/sci/ - Science & Math

[Anonymous No.14867292]

the monty hall program is nonsense
if you eliminate one of the doors, and one of the goats, at that point you're left with one goat, one car, and two doors. that means that both doors have a 1 in 2 chance of having the goat and a 1 in 2 chance of having the car, and the choice to swap to the other door has the same probability of getting either as the choice to stay at the same door.
i don't understand why people pretend as though the 1 in 3 chance from when you first make your choice somehow carries over once one of the doors is revealed to be a goat. the second choice is not the same as the first choice, it's an entirely new choice between two doors, one which contains a goat and the other which contains a car, and there's no reason why one door should have a higher chance than the other of having the car
pic rel summarizes everything that is wrong with how people approach the problem. they assume, for no real reason, that the 2/3 chance of being wrong at the beginning somehow translates over to your second choice, whether or not to switch which door you chose, despite the fact that the second choice only involves two doors and four possible outcomes, rather than the three doors and nine possible outcomes from the first choice

[Anonymous No.14867326]

make it 1000 doors and it becomes obvious

[Anonymous No.14867339]

>>14867292
one way to think about it is that Monty is a kind of force that is unleashed on the doors to do his thing
when you pick your first door you hide it away so monty does not have access to it
then after monty has opened a door, your door is put back into play.

the idea being he would PROBABLY have touched your door had he been able to

[Anonymous No.14867343]

>>14867326

[Anonymous No.14867350]

You are correct if you assume the host doesn't know which one has the goats and what has the Ferrari
But he knows with perfect information and gives you additional information so switching profits

[Anonymous No.14867355]

>>14867350
he does not actually have to know
if he opens the door with the car you just discount the match because it obviously does not matter and you can clearly see it

what *is* necessary is that monty always offers you the choice of swapping.
If monty was adversarial he could offer you the choice only when your door has the car, and refuse to let you switch if you picked the goat.

You have no way of knowing whether monty is adversarial though so it is usually just assumed he is not.

[Anonymous No.14867948]

The essence of the paradox is the host, which by the rules of the games, always needs to reveal a goat that is *not* the players initial choice.

This rule cannot be done with two doors.

[Anonymous No.14867959]

>>14867326
>make it 1000 doors and it becomes obvious

[Anonymous No.14867963]

>>14867292
You realize you can simulate this problem and see for yourself that you suffer from severe mental retardation, right?

[Anonymous No.14868009]

>>14867326
>muh 1000 doors
Not the same ratio anymore fag. That's like saying getting heads in a coin toss is not 1/2 because getting heads in a 1000 coin toss is almost 1.

[Anonymous No.14868192]

>>14868009
it is the same situation
your counter example makes absolutely no sense

[Anonymous No.14868231]

>>14868009
Wdym? You still are left with just 2 doors in the end. What's the issue here?

[Anonymous No.14868233]

>>14867292
Under what circumstances does swapping doors make you lose?

[Anonymous No.14868234]

>>14867292
>probability that a randomly picked point ends up on the x axis is 1/2 because it either is there or not.

[Anonymous No.14868289]

>>14867355
>he does not actually have to know
He does. If he's not guaranteed to open a goat door, the probability is 1/2.

[Anonymous No.14868295]

>>14868289
Only if the host opening the door with a car is a valid game. Obviously it isn't and if he doesn't know which door is which and you reject games where he opens the door with the car you are back at the 1/3 2/3 odds.

[Anonymous No.14868297]

>>14868295
>if he doesn't know which door is which and you reject games where he opens the door with the car you are back at the 1/3 2/3 odds.
Wrong. It's 1/2 then.

[Anonymous No.14868299]

>>14868297
the implicit question is
>does switching when the host shows you a goat increase your chances?

the question is not
>does switching (even when the host shows you a car) increase your chances?

if you think the latter makes more sense I do not know what to tell you

[Anonymous No.14868300]

>>14867292
This should’ve been the first reply: https://youtu.be/n-EAOKuM5Eg

[Anonymous No.14868305]

>>14868297
>>14868299
to expand on it, you are fucked if you are in a situation where monty opens a car door.
There is nothing you can do.
You do not care about this game.

You only care about the games where he shows you a goat door because that is where your choice matters.
The question then is (when your choice matters), does switching increase your odds?
You are only calculating your odds in the goat-was-shown situation
you are not calculating them for the car-was-shown, so you do not count the game as valid if the car was shown.

[Anonymous No.14868308]

>>14868299
>the implicit question is
>>does switching when the host shows you a goat increase your chances?
and that question is ambiguous, because the probability of getting a car after a switch is either 2/3, or 1/2, depending on whether the host knew which door to open, or whether he opened the door randomly.

[Anonymous No.14868309]

>>14868297
Yes sorry, that's correct.
>>14867355
He is correct here. The game would be rejected half the time you didn't pick a car in your original choice which means that the odds are now 1/2. 2 out of 6 possible games you pick a car, and half of the 4 remaining games are invalid.

[Anonymous No.14868311]

>>14868305
run a simulation then. let the host open doors randomly and discard cases where car door was opened. you will get 1/2 probability, not 2/3.

[Anonymous No.14868312]

>>14868308
>and that question is ambiguous, because the probability of getting a car after a switch is either 2/3, or 1/2, depending on whether the host knew which door to open, or whether he opened the door randomly.
no

it is 2/3 because these two situations are equivalent:
>the host knows where the car is and only opens a goat door
>the host can open a car door, but if he does you pretend the game never happened and move on to the next game

there is no ambiguity

[Anonymous No.14868314]

>>14868312
>it is 2/3 because these two situations are equivalent:
they aren't. proof is left as an exercise.

[Anonymous No.14868316]

>>14868314
they are the very same, anon

[Anonymous No.14868318]

>>14867292
>>14868289

[Anonymous No.14868319]

>>14868316
>they are the very same
can you prove it?

[Anonymous No.14868320]

>>14868312
>it is 2/3 because these two situations are equivalent:
They are not equivalent. The rejected games only happen when you don't pick the door with the car so it's not symmetric.

1: Car
2: Goat
3: Goat

1. Pick 1, host opens 2
2. Pick 1, host opens 3
3. Pick 2, host opens 1 [invalid]
4. Pick 2, host opens 3
5. Pick 3, host opens 1 [invalid]
6. Pick 3, host opens 3

You pick car 1/3 of the time
You don't pick the car 2/3 of the time
1/2 of the time you don't pick the car, the game ends prematurely so 1/3 of the time the game is invalid.
1/3 you win
1/3 you lose
1/3 you restart
It's 1/2, when the host is stupid. When the host knows you can treat the unpicked door as a single choice and you have no rejected games so you pick between the chosen door (1 out of 3) or the remaining door (2 out of 3). The host is now a dramatic decoration and is irrelevant. What matters is whether you picked the car originally or not and solely that determines the outcome.

[Anonymous No.14868323]

>>14868319
I think I may have been wrong about it then
the new game is not resampled from the same distribution as the rejected game
oops

[Anonymous No.14868324]

>>14868320
guess the brainlet is me then
thanks

[Anonymous No.14868326]

>>14867292
Here you go, brainlet:
https://www.online-python.com/9KsnZt6RfI
It's empirical fact.
>the second choice is not the same as the first choice, it's an entirely new choice between two doors
Learn what a dependent event in probability is. This is basic shit.
And the host ALWAYS has to show you a goat door. To do otherwise would defeat the purpose of the part where he shows you a door's contents to get you to reconsider your choice since obviously you would always choose to switch if you knew the door he showed you had the 1 car in it.
You can exploit this knowledge of what he ALWAYS must do.
YOUR initial pick is random 1/3.
His reveal to get you to change your mind is not at all random.
If your random initial pick landed on the car, that's only a 1/3 occurrence.
If your random initial pick lands on one of two goats, that's a 2/3 occurrence.
If you switch when your initial pick was the car then you lost (1/3).
If you switch in a goat initial pick, you NECESSARILY win. Because the host ALWAYS shows you the NON-CAR door remaining when you don't pick CAR.
Do you get it yet? Fucking tard.

[Anonymous No.14868388]

>>14867292
You pick a door
>there is a 2/3 chance a goat is there
They reveal one
>there is definitely a goat there
You probably already picked the other goat, so the remaining door probably has a car

[Anonymous No.14868555]

>>14867292
>that means that both doors have a 1 in 2 chance of having the goat
Nope. The door you chose had a 2/3 chance of having a goat. If Monty knows where the goats are then he will always reveal a goat regardless of whether you chose a goat, so the probability you have a goat remains the same, 2/3. Where do you get 1/2 from?

[Anonymous No.14868564]

>>14867355
>he does not actually have to know
He does. If he doesn't know then he is guaranteed to reveal a goat when you have the car, but only has a 1/2 chance of revealing a goat when you have a goat. Sonce you initially had double the chance of having a goat vs the car, it's now an equal chance.

[Anonymous No.14868565]

>>14867292
The problem with the Monty Hall problem is that a goat is a pretty good reward still. If I rolled the goat that's like 50lbs of free, fresh goat meat to roast.

How can a free goat be a bad prize?

[Anonymous No.14868593]

>>14868565
Honestly, I'd much prefer a goat than a car. Goat meat is so fatty, it is delicious. It's also more expensive compared to other common meats.

[Anonymous No.14868985]

The problem is retarded because it makes assumptions which don't matter. It's 50/50 by merit alone of "you play the game once". you either chose right, stayed. chose wrong, stayed. chose right, switched. chose wrong, switched. two total choices. 4 total options. 2 options win. 2 options lose.

anyone who feels the need to run algorithms, much less attempting to "play the game" multiple times, is completely missing the point. the game is played once, and you win or lose once. If the problem was about a videogame version of this gameshow, or otherwise a completely different but repeatable event, then of course "switch is statistically more likely to win more often". but its a goddamn game show. It is almost statistically impossible to become a contestant on a game show more than once, compounding against successfully reaching it to the "end" of the game show where you'd get to play car door goats as the final.

you will play once. you will feel retarded if you lose, no matter how you lose. there are a variety of different ways to feel retardee for losing, but among the worst would probably be "switching because you already know it's statistically more likely you'll win by switching", yet losing anyway.
Then what. You blew it. You blew your only chance cause you listened to some anonymous schizoid on 4chan. and you just have to blame the numbers for your loss as sone kind of self-protection cope, trying to make it seem like it was beyond your power to win even though you picked to switch which [math]\text{should have}[/math] won, yet didn't. Yup, imagine wagering your will to exist and earn as you deserve on muh random numbers and invisible imaginary forces of nature.

you die ashamed.

here's the only tip on how to win the monty hall problem:
goats have value too.

[Anonymous No.14869006]

>>14868985
>Getting the royal flush in poker is 50/50. You either do or you don't.

[Anonymous No.14869013]

>>14869006
Getting aids and cancer is guaranteed for you

[Anonymous No.14869065]

>>14869013
kek

[Anonymous No.14870493]

You have a 1 in 3 chance of choosing the winning door at the START. Which means that there is a 2 in 3 chance that you picked wrong. After the host reveals a goat you still had a 1 in 3 chance of picking correctly since you picked out of 3 doors. That's why it's better to switch.

If you think about it as a 1,000 doors. You have a 1 in 1,000 chance at the start. The host reveals 998 goats. You still only had a 1 in 1,000 chance that your initial pick was correct.

Now if you showed up and mentally decided to pick door 2. And the host then revealed door 1 it's 50-50 because the host COULD have revealed the door your mentally decided on. The fact that the host can't reveal your door is why the odds are against you with your first choice. The host could've revealed 2 and then you have to repick your door, but the host doesn't know that.

[Anonymous No.14870521]

>>14867292
It's because of conditional probability.
It holds if the distribution occurred before the first choice.

[Anonymous No.14870636]

>>14870521
The host is a red herring. Using conditional probability here is like using a chainsaw to open a can of beans.
You either bet on 1 picked door to have the car or bet on 2 picked doors to have the car. You are choosing whether you want to open 2 doors or 1 door for the same reward. The actions of the host are entirely inconsequential.

[Anonymous No.14870641]

[Anonymous No.14870647]

>>14870636
Using contrived this "2 picked doors" logic here is like pulling turds straight out of your rectum and smearing them all over the wall with a finger to do the calculations. It's plainly obvious that you lose by swapping only if you had the car on your first guess (1/3), otherwise you win (2/3).

[Anonymous No.14870649]

>>14870647
Yes, because you either pick 1 door or other 2 doors. You are being offered 1/3 and 2/3 bets for the same reward.

[Anonymous No.14870664]

>>14870649
>because you either pick 1 door or other 2 doors

[Anonymous No.14871027]

The total number of possibilities in this problem - 3. A 3 day long discussion ensues for the n-th time on the subject. Who is retarded?

[Anonymous No.14871074]

>>14868985
So if you only play the lottery once it's 50/50? Amazing.

What is the probability of getting goat 1 vs goat 2? 25%? How does it work?

[Anonymous No.14871263]

>>14867292
Why do you retards always think that the host opening a door retroactively increases the probability that you picked a car from 1/3 to 1/2?

[Anonymous No.14871272]

>>14867292
I think the best way to see it is to rephrase the problem like this:
You pick a door, then Monty tells you that you can choose too exchange your door for the other two doors, known that one of the other doors inevitably contains a goat. Do you take the exchange?

[Anonymous No.14871486]

>>14868326
>And the host ALWAYS has to show you a goat door
Not really

[Anonymous No.14871504]

>>14867292
Write a fucking Python program and verify it yourself, you absolute jackass with the intelligence of a wet sponge

[Anonymous No.14871698]

>>14870647
it's literally the same exact scenario, dumbass faggot

[Anonymous No.14871740]

>>14871698
I can contrive a thousand more obtuse "exact same scenarios" but it doesn't change the fact that your mongoloidal explanation is a lot worse than simply applying conditional probability, let alone a decent explanation like mine.

[Anonymous No.14872584]

50/50 BRO YOU EITHER WIN OR YOU LOSE HOW ARE YOU PEOPLE SO FUCKING DUMB FRFR NO CAP ONG

[Anonymous No.14873846]

you have a 33% chance of losing by switching
not 0%
not <5%
not a negligible chance of losing
33 fucking percent

imagine the feeling of being any of the non-negligible amount of people who unknowingly picked the car door then switched to a goat

[Anonymous No.14873855]

>>14873846
>you have a 33% chance of losing by switching
and 66% chance of winning

[Anonymous No.14873925]

>>14867292
>the monty hall program is nonsense
Just run your own simulation of it, you dumb retard.

[Anonymous No.14873930]

>>14873846
Anon, I don't think the vast majority of people who would switch based on this knowledge would have thought it was a 0% chance of losing. Not sure where you got that from.
The point is it's ALWAYS your best move given what you know in the moment.
You might win by making a bad move instead, but you'd be retarded to pick the bad move for that reason.
It'd be like deciding to aim and shoot at someone in the arm instead of at their center of mass just because shooting at their center of mass is merely the best option and not a 100% foolproof guarantee of success. You might get lucky and do better with an arm shot somehow but it'd make no sense to try that.
All that aside, this game is not real and never even has been real in the way the trick problem is defined. The game it's named after was pretty different in reality.

[Anonymous No.14873943]

>>14871486
>And the host ALWAYS has to show you a goat door
>Not really
Yes really as far as the Monty Hall Problem (not the show) is defined. The host isn't allowed to just not show you any door. And that means when he's forced to show you one door of the two doors you didn't pick he must ALWAYS show you a goat door or else he will effectively commit gameshow suicide and show you the location the car is at.
In the event he shows you the door with the car then obviously you pick the door with the car (which would mean switching anyway).

[Anonymous No.14873951]

>>14868326
Buddy your program and your logic is wrong.
t. Master's in Math

[Anonymous No.14873958]

>>14873951
>>14873943
Where is this previous thread, where I wrote half a page explaining each probability?

Can you guys link it here from an archive?

[Anonymous No.14873974]

>>14873951
Yeah, a shitt ton of highly credentialed academics wrote statements just like yours insisting that solution to the Monty Hall Problem was wrong.
You're not saying anything new. People like you are exactly how the problem became (in)famous to begin with.
Also that almost definitely isn't the anon's program. He just pasted it into a Python interpreter site from the first github link that hosts the code when you google for it.

[Anonymous No.14873985]

>>14873974
You don't go all feisty.
Let's link to that old thread. Or let's analyze together each step in your program.

Or let's analyze logic. But I really don't want to redo any logic since it is long but it is obvious.

Essentially. You either allow RECALCULATE probabilites. Or you don't.
If you allow recalculation:
1) Each door is 1/3 (or 1/100 if 100 doors). Host opens one (or host crosses 98 only leaving 2 doors).
Each probability is recalculated - since only 2 doors are left it is 1/2.
Regardless if you switch or not - 50%

If you can't recalculate probabilities:
1) each door is 1/3 (or 1/100 for 100 doors). Host opens one, you are left with 1/3 or 1/3 (or 1/100 or 1/100)
in either case identical probability.

[Anonymous No.14874005]

>>14873985
I don't think any of the archive sites for 4chan have /sci/ with search enabled anymore.
That said, you're doing what I expected and arguing for a 50%, which is what a shit ton of highly credentialed academics did when the problem first was introduced.
That's why the problem is notorious and famous today: Because it's recognized as having tricked a large number of extremely intelligent academic professionals.
I don't think I, or whoever wrote that program, or the other anon/s are going to succeed at convincing you of this, but I would recommend you check out the historical background with Marilyn vos Savant and all the angry letters from PhDs insisting it must be 50%.

[Anonymous No.14874038]

>>14873985
>Each door is 1/3
>Host opens one
>Each probability is recalculated - since only 2 doors are left it is 1/2.
>Regardless if you switch or not - 50%
I think what you're missing is your choices and the hosts choices aren't the same.
Yours are a random 1/3 initially.
The host knows the answer though.
You know he knows which gives you the ability to get to the 2/3 answer.
If you pick car initially then it doesn't matter what he shows you (will be goat either way). So 1/3 you lose by switching.
If you pick a goat door initially, he will show you the door with the other goat. This is because he can't show you the third door with the car without making himself lose the game.
So that knowledge lets you know that, for the 2/3 times your initial random pick lands on one of the two goats, switching will end up getting you the car due to the host being forced to always eliminate the remaining goat door.

[Anonymous No.14874089]

>>14874038
This is the correct answer

[Anonymous No.14874136]

>>14873846
The only winning move is not to play.

[Anonymous No.14874212]

>>14873985
Let's do a simple simulation of this, shall we
we have doors 1, 2, 3. Door 3 contains the car.
You pick door 1, host shows you door 2 (only other door with a goat), you get the car if you switch
You pick door 2, host shows you door 1 (only other door with a goat), you get the car if you switch
You pick door 3, host shows you door 1 (random door with a goat), you lose the car if you switch
You pick door 3, host shows you door 2 (random door with a goat), you lose the car if you switch

The only way to get 1/2 on the switch is of you consider the last 2 possibilities as equal to the first 2 - i.e. you have a larger than random chance to choose the car on your first choice. Since this obviously does not apply, you can see that switching will get you the car more often than lose it. The probability is not recalculated because which door(s) the host opens is not a random event unless you already chose the car. If you didn't choose the car there is only one door the host can show you, so it's not an independent event. The choices cannot be 1/3 and 1/3 because these do not add up to 1 - a very clear hint that there is something missing.

Yes, if you chose a door AFTER the host has shown you one goat then it's 1/2. But which door the host shows you is bound by your initial choice, which skews the probabilities.

[Anonymous No.14874277]

>>14867292
kek brainlets will always struggle with this, conditional probabilities are the ultimate brainlet filter, even Erdos got filtered by them lmao
>Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result.

[Anonymous No.14874329]

>>14874277
Smart / well educated people get extra-filtered by problems like this because their credentials keep them from doubting themselves and then they keep doubling down on the same mistake they fell for.
At least an average intelligence person has a chance at being humble enough to doubt himself and try to see the perspective he missed initially.

[Anonymous No.14874390]

>>14874329
There are some cognitive biases at play here, yes
>Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.

[Anonymous No.14874466]

>>14867292
Your illustration shows why. It is a collapse of probability.

You pick a door at random. The host knows what is behind each door. Monty picks one of the doors that has a goat (not at random). You had a one third chance of choosing correctly, and a two-thirds chance of choosing incorrectly. With the reveal of the goat, the remaining door has a two-thirds chance of having the car.

[Anonymous No.14874472]

>>14867355
He does have to know, otherwise it is a boring game. If he reveals a car by chance, then there is no point in you swapping doors.

[Anonymous No.14874489]

>>14867355
>If monty was adversarial he could offer you the choice only when your door has the car, and refuse to let you switch if you picked the goat.
The Monty Hall Problem is defined in such a way that he always reveals a door each time.
If he doesn't show a door each time then it's no longer the Monty Hall Problem.

[Anonymous No.14874703]

>>14873958
https://en.wikipedia.org/wiki/Monty_Hall_problem

[Anonymous No.14874904]

>>14873958
search for it here
>>/sci/

[Anonymous No.14874987]

[Anonymous No.14876151]

>>14867292
you are a moron,
monty hall is beyond simple
has nothing to do with probability
and none of the videos on youtube explain it properly

[Anonymous No.14876154]

>>14868318
this is (actually) still not the _truly_ correct answer to the question
while it is factual, it isn't the truest answer that represents what is happening
because what it happening isn't even about probability
it's just comparison
is one bigger than the other
it is much simpler than this

[Anonymous No.14876163]

>>14876154
>>14876151
>lots of claiming, no explaining

[Anonymous No.14876179]

>>14868009
If you keep the door the probability of getting the car is 1/1000 wich are the odds of sniping it from the 1000 doors. If you open the other door you get the car IF AND ONLY IF you have chosen a goat as the first pick

P(getting car changing door)=P(selecting goat door at the start)= 999/1000

[Anonymous No.14876191]

>>14874987
Nice

[Anonymous No.14876221]

>>14867959
If you're doing it by simulation you only need to do it 5 or so times for the 1000 door case to see that it's not 50%, whereas for the 3 door case you would need a larger sample size, like 40ish

[Anonymous No.14876226]

>>14871074
If you're actually on the show there is also the complication of the host possibly not offering the switch each time.

And he's trying to argue that you should maximize utility or minimize regret instead of EV.

[Anonymous No.14876945]

You have a 66% chance of having chosen wrong on the first turn. If you did chose wrong then the only other wrong choice is eliminated before the second turn and you should switch.

[Anonymous No.14878658]

>>14867292
99% likelihood of coin toss landing heads is still 50/50 for whom the coin toss is decisive.

[Anonymous No.14879219]

>>14870641
/thread

[Anonymous No.14881418]

The probability is 1/2. I can prove it very simply. After the host reveals all the goat doors, swap the original contestant out with a new one. This contestant has no knowledge of what the host has just done. Ask this contestant what their chances of winning are. It's 50%.

[Anonymous No.14881422]

>>14881418
>This contestant has no knowledge of what the host has just done
ngmi

[Anonymous No.14881559]

>>14881418
No shit. The original contestant had information that the new contestant lacks. Guess what. If you were to tell the contestant which door the car was behind then they can win 100% of the time, information is important.

[Anonymous No.14881612]

>>14867292
>there are 1,000,000 doors
>anon picks 1 at random
>Monty reveals that behind 999,998 of the doors, there are goats
Monty: do you switch?
anon: actually it's 50:50 and I kinda feel like staying so no I won't switch thanks

[Anonymous No.14881615]

>>14867292
Honestly can't tell if you're joking or just stupid.

[Anonymous No.14881617]

>>14881418
You really think you're saying something here don't you