Goys?

yup

2/3 is the correct answer

This isn't a monty hall problem.

It's 1/3

The second ball drawn has to come from the same box the first ball came from. Therefore the only way to pick a second gold ball is if the first box was chosen at random initially.

This

saged and hidden

I have now realised that only 2/3 boxes have a gold ball. So you'd have to have chosen one of those two initially.

Final answer is 1/2

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1-1/3=2/3

7/256
god its fucking obvious everyone here is a fuckin dropout loser why do i even come to this website

Jk it's 1/2

66% because you picked the box with the gold ball

it's 1/3

there was a 1/3 chance you picked the box with 2 gold balls, that remains unchanged by you drawing one out

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2/3
lrn2conditionalprobability

true but its still 2/3

the possible outcomes are
2 grey (1/3)
100% grey - grey => 1/3 chance
1/1 (1/3
50% grey - gold => 1/6 chance
50% gold - grey => 1/6 chance
2 gold (1/3)
100% gold gold => 1/3 chance

exclude options when you pick grey first

it's 2/3

But that's not the question..

the question is what are the odds you picked the box with 2 gold balls, it just appears that information OP gives you changes that but it really doesn't

this is a famous mathematical riddle called the Monty Hall Problem

You aren't picking the second ball from a box at random, you're picking it from the same box. So only the first (2 gold) and second (1 gold 1 silver) are of interest, the 2 silver box is discarded once we've successfully picked a gold ball (which is implied in the question).

Again, the question is not "what is the chance of drawing two gold balls", it is "what is the chance of drawing a second gold ball from a box when your first draw yielded a gold ball". This is 1/2 - by picking the gold on your first draw, you can only have 1 of 2 boxes, so your next draw will either be another gold (having picked box 1) or a silver (having picked box 2)

1/2 it is

Jesus the state of this board. To all you fucks over thinking this its literally 50/50. 2nd ball in ANY box containing a gold ball will be either gold or silver. 1/2. fuck

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Last bus to literacy camp is departing in 20 seconds.

Retard alert

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It isn't Monty Hall, it is 100% guaranteed that you have picked a box with a gold ball and are going to take a second draw from the very same box, the 3rd box is out of the equation. If you were able to put the ball back and switch boxes you would be right, but you're locked into the box you initially chose here.

it's the same probabilities how can you not see that?

you drawing one gold ball is the same as the Monty Hall host showing you one goat. It doesn't change the underlying probability that your initial choice of box was 1 in 3

You're more likely to have initially picked the gold ball from the box with 2 gold balls, given that it presents more chances for the winning condition (2/3 gold balls vs. 1/3 gold balls)

You just aren't thinking 4th dimensionally

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You dont choose the box. you start with 1 gold ball and the box it came from. the 2 silver box is negligable entirely. it may as well not exist

>you can only have 1 of 2 boxes, so your next draw will either be another gold (having picked box 1) or a silver (having picked box 2)
Exactly. Also you might die tomorrow or you might survive, therefore you must have a 50% chance to die tomorrow. Better tell your loved ones.

Oh fuck, I better tell my loved ones too.

no read the OP image it says you pick one box at random

there were 3 boxes and you picked 1. therefore the odds that box contains 2 gold balls is 1 in 3.

I think its a 2/3 chance

2/3

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>YOU PUT YOUR HAND IN AND PULL OUT A GOLD BALL

LISTEN FUCKO THIS IS WHERE THE QUESTION STARTS. WITH A GOLD BALL IN HAND. THE 2 SILVER BOX DOESNT FUCKING MATTER.

I didn't think of that, that's possible - but it may be arguable that the initial draw has no adherence to probability: it was guaranteed, for the sake of the question, to be a gold ball, so the initial pick was 50/50 between the 1st & 2nd box, and your draw, the only with any element of randomness, will be from one of those two.

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It's not 2/3, the remaining balls from the boxes aren't shared ("from the same box").

You picked the gold ball, either from box 1 (which contains one more gold), or box 2 (which contains one more silver).

50/50

The number of balls irrelevant.

50%

Either you picked the box with 2 gold balls or you picked the box with 1 gold ball. You couldn't have picked the all silver box.

this was my initial thought, but
make me reconsider. So it may not be that you have two options (box 1 or box 2) but 3 (gold A from box 1 gold B from box 1 or box 2), thus you'd have three possible picks of your own, the first two yielding a gold and the third a silver so 2/3 seems to make sense (but I'm a Humanities fag so what do I know)

No, its 2/3rds if you switch boxes. The question says that you pick both balls from the same box, so its 1/2.

>it may as well not exist
>but it may be arguable that the initial draw has no adherence to probability
This is what happens when you ignore the whole "you pick a box at random" part. It says that very deliberately and is an essential part of the riddle. It's not telling you what WILL happen, it's telling you what DID happen after you made a completely probabilistic choice. If you were to reproduce this experiment in real life (and you should give it a try if you're unconvinced since it's so damn easy to do as long as you have some opaque cups and are capable of closing your eyes) you would find that you get the gold ball (or whatever analogous binary-state object you use) 2 out of 3 times.

en.wikipedia.org/wiki/Bertrand's_box_paradox

We're given the fact that we picked a gold ball. Between the two boxes, one of them contains more gold balls and so was more likely to have been the one that was picked from, given this information.

Imagine if the question was reframed, so that one box was 100/100 gold, the second box was 1/100 gold, and the third box was 0/100 gold. If you randomly pick a gold ball from one of the boxes, then you can be almost certain that it was from the first box, because 100/101 gold balls come from that box. Just like how in this problem you can be slightly more certain it came from the first box because it contained 2/3 gold balls.

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if you actually think this after looking at his photo you are really dumb, i'm sorry. I actually thought everyone saying 1/2 was memeing but you sound legit.

You have WORSE odds if you switch boxes, this isnt the monty hall problem, but it does use a similar logic.

that's not what it is though, really there were 6 initial possibilities and you are only showing 3. the overall probability is still 2 in 6.

50%

If you picked a gold ball it means you either picked the 2gold balls box or the 1 gold 1 silver box

this feels like facebook...

P(G2|G1) = P(G1 and G2) / P(G1)
(1/3) / (1/2) = 2/3

It didn't ask for the overall probability, it asked for the probability given that you've already picked a gold ball, which rules out 2/6 of the possibilities from the beginning

I'm about to do some real life trials with identical plastic bags and some bottle caps. I fully expect to get a result of aprox. 50% gold for the second pick.

Turn 360 degrees and walk away.

Can't believe there are brainlets that can't grasp that because you got a golden one first you have double the likelihood to have chosen the 2xgolden box an therefore two thirds
probably nonwhites

2/3

no it doesn't it just appears to, that is the trick

you are not changing anything after you make your initial choice of box so the scrap of additional information is meaningless to the underlying probability

based gimpposter

If you put your hand in the first box, 100%

If you put it in the second, 0%

Chance doesn’t actually exist

YOU FUCKING BRAINLETS I KNOW THIS IS BAIT BUT YOU TRIGGER ME TOO MUCH.

The answer is 2/3. Yes it's counterintuitive, but follow my mathematical proof and you will be convinced, faggot.

We want the probability of picking GG KNOWING that we picked a box with one G ball (we will call this event 1G). We denote this probability by : P(GG|1G). Now, use Bayes theorem (google it, I don't want to prove it here), which tells us :

P(GG|1G) = P(1G|GG)*P(GG)/P(1G)

Then, we have that P(1G|GG)=1 since if we picked the GG box we necessarily got 1 G ball. P(GG) =1/3 because it's simply the probability of picking the GG box without a priori info. Finally, we need P(1G). This can be viewed in two ways. You can consider you have half and half G and S balls, so it's 1/2 to pick a G ball. You can also say that to get a gold ball you have to either pick the GG box (1/3) OR pick the GS box (1/3) AND pick the G ball inside (1/2). Unsurprisingly, we get again P(1G)= 1/3+1/3*1/2 = 1/2.

PLUCK THAT SHIT INTO (1), and we get :

P(GG|1G) = (1/3)/(1/2) = 2/3

Can't believe I spend 10 minutes explaining this shit to brainlet, but at least I can sleep now.

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okay, I'm done trying to explain basic statistics to brainlets, look up conditional probability sometime you fucking mong

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this is monty hall problem but forced to not switch. 1/3

the problem is dude, you do have a priori information. So the brainlet is you here.

en.wikipedia.org/wiki/Bertrand's_box_paradox

everone go home now

Re-read again what I wrote, but slowly

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make them black please. it's easier to run over niggers

This was my understanding but all the circus math going on in here had me second guessing.

Looks like I was right, brainlets

we only made one choice, it doesn't matter that we get more information AFTER we made that choice, it cannot affect the outcome

3/6 balls are golden so it's a coin toss that the first ball you draw any given time you run the experiment will be golden

if you ran the experiment 100 times you would only pick the box with 2 gold balls one third of the time, but the first ball you draw would be gold half the time

the question is not "what are the odds the second ball is golden only of the times the first ball is golden?" it's just "what are the odds on any given experiment both balls are golden?"

its 2/3 you brainlets reeeeeeeeeeeee

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first 10 trials, literally 50%. The ignores are ones where the first pick was silver:

1: ignore
2: silver
3: ignore
4: silver
5: ignore
6: ignore
7: gold
8: ignore
9: gold
10: ignore

also, Im a little retarded re: is right about your odds being worse if you switch, but not about this particular scenario not being 50%.

If you're allowed to eliminate the box with only silver balls you have 2/3rds. If you're not you have one third. If you don't switch you have 1/2. For some reason I assumed that in an alternate version of this question where you switch, you'd eliminate the 100% silver box.

Gonna do 10 more tests and report back.

>the question is not "what are the odds the second ball is golden only of the times the first ball is golden?"
It literally is, Jesus go back to ESL class you fucking pajeet

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I don't understand a single thing about your incoherent ramblings. I defined the probabilities, used equations, arrived at the result using only logical steps.

I don't give a fuck about your "arguments", the equations tell me that you are wrong. Just check the post linked by the other user, apparently it's a famous problem :

There a 3 gold balls you could possibly pick initially.
By that fact alone there are 3 possible outcomes.

You’re not picking a box, that’s not what the question is asking of you.

if im wrong explain how this works
are you literally doing this by hand? holy fucking christ you are a brainlet. There is no way this ins't a troll at this point

jeez fucking code so verbose I hate you

>mfw brainlet try to simulate probabilities IRL
>mfw plenty of people already simulated it
>mfw 2018 and user still can't code

whatever 6/9 is

75% chance. Congrats op you win.

You must be a sub 100 IQ being.

kek

i was trying to be explicit for people that don't know how to code no bully plz

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There are 3 possible states and 2 end up in gold. Therefore 2/3.
It's really interesting neurologically. If I think about states 2/3 seems fine. If I think about boxes 1/2 seema like the obvious answer.

This means that there's a bug in human world modeling hardware. Seems to be genetic and universal.

Nice try, fag, you obviously overloaded the native print function with one that prints wrong answers. Or maybe you got lazy and just photoshopped it. Or maybe I'm wrong and you're being honest after all.

Given that there's three possibilities and you're only honest in one of them, there's a 2/3 chance you're of shit.

1/2

conditional probability. The question is worded differently to Betrands problem

>3/6 balls are golden so it's a coin toss that the first ball you draw any given time you run the experiment will be golden
The problem specifically states a gold ball has already been chosen, that is an initial requirement for the experiement, so the odds of not picking a gold ball initially are not relèvent.

You’re arguement is like saying a coin toss isn’t 50/50 because there’s a chance you don’t have a coin. A coin toss problem assumes you have a coin, this problem assumes you have already pick a gold ball, in both cases you only consider possibilities that have the initial conditions already satisfied.

>1 million samples, his cpu hates him
>uses virgin format method instead of chad % builtin
>values instead of truths for 2 choices
>not spyder ide
you're not going to make it

phd in Physics, astrophysics, and mechanical and electrical engineering
0% is the answer

thank me later

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I modified it with cheatengine, it was 3/4 fag get rekt