/sci/ - Science & Math

[Anonymous No.15297977]

>le probability
What a laugh. People lack common sense and use their flawed "logic".
Just put the best supercomputer to test this "probability" and you will see that it is bullshit.

https://youtu.be/dOQowCeAnRs

I am not denying that the woman is smart, but this is a classical mistake of the so called "smart" people.

The probability of winning will not improve if you switch doors. This is fucking obvious.

[Anonymous No.15297995]

Probability is one of the best midwit filters. When confronted with the Monty Hall problem midwits will either fail to understand it or they will mindlessly repeat some shitty pop sci highschool reasoning (which again demonstrates lack of understanding). Only a true high IQ person can explain the actual issue.

[Anonymous No.15297998]

>>15297977
>>15297995
Being contrarian doesn't make you smart.
Real math/science is about being right, that's all, there's no strings attached.

There is a 2/3 chance of winning if you switch doors and a 1/3 chance of winning if you don't switch doors.

[Anonymous No.15298002]

>>15297995
Are you a true high IQ person?

[Anonymous No.15298003]

>>15297998
Being right for the wrong reasons is worthless. When you're right but can't explain why, then you're no better than someone who is merely guessing.

[Anonymous No.15298005]

>>15298003
>explain why

[Anonymous No.15298006]

>>15297998
Wrong. There is a 2/3 chance of winning if you switch doors and there is a 2/3 chance of winning if you don't.
Being a conformist doesn't make you smart.

[Anonymous No.15298012]

>>15298003
In the original problem here is the solution and explanation:

[x] [o] [x]
There's three doors, one contains an o and two contain an x.

There are three possible "branches" we can take from here, x, o, or the second x, which I'll call x2.

The chance of blindly choosing x or x2 is known as 2/3 probability in math, because there's three choices and 2 of them are x.

Therefore, ignoring whatever happens next, we can confidently say that if I pick a door randomly, there will always be a 2/3 chance I got x, it will never be 1/2 chance, it will always be 2/3 forever, because I am picking right now, between three possible doors.

Then here is the part that midwits don't realize:

The host reveals the other x, and now I only have two doors left, but there are two situations in which he can reveal the other x, because there's two x's, and there's only one situation (out of 3) where I picked o the first time and he revealed an x.

The branches are as follows:

1. I pick x1, and the host reveals x2, I win if I switch.
2. I pick o, and the host reveals either x1, or x2, I lost if I switched. This is trivial and can be considered one scenario only.
3. I pick x2, and the host reveals x1, I win if I switch.

There are only three possible scenarios and in 2 of them switching is better.

THat is called 2/3 probability of switching wins.

As where the initial choice is always 1/3 probability because the branches are liek this:

1. I pick x1 and I stay.
2. I pick o and I stay.
3. I pick x3 and I stay.

Not switching wins 1/3 of the time, switching wins 2/3 of the time.

Also, science simply is what it is, so is math, being right for the wrong reasons is OKAY, Math is only about being right and nothing else matters really.

[Anonymous No.15298016]

>>15297998
>>15298005
You are the contestant in a variant of the Monty Hall problem. You are not made privy to how it differs from the classical Monty Hall problem.

You come on stage and there's three sets of three doors. You go to the first set, pick one, the host opens one you didn't pick, there's a car behind it. You move to the second set of doors and the same thing happens. You move to the third set, pick a door, the host opens a different door and there's a goat. You can stay or switch doors, what do you do?

Does knowing the answer to the classical Monty Hall problem without knowing why it's the answer help you here?

[Anonymous No.15298017]

>>15298006
I guess my poor midwit brain can't comprehend how you add up four thirds to get one

[Anonymous No.15298022]

>>15298012
In your logic there's 4 branches:
1. you choose x and host x2
2. you choose x2 and host x
3. you choose o and host x
4. you choose o and host x2

In 2 cases switching loses, in 2 cases switching wins. Probability 2/4 = 0.5

QED

[Anonymous No.15298023]

>>15298012
>>15298017
Do not disconsider who is doing the math. Math can be right 100% of the time, but the ones doing math usually fucks up. And that is the case here.

Here is how this problem really works:
1. The host reveals one of the doors
2. Now I have 2/3 chances of winning with either doors that have left.

It is not magic. It is obvious. And if the math says otherwise, you can be sure the one doing the math fucked up.

You can prove that by running this test in a supercomputer for a googol times.

[Anonymous No.15298028]

>>15298016
That problem has the same explanation as this one I provided: >>15298012
The only difference is that you added in a stipulation that did not exist in the original problem without understanding that it is inconsequential anyways.

Monty does not open the door with the car, and even if he did, you wouldn't have the choice of switching in those scenarios so all of those cases are completely invalid and make no sense, and shouldn't be included when considering the problem.

[Anonymous No.15298031]

Fun fact: In quantum mechanical Monty Hall switching actually has only 1/2 chance of winning because the host doesn't know which door's wave function to collapse.

[Anonymous No.15298033]

>>15298022
No there's three branches, I already explained why, don't make me do it again.
I'll make it even stupider so you can understand:

1. You choose a goat and he opens another goat.
2. You choose a car (there's only one) and he opens EITHER goat. Switching loses. This is only one scenario because there's only one car.
3. You choose the second goat and he opens the first one.

[Anonymous No.15298036]

>>15298033
>2. You choose a car (there's only one) and he opens EITHER goat. Switching loses. This is only one scenario because there's only one car.
Those are two distinct cases. Don't mix them up. The host has two different choices.

[Anonymous No.15298038]

>>15298016
No but it doesn't hurt either. Anyway, you can't fault people for knowing an answer in one situation but not in another.

[Anonymous No.15298040]

>>15298022
>3. you choose o and host x
>4. you choose o and host x2
If you choose o, the host chooses randomly between x and x2, making either one 50% likely at this point. If you choose either x, the other x is 100% likely. So (branch 3 + branch 4) which represents choosing o has the same probability as either branch 1 or branch 2.

[Anonymous No.15298043]

>>15298036
I acknowledged both cases and then mentioned that it doesn't matter because either way switching still loses in that ONE case. The host has 4 choices but the contestant has 3.

[Anonymous No.15298045]

>>15298040
That can't be right. By that logic total probability is 50% + 50% + 100% = 200% which is more than 100%.

[Anonymous No.15298046]

>>15298043
>The host has 4 choices
There's only 3 doors ... I hope you're just trolling.

[Anonymous No.15298047]

>>15298045
No no no, it's like this:
x: 1/3*1
x2: 1/3*1
0: (1/3*1/2) + (1/3*1/2)

[Anonymous No.15298053]

>>15298031
nah
https://en.wikipedia.org/wiki/Monty_Hall_problem#Quantum_version

[Anonymous No.15298063]

>>15298022
>>15298047
Or, in other words, there are three branches (one for each door) and one of them branches further.

[Anonymous No.15298083]

>>15297977
The chance of always 50/50. You either switch or you don't

[Anonymous No.15298097]

>>15297977
"the best supercomputer"

Bro, you could simulate this problem on a Windows 98 in like 40 lines of code. Why don't you do some science. Post your experiment and results. Then we'll consider your theory. Until then, OP, you are a faggot.

[Anonymous No.15298114]

>To get more data, we wrote a program in C to play the game repeatedly. The program
uses pseudo-random numbers generated by the functions rand and srand [2], available in
standard Unix-based systems. Although the numbers generated are not truly random, they
are sufficiently random for the purposes of this experiment. The program is designed so that
it is easy to change the number of times the game is played and the number of cards used.

>The program was set to generate samples of one million games. In ten runs, the average
percentage of times the switching strategy proved successful was 66.708%, with a standard
deviation of of 0.033%. In a separate set of ten runs, the strategy of always staying with
the original choice succeeded 33.326% of the time with a standard deviation of 0.056%.

https://web.mit.edu/rsi/www/2014/files/MiniSamples/MontyHall.Old.Bad/montymain.pdf

I still need other simulations, other programs, and I need it to run a googol times.

https://youtu.be/4Lb-6rxZxx0
>over the long run
Here. I spotted the problem.
If the person playing the game had multiple chances, it would be correct for her to switch doors. But the person has only one shot. So her chance is 50-50.

https://youtu.be/UgKrQ2ywVfs
The Monty Hall Problem always contemplates multiple choices and not only one.

You /sci/entists need to refine that problem, to reach the proper solution now that you know what is its flaw.

[Anonymous No.15298119]

>>15297977

[Anonymous No.15298130]

>>15298114
Bro you fundamentally don't get how statistics work

Okay, suppose you're on a game show in a computer simulation designed to test the Monty Hall problem and you'll be doing it a million times. Of those million times, there are 667,080 where you will win by switching. You do not know if you are in one of these 667,080 instances or in one of the 332,920 remaining ones. But you can see which number is bigger right? Which one you are more likely to be in if you were randomly dropped into one of the million simulations?

[Anonymous No.15298132]

>>15298119
I was just passing by and I love you for this. If you're the one who did that in excel and it isn't a copy from the millionth time this thread was posted. You are my favorite anon now.

[Anonymous No.15298136]

>>15297977
Why are there always so many of these threads? It's not that hard. For example...
>1 door out of 10,000 has a prize behind it
>You pick one at random
>The host opens 9,998 doors, revealing that they have no prize behind them.
>Do you switch?

The confusion comes in because there is an unwritten rule. The host will not open the door that you select, no matter what is behind it.

[Anonymous No.15298147]

>>15298028
The point is that it's a different problem. You no longer know whether or not Monty knows what's behind a door before he opens it. Understanding the original problem helps in understanding this variant, but if you've only memorized the answer to the original problem then all you can do is say "you aren't asking the question right." Which is what you've done.
>>15298038
>it doesn't hurt either
It could, if it leads you to making the wrong choice.l based on faulty reasoning.

[Anonymous No.15298149]

>>15298114
I want to help you out here but I am also not sure if you're trolling.
>If the person playing the game had multiple chances, it would be correct for her to switch doors. But the person has only one shot. So her chance is 50-50.
Let's go with that to illustrate the problem. Now what are the odds if 1,000 individuals only have one shot? It won't be 50/50, it'll be 2/3 on switch 1/3 on stay. Your reasoning does not work. Probabilities do not change merely because it's a single run.

[Anonymous No.15298151]

>>15298147
>It could, if it leads you to making the wrong choice.l based on faulty reasoning.
What would be the wrong choice here?

[Anonymous No.15298158]

1000 doors, pick one 1/1000
998 dud doors open suddenly, 2 doors left
change for 999/1000

[Anonymous No.15298162]

>>15298151
Not punching the host in the face until he tells you where the car is.

[Anonymous No.15298164]

>>15298119
You would be right, if the person playing this game was playing it more than once. But the person will only play it once, so her chances of winning by switching doors or not is 50-50.

[Anonymous No.15298169]

>>15298164
>But the person will only play it once, so her chances of winning by switching doors or not is 50-50.
It isn't 50-50 because you have information provided by the host. It's called a conditional probability. There's a difference between that and unconditional probability.

[Anonymous No.15298170]

>>15298149
I'm not trolling and I think you actually convinced me.

[Anonymous No.15298172]

>>15298130
The number being bigger still sounds like magic to me. "Branches" does not sound logical.

[Anonymous No.15298175]

>>15298170
>I'm not trolling and I think you actually convinced me.
Oh, cool. No worries. I don't ever mind answering genuine questions, I'm not one of the pricks who sneers going "Buht it's obvius" or anything like that. There's just one or a couple "it's 50-50" trollposters because it's a bit of an in-joke at times.

Usually conditional probability completely fucks people up, since it isn't as straightforward as usual random fractions or single die rolls or coin tosses.

[Anonymous No.15298179]

>>15298175
Motion creates emotion. Remember that.

[Anonymous No.15298193]

>>15297977
You're overlooking the probability, which is to say, the probability of the probable previous probability and its future probability and the probability that it's probable. Take any probability and quadruple it, then what do you have? We think about probable improbability as though the probability is probably improbable when we should be thinking in terms of probability

[Anonymous No.15298200]

>>15297995
>>15297998
>>15298005
>>15298012
>>15298023
>>15298119
>>15298130
>>15298158
>>15298149
You morons dont realise you are on a game show, not a computer simulation. No matter what your idea of probability is, the problem is the same so lets use the frequentist interpretation since the problem is an empirical one.
If you played this scenario an infinite number of times, the probability of winning after switching would approach 2/3, i think everyone agrees. But you arent playing it an infinite number of times. You are playing it once. Probability theory has no basis for one-off events, they either happen or they do. Its closer to 50-50 in this case, maybe not exactly and there is a very slight advantage in switching, but the odds are much closer to 50-50 than you having a 2/3 chance. You cant say you have a 2/3 chance of winning because you dont, you would however win 2/3 of the time were you to play the game an infinite number of times.

The data from the show and shows like it such as deal or no deal support this actually, your probability is of limited help in such situations. Same with gambling rules.

[Anonymous No.15298207]

The Monty Hall problem is super digestible if you just start with 10 doors, then 9 and so on until you hit 3.
Even a middle-aged woman with no education could understand it.

[Anonymous No.15298208]

>>15298200
>on a game show
Actually it is called 'the Cosmic Play'. Everybody who has a body is a character in it, knowingly or not.

[Anonymous No.15298210]

>>15298207
Not if she is American.

[Anonymous No.15298214]

>>15298200
>But you arent playing it an infinite number of times. You are playing it once. Probability theory has no basis for one-off events
Put simply, you are confusing something like descriptive quantitative statistics with abstract probability. Again I am not sure if this is on purpose. One could not "describe" strictly on past observation of a single event the "odds of that event" given a sample size of 1, but that does not apply to abstract examples.

[Anonymous No.15298215]

>>15298200
>Probability theory has no basis for one-off events
lol

lmao

Yeah and maybe when the goat is revealed it's actually the demon Baphomet in disguise come to reap your soul and he bites your cock off. I mean, I'd say the odds of that are generally pretty low, but this is a one-off event, so it's 50-50, he either is or isn't
>Its closer to 50-50 in this case, maybe not exactly and there is a very slight advantage in switching
What the fuck do you base this on, then?

[Anonymous No.15298216]

>>15298210
Even then I believe it simplifies to such an incredible degree that she could manage.

[Anonymous No.15298235]

>>15298215
>What the fuck do you base this on, then?
Either an incredibly clever "inverted reification" ("Thingification"?) fallacy to troll people or... or honestly I have no fucking idea.

[Anonymous No.15298241]

>>15298216
No me mang. If she is American, she will not get it. https://youtu.be/dOQowCeAnRs

[Anonymous No.15298263]

>>15298241
If you just do the exact same setup with 10 doors it's ridiculously easy to see. I'm surprised it took them so long.

[Anonymous No.15298264]

>>15298214
>>15298215
wow brainlets. statistics truly is the great filter. The sample of games you play is small (1) so the CLT is moot. Your SD is so high that you cant trust the long term probability. In such a case as this, there is no real advantage in switching so we say its 50-50. Again, the data supports this. Your theory doesnt apply.

[Anonymous No.15298280]

>>15298264
This is the equivalent to saying if you roll a dice once it’s a fifty-fifty chance kek.

[Anonymous No.15298290]

>>15298207
>>15298210
>>15298216
>>15298241
Sorry but this is such a big *WHOOSH* that I'm starting to wonder if it was even intentional or if you're both stupid

[Anonymous No.15298293]

>>15298280
You either get a six or you don't! It's just when you roll it six times you'll probably only get a six once. For some reason. Because that's what happens with six 50-50 chances in a row.

[Anonymous No.15298296]

>>15298280
>6 sides
retard.

[Anonymous No.15298301]

>>15298296
>3 doors

[Anonymous No.15298317]

>>15298301
2*
I know what you faggots say: "you picked the door when it had a 1/3 chance"
once the door is opened the probability changes. its a completely different scenario. you arent picking 1 from 3, you are picking 1 from 2.

[Anonymous No.15298318]

>>15298264
>wow brainlets. statistics truly is the great filter. The sample of games you play is small (1) so the CLT is moot. Your SD is so high that you cant trust the long term probability.
Again, that is if you were constructing odds ratios from real-world events. You'd be right if it was a mistake to assume a real-world game, for example, wasn't fair according to expectation. Then you'd need to figure out how many runs you'd need to determine if the game were actually fair. Yes, reality can be mroe complicated.

...But we're talking an abstract example assuming a non-rigged game.

[Anonymous No.15298321]

>>15298318
>But we're talking an abstract example assuming a non-rigged game
We arent though, the problem is WWYD?

[Anonymous No.15298323]

>>15298317
>the dice only has two sides after it’s rolled, the one that I wanted and the one I got!

[Anonymous No.15298328]

>>15298321
>We arent though, the problem is WWYD?
Not gamble. Also, yes we are, you're just a twat for assuming people aren't and calling people stupid because YOU chose to change the assumptions. That isn't cleverness. That's miscommunication.

[Anonymous No.15298332]

>>15298323
u might be confusing a die for a coin? try again homo

[Anonymous No.15298334]

Maybe the confusion comes from the fact that the player does not know before if the host is going to open any doors or not.
Like lets say you choose door 1 and host is only going to open empty door if you miss or pick the right one, if those rules would apply then it would be rigged

Like example with 1000 doors, would the host open 998 doors or only 1, or perhaps it even have the factor in place that if you correctly pick car, no doors gets open or otherway around

[Anonymous No.15298357]

>>15298334
>Maybe if the problem was completely different it would be simpler
Shut the fuck up mongoloid.

[Anonymous No.15298361]

>>15298317
I know this is a troll but the probability never "changes". It's so much fucking simpler

Opening the door is a distraction. You're just being asked to bet on whether the car is more likely to be behind the door you chose originally (n) or the set of all doors excluding n. If there are more than 2 doors in the total set, the latter is more likely. That's it.

[Anonymous No.15298377]

>>15298361
Well trolling or not that IS how a lot of people react, by inferring one is arguing the odds "change" in a manner violating statistical independence. A kind of reversion to simpler heuristics that can very easily get one trapped in a misunderstanding. Not just with statistics but with virtually everything. If someone is confused by a given proposition ENOUGH, odds are that very simple "red light time to stop" style cognitive heuristics will just instantly halt any exploration of alternatives. Rather than "Okay I think that's wrong, but let's figure out how to test if it's right".

This is definitely students tend to inevitably come across when they struggle remembering certain things at any given age, due to "just not getting it", and also why "shut up and number crunch" memorization may not help at all. Since it may not resolve that mistaken error detection feeling people can't explain.

[Anonymous No.15298411]

>>15298200
Okay, if there's a thing that has a 50% chance to succeed, and you do it ten thousand times, how many successes would you expect? Five thousand, I'd assume. So now we're reasoning the other way around, starting from the knowledge that there's this thing that, if you did it ten thousand times, would lead to 6,600 successes. What do you reckon the odds are for a single try?

[Anonymous No.15298427]

>>15298200
Your ilk always makes the same mistake. You conflate percentages that serve to denote partitions of a certain whole, with percentages denoting likelihood.

The two outcomes are
{car, goat}
This has the size 2. Meaning, each choices accounts for 50% of possible outputs.

But this doesn't say anything about the likelihood. It's literally not connected. They just use the same notation because, to non-midwits, they are very equivalent, but midwits get confused and make absurd statements like the *likelihood* is 50%.

Imagine an asteroid could impact the Earth, measured every day.
{asteroid_impact, no_asteroid_impact}
Each outcome is 50%. But the likelihood an asteroid hits is not every day 50%.

[Anonymous No.15298428]

>>15298357
Not what i said, there is a 50/50 chance for there to be hidden rules.
Normally monty hall goes like this "pick a door from 3 doors, NOW host opens a door. etc."
but
You have 50% chance to be in a game where a door gets opened even if you land on a car and 50% chance to be in a game where no doors gets opened BEFORE THE QUESTION
Because the NOW is implied only after your choosing "read ""these"" again" from 2 line of text.
>you pick before the rules
>those rules are 50/50

[Anonymous No.15298441]

>>15298428
Are you actually retarded?
The Monty Hall problem has very few rules. Adding, changing or taking away rules completely changes the problem and how it works?
It isn't very complex ruleset-wise to begin with. Who are you even simplifying this for? A chimp?

[Anonymous No.15298446]

>>15298290
I'm the hand up Mona Lisa's skirt. I'm a surprise, bro. They don't see me coming: that's what you're missing.

[Anonymous No.15298449]

>>15298200
>getting struck by lightning once is a 50-50 chance

[Anonymous No.15298453]

>>15298334
>>15298428
Am I having a stroke

[Anonymous No.15298455]

>>15298441
Thats why i was saying "maybe this is the reason for the initial confusions"
And it is

[Anonymous No.15298462]

>>15298455
Now I don't mean this in an offensive way. But, is your first language English?

[Anonymous No.15298470]

>>15298462
>Shut up mongoloid
>Are you actually retarded
>Now I don't mean this in an offensive way.

[Anonymous No.15298478]

>>15298470
I'm insulting his intelligence not his origins.

[Anonymous No.15298496]

/sci/entists... what if the host didn't show what is behind some door? Would still worth to make the switch?

[Anonymous No.15298510]

>>15298496
then it wouldn't make any difference. the whole thing is the host providing information

[Anonymous No.15298524]

>>15298496
That's an interesting question. What matters in the "the host doesn't know either" situation is whether the staff - the people controlling the doors - know. If the host doesn't know, but the staff still know, the staff could always choose to eliminate a goat. If the staff made it known to the player that they eliminate a goat, then the player should obviously still switch even though the host may be unsure of the goat's position.
Personally, I would find the "the host doesn't know either!" situation rather untrustworthy. How do I know the host doesn't actually know? If both the host and staff say they don't know, might they be lying? I would switch doors, purely on the chance that the staff was not properly informed of the new rules of the game, as it leads to the two situations:
A. The staff does their job correctly so switching makes no difference
B. The staff incorrectly eliminated a goat on purpose so switching improves my odds

[Anonymous No.15298618]

>>15298147
its not whether Monty knows or not, (its specified that he does) its whether he could choose to not open the door. If he could, then its reasonable to assume malicious intent on his part, lets assume (you) are like one of the midwids posting here saying its always better to switch, lets see how Monty fucks up your dumb ass:
>you pick a door with a car
Monty knows you will switch so he opens another door to tempt you
>you pick a door with a goat
Monty keeps his mouth shut and you don't know what to do becouse this wasn't part of the simulation
200+ IQ my ass, go make me a sandwich BITCH

[Anonymous No.15298655]

>>15297977
You need to be impaled alive, and maybe once you are being flayed and then burned alive you will be able to understand how fucking stupid you are.

[Anonymous No.15298673]

>>15298524
Ah but have you considered what if the player knows the host doesn't know and the staff knows? Then the Staff must account for the player knowing and must rig it so that switching wins on account of them not switching because if the staff were to know the player would switch then it means the staff would be certain that switching is losing.
This math stuffs sure is complex and psychological.

[Anonymous No.15298972]

>>15298002
I am. The problem is actually very simple. What is the probability that the car is behind door number 1?

Well, the probability is obvious: it either is, or it isn't.

[Anonymous No.15299086]

>>15298972
>Well, the probability is obvious: it either is, or it isn't.
Ha-ha funny set theory joke I bet, right? Just define the element as the set "muh set theory has no definition of a set so they're all equal durrhurrr". Or am I too smart for the "smart person"? In any case that's what you're effectively doing. Communicating badly and not being clever >>15298328
>I am. The problem is actually very simple. What is the probability that the car is behind door number 1?
One element of the sample space is not the set of all elements in the sample space. Yes, it is very simple. Almost like this fallacy describes a certain dishonest behavior you're highlighting about intentionally ignoring other elements of a sample space. https://en.wikipedia.org/wiki/Cherry_picking

[Anonymous No.15299113]

Don't worry guys, I watched vsauce
When the host opens the door he provides a useful target in 2 out of the 3 scenarios, in one scenario he provides a false positive
resulting in a 2 out of 3 chance of winning should you heed his advice

[Anonymous No.15299122]

>>15299113
The pic assumes that you've chosen door one on the left hand side at the first stage of the game

[Anonymous No.15299185]

>>15299122
I'm frankly not sure what it's assuming as it looks like a pretty terrible illustration of the Monty Hall problem, but it doesn't matter which door you pick initially.

[Anonymous No.15299268]

>>15297977
>Just put the best supercomputer to test this "probability" and you will see that it is bullshit.
you don't
lol you don't need a supercomputer for this shit
https://pastebin.com/n91cFvwG
copypaste this into your javascript console and call mhRepeat(1000), you'll get a number around 666

[Anonymous No.15299356]

>>15299185
It matters for the pic to maintain its internal logic

[Anonymous No.15299462]

>>15298524
If the host knows or not, but the one telling the host knows and always eliminates a goat, then it is irrelevant and the host "knows"
In reality, the host doesn't know shit so they can't signal to the player. The host just receives a signal of which door to open. Do you really think the host is more than a pretty face who can speak well?

[Anonymous No.15299812]

>>15297977
Let me take you through an algorithm you can use to test this in C.
Download like 1000 digits of pi or 1000 other true random digits in a text file.
Read them one by one.
Goat position is currentDigit%3.
Choose a door.
Revealed door position is currentDigit%2, you exclude the door you chose.
Stay 100 times, switch 100 times (don't count the times you got it right first try).
Write down results.
You have now simulated this problem in C and got a data set, read it and weep.

[Anonymous No.15299819]

>>15297998
thats not true. Monty NEVER opens the door with the prize. Because he isnt acting at random, the odds are different

[Anonymous No.15299823]

[Anonymous No.15299835]

>>15297977
I know, there's 50% chance of winning all the time, and I haven't needed supercomputer to know that.

[Anonymous No.15300178]

>>15299462
Remembering where you left a fucking car isn't exactly a superhuman feat requiring a team of assistants

[Anonymous No.15301184]

>>15299819
>Monty NEVER opens the door with the prize.
Irrelevant.
Your door has a P of 1/3 from the start. Your unchosen pair have a shared probability of 2/3. When he shows that one of these has a P of 0, the P of the other is therefore 2/3. To confirm this, you cosider that you learned no new information about your door, so it still has a P of 1/3.

https://www.youtube.com/watch?v=T5QYTrDReTo

[Anonymous No.15301524]

>>15298012
This is explained very poorly
I'll break down all possible states as first choice|monty reveal|second choice
x1|x2|o
x1|x2|x1
o|x1|o
o|x1|x2
o|x2|o
o|x2|x1
x2|x1|o
x2|x1|x2
There are four choice paths where you end up with the car:
>Pick goat 1, switch to car
>Pick goat 2, switch to car
>Pick the car, keep the car after seeing goat 1
>Pick the car, keep the car after seeing goat 2
If you originally pick randomly, then you have a 2:3 probability of picking goat 1 or goat 2, and only a 1:3 probability of having picked the car. This means the probability of getting either outcome 3/4 is the same as getting either 1 or 2.

The elegant way to explain it is that when you start the game, for two of the doors your only way to win is by switching. There is only one door where you can keep your choice and win. Therefore, you should always switch regardless of the goat revelation.

[Anonymous No.15301535]

I think the generalized problem makes it obvious.
>You are on a game show, and standing in a hallway with n doors. After you pick a door Monty Hall opens n-2 doors to reveal goats. He then offers to let you switch to the remaining door or stick with your original door.
As n gets larger it should be clear why switching is worth it. Imagine 100 doors of which Monty Hall has opened 98.

[Anonymous No.15301761]

>>15298003
Vaxxtard logic

[Anonymous No.15301770]

It's correct to switch doors because since it's 1/3, you most likely picked the wrong one to begin with.

Imagine if there's 1000 doors and he opens all of them but one, revealing goats. What's more likely - that the other door contains the car, or you luckily picked the right one out of 1000?

[Anonymous No.15301923]

>>15298441
>The Monty Hall problem has very few rules

The original Monty Hall problem didn't specify these rules, and doesn't describe the host's behavior, only that he chooses to reveal a door with a goat behind it and is willing to offer the switch. This is hugely significant, as it means the interpretation of the problem where Monty's behavior is specified was made up later and then subsequently accepted as the 'original', when the original problem was always ambiguous.

[Anonymous No.15302024]

>>15297995
Most of them simply don't understand the Question.

[Anonymous No.15302029]

>>15301923
NTA. Seems pretty clear cut what the intention is as written by Marilyn herself https://web.archive.org/web/20130121183432/http://marilynvossavant.com/game-show-problem/

That includes pointing out that popping in mid-game without prior knowledge (unconditional probability) makes the result a 1/2 choice. That is, explicitly stated, NOT the monty hall problem.

[Anonymous No.15302042]

>>15301923
>>15302029
You can also review the first publication attributed to Steve Selvin here https://www.jstor.org/stable/2683689

This also does not support your interpretation. The whole point stems from reference to the TV show "Let's make a deal". That also excludes your interpretation, as no such variance exists in the show. Or at least does not to my knowledge. Either way, it is also clear from Selvin's setup what the intention is.

[Anonymous No.15302176]

>>15299086
When you pick the door, you guess if the the car is behind that door, or it isn’t. The probabilities are 1/3 and 2/3 respectively. After opening the other goat door, the hosts asks you if you want to choose the door you thought it was behind (1/3) or the one you thought it wasn’t (2/3). It’s simple my schizo friend.

[Anonymous No.15302192]

>>15302176
I think I misidentified you as the "muh 50/50" person due to the statement "it is or it isn't". In which case there's been a misunderstanding, as that was meant for him. I assure you I need no assistance understanding the monty hall problem.

[Anonymous No.15302691]

>>15297977
Here is your common sense:
> Picked doors.
> I am more certain that I chose wrong doors, because there are twice as much wrong doors than the good ones.
> Person shows me other door that is certainly wrong.
> I know that the person would show me the wrong doors independently on whether my first choice was good or wrong.
> Therefore, my belief that I chose the wrong door is not updated.
> Thus I am more certain that the remaining door is the good one.
> I pick the remaining door.

[Anonymous No.15302697]

>>15302691
One might argue that some goats drive better than some cars

[Anonymous No.15302758]

>>15298427
Since there are infinite versions of this planet, it might be that 50% of them are catastrophically destroyed every day. We just wouldn't know. We can only find ourselves in a version of this planet that hasn't been destroyed yet because people on destroyed planets don't find themselves anywhere.

[Anonymous No.15302761]

>>15298005
/thread

[Anonymous No.15303596]

>>15302029
The wording here is by Craig F. Whitaker, and it specifically doesn't specify the reasoning behind the host's behavior beyond that he "knows what’s behind the doors", which doesn't exclude the possibility of things like him being 'evil' and, say, choosing only to reveal a goat in the case that you've chosen a car, or a heap of other possible behaviors.

Marylin claims that "the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance!) Anything else is a different question." but later (in other articles) admits that the wording of the question is meaningfully ambiguous, and that it is her choice to interpret it in this way. In the end, the only people she has harsh criticism for are those who DO interpret it in the same way, but who don't agree with the 2/3 chance to win solution (which IS correct, given that specific interpretation).

>>15302042
The first published instance of the problem is even more interesting, because, while there is meaningful reason to support the 2/3 on switch chance, in this case that probability is conditioned on the host actually being ignorant of the implications of the player's request to switch, which is completely reasonable, but it's interesting at least that the (unlikely, but still possible) possibility of the host actually understanding the probabilities involved, yet lying about this and accepting the switch only on the condition that he's aware that this will result in the guest losing (since he could point out the advantage and refuse otherwise), is still at least not ruled out in this scenario.

Monty explicit discusses this possibility of deal making and the practical implications of it which Marylin, and they both agree that, in cases where the host can make free decisions like this, you can't guarantee the 2/3 chance on switch.

[Anonymous No.15305165]

>>15298022
Three branches:
OXX
XOX
XXO.
Nothing more.

[Anonymous No.15306989]

>>15297977
Monty Hall is the easiest shit ever if you just assume there are 100 doors instead of three. You pick one at random, moderator opens all the ones with goats. This leaves two doors unopened. - the one you picked and the one moderator didn't open. Will you witch your choice? Of course you fucking will

[Anonymous No.15307078]

>>15297977
Didn't mythbusters statistically prove that monty hall logic is correct?

[Anonymous No.15307340]

>>15298022
Your four cases don't have equal probability. Branch 1 and 2 each have a 1/3 chance of occurring and branch 3 and 4 each have a 1/6 chance of occuring.

[Anonymous No.15307477]

>>15297977
People are confused by chances.
Suppose that the car is behind door 1. Then the chances that the car is behind door 1 is 100%, and the chances that the car is behind doors 2-3 is 0%. Chances where the car actually is do not change when any doors are opened, because it was arranged beforehand.
The chances this problem is asking is "the chance you guessed the door correctly based on available information". When the door with goat (it's always goat, because host knows where the car is and never opens a door with a car) is opened, both "available information" and "available options" change. Switching the door is chance 2/3 is specifically because you know stuff you didn't know before. Suppose stranger wandered in and he didn't know which door you chose before the door is opened. The chance for him based on available information is 1/2, regardless which door he chooses.
Suppose there's a third player, which have insider information. Then the chance for him to guess correctly is 100%. Because of available information.

[Anonymous No.15307488]

Why do people still make threads about this, we're they caboose clobbered their intuition was wrong or is it deranged contrarianism?

[Anonymous No.15307493]

have 10 doors, 1 winner, 9 losers
pick one
remove 8 losing doors, leaving the one you picked and alone other door
would you switch to the other door, or keep your first pick?

now repeat procedure w/ only 3 doors

[Anonymous No.15307509]

>>15298033
>You choose a car (there's only one) and he opens EITHER goat.
Imagine NPC who already won, but switches the door because he feels smart and "knows about Monty Hall problem".

[Anonymous No.15307892]

>>15298208
kek
>mfw it's true

[Anonymous No.15307945]

>>15297977
i watched a minute of this. its seems very obvious that the probability would change. if you stick with your original pick, it's a 1 in 3 chance of being right. if you switch, its a 1 in 2 chance. so you have a ~17% higher chance by switching

[Anonymous No.15307947]

>I got even more disturbed when I told the problem to the late Paul Erdös, one of the most famous mathematicians of the century, when he visited my home in 1995.
>Erdös was considered by number theorists as one of the greatest experts in probability theory. In a conversation about the use of probability theory in decision making, I mentioned the goats and Cadillac problem and the answer to Erdös, fulling expecting us to move onto the next subject. But, to my surprise, Erdös said, “No, that is impossible, it should make no difference.”
>I mentioned Bayes, and showed Erdös the decision tree solution I used in my undergraduate course. I reminded him that probability is not a fixed, static thing; it changes as time goes by. To my amazement this didn’t convince him. He wanted a straightforward explanation with no decision trees. I gave up at this point, because I have no common sense explanation. I came to the conclusion
that unless your are educated in using decision trees, and know how to apply the real-world Bayes theorem, it is hopeless to understand the solution.
>So I told Erdös, “You don’t know about decision trees so you can’t understand the solution. Put on your earphones, listen to your music, and stop bothering me.” (When Erdös appeared in my house, the first thing he did was unpack his radio and start listening to classical music. The radio blasted from 5:00 am to midnight. He didn’t seem to be able to live without it.)
>[...]
> I ran the program, without the pictures, 100,000 times and found that if I do not switch, the host will smile about 2/3 of the cases. But if I do switch, he will be crying 2/3 of the cases.
>Erdös objected that he still did not understand the reason why, but was reluctantly convinced that I was right. A few days after he left, he telephoned to say that Ron Graham of AT&T explained to him the reasoning behind the answer and that now he understood. He proceeded to tell me the reasoning but I couldn’t fathom his explanation.

[Anonymous No.15308026]

>>15307945
so i was a bit wrong, and it makes sense why. the possible outcomes:

chose car - revealed goat 1
chose car - revealed goat 2
chose goat 1 - revealed goat 2
chose goat 2 - revealed goat 1

so while the probability of the second door having the car is 1/2 after the reveal, there is a 1/3rd chance of choosing the car from the beginning. a 2/3rd chance of choosing a goat. which means there is a 1/3rd (33.33%) higher chance that you picked a goat on your first go, meaning by switching you are giving yourself a 33.33% higher chance of ending with the preferable 50-50 outcome after the reveal

[Anonymous No.15308443]

>>15307509
Imagine being you thinking he's smart and contrarian and losing. Which is more likely.

[Anonymous No.15308450]

>>15308026
That's not how any of this works.

[Anonymous No.15308546]

>>15308443
You're incoherent

[Anonymous No.15309456]

>>15308546
Figures that you can't read since you can't do maths either.

[Anonymous No.15309513]

>>15303596
>which doesn't exclude the possibility of things like him being 'evil' and, say, choosing only to reveal a goat in the case that you've chosen a car, or a heap of other possible behaviors.
No problem however phrased can exclude being twisted. That does not give you license to misrepresent things, nor after being corrected claim "Oh but if you completely ignore context and pretend it's otherwise I'm right". Yeah, and I suppose after someone insists they've never beaten their wife you think it's fine to conclude "So you admit to beating your children".
>Monty explicit discusses this possibility of deal making and the practical implications of it which Marylin, and they both agree that, in cases where the host can make free decisions like this, you can't guarantee the 2/3 chance on switch.
Nobody would disagree with that. That is not the problem. Obviously the calculation for a "fair die" does not guarantee getting fair results from an unfair die.

Alluding to acknowledged alterations of the problem doesn't help your case either. On the contrary, since you're fully aware those are alterations to the problem you're fully aware people were right in calling you out on altering it. I'm not saying "don't think creatively", I am saying "stop being a jerk about it".

[Anonymous No.15309566]

>>15309513
No, what I'm saying is, quite simply, is the problem simply does not give the information claimed, specifying only HOW the host acted in this instance and not WHY (which would make it clear what his actions are conditioned on, which is necessary for the construction of a decision tree). So it is dishonest to pretend that the problem given has this definite answer, when someone shoved onto such a gameshow, with no more knowledge of the rules than those given in the actual statement, would have no assurance that switching was better, and claiming otherwise is just a misapplication of probability, at least without further clarification.

Otherwise, how would you actually specify a scenario where you are on a "make a deal" show, where the host performs these same actions, but you simply do not have any knowledge as to whether there are any rules which would determine if he MUST do something like reveal the goat and offer the switch?

Without such knowledge, there simply isn't any reason to assume he would always do such a thing just because he happens to be doing it in your case, which is all the problem statement is saying.

[Anonymous No.15309587]

>>15309566
I don't know what you're not understanding but I've gone from thinking it's trolling to being genuinely concerned.
1. All language is ambiguous, it is simply a matter of degree. It is not "dishonest" to not assume an alternative not given.
2. On the contrary, in ordinary interaction it would be considered dishonest to do that. It's like you live in mirror world.
3. https://en.wikipedia.org/wiki/All_models_are_wrong
You're not breaking new ground here. You just fundamentally are not grasping a social nuance, the communicative necessity, by hyperfocusing on something more engaging for you.

Again, it's fine if you like the more complex problem. But you really need to understand that I do understand what you're saying, but you're not understanding what I'm saying. That would get a whole lot of people really pissed off with you and I'd rather you realize why that would be, than think people are otherwise unfair or stupid.

[Anonymous No.15310234]

>>15298016
>You can stay or switch doors, what do you do?
You toss a coin. Would up the chance from 1/3 to 1/2.

[Anonymous No.15310605]

>>15297977
Everything has a 50% chance of happening because either it happens or it doesn't.

[Anonymous No.15310638]

>>15310605
Everything has a 1/50 chance of happening because either it happens or it doesn't.

[Anonymous No.15311388]

>>15297977
Consider the situation where you always switch. Your first pick has a 1/3 chance to be the car and 2/3 chance to be a goat, obviously. The host opens a goat-door after your choice. If you originally picked the car and switch, you are guaranteed to switch to the remaining goat. And if you originally picked a goat and switch, you are guaranteed to switch to car.

But your original chance of picking a goat was 2/3. So if you switch, you have that chance (2/3 * 1) of getting a car. Thus, you should always switch. If you don't switch, you just have your original chance of getting the car, which is 1/3.

[Anonymous No.15311410]

Probability is simply the application of artificial 4 dimensional properties to inert equations. It reveals no truth, but a strong enough illusion of it to keep gamblers buying lotto. It is a masturbatory calculation that ultimately serves no reliable purpose and should be stricken from the curriculum.

[Anonymous No.15311431]

>>15311388
Unless you have X-ray vision. Then don't switch if you see the car behind the door.

[Anonymous No.15311451]

>>15298005
Realizing that a large portion of people genuinely can't get this is when I accepted that I really am just smarter than the vast majority of people and their opinions on any topic are just about worthless.

[Anonymous No.15311506]

>>15311410
People who know probability do not play the lotto

[Anonymous No.15313257]

>>15311431
I simply tell the staff in the back to move the goat. You lose superman

[Anonymous No.15314345]

>>15297977
Ok this took me like ten minutes to wrap my head around, so let me explain it for anybody who doesn't understand:

Basically, the likelihood that the host has deliberately shrunk the pool of potential doors that hold the car down to *the* door that potentially holds the car is more likely to be the door that has the car than the one that you arbitrarily picked. This is the case because not only does the host have more information than you (thus it can be assumed that his actions have more meaning and potential behind them), but he *can't* give away the door that has the car. Thus he is inadvertedly giving high priority status to the door that remains closed. It helps to think about probability in the context of this game as an even distribution of potential that you have to assume is perfectly even in the beginning because you have zero information to guide your decision. However, as soon as the host eliminates any of the doors in his hand, this is a strong indication of which doors are more likely to hold value. It's quite an abstract concept, but watch the video. They demonstrate this idea with 100 doors. You pick one, the host has 99. The host then eliminates 98 of his 99. Human instinct will hopefully kick in and show you an intuitive understanding that it is far more likely for the door he deliberately didn't eliminate has the car than the door you crbitrarily chose.

[Anonymous No.15314353]

>>15314345
Oh I forgot to add a detail about the distribution of potential. You can think of potential in this case as mass that attracts towards possible options in different bodies. The potential of your door is a planet and the potential of their doors are a cluster of planets. They consolidate all the potential of their doors into just one, concentrating the gravity of their doors' potentials into one body. Like planets crashing into eachother due to gravitational pull. Again, this deliberate shrinking of the scope puts increased potential in the one door they did not eliminate, ***giving stronger reason to think the one door they thought was valuable enough to save is the door with the car than the one door you chose arbitrarily***.

[Anonymous No.15314357]

>>15314353
The planet analogy might have complicated things, so just focus on the words in asterisks at the end. It's as much a probability scenario as it is a psychology one.

[Anonymous No.15316339]

>>15306989
Nah, I’m sticking.

[Anonymous No.15316348]

>>15314345
But I know that he is either going to open one door with a goat, or 98 with a goat. Whether I chose correctly the car or not doesn’t change that. Effectively, the Monty Hall problem is misleading, because if, as the classic version suggests, he always opens a door with a goat, the first stage is effectively meaningless. There is only ever the choice between two doors.

[Anonymous No.15316367]

>>15316348
You're dumb, the first stage, well, sets the stage.
Monty Hall opening one of the fail doors moves you in to a scenario where there are 2 possibilities: either you picked the right door the first time around and the other unopened door has a goat, or you picked the wrong door the first time around and the other door has the car. You had a 2/3 chance of picking the wrong door the first time around, meaning you have a 2/3 chance of being in the latter scenario where you should switch doors.
I've gotta say, the real monty hall problem is trying to explain it in a way brainlets and midwits will actually understand.

[Anonymous No.15316376]

>>15316367
Nah, the first stage is meaningless. You know that Monty will get rid of a goat, it makes no difference. Real brainlets like you throw in talk about probability to make it seem as though you understand, but you all just regurgitate the same tired arguments. And guess what - being able to repeat something you have heard doesn’t mean jack shit. Seriously, the Monty Hall problem is the best way to detect redditors like you who like to think that you are in a club of special intelligent snowlflakes, when really, just like von Savant, the most you will ever achieve is publishing books of brainteasers.

[Anonymous No.15316381]

>>15316376
you don't have to be so mad that I'm smarter than you, anon. No I'm not going to give you the benefit of the doubt that you're trolling. You just made a really stupid, bitter post lashing out at someone for being smarter than you, and it was 100% in earnest and you meant every word. You're genuinely this stupid, and genuinely angry about it. I don't pity you

[Anonymous No.15316386]

>>15316381
Then explain it, rather than malding. The first stage is meaningless - disprove it.

[Anonymous No.15317290]

>>15298005
Makes no sense. Your initial selection is 1/3 odds. No additional information or anything that happens afterwards can modify this. After the door is opened it's just the same thing again, except with odds of 1/2. Revealing the goat is the same as removing a door and trying again, as no matter your choice, a door can be opened to show a goat. Whether you're right or wrong. You're not shifting into another space where it was previously 2/3, nothing has changed. The odds you're right just become 1/2.

[Anonymous No.15317300]

>>15316376
>You know that Monty will get rid of a goat, it makes no difference.
The fact that you know he will always reveal a goat is exactly what makes the difference. There are two doors for him to choose if you picked right and only one if you picked wrong.

[Jackie Hurst No.15317308]

>>15317300
>two doors
two goats* for him to choose from.<div class="xa23b"><span class="xa23t"></span><span class="xa23i"></span></div>

[Anonymous No.15317311]

>>15317308
OH MY GOD, my own post got the GPT-4chan tag.

It must just be automatically applied randomly, or every n posts or something, thank god it's not actual bots.

[Anonymous No.15317419]

What does it mean when I intuitively knew the answer based on mathematics but I rejected it for not making sense?

[Anonymous No.15317523]

>>15317290
Odds that you're right depend on your knowledge. If you discard your knowledge, chance lowers from 2/3 to 1/2. If you have inside info, your odds are 1.

[Anonymous No.15317664]

he was always going to open the door though

[Anonymous No.15317666]

>>15311506
speak for yourself

[Terrence Wyatt No.15318399]

>>15317419
If headpats and good boys don't work, it means you need more vaccines.<div class="xa23b"><span class="xa23t"></span><span class="xa23i"></span></div>

[Anonymous No.15319910]

>>15297977
You can put a toaster to the task and see that it's true, your chosen door still has a 1/3rd chance of being the correct one when the goat is revealed and you're asked to swap

[Anonymous No.15319996]

>>15318399
Appropriately enough this post could well be bot-generated