/sci/ - Science & Math

[Anonymous No.15663921]

is this the matter of fundamental perception of reality?
it appears there are two schools of thought
is this shit quantum?

[bodhi No.15663938]

>>15663921
bitch speak english

[Anonymous No.15663953]

>>15663938
why speak inferior language?

[Anonymous No.15664007]

Why do people think "if there were 100 doors then switching is beneficial". Yeah of course IN THAT CASE switching gives you better odds. But the original question is about 2 doors not 100. With two doors, there is no difference in odds, but with each door added, switching becomes more and more beneficial.

[Anonymous No.15664051]

>>15663921
Statistic gives you info about sets greater than 1
When you only have 1 try its 50/50
It either does or it doesn't does

[Anonymous No.15664058]

>>15664007
There are three doors retard.

[Anonymous No.15664080]

>>15664007
still with either 3 or billion doors you are down to 50:50

[Anonymous No.15664447]

>>15663921
>it appears there are two schools of thought

Yes, the one before and the one after taking an intro to probability course

[Anonymous No.15665514]

>>15664447
>Scenario 1: you pick the winning door first. You are shown one of the two goats. You switch. You lose.
>Scenario 2: you pick the first losing door. You are shown the other goat. You switch. You win.
>Scenario 3: you pick the other losing door. You are shown the other goat. You switch. You win.
You win 2 out of 3 times by switching.

[Anonymous No.15665601]

>>15664080
There's only 18 possible scenarios, so you can easily list each one out and prove it to youself. If you don't switch, you win in 6 scenarios. If you switch, you win in 12 scenarios.

[Anonymous No.15665697]

people say you had a 1/3 chance of picking the car in the first choice so that must mean it was a 2/3 chance of being wrong so you're more likely to win by switching,
but you also had a 1/3 chance of picking a specific goat, like if the goats had names Ray and Charles.

[car] [ray] [charles]

the game doesn't care about your initial pick, it will always reveal either [ray] or [charles] and ask if you want to pick again between what is left over.

you had a 1/3 chance of picking the car
you had a 1/3 chance of picking ray
you had a 1/3 chance of picking charles

if charles is revealed, then the game becomes:

you had a 1/3 chance of picking the car
you had a 1/3 chance of picking ray

and since theres only 2 options left, the odds change:

you have a 1/2 chance of having picked the car
you have a 1/2 chance of having picked ray

you did not pick both ray and charles if you picked either ray or charles initially. ray and charles together are 2/3, and the car is 1/3, but ray doesn't become two goats if charles is removed, and neither does charles become two goats if ray is removed.

[Anonymous No.15665712]

>>15663921
It's a matter of there being a certain subclass of "people" who aren't smart enough to tackle any non-trivial problem, but who desperately want to be "intellectuals", so they obsess over trivial kindergarten problems trying to come up with some new hot take, which inevitably ends up being retarded and wrong because the correct solutions to trivial problems tend to be trivial.

[Anonymous No.15665748]

>>15663921
Choice issue, separate the two choices from each other and evaluate them, then ask what is the question desiring you to evaluate.

Your first choice is 1/3. One element removed (non-goat), the next choice you make is 1/2.

Would adding a new door change your odds of getting a goat?

[Anonymous No.15665752]

>>15665748
>Your first choice is 1/3. One element removed (non-goat), the next choice you make is 1/2.
Perfect example of: >>15665712

[Anonymous No.15665762]

>>15665752
Hahaha u maddddd xD

[Anonymous No.15665764]

>>15665748
>>15665762
>>>/r/teenagers
Go back.

[Anonymous No.15665769]

>>15665764
This is like some kind of self worth issue for you bro the irony LOOL

[Anonymous No.15665773]

>>15665752
you as the player do not know whether or not you have picked from the grouping of [1/3] or the grouping of [2/3] until the end of the game at the final reveal of all doors; provided you believe in the grouping division of [1/3] = car, and [2/3] = two goats.

there is a potential game of
[car: left door] [goat: middle door] [goat: right door]
in one playable version, the left door is chosen, and the middle door is revealed to be a goat.
in another playable version, the left door is chosen and the right door is revealed to be a goat.
in another, the middle door is chosen, and the right door revealed to be a goat.
in another yet, the the right door is chosen, and the middle door is revealed to be a goat.
4 playable versions.
in two of those versions, 2/4, if you picked the left door, and stay, you win the car.
in the other two of those versions, 2/4, if you picked either the middle or right door, and switch, you win the car.

2/4 win by staying
2/4 win by switching

[Anonymous No.15665775]

>>15665773
Please consult >>15665712 and perform some introspection.

[Anonymous No.15665781]

>>15665775
>claim others are retarded
>count 4 different things as only 3 things

[Anonymous No.15665783]

>>15665781
And the mongoloid just keeps going, getting increasingly less coherent with every post.

[Jealous No.15665784]

The problem was that the statistics on the show didn't match the provably correct prediction.

[Anonymous No.15665787]

>>15665783
This is bizarre... Are you actually just describing your own post as you're making it?
My dude, put the spaghetti back in your pockets and gtfo.

[Anonymous No.15665789]

>>15665787
See >>15665712 and >>15665764.

[Anonymous No.15666044]

yes because the goat hears the first door opening and is trained to move, this is a typical trick employed at carnivals to fleece people.

[Anonymous No.15666066]

>>15665697
>if charles is revealed, then the game becomes:
>you had a 1/3 chance of picking the car
>you had a 1/3 chance of picking ray
>and since theres only 2 options left, the odds change:
>you have a 1/2 chance of having picked the car
>you have a 1/2 chance of having picked ray
That's not how it works nigga

[Anonymous No.15666074]

>>15665773
>in one playable version, the left door is chosen, and the middle door is revealed to be a goat.
>in another playable version, the left door is chosen and the right door is revealed to be a goat.
>in another, the middle door is chosen, and the right door revealed to be a goat.
>in another yet, the the right door is chosen, and the middle door is revealed to be a goat.
In the first two scenarios, the host selects randomly between the goats, therefore these are both half as likely as the latter two scenarios, in which the host's choice is constrained by his inability to reveal the car. You will find that this makes up for the discrepancy between your proposed solution and the commonly accepted answer.

[Anonymous No.15666184]

>>15666044
Did they also train it to drive

[Anonymous No.15666297]

>>15666074
so

in scenarios where the host is constrained and may only reveal one specific goat, switching is guaranteed to win.

but in scenarios where the host is unconstrained and may reveal one of either goat, staying is guaranteed to win.

how do you feel to determine which scenarios happen, or how often they happen?

[Anonymous No.15666345]

>>15666297

! = picked door
@ = host reveals door
c = car
ga = goatA
gb = goatB

constrained:
[!ga] [@gb] [c]
[@ga] [!gb] [c]
[!gb] [@ga] [c]
[@gb] [!ga] [c]
[c] [!ga] [@gb]
[c] [@ga] [!gb]
[c] [!gb] [@ga]
[c] [@gb] [!ga]
[!ga] [c] [@gb]
[@ga] [c] [!gb]
[!gb] [c] [@ga]
[@gb] [c] [!ga]
>12 scenarios guaranteed to win by switching; host is constrained and may only reveal 1 specific goat, the goat you didn't pick


unconstrained:
[@ga] [gb] [!c]
[ga] [@gb] [!c]
[@gb] [ga] [!c]
[gb] [@ga] [!c]
[!c] [@ga] [gb]
[!c] [ga] [@gb]
[!c] [@gb] [ga]
[!c] [gb] [@ga]
[@ga] [!c] [gb]
[ga] [!c] [@gb]
[@gb] [!c] [ga]
[gb] [!c] [@ga]

>12 scenarios guaranteed to win by staying; host is unconstrained and may choose to reveal either specific goat, because you didn't pick either goat

50/50

[Anonymous No.15666427]

>>15666345

[Anonymous No.15666436]

>>15664007
The point of going to a 100 doors is that it makes the problem more intuitive: Is it more likely that I picked the correct door out of 100 first try, or is the only remaining other door the correct one.
With 3 door it's still the same, except with fewer doors

[Anonymous No.15666441]

Well, which one is it?

[Anonymous No.15666477]

>>15666436
you believe that changing the game rules wouldn't change the outcome. you think if you pick a door out of 100 doors, you're more likely to have not picked the card door among 99 goat doors, and that despite fundamentally changing the game rules with 100 options rather than 3 options, a standard rule will apply and you will get the choice between staying with the door your picked or picking whatever remaining door the host didn't pick after the host reveals 98 goat doors.

this is retarded, and it's unnecessary, because a game like this already exists called Deal or No Deal,. At the beginning of the game, the player choose from 50 briefcases, and all briefcases have an amount of money between $1 to $1,000,000. Is it likely you picked the one briefcase with $1,000,000? no. But it's also unlikely you picked the one briefcase with $1.
The point of the game is to choose "Deal" or "No Deal", where the host will make a monetary offer for what is potentially in your briefcase. The game doesn't just have 1 positive outcome though, even $10,000 might be a lot to win on a gameshow, so going into the game with low expectations can lead to better outcomes. You could have picked the $100,000 case, for example.
One by one, the remaining briefcases are revealed and removed. In such a scenario where there may be, 4 cases remaining, the $1,000,000, the $500, the $100, and the $1, a deal from the host might be $100,000. The player has the choice to take that money and end the game, or continue playing and revealing the remaining cases. So, to clarify, the player has to choose to stay with their picked briefcase all the way through, aiming to either reveal that their briefcase has $1,000,000, or aiming to attempt dealing out at a high reward value.
If the player in this example declined the $100,000 deal, and on the next turn, a case were revealed to contain $1,000,000, that's it, basically game over. The host will offer $200 instead.

point is, you changed the game.

[Anonymous No.15666483]

>>15666477
>you believe that changing the game rules wouldn't change the outcome
no. I deliberately tweaked the parameters of the game to make it easier to explain
I'd gladly reply to the rest of your post, but it's way to rambly to comprehend what you ant to tell me, sorry

[Anonymous No.15666647]

A: [c] [g1] [g2]
B: [c] [g2] [g1]
C: [g1] [c] [g2]
D: [g2] [c] [g1]
E: [g1] [g2] [c]
F: [g2] [g1] [c]
6 initial game starting conditions, no more, no less.
Game arrangements A-F are all the ways to arrange the two goats and the car behind left, middle, and right doors.

(player choice)
{host reveal choice}
A1: [(c)] [{g1}] [g2] : stay
A2: [(c)] [g1] [{g2}] : stay
A3: [c] [(g1)] [{g2}] : switch
A4: [c] [{g1}] [(g1)] : switch

B1: [(c)] [{g2}] [g1] : stay
B2: [(c)] [g2] [{g1}] : stay
B3: [c] [(g1)] [{g2}] : switch
B4: [c] [{g1}] [(g2)] : switch

C1: [(g1)] [c] [{g2}] : switch
C2: [{g1}] [(c)] [g2] : stay
C3: [g1] [(c)] [{g2}] : stay
C4: [{g1}] [c] [(g2)] : switch

D1: [(g2)] [c] [{g1}] : switch
D2: [g2] [(c)] [{g1}] : stay
D3: [{g2}] [(c)] [g1] : stay
D4: [{g2}] [c] [(g1)] : switch

E1: [(g1)] [{g2}] [c] : switch
E2: [{g1}] [(g2)] [c] : switch
E3: [{g1}] [g2] [(c)] : stay
E4: [g1] [{g2}] [(c)] : stay

F1: [(g2)] [{g1}] [c] : switch
F2: [{g2}] [(g1)] [c] : switch
F3: [{g2}] [g1] [(c)] : stay
F4: [g2] [{g2}] [(c)] : stay

from the initial 6 potential starting conditions, there are 24 potential game outcomes based on which door you pick, and which door the host reveals.
12/24 win by switching
12/24 win by staying

If you believe there is somehow a cheat here, that somehow games A1 and A2 are supposed to be consolidated into a single outcome, then we must use some certain logic.
A1: [(c)] [{g1}] [g2]
A2: [(c)] [g1] [{g2}]
A3: [c] [(g1)] [{g2}]
A4: [c] [{g1}] [(g1)]
It is not correct to say that A1 and A2 are equally likely to occur and should be consolidated as a single outcome, because any other game is also equally likely to occur, such as F3 and D4. All 24 game outcomes are equally likely, 1/24

It is also not correct to say that A1 and A2 aren't different, or else B-games aren't different than A-games.

[Anonymous No.15666779]

>>15664007
lolol you are so fucking stupid
You would fail any math course.

[Anonymous No.15666839]

There are four outcomes when you happen to pick a door with the car behind it.
>pick car, host reveals Goat A; you stay; Host reveals Goat B; you win
>pick car, host reveals Goat B; you stay; Host reveals Goat A; you win
>pick car, host reveals Goat A; you swap; Host reveals Car; you lose
>pick car, host reveals Goat B: you swap, Host reveals Car, you lose

There are two outcomes when you happen to pick a door with Goat A behind it
>pick Goat A; host reveals Goat B; you stay; Host reveals car; you lose
>pick Goat A; host reveals Goat B; you swap; Host reveals Goat A; you win

There are two outcomes when you happen to pick a door with Goat B behind it
>pick Goat B; host reveals Goat A; you stay; Host reveals car; you lose
>pick Goat B; host reveals Goat A; you swap; Host reveals car; you win

so there are 4 outcomes if you pick the door with the car behind it, and 4 different outcomes if you pick a door with a goat behind it. 8 total outcomes.
In these 8 outcomes, 4 outcomes are from staying. Of the 4 staying outcomes, 2 choices to stay lead to losing, and 2 choices to stay lead to winning. 4 outcomes are from swapping. Of the 4 swapping outcomes, 2 choices to swap lead to losing, and 2 choices to swap lead to winning.

So there's a 1/3rd chance of 4 outcomes, a 1/3rd chance of 2 outcomes, and another 1/3rd chance of 2 outcomes. Unknowingly picking the car leads to 4 outcomes, while unknowingly picking EITHER goat leads to 2 outcomes (the case scenarios for each goat already describe what happens to the other goat).

if you pick the car, there are 4 outcomes
>stay, goatA revealed, win car
>stay, goatB revealed, win car
>swap, car revealed, win goatA; lose car
>swap, car revealed, win goatB; lose car
if you don't pick the car, there are 2 outcomes
>stay, car revealed, win your goat; lose car
>swap; your goat revealed; win car
33% of the time, you have two opportunities to win by staying
66% of the time, you have one opportunity to win by switching

50/50

[Anonymous No.15667091]

>>15666297
>in scenarios where the host is constrained and may only reveal one specific goat, switching is guaranteed to win.
>but in scenarios where the host is unconstrained and may reveal one of either goat, staying is guaranteed to win.
>how do you feel to determine which scenarios happen, or how often they happen?
Well one of those happens two out of three times, so

[Anonymous No.15667106]

>>15666647
Bitch nigga the key here is that A1 and A2, if treated as separate outcomes, are both HALF AS LIKELY as A3 or A4. And that goes the same for B through F because they are all functionally identical to A and you could've saved yourself a lot of autistic busywork realising that.

The fundamental error people tend to make in Monty Hall threads is to assume that probability is equally distributed among all outcomes, but the point is it's not. Whether you say "you either get it or you don't", "there's only two doors so it's 50/50", or... all this bullshit, they all rest upon this same flawed assumption.

[Anonymous No.15667112]

>>15666839
>33% of the time, you have two opportunities to win by staying
lmao

[Anonymous No.15667171]

>1/3 chance initially
>okay actually it was 1/2 because we wanted to reveal a door.
>50% chance my door is correct right now.
there is no logic to this. If my chance is already 50% then switching my door would still be 50%.

Who is the brainlet that came up with this shit

[Anonymous No.15667178]

>>15667171
>Who is the brainlet that came up with this shit
You

[Anonymous No.15667182]

>>15667178
this is retarded and not in any sense logical. In fact I will say this is just shit for superstitious retards. With three doors there are only TWO possible states after a door is revealed.

[Anonymous No.15667203]

>>15667182
A weighted coin also has two possible states, exactly as many as a fair one.

[Anonymous No.15667208]

if we know the rules then there will always be a ratio of one car door to a number of goat doors that matter in choice equal to (total doors - 2). There will always be a situation where a goat is revealed so one door will always be moot.

[Anonymous No.15667236]

>>15667208
we also know the host will never open the door of the car door so if we happen to choose the car door then we know its either our door or the other door.

[Anonymous No.15667263]

this is why it isn't 50%.

[Anonymous No.15667489]

[Anonymous No.15667525]

>>15665697
>is this the matter of fundamental perception of reality?
no
>it appears there are two schools of thought
no, some people are just wrong
>is this shit quantum?
no
there is nothing weird or spooky about this problem. it is counterintuitive to people whose intuitions about probability are built on the way an idiot woman taught them in middle school. that is because they were taught badly, not because it's a hard problem. literal pigeons can figure it out given the chance to experiment. humans have trouble because they are told it's a math problem and they've had the ability to learn math by experimentation tortured out of them.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3086893/

[Anonymous No.15667530]

>>15667525
did not meant to link that post

[Anonymous No.15667789]

>>15667091
so, two out of three times, regardless if you picked the left door, the middle door, or the right door. So the quality of "2/3", without knowing what is behind each door, is applied to each door.
[ -A-] [ -B-] [ -C-]
[ 2/3] [ 2/3] [ 2/3]
like this.

so, if you pick A, two out of three times, you've constrained the host into only being able to reveal one goat.
and, if you pick B, two out of three times, you've constrained the host into only being able to reveal one goat.
but, if you pick C, two out of three times, you've constrained the host into only being able to reveal one goat.

How strange, I thought there were only 2 goats in the problem, and each goat were behind 1 door respectively. There was also supposed to be a car in there somewhere.
Instead, your logic suggests there are 2 goats behind every door.

>>15667106
> A1: [(c)] [{g1}] [g2]
> A2: [(c)] [g1] [{g2}]
> A3: [c] [(g1)] [{g2}]
> A4: [c] [{g1}] [(g1)]
You believe the weight of A1 and A2 are half as much as the weight of A3/A4. A1+A2 = A3; or A1+A2 = A4.
Why?
All choices, all moves, in any specific game, must add up towards a single game; so all choices, all moves, must add up to 1.
[math] \frac{x}{n} + \frac{y}{n} = \frac{n}{n} ; (x+y=n) [/math]

So lets slightly alter the problem, and say the car is a Quantum Car. Let's say Goat 1 is wearing a red sweater, and Goat 2 is wearing a blue sweater, and depending on which goat the host reveals first, will change the revealed color of the car. If the red goat is revealed by the host, the car will be red. if the blue goat is revealed, the car will be blue. Are these the same outcomes? It's the same exact model of car, regardless of anyone's opinion that "red paint makes it go faster".

> A1: [(c)] [{gR}] [gB] : win red car by staying
> A2: [(c)] [gR] [{gB}] : win blue car by staying
> A3: [c] [(gR)] [{gB}] : win blue car by switching
> A4: [c] [{gR}] [(gB)] : win red car by switching

A1+A2 = A3+A4

[Anonymous No.15667865]

>>15667789
adhd, the post

[Anonymous No.15667986]

>>15667789
in the above example, the quantum car was meant to represent that the host's choice has the capacity to matter; and could have just as well been more realistically visualized as a team of painters hiding behind the car door ready to paint the car at a moment's notice.

A1 has a 16.666% chance of happening
A2 has a 16.666% chance of happening
A3 has a 33.333% chance of happening
A4 has a 33.333% chance of happening
if the desire to win was only a red car, that from a personal perspective the blue car was equivalent to a goat, then staying wins the red car 16% of the time, and switching wins the red car 33% of the time, doubling the odds of winning over staying.

this is because, 33% of the time, you will pick the blue goat, causing the host to reveal red.

33% of the time, you will pick the car, but 16/33 of that it could be a red car, and the other 16/33 of that could be the blue car.

so the quantum car needs to be more quantum. an advantage for initially picking the car door means the car's color will instead be influenced by personal desire, rather than the host's revealed goat color, but still only between red and blue color choices. If the goal was still to acquire a red car, and a blue car was less valuable, then:
> A1: [(c)] [{gR}] [gB] : win red car by staying, personal color influence
> A2: [(c)] [gR] [{gB}] : win red car by staying, personal color influence
> A3: [c] [(gR)] [{gB}] : win blue car by switching, host color influence
> A4: [c] [{gR}] [(gB)] : win red car by switching, host color influence
now,
33% of the time, you win a red car by staying
33% of the time, you win a red car by switching
33% of the time, you win a blue car by switching
In this example, the quantum car not only represents that the host's choice matters, but so does the player's choice, that the player gains an advantage by being self-determined with being initially correct rather than easily manipulated into changing their choice.

[Anonymous No.15668388]

>>15667525
>it is counterintuitive to people whose intuitions about probability are built on the way an idiot woman taught them in middle school
Remember: it was a woman who got the correct answer where many men failed

[Anonymous No.15668390]

>>15667789
What the fuck is this schizo shit lmao

[Anonymous No.15668403]

>>15667789
>so, two out of three times, regardless if you picked the left door, the middle door, or the right door. So the quality of "2/3", without knowing what is behind each door, is applied to each door.
>[ -A-] [ -B-] [ -C-]
>[ 2/3] [ 2/3] [ 2/3]
>like this.
Where the fuck do you think that probability comes from in the first place? It's 2/3 because there are two goats and three doors. Simple as. So yeah, each door is always 66.66% likely to have a goat behind it rather than a car. That is in fact the logic that leads to this probability.

>You believe the weight of A1 and A2 are half as much as the weight of A3/A4. A1+A2 = A3; or A1+A2 = A4.
>Why?
Because if you select the car initially, the host will randomly select either goat, with a probablity of 1/3*1/2 for each, but if you select a goat, the host's hand is forced, giving a 1/3*1 probability of the other goat being revealed.

[Anonymous No.15668569]

>>15664007
I can't tell if this is bait.

[Anonymous No.15668572]

>>15667789
>Instead, your logic suggests there are 2 goats behind every door.
>2/3 = 200%

[Anonymous No.15668710]

>>15665697
>and since theres only 2 options left, the odds change:
>you have a 1/2 chance of having picked the car
>you have a 1/2 chance of having picked ray
No this is not true. You had 1/3 chance of car and 2/3 chance of Ray. Just because they're two doesn't mean it's 50-50; the goat absorbs the probability of the other one because if you draw the probability tree they lead to the same result.

[Anonymous No.15668748]

>>15668388
both parts of the noun phrase "idiot woman" are important to the meaning of the sentence. "idiot" is not meant to be read as a redundant intensifier, it adds information

[Anonymous No.15668771]

>>15663921
purposefully badly phrased problem to look smart and smug

[Anonymous No.15668816]

>>15668388
>anon1 specifically says "idiot woman"
>anon2 defends women by assuming the phrase "idiot woman" is redundant
Accidental sexism from white knights is always hilarious.

[Anonymous No.15668832]

>>15668748
>>15668816
Why the explicit assumption that the idiot is a woman? What does it add to our understanding of the situation? Would an idiot man have taught probability correctly in middle school, and is that not an implication that an idiot man is still somehow less of an idiot solely for being a man? In short, aren't you being disingenuous?

[Anonymous No.15668848]

>>15668832
There are far more female teachers than male teachers beneath the high school level. If you guess that a woman taught a middle school or elementary school class you'll be correct an overwhelming majority of the time.

[Anonymous No.15668854]

>>15668832
>aren't you being disingenuous?
Project harder.

[Anonymous No.15668855]

>>15668832
>Why the explicit assumption that the idiot is a woman?
two reasons:
1. most middle school math teachers are, in fact, women.
2. women tend to be worse at teaching things at the limit of their own understanding, because they perceive their authority to be fragile and consider being corrected by a student to be a threat.
men are more accepting of arbitrary formal hierarchy: i had a couple male math teachers who were dumber than me in high school and whom i corrected a couple of times when they were teaching something incorrectly, and they were fine with it because they knew their formal authority over the students didn't depend on knowing every bit of the material better than every student.
being an idiot isn't actually sufficient to totally botch teaching simple math. you also have to be insecure in your authority such that when a student correctly points out your mistake you view it as a challenge, and take actions to discourage students from correcting misunderstandings. that is not usually (but not exclusively, of course) a woman thing.

[Anonymous No.15668875]

>>15668855
correction, that is *usually (but not exclusively, of course) a woman thing.

[Anonymous No.15668887]

>>15668855
A
>>15668848
C
>>15668854
F apply yourself