/sci/ - Science & Math

[Anonymous No.10685430]

/sci/ how is this supposed to be 2/3 chance? I cannot wrap my mind around the reasoning of it. 1/2 makes complete sense.

[Anonymous No.10685440]

lol nigga just count the outcomes
you want a MAN'S problem?

[Anonymous No.10685442]

three possible outcomes. two of those outcomes are gold balls. note you are pulling from the same box which you already pulled a gold ball from.

[Anonymous No.10685453]

>>10685442
What I'm thinking is that you already pulled a gold ball, as stated. You don't know which you pulled a ball from. You only know that it isn't the silver box. Now that you already pulled a gold ball from one of the boxes, you either pulled it from the mixed box, where you will get a silver, or you pulled it from the pure box, where you will get a gold. The gold ball was already pulled so why should it matter whether or not you pulled from the mixed or pure box? So 50% makes the most sense because you'll either pull a silver ball or a gold ball next.

[Anonymous No.10685458]

>>10685453
Imagine the middle box has 1000 silver and 1 gold

You just picked a gold ball. Which box did you likely put your hand into?

[Anonymous No.10685465]

>>10685458
I don't understand why it matters which box I more likely put my hand into. They are separate boxes, yes, but I pulled a gold from one already.

[Anonymous No.10685466]

>>10685453
we are not counting boxes. we are counting balls. there are three balls remaining you could pull.

[Anonymous No.10685467]

>>10685466
I have to pull from the same box though.

[Anonymous No.10685472]

>>10685465
you just picked a gold ball mate. Now you're going to guess what color the next ball will be.

Is it more likely you picked that gold ball from the gold ball box, or you somehow managed to pick the 1 gold ball in the sea of silver balls in the middle box?

Take it to the extreme, let's say the middle box has a million silver, and 1 gold

[Anonymous No.10685473]

>>10685440
36cm[math]^2[/math]

[Anonymous No.10685478]

>>10685430
It's always helpful to exaggerate things.
Now assume that each box contains 1000 balls, one only contains golden balls and the other still only contains one gold ball.

Would you still say that there's a 50/50 chance once you pick a gold ball?

[Anonymous No.10685479]

>>10685467
you don't yet know which box you are pulling from because you chose at random. there are two boxes from which you could have pulled a gold ball. therefore there are three remaining balls which you could pull. of those three two are gold.

[Anonymous No.10685481]

>>10685430
Increase the number of boxes and gold balls and it makes sense. For instance, suppose you had 1000 boxes; 998 of them contain 2 gold balls, 1 contains 2 silver balls and 1 contains a gold and a silver ball. Now if you pick a box at random, and you find that you selected a gold ball do you think the odds of the second ball being gold is 1/2 or 998/1000?

[Anonymous No.10685482]

>>10685473

[Anonymous No.10685483]

>>10685472
I already picked a gold. I am limited to the one box. I either picked a gold and will now pick a silver, or I picked a gold and will pick a gold. I don't know why adding more silver balls does anything. I already picked a gold, and since I'm limited to that box I'll inevitably pick a silver or a gold ball.

[Anonymous No.10685486]

>>10685473
explanation?

[Anonymous No.10685492]

>>10685483
mate, if you just picked a gold ball, which box is your hand likely in? You have double the chance of picking a gold ball from the first box as from the second. So two times out of three your next choice will be a gold ball

Think about it like this, there are three gold balls and each one has a pair.
Gold ball #1 is paired with a gold
Gold ball #2 is paired with a gold
Gold ball #3 is paired with a silver

so if you just picked a gold ball, two times out of three it's pair will be a gold

[Anonymous No.10685496]

>>10685467

you KNOW that your box is not the one with two silver balls.

you DON'T KNOW which of the two boxes that have gold balls you picked from.

so you can ELIMINATE the silver box, but you can't ELIMINATE the other two.

So WHICH box of the two that are left is irrelevant.
So there is NO DIFFERENCE between ONE box with three balls left and TWO boxes one with one ball and the other with two OR even THREE separate boxes with one ball in each.

The boxes or the ORDER and ARRANGEMENT of the balls is only relevant in eliminating the two silver balls.

[Anonymous No.10685503]

>>10685483
-I already picked a gold
-inevitably

what do you know since you already pick a gold?

not thinking in terms of inevitably. its like saying I have a million gold balls and 1 silver ball what is the probability I will choose a gold ball? if you respond 50% and justify well inevitably ill pick one or the other it doesnt make sense logically

[Anonymous No.10685509]

>>10685496

The ARRANGEMENT of balls is only important for the elimination that the knowledge of pulling a gold ball out gives you by deduction. After that the BOXES don't change the probablility of picking a gold ball out of three balls when two are gold and one is silver ...

That the ARRANGEMENT information allowed you to compute.

Permutation is not combination, and Combination is not Permutation. The problem changed from one where order mattered to one where order does not matter.

[Anonymous No.10685515]

>>10685430
The probability is 1 in 2. The box with the two silver balls is eliminated so now there's 1 in 2 chances of this being the box that has two gold balls or a silver and a gold ball. You already have a gold ball so the next one will be either a gold ball or a silver ball.

[Anonymous No.10685518]

>>10685515

wrong, it's 2 / 3 you have just failed prob / stats. Please try again next year.

[Anonymous No.10685528]

>>10685492
>>10685503
>>10685492
I don't know which box I picked from. In both scenarios, if I pick gold, since there is one gold ball in the two boxes I will either get a silver or gold when further picking from the box. Also if there were 1 million gold and 1 silver that would be a completely different story, because in this scenario there are two balls in each box.

[Anonymous No.10685529]

>>10685518
You pick a gold ball so you know for sure the box with 2 silver balls is out of the equation. There are two possible boxes that you pickes on and there is one ball remaining. It's either gold or silver. I don't understand how it's 2/3.

[Anonymous No.10685532]

>>10685515
You're treating it as if you are just presented with a box which has either a gold or a silver ball in it, but that is not the case. Drawing the first ball tells you more than that. It not only allows you eliminate the all-silver box, but also conclude that your box is more likely to be the all-gold box. Twice as likely as the gold-silver box, in fact. And when you have the all-gold box, your next ball will be gold as well. So that's the same probability.

[Anonymous No.10685536]

>>10685528
>Also if there were 1 million gold and 1 silver that would be a completely different story, because in this scenario there are two balls in each box.
It wouldn't be completely different. You're able to recognise that a million is more than one, surely you also know that two is more than one?

[Anonymous No.10685537]

>>10685503
>>10685515

see >>10685496 >>10685509

Bayesian problems are one where the probability changes because of new information. So think of it this way.

If I have a box of 1000 chickens where one is alive and the others are dead, the probability of picking a live chicken is 1 in 1000.

However if I told you I removed all of the dead chickens, the probability now of picking a live chicken is 100% because that is the only chicken left.

you started off with 6 balls, 3 silver and 3 gold,

But you used the arrangement of boxes to deduce that by picking a gold ball (new information) that the population size has been changed. Now there are only 3 balls left, two of them gold, one silver. Hence a 2/3 probability.

All Bayesian problems Play with the adjustment of the population size.
In the Monty Hall problem, Monty removes one of the choices changing the population size and changing the problem.

[Anonymous No.10685541]

>>10685536
It's asking the chance that I pick another gold ball. In the 1 gold 1 million silver scenario it doesn't matter because the gold ball was already picked and in one box there are no more silver balls. If there were 1 million gold the other box would have an extremely low chance of picking any silver at all after I picked 1 gold.

[Anonymous No.10685544]

>>10685528
well done troll thread

[Anonymous No.10685545]

>>10685544
I'm not trolling

[Anonymous No.10685553]

>>10685529

Label the balls, you are holding 1 out of 3 possible gold balls, 2 of them are paired with another gold ball and one is paired with silver ball.

This problem is hard and counter-intuitive, but read the thread. Many anons have explained it before me

[Anonymous No.10685554]

>>10685541
I accidentally mistyped. I meant in the 1 gold 1 million silver there are no more gold balls in one box.

[Anonymous No.10685555]

>>10685430
Since the probability of choosing the gold ball out of the first box is higher, you expect that you picked a gold ball/speck out of the first box.

[Anonymous No.10685558]

>>10685541
See, that's not true. It does matter, because it is statistically very improbable that you managed to pick the gold ball from that box on your first attempt, as opposed to the other box, which has a million gold balls and it would be very improbable for you not to have gotten it. Therefore, at a million to one odds, your next ball is also probably gold.

[Anonymous No.10685561]

>>10685555
It's not asking the probability that I pulled from a specific box, it's asking the probability that I grab another gold ball. In this case it's 50% because I either get a silver piece of dust after picking from the singular gold piece (since it said I picked a gold ball) or I pick one of the many millions of gold specks from the gold speck box.

[Anonymous No.10685562]

>>10685561
>It's not asking the probability that I pulled from a specific box, it's asking the probability that I grab another gold ball.
True enlightenment is to realise that those are one and the same.

[Anonymous No.10685564]

>>10685562
No it isn't. Those are different questions.
Obviously if it is asking the chance that I pulled from a specific box it would be 2/3rd. But it's asking me the chance that I pull another gold ball, which it only makes sense if it's 1/2.

[Anonymous No.10685566]

>>10685555
Quads of fail. There are few enough discrete outcomes here that you can count em on your damn fingers. Quit being a bitch.

[Anonymous No.10685567]

>>10685561
33% of the time you will choose the first box. Therefore, 33% of the time you will choose a gold speck. Do you really think that you have a 16.5% chance of pulling the gold speck out of the middle box every time?

[Anonymous No.10685569]

>>10685567
It doesn't matter. I already picked a box and got gold. Why does it matter which box I chose out of?

[Anonymous No.10685571]

>>10685564
>No it isn't. Those are different questions.
They're not. They're the exact same question. If you pulled from the all-gold box, your next ball will be gold. If you pulled from the gold-silver box, your next ball will be silver. If you know how likely it is which box you pulled from, you know how likely it is that your next ball will be gold.

[Anonymous No.10685572]

Once again we have a thread that proves beyond a shadow of a doubt that people who get the wrong answer are either stupid or evil or silent, justifying killing those who get the wrong answer regardless of the reason, just like nature does...

[Anonymous No.10685573]

>>10685569
>Why does it matter which box I chose out of?
Because that determines what the next draw will be.

[Anonymous No.10685579]

>>10685569
If you pick a gold speck, your immediate reaction should be “I probably chose this gold speck out of the first box” because your chance of picking that gold speck out of the second box is close to zero. If you choose a gold speck, and think that your next speck will be silver half of the time, then you’re admitting that you think you will pull that gold speck out of that second box 16.5% of the time, which is obviously highly improbable.

[Anonymous No.10685580]

>>10685571
>If you pulled from the all-gold box, your next ball will be gold. If you pulled from the gold-silver box, your next ball will be silver.
I already pulled a gold, so what you just said is why it's 50%

[Anonymous No.10685583]

>>10685580
>I already pulled a gold
And what were the odds of that?

[Anonymous No.10685586]

>>10685583
100% because the damn problem asked for the probability given that you already drew a gold you illiterate fucking

[Anonymous No.10685587]

>>10685579
Yes, I would probably say that. But I already grabbed the gold so I'm basing the probability off of what will happen then. I have a gold ball, so the chances of me either grabbing the one gold ball in one box or one of the millions of silvers in the other box are the same.

[Anonymous No.10685590]

>>10685583
It doesn't matter how low or high, the gold ball is out and now I am left with the chances of choosing one more. It isn't like I chose out of one bag where there is three gold and 1 million silver.

[Anonymous No.10685596]

>>10685532
That's retarded. Just because I already drew a gold ball doesn't increase my chances of drawing another one.

[Anonymous No.10685603]

>>10685587
I’m getting tired of this.

Let’s run 900 trials for OP’s pic

Each box is picked 300 times. Therefore you pick a gold ball out of the first box 300 times, and 150 times out of the second box. So, out of 900 trials, if you pick a gold ball, you know that this is twice as likely the first box (300:150).

300/450 = 2/3

[Anonymous No.10685604]

>>10685586
Oh, I'm afraid that's entirely wrong. See, when it tells you something happened, and something else happening is entirely contingent on that, and it hasn't happened yet, you can use the probability of the first thing happening to determine the probability of the second thing happening. If X, then Y -> Y has the same probability as X. To put it another way, you're wrong to think of the first event as merely "you have gold ball now". You have to consider that you having a gold ball means either one of two scenarios took place: you drew the gold ball from the gold-gold box (2/3 chance) or you drew the gold ball from the gold-silver box (1/3). Those are your events. And the next event is entirely contingent on it. That's conditional probability. So yes,
>>10685590
it matters, it is exactly like that and
>>10685596
it does

[Anonymous No.10685607]

>>10685603
>>10685604
Is there any way to empirically prove that this is true or is it mental masturbation?

[Anonymous No.10685612]

>>10685596
All combinations:

First draw - Second draw
1) Gold - Gold
2) Gold - Gold
3) Gold - Silver
4) Silver - Gold
5) Silver - Silver
6) Silver - Silver

This represents every single combination of draws, all have an equal probability of occurring.
However we now that we drew a gold ball first, so eliminate all combinations that include a silver draw first.

We are left with the following combinations:
1) Gold - Gold
2) Gold - Gold
3) Gold - Silver

How many combinations result in a silver ball on the second draw? 1
How many combinations result in a gold ball on the second draw? 2
How many combinations total? 3

Therefore the probability of picking a silver next is: 1/3
The probability of picking a gold next is 2/3

It could not get more explicit than this.

[Anonymous No.10685614]

>>10685607
Yes, just do it a couple hundred times and record the results. Then prostrate yourself before us because we used maths to figure it out instead and you realise you could've saved yourself the trouble if you'd only understood.

[Anonymous No.10685615]

>>10685603
And now let’s run a simulation for >>10685555
Let’s assume the probability of picking the gold speck out of the middle box is 1 in 1,000,000. Let’s run the simulation 9,000,000 times. Each box is picked 3,000,000 times. Therefore, the first box yields a gold speck 3,000,000 times, and the second box yields a gold speck 3 times. Out of all the possible times you pick a gold speck first (3,000,003), 3,000,000 of those picks came from the first box. If you pick a gold speck, you better bet that you picked the gold box. Therefore you expect to pick a gold speck next.

[Anonymous No.10685617]

>>10685564
Look, you don't have a 50% chance of drawing two gold balls from the gold-silver box, do you? You can only do that if you got the gold-gold box. And you don't have a 50% chance of getting that, either. So 50% just doesn't enter into it.

[Anonymous No.10685623]

>>10685617
People act like it's the same as which box you were more likely to choose out of. If I choose a gold, then it is obviously more likely to have come out of the gold box. There are two golds that could be there but only one silver that could be there. So 2/3rd. I already chose gold and it's impossible for me to tell which one I chose it from. Therefore it only makes sense to say that I either draw a gold one next, because I chose the pure box, or I draw a silver one next because I chose the mixed box. So 50%

[Anonymous No.10685631]

>>10685623
See >>10685615

I would like to run a real-world simulation and bet some money. If you pick a gold speck, I will bet $100,000 that the next speck is gold. If the next speck is silver, I will give you $1,000,000. We can run this however many times you like. Are you in?

[Anonymous No.10685635]

>>10685623
your answer would make sense if there was a game show host, or some neutral arbiter who handed you a gold ball from the box once you selected the box. But if you're just picking a ball randomly you are wrong

[Anonymous No.10685639]

>>10685623
Wait a minute. Same as who I'm replying to. I grabbed three pens and a pencil off my desk. I thought to myself, "So here is a gold I drew from the gold drawer. There will be a gold coming next, but since I don't know the box it's coming from it could also be the silver. There are more golds so initially I have a 3/4th chance of choosing gold, and a 50% chance of choosing the all gold box or mixed box, when I choose the mixed box I will be choosing a silver next but I don't know that as the picker. Since there is one with all gold I'm probably going to choose from the one with all gold. When I make an initial decision and I get gold, as would probably happen, I don't know whether it came from the box with gold or silver. However, since one box has two gold in it and another has one gold and one silver in it, when I chose that gold instead of a silver it makes it more likely that the next one is gold. Kind of like Monty Hall where when they reveal a door instead of the other one it's more likely to be in that unrevealed door." So am I getting it and team 2/3rd now? I started at the pens and pencils for 3 minutes and cognitive dissonance started to kick in.

[Anonymous No.10685647]

>>10685639
Basically what I'm thinking is if the box I chose out of has a greater chance of being all gold, since picking a gold out of a gold bag is more likely than picking a gold out of a gold and silver bag, then it's also more likely I am going to pick a gold. Is that right?

[Anonymous No.10685653]

>>10685647
Congratulations

[Fan No.10685675]

You have picked a yellow ball.

- With a probability 2/3 you picked a ball from the box which has 2 balls (because there are 2 yellow balls over 3 in this case). In this case the second ball will be yellow too.

- With a probability 1/3 you have picked the yellow ball which is alone in its box. In this case the second ball will be grey.

Thus you the second ball will be yellow with a probability 2/3.

Btw, yere are you from? I am from France.

Bye

[Anonymous No.10685683]

>>10685675
I already figured it out, I am also American.
At first I realized that if I choose a gold ball it will more likely be gold. If I choose a drawer with gold in it it's more likely to be gold only because choosing a gold from two gold balls is 100% while choosing a gold from a gold and silver is 50%, and choosing a gold from two silvers is 0%

[Anonymous No.10685684]

>>10685675
Salût

[Fan No.10685696]

What do you mean by "I already figured it out"

Have a nice evening

[Anonymous No.10685703]

>>10685696
I figured a while ago and was posting about it.
I hope you have a nice evening to

[Anonymous No.10685710]

>>10685486
I made the inside angles 90 degrees, converted the outside to a trapezoid with the top and bottom lines having the same length, got my answer, shrugged and said close enough.

[Anonymous No.10685719]

>>10685478
There's a 50/50 chance that box is all gold or all silver

[Anonymous No.10685746]

>>10685442
>you already pulled a gold ball from
Cripes people. Learn to explain this problem in non-tautological terms.

OP, it works because in BOTH cases when drawing from the gold box will you draw a gold ball next, but in only ONE case from the gold/grey box will you be able to draw a gold ball as the second ball drawn.

[Anonymous No.10685751]

>>10685647
>>10685746
I'm the OP and I already got it. I understand why I thought it was 1/2 and I now understand why it's 2/3rd

[Anonymous No.10685766]

>>10685751
It's usually 2/3 chance, but if you draw from the silver box there's a 50% chance it would be silver. So while it's usually a 2/3 chance, if you drew a gold ball there's a 75% chance it's from the gold box. Cheers.

[Anonymous No.10685777]

>>10685751
At 64 replies I'd already presumed this, but I noted a flaw in the first reply that—if corrected—should hopefully make future versions of these threads shorter.

[Anonymous No.10685814]

>this thread again

[Anonymous No.10685841]

>>10685814
Ignores the configuration of possible gold-drawing samples from the first all-gold box.

[Anonymous No.10685884]

There are two doors, one with nothing and one with money. You are given the choice which door to open to collect your prize but you don't know which contains which. You pick one of the two at random and then the door that you didn't pick is revealed to contain nothing. You are then given the choice whether to stay with your current choice or switch. What is the probability that you are clinically retarded?

[Anonymous No.10685906]

>>10685612
you are over counting. gold-gold and gold- gold are indistinguishable from each other

[Anonymous No.10685928]

>>10685906
The reason the two gold (or two silver) draws are distinguishable is because the balls are all pre-quantized before the experiment begins. If it were some kind of macroquantum system where drawing either of the "two" balls was physically equivalent to grabbing the first, then the quantization would collapse and there would only be the quantum probability that the box-second-draw system was in a superposition of containing a gold or silver ball.

Since the quantization is pre-material, we already know that there are two distinct gold balls to draw, and not one quantum duplicating gold (or silver) ball.

That is, drawing either ball will cause the other ball to be drawn. There are two possible draws (per box).

[Anonymous No.10685933]

>>10685430
my coding skills are piss poor, and copy paste will look like shit, but anyways this is c++, results are %66 gold if firstball is gold.
#include <iostream>
#include <windows.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

int main(int argc, char* argv[]) {
int firstgoldtracker = 0;
int secondgoldtracker = 0;

int i;

int silver = 0;
int gold = 1;

int boxa = 0; // silver-silver
int boxb = 1; // gold-silver
int boxc = 2; // gold-gold


int boxchoice;

for (i=0;i<=100;i++)
{

srand (time(NULL)+i*2);

int firstball = 2;
int secondball = 2;

boxchoice = rand() % 3;

if (boxchoice == boxb)
{
srand (time(NULL)+(i*3));
firstball = rand() % 2;
}
if (boxchoice == boxc)
{
firstball = 1;
}

if (firstball == gold)
{
firstgoldtracker += 1;
secondball = boxchoice-1;
}


if (secondball == gold)
{
secondgoldtracker += 1;
}

}
std::cout << firstgoldtracker << " " << secondgoldtracker << std::endl;

return 0;
}

[Anonymous No.10685942]

>>10685933

[Anonymous No.10685949]

[Anonymous No.10685965]

>>10685440
how to solve? assuming that the sections aren't to scale

[Anonymous No.10685968]

>>10685482
wrong

[Anonymous No.10686045]

>>10685529
If you got a gold ball, you're much more likely to have gotten it from the box with 2 balls than the box with one.

[Anonymous No.10686051]

>>10685430
I take all the gold and silver balls I can. Silver and gold are only going to become moee valuable as world demand rises and robotics bexome moee popular.

[Anonymous No.10686077]

>>10685440
68/3 \approx 22.7

[Anonymous No.10686093]

>>10685430
You picked a gold ball.
Therefore you are in B1 or B2.
There were 3G 1S across B1 and B2.
Give that you picked 1 gold, you have 2G 1S left in your possible draws.
2 in 3.

Just imagine the other ball in the box is any of the available options, because the knowledge of which box B1 or B2 you have is unavailable to you.

[Anonymous No.10686121]

>>10685814
>Leaving you with 2 possibilities
Yeah one possibility with a 2/3 chance and one with 1/3

[Anonymous No.10686137]

>>10685884
THe solution is to stay with the same door. This is because the conditional probability that the other door contains the prize is affected by opening the empty door. About 4/5 of the time the prize will turn out to be behind the empty door.

[Anonymous No.10686144]

>>10685884
THe solution is to stay with the same door. This is because the conditional probability that the other door contains the prize is affected by opening the empty door. About 4/5 of the time the prize will turn out to be behind the empty door.

Inb4
>the other door has 100% chance of having the prize

[Anonymous No.10686164]

You pick up a gold ball. There's a 2/3 chance that the gold ball is in the gold box. done

[Anonymous No.10686177]

[Anonymous No.10686371]

>>10685777
So correct it.

[Anonymous No.10686382]

>>10685884
Pic related is interesting.

[Anonymous No.10686657]

>>10685928
So is the reason I intuitively get all of these wrong that I did quantum mechanics before I did any probability?

[Anonymous No.10686699]

>>10686657
It's possible it mal-adjusted your intuitive sense of object quantization a little, but nothing you can't overcome. Especially if you've already identified the problem.

[Anonymous No.10686735]

>>10685430
Imagine you have 3 boxes:
1. With billion golden balls
2. With 999999999 silver balls and 1 golden
3. With billion silver balls
You pick a ball and it’s golden. Do you think you’re so lucky that you picked 1 golden ball from box 2? Do you think there equal probabilities that you’ve chosen box 1 or 2?

[Anonymous No.10686749]

>>10686735
in that scenario if I picked a gold ball, there's no way in hell I'd believe it came from box 2. I mean, what are the chances

[Anonymous No.10686752]

you dont understand the definition of probability

[Anonymous No.10686755]

>>10686749
>what are the chances
Now apply this logic to 1 golden/1silver case. Clearly to avoid paradoxes chances should decrease if number of silver balls grows compared to number of golden balls. There is no way there is 50/50 chance between box 1 and 2.

[Anonymous No.10686766]

>>10686735
in that scene it makes no difference if box #1 has only two golden balls

[Anonymous No.10686775]

>>10686766
It does. A box with only two golden balls would weight less.

[Anonymous No.10686785]

>>10686755
I think it's best explained here: >>10685492

[Anonymous No.10686797]

Because the exercise is in thinking past bull-shit pics. First of all, the area view of the problem actually looks like the top row in this pic, secondly, the lower 1 is all you're dealing with at that point.

[Anonymous No.10686818]

Though, if you wanted to be a smartass, you could say:3 in 4, since the boxes are clearly refilled with the same color balls in some way.

>But muh words over illustrations!

True legally, but not intellectually. The pic was presented with the problem in total.

[Anonymous No.10686868]

>>10685555
I choosed the box and got a gold spark from it, always. Description is just confusing. I pick the BOX, not a ball/dust, so I either pick a box full of silver or full of gold.

[Anonymous No.10686881]

Absolute state of /sci/

1.
You have picked a gold ball, which maybe any of 3 gold balls in the 2 boxes.

2.
If it was gold ball #1 in box #1, then the second ball in the box is also golden.

If it was gold ball #2 in box #1, then the second ball in the box is also golden.

If it was gold ball #3 in box #2, then the second ball in the box is silver.

3.
Two outcomes in gold, one in silver. 2/3 end of story, for fuck's sake.

[Anonymous No.10686915]

Screw it go get three boxes, three silver balls, and three gold balls and figure it out.

[Anonymous No.10686978]

>>10685623
>Literally on the cusp of grasping it but then going for 50-50 anyway because "you either have it or you don't"
Has to be a troll

[Anonymous No.10686983]

Hey /sci/, I got two balls you can grab! Haha, got em

[Anonymous No.10686990]

>>10686818
I don't think legality enters into it. But intellectually speaking you should be able to grasp that the illustration depicts the initial situation before any ball is drawn.