>>8347660

>crypto intellectuals

Attached: 1521080196037.jpg (701x576, 129K)

Other urls found in this thread:

en.wikipedia.org/wiki/Bertrand's_box_paradox
onlinegdb.com/SJ7-knDFz
onlinegdb.com/r1kDI2vKf
twitter.com/NSFWRedditGif

fucked up the link lmao

1/3

its 2/3 user

Not this shit again
Last time was over 300 posts of people trying to explain conditional probability to morons who thought the answer was 1/3rd

Attached: 220px-Einstein-formal_portrait-35.jpg (220x285, 16K)

yea i fat fingered it, sorry

am I dumb for thinking its 50%?

just a little
but you're better off than anyone who said 1/3

No. You know that there's a gold ball in the box, so it can't be box 3. It's a 50/50 shot you're in the goldibocks, from there. It's only at this point that the question is posed

>grab a box
>pull a gold ball out of box
>I now know that I either have the box with two gold balls or the one with both kinds.
Sorry how could it be anything other than 50%

Attached: 1354896339846.jpg (500x490, 71K)

embarrassing

jsut take the total number of gold balls and divide it by the ones that are left 2/3. it's not 50% that's not ho w it works

here's your math homework

Attached: 1521090829157.png (701x576, 264K)

Ah. I feel stupid for thinking 50/50 now. I understand why I had the thought process so I am not that mad though.

Because sometimes there can be two choices with unequal odds.

This is one of those tomes.

2/3

Attached: bertrandBox.png (728x441, 14K)

No, this is wrong. It's absolutely 50%. Once you've pulled the first gold ball out, you are in a state where the box has just one other ball.

That other ball is either a gold ball, or a silver ball. It's 50%.

The 2/3 answer assumes you would be able to reach into the other box as well, but that's not how the problem was phrased: "What is the probability that the next ball you take FROM THE SAME BOX will also be gold?"

What the fuck. You removed a critical aspect of the original question.
You should be. He removed the part that said "from the same box" ffs.
The answer is 1 in 2.

as in any good con
you're fooled in the very beginning
"grab a box" isn't as simple as you think

You're right. The other answers are just trying to confuse you with their convoluted bullshit

2/3

Attached: bertrandBox_diagram.jpg (125x118, 3K)

Check this out. You have 1 box with 999 silver balls and 1 gold ball. And another box with 1000 gold balls. You close your eyes and pull out a gold ball from one of the boxes, you don't know which one. What's the chance of pulling another gold ball.

Exactly. You have to reach into the SAME box, and the question is posed at the point of only 2 possible outcomes. Anything that happened previously is irrelevant (think gamblers fallacy)

Except you don't know what box you picked.

If you picked from the first box you'll have a 100% chance

If you picked from the second box you'll realistically have a 0% chance

But there's a 2/3 chance to pick the first box, so the answer is 2/3.

Not enough information. Do I have to reach into the same box again?

Yeah.

>but there's a 2/3 chance to pick the first box
how, exactly
each box has a 1/3 chance of being picked at random lol

50% because there's only two boxes

next

what part of
"next ball you take from the same box"
was hard to understand?

One in two, then. I've either already depleted the box of gold (zero chance of picking another) or will definitely pick another, assuming the second box has 1000 gold balls and nothing else in it.

If you have to reach into the same box again, it's 50%.

Either you grabbed the 1 gold ball, so the remaining 999 balls are going to be silver, or:

You reached into the box with gold balls, so the remaining 999 balls are going to be silver.

50% chance.

Because ANY time you pick a silver ball it doesn't count. You HAVE to pick a gold ball. And the only way to do that is either the double gold or split box. And you have a higher chance of picking gold from the double gold box.
See

reread his post, he didn't specify like OP did. I was just being specific.

A person first picks a box from the 3 and then draws both balls from it one at a time. Let E be the event that the second ball he draws is gold and G be the event that the first ball he draws is gold. We are looking for the probability

P(E|G)=P(E,G)/P(G)

P(E,G) = 1/3 because it would mean that the person chose the first box out of 3.
P(G) = 1/3*1+1/3*1/2 = 1/2. If the person chose the first box then it is 100% chance that 1st ball is gold. If the person chose the 2nd box then its 50% chance the first ball is gold.

so the final answer is (1/3)/(1/2)=2/3

FAIL, sorry, I meant gold. Let me try that again:

Either you grabbed the 1 gold ball, so the remaining 999 balls are going to be silver, or:

You reached into the box with gold balls, so the remaining 999 balls are going to be gold.

50% chance.

There's no "have to" anywhere. The problem specifies the following given information:
1.) you have already picked out a gold ball
2.) you must pick from the same box
There is no rule that says you HAVE to pick a gold ball.

Frequentist approach:

ok, say we do this 1500 times

500 times you pick from SS
500 times you pick from GS
500 times you pick from GG

SS: 0 favourable
GS: 250 favourable
GG: 500 favourable
--------------------------------------------
250+500= 750 favourable ("It's a gold ball")

500 of those 750 times you have locked into the GG box.
500/750 = 2/3

Except that's fucking wrong lol. You (the person solving the problem) haven't chosen the first ball. The ball has already been chosen for you.

There are 3 gold balls I could have picked and 2 gold balls remaining to pick. The odds should therefore be 3/2 I think.

No, read the OP again. Your math is assuming you have a chance to pick from the box with two silver balls in it.

winner winner chicken dinner

You...take a ball...It's a gold ball.

lrn2read

Right. But the only way to pick a gold ball is to choose from one of the boxes. What has a higher chance: picking a gold ball from the all gold ball box or the 1:999 chance box. If you pick a silver then you try to again until you get a gold ball. If you pick the 1:999 box you have a 1 in 1000 chance of getting a gold ball. If you pick the other box it's 100%. So we can see if you DO manage to find a gold ball, it's really likely that it came from the all gold box. And if you pick another ball from the SAME box you got the first one from, it's really likely to be gold since it's likely you got the first gold from the all gold box.

Attached: lkje1prff1j01.jpg (660x315, 31K)

This guy gets it!

I don't think you understand. The question clearly states that you already picked a box that has a gold ball, But you don't know which box. So we have to find the probability of picking the box with the gold ball.

The probability of picking the first box with the gold ball is 2/3

The probability of picking the second box with the gold ball is 1/3

The probability of picking the third box with a gold ball is 0/3

The probability that the second ball in the first box is gold is 100%

The probability that the second ball in the second box is gold is 0%

The probability that the second ball in the third box is gold is 0%

Therefore:

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

If we knew which box we first picked, it'd either be 0% or 100%.

>i'm right but only if i make up a totally different problem than the one in the op
Are ALL 50/50 fags just stupid to begin with? I always want to give benefit of the doubt since even the smartest people can be fooled by unintuitive problems, but for some reason they inevitably back themselves into a corner with increasingly ridiculous reasoning to get their wrong answers.

Y'all are hopeless idiots. It's 2/3. True, you either have Box A in which case the other ball is gold, or you have Box B in which case the other ball is silver. But, the probability that you have Box A is higher, and that is why the answer is not 50%.

Here I was thinking Veeky Forums was supposed to intelligent. I got taunted on other boards saying I'd never be smart enough to be allowed to post on Veeky Forums or Veeky Forums. Low IQ they said. Way too low they said. And now I see you "high IQ" individuals can't even figure out this simple math problem? Sad!

Wait a fucking second. After we pick the first ball, do we put it back in the box or is it assumed to have been removed?

Yes, when you look at the whole situation.
Once you have that under control, you can move on and inspect the subset of when the first ball taken was gold.
That's why the favorables are 250+500=750
(and not 500+500)

Typically questions like these it's assumed that you don't put it back in, it's removed from the equation.

Except that is irrelevant. Your first probability of picking a box with a gold ball in it is 100%: The question is only asking for the cases where a gold ball was picked on the first pick.

Since a gold ball was picked (as stated in the problem), that probability doesn't matter.

From there it's simple to see it's 50%.

en.wikipedia.org/wiki/Bertrand's_box_paradox

>the probability that you have Box A is higher

No it isn't. Gambler's fallacy

See and

>the probability of picking the first box with the gold ball is 2/3
No, it isn't. The probability of picking either the first or second box with either one or two gold balls in it is 100%, because it has ALREADY BEEN DONE BEFORE THE PROBLEM HAS STARTED.

>>It's 2/3. True, you either have Box A in which case the other ball is gold, or you have Box B in which case the other ball is silver. But, the probability that you have Box A is higher, and that is why the answer is not 50%.

I think this is the best explanation I've heard yet.

based retards

It's easy to do this experiment yourself if you don't believe the math:

Get 4 marbles or coins or something of identical shape but different color.

Then mix them up, take two into each hand, and set one down. If the one you set down was a gold one, check the color of the remaining one in that hand.

Mark down if it's silver or gold. Repeat 30 times. You'll see you get more gold answers than silver. Science!

the problem OP posted stated the first ball will always be gold (It's a gold ball). As such the possibility of picking a silver ball first is always 0%, no matter which box he picks from at first. If you always pick a gold ball you either picked one of two golds from Box A or the only gold ball from Box B.
Much like how you can't pick Box C because it is has only silver balls and therefore no gold ball to pick you can't pick the silver ball from Box B first as it is not gold.
50%.

75%, this is 4th grade shit.

You had a 100% chance to choose a golden ball, you didn't choose a box that has a golden ball, therefore implying probability of getting which box with the golden ball.

2/3.

The problem states the first ball will always be gold. Box B could have 1 gold and 999999 silver balls but the probability of picking a gold ball first from said box would be 100%.
The use of random selection of balls at first refers to the random selection of gold balls within a box.

tried this and got 22 gold, 8 silver.

2/3

Because you don't have probability defying abilities like the OP is assumed to and is capable of choosing a silver ball first. Remove all the silver balls and boxes without gold balls (what OP has a chance of picking) and try again.

>>Remove all the silver balls and boxes without gold balls

Probability of choosing a gold ball: 100%
10/10 would chance again

New problem:

There are two boxes.
Each box has 10 cupcakes in it.
In one box, 9 of the cupcakes contain small bombs that will explode exactly 10 minutes after consumption, spilling out your guts in front of you.
In another box, only one cupcake has a small bomb.
It is impossible to tell the difference between a normal and loaded cupcake until eaten.
The boxes are unlabeled and it is impossible to tell which is which.

You pick a box at random and open it. The cupcakes all look the same to you, so you pick a random one and eat it. The cake is fresh and moist, and the cream cheese frosting is perfect, but the sprinkles are stale and the thought enters your mind that you might have preferred it without them.

5 minutes pass. So far so good, but you won't know for sure what will happen until another 5 go by. As you anxiously await your fate, you notice that the door to the chapel, where you're enjoying your dessert, is unlocked! All of your family members should be fast asleep in the guesthouse, but you suspect your chubby nephew likes to take some late night strolls so you take action, just in case. The door locks with a satisfying *click* and you briefly consider that it would be quite a mystery if you ended up dead here in this locked room. Amused by your little fantasy, you notice that 10 minutes have finally passed and you are somehow still whole.

You sit back down. One cupcake is not enough for you; you knew this from the start. But you also don't want to eat any small bombs. You contemplate for a good minute before hunger forces you to act. You open the OTHER box and eat a cupcake from it. Your reasoning is as follows:

>I just eliminated a safe cupcake from the open box. This reduces the chance of another safe cupcake coming from that same box. It may even be that I ate the ONLY safe cupcake in the box! At least if I eat from the other box, I must have some chance of picking a safe cupcake.

What are the odds that your family wakes up to a locked room mystery?

Attached: cupcake bomb.jpg (960x1122, 178K)

Yes the probability of choosing a gold ball first is 100% as it clearly says here.

>It's a gold ball

It is
>I flip a coin. What are the odds of me flipping heads twice in a row
vs.
>I flip a coin. It is heads. What are the odds of me flipping heads twice in a row.

>What are the odds

Pretty good odds... You should have eaten from the same box again.

(since you didn't asplode the first time, you probably picked from the box with only 1 bomb. so eat more from that box!)

I just mean that you said to remove all silver balls, and boxes with no gold balls. That leaves one box with two (gold) balls, and 1 box with 1 (gold) ball.

It's impossible for the second pick to be a silver ball, since you specified there aren't any.

Only remove them for the first pick. If you want to put the silver one back in the box with the single gold to represent the second pick you can but it should be obvious at that point what the next ball you will pick given the number of gold balls left.

If you picked the dangerous box first you have a 90% chance of living. If you picked the safe box first you have a 10% chance of living.
Given that you lived long enough to know you ate a normal cupcake both scenarios are equally likely so you have a (10+90)/2 = 50% chance of living.

>Only remove them for the first pick.

In that case that's exactly what was described in

stage #1: 3 ways to have picked the first gold
stage #2: Next ball, 1 silver, 2 golds

That includes the possibility of choosing a silver ball first, therefore it's a different problem.

No it doesn't. Read again:
>If the one you set down was a gold one, check the color of the remaining one in that hand.
You don't mark anything down if you pick silver first.

No it doesn't, if you read what's written:

>>If the one you set down was a gold one, check the color of the remaining one in that hand. Mark down if it's silver or gold.

The check is only done if the one you picked was gold. In other words, it does not include the possibility of choosing a silver ball first.

>If the one you set down was a gold one
>If
So there are times where he selects a Gold-Silver pair but picks silver first. Which is impossible given that he always picks gold first.

Here's the problem explained for retards

>You saw the three boxes and noted the balls and their colors in the three boxes
>You pick out a ball from a random box that you don't know about
>It's a gold ball
>You remember that only two boxes have gold balls
>The first box has two out of three gold balls, meaning you had a 2/3 chance of picking this box
>The second box only has one gold ball, meaning you had a 1/3 chance of picking this box
>You are asked to guess the probability picking ANOTHER gold ball
>Well if you picked the first box, you have a 1/1 chance of picking a gold ball
>If you picked the second box, you'd have a 0/1 chance of picking another gold ball
>So a 2/3 * 1 = 2/3 chance of picking a gold ball from the first box and a gold ball again
>And 1/3 * 0/1 = 0/3 chance of picking a gold ball from the second box and a gold ball again
>Add the probabilities up: (2/3 + 0/3) = 2/3 chance of picking a gold ball on the second try, given that you chose a gold ball on the first try

Was that so difficult?

>So there are times where he selects a Gold-Silver pair but picks silver first.
Yes, but fortunately they are not included in the results of the experiment.

Your obsession with probability-defying powers is getting tiresome. You're lucky you didn't have to eat that cupcake because it would have been extremely likely that you picked an exploding one the second time around

Read the OP more carefully next time. It's not telling you that you have magic gold picking powers. It even specifically says that you cannot look inside the boxes. It's just telling you what happened. You picked a box at random. You picked a ball at random. The ball happened to be gold.

OP is not predicting what will happen, it's telling you what already happened and asking you to use this information to figure out the probability of what will happen next. The fact that you RANDOMLY picked a gold ball tells you something about the box you picked (and the boxes you didn't pick), much in the same way that Monty opening a door that he knows contains a goat gives you information about the doors he chose not to open.

It's more likely that you picked it out of the first box than the second. Since it's more likely that you picked it from the first box it's not 50%.

onlinegdb.com/SJ7-knDFz
>Events: 3334921
>Out of: 5002330
>Raito: 0.666674

I understand what you 2/3-ers are saying, but as the problem states, you pick the box first, so it's irrelevant what the number of balls in the box are.

If for example we have 2 boxes where one has 1000 golden balls and the other has 1 golden and 999 silver balls, you first pick one of those two boxes at random, and then the invisible hand of fate gives you a golden ball from either of them. So now you either have a box of 999 golden balls or 999 silver balls.
We're not counting the number of balls at all, we're counting the likelyhood that the next ball in the box which you picked randomly is a gold one. And the likelyhood can only be 50/50 because there are only 2 boxes. If you picked the first box, you're 100% going to get the golden ball, if you picked the second box it's 100% going to be a silver ball.

The "you put your hand in and take a ball from that box at random" is a red herring, because it's already decided that you get a golden ball no matter which box you picked.

I hope this clears up my thinking.

Alright let me propose a similar but clearer problem.
You have 2 chickens, one of them lays golden eggs, the other one lays regular eggs, but it has a golden egg stuck up its butthole. You pick a random chicken and it lays a golden egg, what's the probability that the next egg is going to be golden?

After you've picked the gold ball, one of two things have happened and you don't know which one. When you've picked a gold ball, it's more likely that the outcome of the first choice was that you picked the first one.
2/3rds of the time when you pick up a gold ball it will be because you picked the first box.

How about three chickens, one lays only golden eggs, one lays golden eggs 50% of the time, and the other lays no golden eggs.
You pick up a golden egg from one of the nests but you don't know which chicken lays in that nest. What is the probability that the next egg the chicken lays is going to be golden?

If you made the first choice many times you'd have 50% times when you got a gold ball and 2/3 of those times would be after choosing the first box.

The code says you're wrong.
onlinegdb.com/r1kDI2vKf
>Events: 50000222
>Out of: 50050267
>Raito: 0.999000105234

Same guy here.
I think I understand what the issue is.
The 2/3 argument would be that you have a golden egg laying chicken, and a chicken who lays golden eggs 50% of the time and regular eggs 50% of the time.
If you picked a chicken at random and it lays a golden egg, then of course there's a 2/3 possibility that you have a 100% golden egg laying chicken.

But the thing is that golden egg laying is predetermined for both chickens as they are presented to you. The word "randomly" threw people off.

Imagine that it's the same chickens, but the regular egg laying one lays golden and regular eggs cyclically.
You pick a chicken, but it has already laid its golden egg.
The night before, one of the chickens laid a regular egg, but the chicken peddler closed shop that day because he wants both chickens to have laid golden eggs, because then people might buy the cyclical chicken, which is valued less.
You simply have a choice between 2 chickens with a golden egg beside them, you don't get meaningful info from this.

You assume that the precedent doesn't have external agency involved, as the problem states, we know that you were going to pick up a golden egg from the nest anyway, all regular eggs are discarded, the crooked chicken peddler exchanges all regular eggs in all the nests with golden ones.

You don't actually get to pick a ball though, it's an illusion. You pick the box.

>The word "randomly" threw people off.
It would be more accurate to say that people who came up with the incorrect 1/2 answer failed to comprehend the premise of the question and would rather make up a completely new question that fits their solution.

The answer to the OP's question as written is undeniably 2/3.

If you assume you have a magic gold tractor beam that always picks a gold ball first in any box that has at least one, then the answer is undeniably 1/2. Feel free to turn that corrupted version of the question into a fun image and see how many people you can bait with it in a new thread. It would actually be pretty interesting: I bet some will skim it and try to answer 2/3 because they think it's something they already solved.

You don't pick the ball!
You pick a box, and a golden ball gets magnetically pulled out of it.
What the premise is that SS gets completely discarded because there are no golden balls to take out of them first.
If you pick G(1000xS) then the G gets deducted, you have 0 chance to pull a G from it now.
If you pick 1000xG, a G gets deducted, you have 999xG left, nothing else, you have 100% chance to pull a G.
It entirely depends on which box you picked.

Yeah but there's a 2/3 chance the first gold ball came from the box with two gold balls

>The "you put your hand in and take a ball from that box at random" is a red herring, because it's already decided that you get a golden ball no matter which box you picked.
Rephrasing into an equivalent problem:
I'm going to hand you a box.
0.1% of the time, it's going to have a silver ball in it.
99.9% of the time, it's going to have a gold ball in it.

Now I hand you the box and you claim the chance of the ball being silver is 50% because you either have it or you don't?
lol

Well that's exactly how my mind comprehends the situation, your own agency in choosing a ball gets taken away when it's directed that you get a golden ball regardless. Thus for me the "you randomly get a golden ball" is contradictory any element of randomness is taken away at that level when there are conditions of result for the entire set up of the problem.

let's tweak to almost(***) match your example:

Frequentist approach:

ok, say we do this 3000 times

1000 times you pick from SS
1000 times you pick from G(999xS) (***)999 instead of 1000 silvers makes math easier
1000 times you pick from 1000xG

SS: 0 favourable
G(999xS): 1 favourable
1000xG: 1000 favourable
--------------------------------------------
1+1000 = 1001 favourable ("It's a gold ball")

1000 of those 1001 times you have locked into the 1000xG box.
1000/1001 = [math]0. \overline{999000}[/math]

1/2 m8

That's not equivalent at all.
First of all, the boxes don't generate balls, each one has a distinct number of balls, second, there are two boxes.

Honestly I think it's just a simple case of misreading the OP. It's not saying "you will get a gold ball if you pick one randomly." It's just saying "you picked one randomly and it was a gold ball."

You don't need to add extra shit to the question, there's no crooked ball fondlers involved switching up the boxes. No color ball was preferentially selected.

Hell, if you're so concerned about agency let's go all the way. Forget about you. Jimmy picked a random ball. Jimmy used the random.org app on his smartphone to pick a random box and then again to pick a random ball. He holds up the ball he picked.

If the ball he holds up happens to be gold, what is the probability the remaining ball in the box is gold?

If the ball he holds up happens to be silver, what is the probability the remaining ball in the box is silver?

The answer is 2/3 in both cases. All OP does is put you in the position of Jimmy after he lifted up that gold ball. If you're saying there's no information to be gained about the box Jimmy picked by looking at the color of the ball he picked, then I don't know what else to say. You might just have trouble with hypothetical scenarios in the first place.

But that's what it is.
You have magical fingers that find you the golden ball first every time.