's problem isn't the monty hall, nor is it bertrands box. Unbelievable that you don't comprehend this.

• 9 months ago
Anonymous

It is Bertrand's Box, in fact. And Bertrand's Box is furthermore similar enough to Monty Hall that some of the same conclusions carry over. Especially when someone is foolish enough to reason that two choices = 50-50. If you understand the Monty Hall problem (or weighted coins for that matter) you should know what a laughing stock you are.

No, the condition here [...] is
>next ball you take from any box?
It doesn't matter what box the balls are in. If you pick 1 of 5 balls at random from 3 boxes it's the same as picking 1 of 5 balls at random from 1 box (or from 5, etc.).

If the balls remain where they are then it necessarily follows that you must pick a box first in order to pick a ball from it.

https://i.imgur.com/N7o007Z.jpg

You pick a ball at random. What is the probability that you pick the ball in the box?

I get that the answer you're fishing for is 1/3 but how do you "pick" a ball and then learn it happens to be in a box? Are the balls numbered and we're required to state a number? If you're going to come up with an unusual selection process you can make the answer anything. But if we assume the selection process remains unchanged, you still have to pick a box first to pick a ball from it. Again, you're relying on semantic ambiguity to impose an idiosyncratic interpretation that ignores the materiality of the problem posed.

• 9 months ago
Anonymous

>If the balls remain where they are then it necessarily follows that you must pick a box first in order to pick a ball from it.
Lol, no. 5 pool balls on a table, 1-5. Is it necessary to pick "prime" first before you choose the 2 ball?

• 9 months ago
Anonymous

Oh, now they're on a table? Sure, let's just change everything according to your whims and then declare you the smartest person in the thread.

Actually the answer is 3/6 because the ball in your hand is also originally "from" a box and therefore still eligible :^) I am very smart, you see

>It is Bertrand's Box, in fact.
wrong.

I haven't been wrong about a single thing once in this thread (except above where I am being sarcastic, if someone wants to exploit *that* amibguity).

• 9 months ago
Anonymous

>I haven't been wrong about a single thing once in this thread
you're the same kind of egotistical jackass who's never admitted to being wrong once in his life.

• 9 months ago
Anonymous

Wrong again. I have been wrong many times before. That is why I can confidently state this is not one of them. Perhaps you should have a look in the mirror. After all, the only argument you seem capable of making is "no you're wrong".

• 9 months ago
Anonymous

>I have been wrong many times before.
prove it. tell us about a time you were wrong and humbly admitted it. bonus points if it's a mathematical statement you were wrong about

• 9 months ago
Anonymous

I don't understand your point. Label 5 balls 1-5. Pick one at random. Why do you imagine putting them into some array of boxes is any different than classifying them as prime or imprime?

• 9 months ago
Anonymous

Because, dear Anon, you're changing the problem to force the answer you want.
The original problem clearly implicitly defines "picking" as "reaching into a box and taking out a ball in order to identify it". So if you ask about the probability of taking a silver ball from any box, obviously people will tell you the probability of obtaining a silver ball following that same procedure. How else would you do it, when the balls are in boxes? Of course when you idiosyncratically redefine the word "pick" to instead refer to, I don't know, mentally assigning special status to one particular identifiable ball while leaving it undisturbed in its box, you will get a different answer. But that doesn't make you clever. It makes you a poor communicator. You're asking people to think about the existing problem yet expecting them to ignore part of the problem as stated.

>I have been wrong many times before.
prove it. tell us about a time you were wrong and humbly admitted it. bonus points if it's a mathematical statement you were wrong about

For one, I was wrong in assuming good faith in this conversation.

• 9 months ago
Anonymous

I don't disagree with what the OP says. We've been talking about a different problem for close to 48 hours. Why are you so obsessed with OP that you can't dislodge your own head from his butthole long enough to see that I agree with your opinion of how his butthole smells to you?

• 9 months ago
Anonymous

Anon, dearest, sweetheart, your "different problem" asks about picking a random ball from any box, given the same conditions as the OP problem. Note: the conditions stated in the OP still apply. Note also: the balls being in boxes is among them and is also acknowledged in the "different" problem. If you want the problem to be "what are the odds of picking a random ball from among five balls" then just ask that, but then leave out the boxes, because they do change things, even if you don't realise it. If you really don't disagree with the OP, you shouldn't disagree with me either.

• 9 months ago
Anonymous

five balls on a pool table
(1)(2)(3)(4)(5)
five balls in boxes <> mean = 2.5
[ (1)(2) ] [ (3)(4)(5) ]
five balls if prime
(1) (2)(3) (4) (5)

now pick a ball from each. how is it different in each case?

• 9 months ago
Anonymous

lol you wiener you put three balls in one box
Also, you can't just pretend we have x-ray vision here. You can say "there are five balls, pick one" but that is because you are only envisioning five balls in the abstract, and refusing to consider the implications of them being in boxes. You cannot "pick" a ball that you cannot physically access. You can't say "that one" and point to an opaque box and count it as picking a ball. Picking a ball means taking it from a box, and that places constraints on your choice.

• 9 months ago
Anonymous

> lol you wiener you put three balls in one box
Yes, If you distribute {1..5} by floor(x/2.5) then there will be 3 balls in the same box. Lol. It's how math works.
> a ball that you cannot physically access
> physically access
Just leaving this here.

• 9 months ago
Anonymous

>Yes, If you distribute {1..5} by floor(x/2.5) then there will be 3 balls in the same box.
Sure. Why are we doing that? You don't even grasp the original "different" problem and you already insist on moving on to different "different" problems.

• 9 months ago
Anonymous

Honestly that sounds like you criticizing other people for your own stupidity.

• 9 months ago
Anonymous

>Honestly that sounds like you criticizing other people for your own stupidity.
You misspelled "their"

Which one are you referring to now?

• 9 months ago
Anonymous

Lol it's pretty easy. Pick a ball 1-5. Doesn't matter if 235 are prime. Doesn't matter if 12 are less than x/2.

• 9 months ago
Anonymous

What are you babbling about, man

• 9 months ago
Anonymous

moron lol.

• 9 months ago
Anonymous

You said "read the original problem". Now you're saying "Pick a ball 1-5". No one was talking about that until you brought up pool tables. And what is
>Doesn't matter if 235 are prime. Doesn't matter if 12 are less than x/2.
even supposed to *mean*? Which 235? Which 12? Which x/2? What does any of this have to do with the problem?

• 9 months ago
Anonymous

Are you actually moronic?

• 9 months ago
Anonymous

No, I'm pretty sure you are, because any attempt you've made to explain yourself is gibberish.

• 9 months ago
Anonymous

>We've been talking about a different problem for close to 48 hours.

Not him but what if the question were "next ball you take from any box"?

>Not him but what if the question were "next ball you take from any box"?
>Posted not even 24 hours ago
>OP isn't even 48 hours old
I'm not taking math advice from you

• 9 months ago
Anonymous

You sound moronic. Good luck in life.

• 9 months ago
Anonymous

You sound defeated.

• 9 months ago
Anonymous

>I'm not taking math advice from you
Yes, very defeated.

• 9 months ago
Anonymous

>It is Bertrand's Box, in fact.
wrong.

• 9 months ago
Anonymous

>you're relying on semantic ambiguity to impose an idiosyncratic interpretation that ignores the materiality of the problem posed.
Didn’t know they made spergs that were this autistic. Updoots and reddit gold for you my good sir.

Take a ball. All three of them are filled with explosive powder, but only one has a working motion sensor hooked up to a primer. What is the probability that you fricking explode?

• 9 months ago
Anonymous

It's 100% because you're a malicious liar and actually just operating a remote kill switch. Oh, that's not stated in the problem? Well it's what we're talking about now. So if you were thinking 1/3, ha, wrong b***h. Outplayed.

• 9 months ago
Anonymous

No, the condition here

Not him but what if the question were "next ball you take from any box"?

is
>next ball you take from any box?
It doesn't matter what box the balls are in. If you pick 1 of 5 balls at random from 3 boxes it's the same as picking 1 of 5 balls at random from 1 box (or from 5, etc.).

20. 9 months ago
Anonymous

>mfers be thinking the hobbit is a history book because it says "in a hole in the ground there lived a hobbit"

21. 9 months ago
Anonymous

what if the silver ball in box 2 is actually able to change states depending on if it was selected first or not? If it is selected first, it is gold. If it is selected second, it is silver. Is the answer now 1/2?

• 9 months ago
Anonymous

Yes, because what you said is logically inconsistent so the answer can be anything

22. 9 months ago
Anonymous

Yes because it learnt from 10 years of morons trying to solve this question. Its not Artificial general intelligence is a really good copy cat

23. 9 months ago
Anonymous

You pick a ball at random. What is the probability that you pick the ball in the box?

• 9 months ago
Anonymous

1/3 per jpg

• 9 months ago
Anonymous

depends if you can open the box

24. 9 months ago
Anonymous

Latex test [math] frac{a}{b} [/math]

• 9 months ago
Anonymous

[math]kfrac{y}{s}cdots[/math]

25. 9 months ago
Anonymous

• 9 months ago
Anonymous

Almost nobody can understand this gibberish. You should write it in a non-weird language.

26. 9 months ago
Anonymous

You are responding to a bot.

27. 9 months ago
Anonymous

It's 1/3.

For those who claim it's 1/2, please point out where your reasoning diverges from the following:

Let there be 198 people, each of whom have 3 boxes {GG, GS, SS} in front of them. Everyone picks a box at random. 66 pick the first box, 66 the second, and 66 the third. We eliminate those who picked the third box.

132 people left. 66 picked the first box, and 66 picked the second box. They take out a ball at random. All 66 who picked the first box take out a gold ball, while 33 of those who picked the second box take out a gold ball. We eliminate the 33 who picked the second box and then took out a silver ball. This is where the set up of the question leaves us.

Now with 99 people left, 66 who picked the first box and took out a gold ball, and 33 who picked the second box and took out a gold ball, we are asked about the probability that the next ball they take out will be silver. All 66 who picked the first box and took out a gold ball take out the remaining gold ball, and all 33 who picked the second box and took out a gold ball take out the remaining silver ball, so 1/3rd of those who picked a box at random, then took out a ball at random, and that ball turned out to be gold, will take out a silver ball next.

QED.

Note: You can interpret the 198 people as one person doing the procedure 198 times instead, with the same results. For the sake of simplicity, the outcomes of all chance samplings are stipulated to conform to the expected value as the number of trials approaches infinity.

• 9 months ago
Anonymous

>For those who claim it's 1/2, please point out where your reasoning diverges from the following:
>Let there be 198 people
I'm one person, not 198 people.

• 9 months ago
Anonymous

>I'm one person, not 198 people.
This is just a didactic device to demonstrate the probabilities in more concrete terms. The answer does not depend on how you choose to illustrate it, as long as the probabilitic reasoning involved is equivalent.
It is just a restatement of the demonstration in a sentence that closely mirrors the language in the original problem statement, to make explicit how the solution offered is indeed the solution to the problem posed in the OP. Can you point out a specific disagreement with the reasoning?

So you pick 1 out of 3 boxes.

You choose both the balls in that box, and you're about to confirm which other 2 boxes are left through elimination. What's the chances that this specific box is the one with the gold and the silver? 1/3. But what is the chance then, that after you pick gold, you confirm it's silver. There are 3 boxes. You know you got gold. So it's either the box with gold and silver or two golds. How likely is it that out of two gold source boxes, you picked the one with the silver. It 50%

>But what is the chance then, that after you pick gold, you confirm it's silver. There are 3 boxes. You know you got gold. So it's either the box with gold and silver or two golds. How likely is it that out of two gold source boxes, you picked the one with the silver. It 50%
No, it's 1/3rd, because having picked gold, you the box you picked is more likely to be the one with 2 golds than the one with 1 gold.
>wrong. per the problem, all 66 people who selected the 2nd box got gold.
Wrong. The problem says you take a ball at random and get gold, i.e. you did in fact get gold, but randomly, not with certainty. That you did in fact get gold does not alter the probabilites. The answer of 1/2 would be correct if the problem stated that you non-randomly, with certainty, take out gold, but as per the plain language in the problem statement, you just happened to take out gold randomly.

• 9 months ago
Anonymous

>Can you point out a specific disagreement with the reasoning?
see

>while 33 of those who picked the second box take out a gold ball.
wrong. per the problem, all 66 people who selected the 2nd box got gold.

• 9 months ago
Anonymous

"Per the problem" as interpreted by you, all 198 people should've gotten gold, and you know that's impossible.

• 9 months ago
Anonymous

what the frick are you babbling about? i told you exactly where you were wrong. 66 chose box 1, 66 chose box 2. the sample size got reduced to all 132 of these people choosing gold. half of them are in box 2.

• 9 months ago
Anonymous

you're a moron; no one is talking about that.

• 9 months ago
Anonymous

>tell me where you disagree
>n-n-noooooo you're a moron!
you're embarrassing yourself

• 9 months ago
Anonymous

> moron. pick a ball 1-5, does primality constrict your choice?

• 9 months ago
Anonymous

"Per the problem" as interpreted by you, all 198 people should've gotten gold, and you know that's impossible.

No, because the problem statement uses the word randomly, we have to model the outcomes probabilistically. The fact that we are supposed to assume this or that outcome is just an indication of where in the probability tree we are located when the final question is asked. The language explicitly repudiates the interpretation that the outcome at any one sampling point was non-random, but that is what you have to stipulate in order to get the answer of 1/2.

• 9 months ago
Anonymous

>roll two dice, randomly, and sequentially
>the first comes up six
>what's the probability the 2nd comes up six?
you should be able to answer this. the first die being random is meaningless, because the result is six.

• 9 months ago
Anonymous

No one cares about your hypothetical because it has no value and you're an annoying butthole for pretending it does.

• 9 months ago
Anonymous

do you understand the analogy and why it's relevant? w

• 9 months ago
Anonymous

The question (essentially) is can you choose 1 of 5 balls randomly if 1 ball is in 1 box and the other 4 balls are also in 1 box.

• 9 months ago
Anonymous

That would be an accurate analogue of the problem if we make the following modifications. You have two dice. One of them has one face with a six on it, and the other has three faces with a six on it. You roll them randomly and sequentially etc.

• 9 months ago
Anonymous

no modifications necessary. the probability distribution is irrelevant.

• 9 months ago
Anonymous

>the probability distribution is irrelevant.
Why would it be?

• 9 months ago
Anonymous

idk, you're the person trying to modify it.

• 9 months ago
Anonymous

If you know one outcome is likelier than the other, and also the outcome of the second trial is dependent on the outcome of the first, how can you say the probability distribution is irrelevant?

• 9 months ago
Anonymous

here let me help you. the probability distribution is that you have 100% chance to pick the gold from box 2.

• 9 months ago
Anonymous

You don't, sorry. And again, it's easy to demonstrate the error: if you say getting a gold ball is always forced then you're saying we'd get a gold ball even from the all-silver box. Because it's guaranteed no matter the box. So then your probabilities are even more fricked and more impossible, because you're assuming contradictory things.

• 9 months ago
Anonymous

if you make an assumption and get a contradictory result, throw the assumption out: namely, you cannot have selected the all silver box. this isn't complicated.

• 9 months ago
Anonymous

>you cannot have selected the all silver box
So then they must have been forced not to, so no one selected it. See? In fact, how do you know they weren't simply all forced to select the all-gold box in order to force their random pick from it to be gold? If you're making unfounded assumptions like this, you're completely lost, because you can speculate anything.

• 9 months ago
Anonymous

Of course the result here is 6, but your problem is not analogous to the problem posed by OP. Say you have two dice, one 6-sided and one 12-sided. You randomly pick one die and roll. The random result turns out to be 6. What is the probability that the next roll will be 6?

• 9 months ago
Anonymous

ah but you see fine lad, the first die could have come up as anything but a six. therefore the probability to roll two sixes sequentially is 1/36!!!

that's what you're arguing with op's problem, but it's a stupid argument because it's analogous to saying you can potentially pick the silver ball first out of the 2nd box. that's explicitly forbidden. so all 66 of the 198 people who chose box 2 must have selected gold, per the problem.

do you understand now?

• 9 months ago
Anonymous

You're not being very nice to the people who see through your bullshit, are you? lol

• 9 months ago
Anonymous

>i got irrefutably pwned so i'm just going to accuse him of saying bullshit
not an argument. i accept your concession.

• 9 months ago
Anonymous

> The question (essentially) is can you choose 1 of 5 balls randomly if 1 ball is in 1 box and the other 4 balls are also in 1 box.

• 9 months ago
Anonymous

honestly i don't even know what side you're one (1/3-tards or 1/2-chads), but you're easily the dumbest person in this thread.

• 9 months ago
Anonymous

No, you misunderstand the original question, and you didn't answer mine either. As I explained before, the stipulation that you picked gold randomly is an indication of where you are in the probability tree and does not change the probability tree itself. What would change the probability tree is saying that after randomly picking the box, you non-randomly take out a gold ball, because only then would you be equally likely to have picked box GG or GS at the final decision point. But the problem specifically states that you randomly take out a gold ball, and so you are twice as likely to have picked box 1, which has double the number of gold balls in it, just like it is more likely that you picked the 6-sided die in the example I gave.

• 9 months ago
Barkun: Demi-GodofSCI

Were you educated in America?

• 9 months ago
Anonymous

So, no response after I make your misunderstanding explicit, and even go so far as to explain the erroneous model you are using to come to your conclusion. I understand your position perfectly; it just contradicts the plain language in the OP, which states that the gold ball is taken out randomly, i.e. in accordance with the original probability tree, and not the modified one you use to get the incorrect answer.

• 9 months ago
Anonymous

i'm assuming you meant to respond to me and not that dude. look, when you pick an empty door in the monty hall, i don't care how you came to that random conclusion. the probability you picked empty door 1 could be 90% or 10% or 15%. it's irrelevant; what matters is that you were forced to pick an empty door. in OP's problem, the word "randomly" doesn't matter, you were forced. 100% randomness exists, and if you don't believe me write a quick code to justify it to yourself.

• 9 months ago
Anonymous

>the word "randomly" doesn't matter, you were forced
Amazing how you people will say that the literal words of the problem don't matter because they conflict with your assumption

• 9 months ago
Anonymous

redundant words can exist in math problems, dunce. if i tell you that 2 is greater than or equal to 1, i am being truthful and accurate. but the "equal to" is irrelevant and doesn't matter.

• 9 months ago
Anonymous

Redundancies aren't contradictions, the way "random forced outcome" is

if you make an assumption and get a contradictory result, throw the assumption out: namely, you cannot have selected the all silver box. this isn't complicated.

Right, and if you don't assume "random" when it says "not random" there's nothing else to throw out. Because just as you cannot get gold from the all-silver box, you also cannot get gold every time from the gold-silver box.

let me clarify. when i say random, i mean many outcomes are possible. when i say forced, i mean a specific outcome must occur. a die can randomly show up on any face. but when i tell you that it landed on 6, the outcomes 1,2,3,4,5, are strictly forbidden, and it's no a probability 1/6 that a 6 shows up, rather it's 100%.

And here you are conflating things again. Yes, the six is there. But it had a 1/6 chance of occurring. You seem to acknowledge this and yet you insist that if we say "it landed on a six" then that outcome had to be forced. That's a contradiction. You can't say other outcomes were possible when they were actually not possible. You can't get around the contradiction here.

• 9 months ago
Anonymous

you're pretty dense, aren't you? if i tell you i have cancer, the probability i have cancer is 100%. it doesn't matter what cancer i have (all of which have different probabilities to get). all that matters is i tell you i have cancer, and now that's set in stone (all the probability functions collapse).

whether the gold ball from box 2 was selected randomly or not is irrelevant, and if it was random its probability distribution is also completely irrelevant because gold was gotten, period.

therefore, if box 2 was randomly chosen, getting the gold ball was forced. likewise if box 1 was randomly chosen, a gold ball was forced. therefore the probability i pull a silver ball next is 50/50.

perhaps an analogy will help. consider the monty hall problem, where you can pick any door. the host reveals an empty door, and asks you if you want to switch: what are the odds you switch to the prize? 2/3. recognize, however, that there's a finite probability that the first door you picked was the prize (1/3).

now let's modify the problem. i restrict your first guess in the monty hall problem to have picked an empty door. it could be either one, i don't care. now, as the host i reveal an empty door. should you switch? yes, because now you have a 100% chance to pick the prize.

do you understand how forcing a choice modifies the problem?

• 9 months ago
Anonymous

>therefore, if box 2 was randomly chosen, getting the gold ball was forced.
Wrong conclusion and precisely where you go astray. Again, this is just like saying that if you rolled six it had to have been forced because it is stipulated that you rolled six. Forget about a second trial, because you fail to understand that the first trial is random, and if we can't build on that knowledge, we can't move on.

• 9 months ago
Anonymous

anon, the first trial can be both random and forced. it being forced nullifies the random aspect of it.

• 9 months ago
Anonymous

Ah yes, just as how the box itself can be both round and square, because it being round nullifies it being square.

• 9 months ago
Anonymous

let me clarify. when i say random, i mean many outcomes are possible. when i say forced, i mean a specific outcome must occur. a die can randomly show up on any face. but when i tell you that it landed on 6, the outcomes 1,2,3,4,5, are strictly forbidden, and it's no a probability 1/6 that a 6 shows up, rather it's 100%.

• 9 months ago
Anonymous

you randomly pick a number out of a box it is a 6, you randomly pick another number out of a box it is 6, you repeat this 100 times. What is the probability that on the 101th attempt you pick out a 6.

according to your moronic logic it would be 1/6 and the other 100 trials are completely irrelevant to what is being asked. you can make that assumption if you want but you should know that it makes you moronic

• 9 months ago
Anonymous

>you randomly pick a number out of a box it is a 6, you randomly pick another number out of a box it is 6, you repeat this 100 times. What is the probability that on the 101th attempt you pick out a 6.
>according to your moronic logic it would be 1/6
no, based on my reasoning it'd be 100%. it seems you still don't understand my argument... me, and the other 1/2-chads are the ones who are using prior knowledge of the problem (namely, that a gold was forced). you're the one neglecting this and other prior knowledge.

• 9 months ago
Anonymous

whatever you say chud, good luck to your future endeavors in life, you will be very successful, I can tell just from the way you write and think about probabilities that you are destined for great success. it is 100% guaranteed

• 9 months ago
Anonymous

i'd bet good money i'm already more successful than you. are you tenured? 🙂

>using prior knowledge of the problem (namely, that a gold was forced
Faulty assumption that because an outcome occurred that it was forced, which is not stated and directly contradicts what is stated.

it is stated.

• 9 months ago
Anonymous

What is stated is that the first pick is random. What is not stated is that it is forced. And considering "forced" would contradict "random" you shouldn't assume it. There's really no way to make this any easier. You are blatantly and intentionally misreading the problem.

• 9 months ago
Anonymous

>it is stated.
To my mind, the straightforward interpretation of the statement

>You put your hand in and take a ball from that box at random. It's a gold ball.

is as an instruction on which branch of the probability tree we are to consider. In the GS box, you will randomly take out a gold ball 50% of the time, and it is the G branch of the probability tree that we are to consider when answering the final question.

Now, you want to say that no, what this is actually saying is that the probability distribution at this point is changed to 100% certainty of picking a gold ball. But in order to accord with your interpretation (taking into account that you cannot draw a gold ball from box SS) the sentence would have to read something like this:

>>You put your hand in and take out a ball from that box. If there is a gold ball in the box, you take out the gold ball.

This reformulated sentence clearly informs us about ther the probability distribution. Now either GG or GS will always yield a gold ball, and SS will not. The S in GS is "masked" in the first draw, and becomes available in the second. On this formulation, everyone would reach the same conclusion as you.

So given that it is very straightforward to reformulate the sentence such as to clearly convey to the rest of us the meaning you are reading into it, and given that your contention is that your reading is the natural one, how would you reformulate the sentence to render our interpretation the natural one? Or are you claiming it's just ambiguous, and either reading is legitimate?

• 9 months ago
Anonymous

no, i'm saying there is no conditional at all. the fact you have to reinterpret my approach as a conditional is where your fundamental misunderstanding is coming from.

>saying "it is gold" is not the same as saying "if it is gold"
Yes it is, considering this is all in the context of a hypothetical scenario. It is exactly the same as saying that.
>you both misunderstand what it means for a decision to be forced. box 3 is forbidden, so it's irrelevant
It's only forbidden because the contents are forbidden. Why say that the box has to be selected randomly (but only from among the boxes that potentially could yield the correct result) but the next pick has to be forced, when it could well be that the first pick is first and the second random? Who knows? Fricking no one because you're making stipulations that aren't stated so it could be anything. We all get how it works if you make your wrong assumption. What you don't get is why it's wrong to assume so. And literal fricking language won't convince you, so what will?
>it doesn't matter how the gold ball randomly got selected--it was selected.
It does matter and that's precisely why this problem resembles Monty Hall. The host always opens a goat door. But which door he is therefore able to open is constrained by your first choice.

one is a conditional, the other is not.
>Why say that the box has to be selected randomly
because you could have box 1 or box 2. the 3rd box is irrelevant.
>the next pick has to be forced
because you're told what your result is: it's a gold ball.
> resembles Monty Hall.
i already addressed this. if i force you to initially pick an empty door, then you have a 100% chance to get the prize by switching when the host reveals the 2nd empty door. forcing your choice changes the problem.

• 9 months ago
Anonymous

>because you could have box 1 or box 2.
No, you couldn't have had box 2, because your box is forced to always yield gold, and box 2 does not do that with a random pick. Only box 1. Ergo, we selected box 1.

Point is this:
Semantically speaking, saying "hypothetically, it is a gold ball" and "if it is a gold ball" are equivalent.

• 9 months ago
Anonymous

> you couldn't have had box 2, because your box is forced to always yield gold
lol you're a fricking idiot.

>no, i'm saying there is no conditional at all. the fact you have to reinterpret my approach as a conditional is where your fundamental misunderstanding is coming from.
I reject this. I offered a reformulation that yields your interpretation. I'm couching it in conditional terms because the conditional directly points at the probability distribution, which we have to alter in order to make your interpretation into the straightforward, natural one. But given that you are unable to understand the original sentence, it is perhaps no surprise that you don't understand how to reformulate the sentence such as to communicate what you're reading into it either.

You haven't answered my question: How would you reformulate the sentence such that in your own way of reading it, it would express what we take it to express?

i don't need to reformulate it, because i'm solving it as phrased. if you're asking how to remove the ambiguity, i guess remove the word randomly. though i doubt that would make people like you change your understanding of the problem.

Do you believe the OP should've been written like this?
>If there were three boxes, and if each box contained two balls, and if one box contained two gold balls, and if another box contained two silver balls, and if the final box contained one gold ball and one silver ball, and if you then picked a box at random and if you then put your hand in and if you took a ball from that box and if it were random and if it were a gold ball... then what would be the probability that the next ball you take from the same box will be silver?
Do you think that's a different problem? Do you believe that's unambiguously hypothetical in a way the OP isn't? Is English not your first language?

are you high?

• 9 months ago
Anonymous

>lol you're a fricking idiot.
As if you'd know. You see one every day in the mirror and fail to realise.
>if you're asking how to remove the ambiguity, i guess remove the word randomly.
See, now you're literally just saying "change the problem" and you're still not seeing you're changing the problem.
>are you high?
I used "if" like you insisted. Now it's conditional.

• 9 months ago
Anonymous

oh, i see. you have three different personalities. take your meds.

• 9 months ago
Anonymous

Anon, am I obliged to politely respond only to those parts of your post you addressed to me? If you say something moronic to someone else, am I to bite my tongue and hope you'll grace me with like idiocy so that I may comment on it? How do conversations arise on BOT at all, if we're only to speak when spoken to?

• 9 months ago
Anonymous

i'm trying to keep track of who is whom this way i know which morons to insult.

• 9 months ago
Anonymous

Anon, am I obliged to politely respond only to those parts of your post you addressed to me? If you say something moronic to someone else, am I to bite my tongue and hope you'll grace me with like idiocy so that I may comment on it? How do conversations arise on BOT at all, if we're only to speak when spoken to?

That said, two of those three posts were mine, so it's basically 50-50, right?

• 9 months ago
Anonymous

>i guess remove the word randomly. though i doubt that would make people like you change your understanding of the problem
Good, we're making progress. The word random explicity reaffirms the probabilistic nature of the draw, so it repudiates your interpretation. Removing it would indeed make your interpretation more reasonable, especially considering that there would then be an implied contrast to the nature of the choice in the case of the boxes, which would still be explicity random.

You haven't answered my question though: You suggested a change that would make your reading more plausible to us, but what I'm asking for is a change that would make our reading more plausible to you. You're not helping your case by struggling to understand simple English. Or maybe you're not answering because the statement is already very clear, and our reading is the straightforward, obvious and natural one, and any reformulation would just draw attention to and reinforce that fact?

• 9 months ago
Anonymous

if you agree that the phrasing is ambiguous, then idk why your knickers are in a knot over someone getting a different answer than you. autism?

• 9 months ago
Anonymous

He clearly doesn't agree. He and I both think it's unambiguous as it is, and your tortured misreading of it has to blatantly ignore what it actually says. Just as you keep ignoring what he is actually asking you.

• 9 months ago
Anonymous

You sound like you're the balls-moron. Is this true? You have the same low-IQ identifiers. No, I won't tell you what they are.

• 9 months ago
Anonymous

>the balls-moron
m8 I don't know what to tell you, this entire thread is about balls, we're all talking about balls here

• 9 months ago
Anonymous

>No, I won't tell you what they are.
Every single time ITT some idiot has gone "I know something you don't but I'm not telling tee hee hee" they inevitably turned out to be pretentious clowns with delusions of competence. I dunno, I'm starting to think that outcome is forced.

• 9 months ago
Anonymous
• 9 months ago
Anonymous

The problem is that your interpretation is precluded by the wording as it stands. It is perfectly legitimate, even interersting, to discuss modifications of the original question, but clearly misinterpreting the question and then getting defensive about it tends to make people want to press you on the point until you admit it. Autism might well play a role in this, but it takes two to tango, ya know?

• 9 months ago
Anonymous

are you asking me to dance with you? because i'm not saying no. whether or not you get eye contact will depend on your hygiene

• 9 months ago
Anonymous

>no, i'm saying there is no conditional at all. the fact you have to reinterpret my approach as a conditional is where your fundamental misunderstanding is coming from.
I reject this. I offered a reformulation that yields your interpretation. I'm couching it in conditional terms because the conditional directly points at the probability distribution, which we have to alter in order to make your interpretation into the straightforward, natural one. But given that you are unable to understand the original sentence, it is perhaps no surprise that you don't understand how to reformulate the sentence such as to communicate what you're reading into it either.

You haven't answered my question: How would you reformulate the sentence such that in your own way of reading it, it would express what we take it to express?

• 9 months ago
Anonymous

Do you believe the OP should've been written like this?
>If there were three boxes, and if each box contained two balls, and if one box contained two gold balls, and if another box contained two silver balls, and if the final box contained one gold ball and one silver ball, and if you then picked a box at random and if you then put your hand in and if you took a ball from that box and if it were random and if it were a gold ball... then what would be the probability that the next ball you take from the same box will be silver?
Do you think that's a different problem? Do you believe that's unambiguously hypothetical in a way the OP isn't? Is English not your first language?

• 9 months ago
Anonymous

>are you tenured? 🙂
r u joking? is this really the state of tenured professors? LARPing as successful morons on public internet forums for incels?

to answer your question, i don't have tenure because it never occurred to me that would be something i should care about. that either makes me a lot smarter than you or a lot dumber. what are the odds, in your opinion, whether it's one or the other?

• 9 months ago
Anonymous

you either are or you aren't. 50%

• 9 months ago
Anonymous

so then that measure of success, tenured professorship, is entirely meaningless on a public and anonymous internet forum for incels. if the odds are 50/50 then it would not have occurred to me seek that distinction as some kind of meaningful measurement of success

the answer tho, to the problem stated, is 1/3

• 9 months ago
Anonymous

50/50. one of us is tenured, and that's me. therefore you're the other 50%.

• 9 months ago
Anonymous

the question was one of intelligence and not tenure but you misinterpreted what was intended. don't worry, it happens to the best of us

• 9 months ago
Anonymous

• 9 months ago
Anonymous

>using prior knowledge of the problem (namely, that a gold was forced
Faulty assumption that because an outcome occurred that it was forced, which is not stated and directly contradicts what is stated.

• 9 months ago
Anonymous

All right, so, the thing is, you are supposed to use prior knowledge. But that prior knowledge is properly how likely something was to happen, and not just that it happened and assuming from there that it was forced.

• 9 months ago
Anonymous

i'm assuming you meant to respond to me and not that dude. look, when you pick an empty door in the monty hall, i don't care how you came to that random conclusion. the probability you picked empty door 1 could be 90% or 10% or 15%. it's irrelevant; what matters is that you were forced to pick an empty door. in OP's problem, the word "randomly" doesn't matter, you were forced. 100% randomness exists, and if you don't believe me write a quick code to justify it to yourself.

You keep trying to change the probability distribution despite the problem explicity leaving the probability distribution intact by using the word randomly. What you are mistakenly interpreting as asking you to "forcing the choice" in the sense of suspending the probabilistic nature of the scenario at a certain point is actually just a specification of the location on the probability tree. Please try to absorb this point before producing more analogies, especially if you're not going to address my correction of your false analogies. Your monthy hall reformulation suffers from the exact same error as the dice roll analogy. You seem to think I'm committing the gambler's fallacy, but I have already shown in my response to your dice example that this is not what I'm doing.

• 9 months ago
Anonymous

i'm not changing anything, i'm saying the probability distribution is irrelevant. the gold ball was picked, full stop. you showed via YOUR dice example that you're not committing the gambler's fallacy, but that's a different problem. in MY dice example, which is more reflective of the op, your analysis is wrong.

What is stated is that the first pick is random. What is not stated is that it is forced. And considering "forced" would contradict "random" you shouldn't assume it. There's really no way to make this any easier. You are blatantly and intentionally misreading the problem.

>What is not stated is that it is forced.
wrong. saying "it is gold" is not the same as saying "if it is gold". no conditional means it's forced. this isn't a complicated concept. if i say you have a 100% chance to die, it's forced that you die even if the way in which you die is random.

If you interpret "its a gold ball" as to force a random selection to always be one outcome. Then by that logic, the third box should also give gold via quantum teleportation, since that is technically something that can happen by random chance.

>you cannot have selected the all silver box
So then they must have been forced not to, so no one selected it. See? In fact, how do you know they weren't simply all forced to select the all-gold box in order to force their random pick from it to be gold? If you're making unfounded assumptions like this, you're completely lost, because you can speculate anything.

you both misunderstand what it means for a decision to be forced. box 3 is forbidden, so it's irrelevant. the problem is equivalent to there being two boxes, GG and GS, randomly picked. out of 100 people, 50 pick box 1 and 50 pick box 2.

of the 50 who picked box 1, all 50 selected gold. of the 50 who picked box 2, it's forbidden for any of them to have picked silver per the problem's wording (they picked gold). it doesn't matter how they randomly found themselves to pick gold. maybe they picked up a silver a million times, threw it back in, and finally got a gold ball. maybe a host looks into the box, picks the gold ball out, and gives it to them. it doesn't matter how the gold ball randomly got selected--it was selected.

so, of these 100 people, how many of them when given the 2nd ball of their respective box receive a silver? the correct answer is that all 50 who ended up with the 2nd box, that is, 50%.

• 9 months ago
Anonymous

>saying "it is gold" is not the same as saying "if it is gold"
Yes it is, considering this is all in the context of a hypothetical scenario. It is exactly the same as saying that.
>you both misunderstand what it means for a decision to be forced. box 3 is forbidden, so it's irrelevant
It's only forbidden because the contents are forbidden. Why say that the box has to be selected randomly (but only from among the boxes that potentially could yield the correct result) but the next pick has to be forced, when it could well be that the first pick is first and the second random? Who knows? Fricking no one because you're making stipulations that aren't stated so it could be anything. We all get how it works if you make your wrong assumption. What you don't get is why it's wrong to assume so. And literal fricking language won't convince you, so what will?
>it doesn't matter how the gold ball randomly got selected--it was selected.
It does matter and that's precisely why this problem resembles Monty Hall. The host always opens a goat door. But which door he is therefore able to open is constrained by your first choice.

• 9 months ago
Anonymous

>the first pick is first
first pick is forced*

• 9 months ago
Anonymous

Who cares about any of this obvious moron shit? All you're saying is that you're an obvious moron. Good for you? Congratulations? Lol what

• 9 months ago
Anonymous

>Now with 99 people left, 66 who picked the first box and took out a gold ball, and 33 who picked the second box and took out a gold ball, we are asked about the probability that the next ball they take out will be silver. All 66 who picked the first box and took out a gold ball take out the remaining gold ball, and all 33 who picked the second box and took out a gold ball take out the remaining silver ball, so 1/3rd of those who picked a box at random, then took out a ball at random, and that ball turned out to be gold, will take out a silver ball next.

• 9 months ago
Anonymous

Not him, but I'm pretty sure you're actually a moron so why should anyone care lol.?

• 9 months ago
Anonymous

Because we're right. This anon explains it.

>while 33 of those who picked the second box take out a gold ball.
wrong. per the problem, all 66 people who selected the 2nd box got gold.

• 9 months ago
Barkun: Demi-GodofSCI

So you pick 1 out of 3 boxes.

You choose both the balls in that box, and you're about to confirm which other 2 boxes are left through elimination. What's the chances that this specific box is the one with the gold and the silver? 1/3. But what is the chance then, that after you pick gold, you confirm it's silver. There are 3 boxes. You know you got gold. So it's either the box with gold and silver or two golds. How likely is it that out of two gold source boxes, you picked the one with the silver. It 50%

• 9 months ago
Anonymous

>while 33 of those who picked the second box take out a gold ball.
wrong. per the problem, all 66 people who selected the 2nd box got gold.

• 9 months ago
Anonymous

If you interpret "its a gold ball" as to force a random selection to always be one outcome. Then by that logic, the third box should also give gold via quantum teleportation, since that is technically something that can happen by random chance.

28. 9 months ago
Anonymous

Solution in PowerShell. The answer is still 1/3

• 9 months ago
Anonymous

Wrong problem.

• 9 months ago
Anonymous

It is the correct problem but morons in this thread are either unable or unwilling to actually write out formally what is going on. Once you do it becomes obvious there is only 1 correct way to interpret what is being presented

• 9 months ago
Anonymous

Incorrect. This anon is right.

This is the boy or girl paradox with with the supporting info being the older child is a girl (the first ball is gold). The answer is 50/50. It's the first question on the wiki article.

There are 4 ways to draw the balls: gg, gs, sg, ss. We can eliminate the last two because we know the "oldest" child is gold. That leaves a 50/50 chance the other "child" is also the same.

• 9 months ago
Anonymous

No, you're just an idiot who no one cares about.

• 9 months ago
Anonymous

So why do you keep responding angrily?

• 9 months ago
Anonymous

But why are you so angry about it?

• 9 months ago
Anonymous

>no u

• 9 months ago
Anonymous

Anger writ large.

• 9 months ago
Anonymous

More verbose than Ruby tho

29. 9 months ago
Anonymous

Our little AI is still a baby trying to learn to walk.

30. 9 months ago
Anonymous

This is the boy or girl paradox with with the supporting info being the older child is a girl (the first ball is gold). The answer is 50/50. It's the first question on the wiki article.

There are 4 ways to draw the balls: gg, gs, sg, ss. We can eliminate the last two because we know the "oldest" child is gold. That leaves a 50/50 chance the other "child" is also the same.

• 9 months ago
Anonymous

Boy/ girl paradox is moronic. have a nice day literally with a literal knife you fricking idiot.

• 9 months ago
Anonymous

It's Bertrand's Box you mong

• 9 months ago
Anonymous

• 9 months ago
Anonymous

The setup for the problem diverges from Bertrand's Box in a key way by removing the randomness of the first pick. This eliminates all of what [...] said because both gold-box and mixed-box choosers have a 100% chance of picking the same color of ball.

It's not even ambiguous. It literally tells you the first pick is random in so many words. You are simply wrong.

• 9 months ago
Anonymous

• 9 months ago
Anonymous

You literally admit you're a moron and are proud of being a moron lol.

• 9 months ago
Anonymous

The setup for the problem diverges from Bertrand's Box in a key way by removing the randomness of the first pick. This eliminates all of what

It's 1/3.

For those who claim it's 1/2, please point out where your reasoning diverges from the following:

Let there be 198 people, each of whom have 3 boxes {GG, GS, SS} in front of them. Everyone picks a box at random. 66 pick the first box, 66 the second, and 66 the third. We eliminate those who picked the third box.

132 people left. 66 picked the first box, and 66 picked the second box. They take out a ball at random. All 66 who picked the first box take out a gold ball, while 33 of those who picked the second box take out a gold ball. We eliminate the 33 who picked the second box and then took out a silver ball. This is where the set up of the question leaves us.

Now with 99 people left, 66 who picked the first box and took out a gold ball, and 33 who picked the second box and took out a gold ball, we are asked about the probability that the next ball they take out will be silver. All 66 who picked the first box and took out a gold ball take out the remaining gold ball, and all 33 who picked the second box and took out a gold ball take out the remaining silver ball, so 1/3rd of those who picked a box at random, then took out a ball at random, and that ball turned out to be gold, will take out a silver ball next.

QED.

Note: You can interpret the 198 people as one person doing the procedure 198 times instead, with the same results. For the sake of simplicity, the outcomes of all chance samplings are stipulated to conform to the expected value as the number of trials approaches infinity.

said because both gold-box and mixed-box choosers have a 100% chance of picking the same color of ball.

• 9 months ago
Anonymous

no. see here

>while 33 of those who picked the second box take out a gold ball.
wrong. per the problem, all 66 people who selected the 2nd box got gold.

31. 9 months ago
Anonymous

Many morons ITT. Good luck being so fricking stupid lol.

32. 9 months ago
Anonymous

>You roll a six-sided die. You got a six. What were the odds of that?
>Durrr 100% because you just said I already got it

• 9 months ago
Anonymous

wrong. more like
>you roll two six sided dice, one after the other
>the first die comes up six
>what is the probability you roll double six?
you would say 1/36 (the 1/3-tards), when the real answer is 1/6 (us, the 1/2-chads)

• 9 months ago
Anonymous

No, Anon. These are not independent trials. You are picking from the same box. And, here is where we'll have to abandon the die-metaphor: the odds are weighted.

• 9 months ago
Anonymous

>the odds are weighted.
no. when you say the odds are weighted, you're assuming the possibility of picking the silver first from the 2nd box. that's explicitly forbidden per the problem.

• 9 months ago
Anonymous

You conflate something happening with something needing to have happened.

• 9 months ago
Anonymous

you're one conflating that. bertrand's box is something needing to happen (conditional if). this problem (op) is one where something did happen (not condition). it changes the probability, and the 1/2-chads like me are the ones not conflating the two.

• 9 months ago
Anonymous

moron. pick a ball 1-5, does primality constrict your choice?

• 9 months ago
Anonymous

moron. pick a ball 1-5, does primality constrict your choice?

Jesus why are you all SO FRICKING STUPID

• 9 months ago
Anonymous

thanks for admitting you're a moron
now pick a ball 1-5
> does primality constrict your choice?

• 9 months ago
Anonymous

This is a fricking clusterfrick but OP is 1/3 and the other problem is 5/9 and your bullshit is 1/5 and I don't know what you're raving about or why you think it's relevant

• 9 months ago
Anonymous

No, the other problem is not 5/9.

• 9 months ago
Anonymous

It is, and if you understand why the first is 1/3, you should know why the second is 5/9, because it works the same way.

• 9 months ago
Anonymous

Absurd.

• 9 months ago
Anonymous

Anon, there's a reason the problem doesn't just say "you pick a ball from a pile of balls, half of which are gold and the other half are silver". Precisely the reason it works out the way it does is because of the boxes. I don't get how you understand this fact in the one case and treat it as a pile of balls in the other.

• 9 months ago
Anonymous

Maybe because it says it in the first case and says the opposite in the second case..? There's no way you're really this stupid. You must be pretending to be stupid. Why though?

• 9 months ago
Anonymous

>Maybe because it says it in the first case and says the opposite in the second case..?
What does this MEAN lol

• 9 months ago
Anonymous

Absurd.

All right, I was genuinely puzzled at how you seemed to understand Bertrand's Box and yet drop the ball (lol) when it came to a slightly modified version of it. But then I realised: you don't actually understand it at all. You just happen to get the right answer if you apply your methodology to the standard Bertrand's Box scenario.

What you're doing is this: you have two boxes that your ball could have come from, with three balls between them, and from those three balls, only one is the correct one, hence 1/3. Just happens to work out to the right answer, but the underlying logic is entirely wrong. You're not dealing with two boxes, for starters. You're dealing with three at first, and then with one. You're trying to narrow down which box it is, and you know it has to be one of two (that's where the two comes in), but not with equal likelihood. The gold ball is twice as likely to have come from the all-gold box, and if you got it from the all-gold box, then you won't get silver.

Your mistake becomes obvious if we modify the problem by putting the gold ball back into the box it came from and then picking from the same box again. You would reason that there are now two "possible" boxes with four balls between them, and still one correct ball, for a total probability of 1/4. But actually it works like this: if you got the gold ball from the all-gold box (probability of 2/3) and put it back, the probability of getting silver from that box remains unchanged at zero. However, if you had the gold-silver box (probability of 1/3), your chances are halved, because instead of a guaranteed silver you now have 50% chance of randomly getting the same ball again. This means that, if you put the ball back and draw from the same box again, you have a total chance of only 1/6 of getting the silver after getting the gold.

We could've resolevd this ages ago if you'd explained yourself instead of insisting "yes yes I know Bertrand's box and you're all moronic"

• 9 months ago
Anonymous

you're a fricking moron. you roll a die, it comes up 6 so you erase the six randomly and then roll again. that's the correct analog of the problem statement

• 9 months ago
Anonymous

>it comes up 6 so you erase the six randomly and then roll again.
lol wtf. you have such a web of nonsense in your head to justify your 1/3 result (because you think this is bertrand's box, even though it isn't) that you're making up random shit.

33. 9 months ago
Barkun: Demi-GodofSCI

It's a red herring.

34. 9 months ago
Anonymous

3 boxes in total

2 balls per box

6 balls in total in the three boxes

one box - 2 gold balls

one box - 2 silver balls

one box - one gold ball and one silver ball

three silver balls in total out of 6 balls

The answer has to be 2/3 - 66.6%

Remember we know that there is three boxes, two balls each. We know the color of balls in each box from the picture. 2/3 chance of picking a silver.

35. 9 months ago
Anonymous

36. 9 months ago
Anonymous

Here you go, morons:
The probability the other coin is silver is 0.33.
The tard-tier python script (pic) proves it.

• 9 months ago
Anonymous

code is wrong, it permits you to choose silver 1st from the 2nd box. forbidden per problem.

• 9 months ago
Anonymous

Nobody has yet been able to explain how its logically possible to make a selection both randomly and not randomly at the same time.

• 9 months ago
Anonymous

already did, you just ignored it.

• 9 months ago
Anonymous

Nope, you can't read. It only considers cases where the randomly selected ball is gold (as per the problem).

37. 9 months ago
Anonymous

You are all morons.

6 equally likely outcomes:

Gg gs ss
gG gs ss
gg Gs ss
gg gS ss
gg gs Ss
gg gs sS

First observation (G) limits us to the first 3, still equally likely. Only one gives S as next draw.

1/3 it is.