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/sci/ - Science & Math

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9379407 No.9379407 [Reply] [Original] [archived.moe]

The goat either IS or ISN'T behind the door. If you take out the choosing door part, it's clearly 50% chance. You can't refute this.

If you flip a coin, it's heads or tails.
If you pick a card it's red or blue.

Get your shit straight!

>> No.9379410

suppose the guy offered u 100 doors u pick one and he opens 98.
do u switch?

>> No.9379425

Absolutely. The probability when you picked the first door was 1/100 and the probability when you pick the second door regardless of whether or not you were actually correct in your initial guess is 1/2. Is this not obvious?

>> No.9379431

The non autistic answer is that it doesjt matter because you already have one of the last two doors selected. Your odds are the same.

>> No.9379438

it's non autistic and also wrong

>> No.9379441

when you first pick a door, the chance of the car being behind the other two doors combined is 66%. When the host reveals the goat behind of those two doors, the chance of there being a car in the one that's left is still 66%.

>> No.9379442

Look at it this way, if you stick with your original door you have 99/100 chance of being wrong because there were 100 doors when you selected. If you swap you have a 1/2 chance of being wrong because you are literally told one of these two doors still has the prize.

>> No.9379446

No, the other doors have been opened, so if you stick with your original choice, you know the goat is behind 1 of 2 of those doors, so your odds are 1 in 2

>> No.9379448

This is actually testably false. Our intuitions about probabilities are just bad.

>> No.9379453

what makes this not the intuitive answer is the fact that it could be either door

>> No.9379459

I'm a brainlet myself but you 50/50 fags take that shit to a whole new level.

It's like you are a child. You see 2 doors, it must be 50/50.

Of course it is a 50/50 chance if you take out all the stipulations and variables relating to the door. But that is missing the point on purpose

>> No.9379462
File: 50 KB, 374x382, monty.png [View same] [iqdb] [saucenao] [google] [report]


>> No.9379491

What are you even saying? I'm confused as to which side you're arguing for because you're not doing a good job at explaining youself

>> No.9379498

then you can't read, his post is completely clear.

>> No.9379512

i'm baffled at how /sci/ will still argue about a problem with a solution so easy to verify with a computer and any programming language. i get it that the problem confused even mathematicians when it was published because it's very non intuitive, but reading some of the replies here i can't believe i'm on the science board

>> No.9379525

your post is literally "im so smart look at me ooohhh"
even worse than the idiots here desu

>> No.9379529

if you think understanding this problem is being smart, that says more about you than about me

>> No.9379540

>hurrr hurrrr don't call me out you're dumb
you're a stuck up faggot, and your posts are useless trash which contribute to nothing.

>> No.9379546

I was baiting. I just wanted to see the retarded 50/50 man explanations. They are usually the quality of my original post

>> No.9379560

There are three doors. The game master will shoot you in the head until you are dead and then open a door at random. If there is no goat behind that door he will open another door at random to get that goat. He then proceeds to murder that goat and grill its meat for dinner.

What is the probability of your death?

>> No.9379565

50/50 either it happens or it doesn't

>> No.9379568

the door that you did choose could have been one of the 98 doors opened, there is no difference between the rest and the one you pick. Probability (1/100), but if you choose the other, the chance of it be the right is the the one you pick in a hundred less de 98 eliminated (1/100-98)=1/2

>> No.9379586


/ \
/ \
/ \
/ \
/ \
1/100 99/100
/ \
/ \
/ \
/\ /\
/ \ / \
/ \ / \
1/2 1/2 1/2 1/2
/ \ / \
/ \ / \
1/200 1/200 99/200 99/200

>> No.9379599

.../ \
.../ \
../\ /\
../\ /\
./ \ / \
./ \ / \
./\ /\ /\
./\ /\ /\
/ \ /\ / \
/ \ /\ / \
/ \ /\ / \
. ||
. ||

>> No.9379600

They can't perceive the problem in their head, even though it has been proven to be 2/3.

Then they use mental gymnastics to fool themselves that they are actually the ones who are right

>> No.9379609

The anon you're replying to isn't agreeing with you, it has been proven to be 1/2.

>> No.9379614

The probability of winning regardless of choice by playing the game a single time is 50%, yes. You either win or you lose.

If you played multiple times though, you have a higher probability of getting more wins than losses if you always switch your choice.

>> No.9379620

You should always switch

>> No.9379622

Thanks for the spacer tip.
.........1/100 99/100
1/200 1/200 99/200 99/200

>> No.9379624

Yeah I know, why are you telling me?

>> No.9379632

>i can't believe i'm on the science board
You aren't on a science board

>> No.9379636

By who?

>> No.9379648

R gets 1/200 + 99/200 = 50%
(IF you choose at random at the end)

But that was never the question. The trick to this fucking question is, that it asks the probability of a fixed strategy (Always Switch) OR (Always Stay) which skews the probability.

(It's like: How many rectangles are there? --> Cue everyone taking it as "How many rectangular tiles are in this image?" and then complaining when people getting it wrong. Ask clearer questions.)

>> No.9379692 [DELETED] 

That's true, the random process is just the initial pick. Sorry.

.........1/100 99/100

Afterwards, only two doors will be left, one W and one R, so that the operation of switching is effectively an inversion of the result (R->W and W->R), so that you flip the initial probabilities at arriving at the results, which yields the mentioned 2/3 of getting the right door in the 3 door case, as mentioned before.

.........1/100 99/100
........|....................|....IF FLIP
....... v...................v

>> No.9379705

That's true, the random process is just the initial pick. Sorry.

.........1/100 99/100

Afterwards, only two doors will be left, one W and one R, so that the operation of switching doors is effectively an inversion of the result (R->W and W->R), which yields the mentioned 2/3 of getting the right door in the 3 door case.

.........1/100 99/100
........|....................|....IF FLIP
....... v...................v

>> No.9379841

Wow - sorry to hear about your low level of intelligence, OP. Maybe don't go around exposing yourself as a dummy.

>> No.9380593

The probability is obviously -1/12

>> No.9380649

Why don't you just stop posting shit and go one with fucking the goat?

>> No.9382307

I wonder if there are actually people out there whose intuition would lead them to believe this.

>> No.9382665

Are there still people who legit don't understand?

>> No.9382667
File: 7 KB, 420x420, b36.png [View same] [iqdb] [saucenao] [google] [report]

Cause you are retarded.

>> No.9382701

>you pick a goat or a car
>he reveals a goat
>if you picked a goat, switching will get you a car
>if you picked a car, switching will get you a goat
>you had a bigger chance to pick a goat at the beginning
therefore you switch, because you are more likely in the situation where the unpicked door is a car

literally as simple as this you fucking mongoloids

>> No.9382713

>still replying to b8 threads


>> No.9382749

Well. The probability for getting a car is 50% IF you pick at random all the time.

The question does not really highlight obviously that it asks for the probability of a different strategy (Always switch) or (Always stay).

>> No.9382758

If you switch you will win in 2 of 3 cases (because you will always switch from a goat to a car) and loose in 1 of 3 cases (because you picked the car at the start and always switch to a goat).

>> No.9383866

what is the practical application of this?

>> No.9384077

none, also most retards argue without using the Bayes theorem

>> No.9384104

>what is the practical application of this?

the entire question is a practical application of probability, what kind of stupid Q&A is this?

>> No.9384105

a tv show

>> No.9384117

the rules of the TV show didn't allow for the contestant to change doors after the host starts opening them.

>> No.9384127
File: 292 KB, 736x1249, monk.jpg [View same] [iqdb] [saucenao] [google] [report]


the goat ain't real

>> No.9384130

heres a novel thought: the tv show is already fictional and you can therefore make up a similar one in which the rules do allow for the contestant to change doors after the host starts opening them

or, you could make a real tv show with these rules

>> No.9384137

>practical application
these threads get worse every year

>> No.9384365

If you flip a weighted coin, it is indeed heads or tails.
If you pick a card from a deck that has 51 blue cards and one red, it is indeed red or blue.
But the odds are not equally distributed.

>> No.9384377

It is based on a real tv show, Let's Make a Deal, hosted by Monty Hall. Hence Monty Hall problem.

>> No.9384386
File: 10 KB, 189x267, 1512687267915.jpg [View same] [iqdb] [saucenao] [google] [report]

> If you take out the choosing door part, it's clearly 50% chance...


>> No.9384451

no they arn't everyone understands shit like this intuitively the people that don't are dumb or trolling and hence dumb.
You may be new but this is the not the first Monty Hall thread and it won't be the last

>> No.9384467

you pick goat = you win
probability of picking goat = probability of winning = 2/3
close this thread

>> No.9384551
File: 302 KB, 769x588, 8uyKclo.png [View same] [iqdb] [saucenao] [google] [report]

>pick a door
>another door has a goat
>now you have two doors with a 50/50 goat or car
>for some fucking reason you're supposed to be more likely to win if you switch doors?
Mathfags are so fucking retarded.

>> No.9384589

>Demonstrably true probability phenomenon.
>"Hurr... stupid mathfags be retards."

For those of you who need pictures and scary looking hags to explain shit to prevent your head from deflating: https://www.youtube.com/watch?v=4Lb-6rxZxx0

>> No.9384704

>>for some fucking reason you're supposed to be more likely to win if you switch doors?
If you choose to ALWAYS switch doors.

Try imagining:
There are 99 doors. One single door has a car behind it.
If you choose a door the GM will open 97 doors that don't have a car behind them. You can then choose to switch to the remaining door.

What is the probability that you win the car, if you ALWAYS switch.

It's 1/99 * 0 + 98/99 * 1 = ~98,98%

Why? If you have chosen the 1 door out of 99 you will win 0% of the time if you switch. If you have chosen one of the 98 doors without a car you will win 100% of the time if you switch.

>> No.9384716

The chance of you getting a goat is actually 100% because he places the car after you've chosen if you'll switch or not.

>> No.9384872

this is good thank you

>> No.9385758

I wrote up a python(3.6) script for you anon.
1/3 :

import random as rnd

def mghtnotexist(changeChoice):

D = [0,0,0,0,0]

w = rnd.randint(0,4) #randomly select door as winner

D[w] = 1 # set random element in D to 1 which is the winnig thing.

choice = rnd.randint(0,4)
# print (D)
# print(choice)
#remove all elements besides for the choice and one other.
DTwoChoices = []
del D[choice]

iterations = len(D)-1
for i in range(iterations):
s = rnd.randint(0,len(D)-1)
del D

# print("We are going throught the iterations at ", i, "we randomly selected the index ", s, "and D is now: ", D)

# print(DTwoChoices)
if changeChoice == 1:
return DTwoChoices[1]
return DTwoChoices[0]

>> No.9385760


def tochooseornottochoose(changeChoice):

D = [0,0,0,0,0]

w = rnd.randint(0,4) #randomly select door as winner

D[w] = 1 # set random element in D to 1 which is the winnig thing.

choice = rnd.randint(0,4)

#remove all elements besides for the choice and one other.
DTwoChoices = []
if w == choice:


if changeChoice == 1:
return DTwoChoices[1]
return DTwoChoices[0]

>> No.9385763


winningrecords_dontchange = []
winningrecords_change = []
winningrecords_notgaranteed_dontchange = []
winningrecords_notgaranteed_change = []
for i in range(1000000):

winningrecords_notgaranteed_dontchange_Average = sum(winningrecords_notgaranteed_dontchange)/1000000
winningrecords_notgaranteed_change_Average = sum(winningrecords_notgaranteed_change)/1000000

print("Chances of winning without changing: ", winningrecords_notgaranteed_dontchange_Average)
print("Chances of winning with changing: ", winningrecords_notgaranteed_change_Average)

dontchangeAverage = sum(winningrecords_dontchange)/1000000
changeAverage = sum(winningrecords_change)/1000000
print("Chances of winning without changing: ", dontchangeAverage)
print("Chances of winning with changing: ", changeAverage)

# Not Garanteed one will win:
# Chances of winning without changing: 0.200017
# Chances of winning with changing: 0.199422

# Garanteed one will win:
# Chances of winning without changing: 0.200583
# Chances of winning with changing: 0.799636

>> No.9385780


These results illustrate two things.

First, that when we are guaranteed that one of the choices is a 'winner', then what is essentially happening is that the rest of the choices are "squished" result into just the correct one. It is unlikely you randomly picked the correct one. This explains why changing your choice is much better since your first choice was out of the whole set but the second choice is a "squished" set.

Second is that when we are not guaranteed one of the choices is correct it resembles our intuitive reasoning. That is, if we take out our choice from the set of possibilities, and randomly select another one, they are both equally as likely. Thus changing one's choice is pointless.

Hope this helps anon!

>> No.9385812

Son. If you need whitespace: code-BB-Tags exist.

(Look into the "normal" posting forms on subboards of 4chan to look if there are special rules or tags for that board or ability to upload more or different files.)

>> No.9385818

>having to code simulations to understand a simple problem than any 5 year old can solve
The absolute state of /sci/

>> No.9385853

Five year old would get tripped up by shoddily written questions that don't really highlight that the problem is asking "Imagine if contestant has decided that he will ALWAYS switch.".

The problem with this question lies mainly with shitty question authors.

I mean look at this shit!
>Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

>Is it to your advantage to switch your choice?
If you decide randomly on the spot instead of choosing a strategy like "always switch", the probability of this is actually 50/50.

THIS is the bullshit this gotcha question hinges upon.

>How many rectangular tiles do you see in this image?
>How many different rectangles could you draw on these rectangular grid lines?
>How many rectangles do you see?

Those questions hinge on different "normal" interpretations of what was actually asked.

>> No.9385876

You have a 2/3 chance to pick a losing door.

One losing door is always removed after picking, leaving only the winning door if you picked a losing one initially.

Picking a losing door and switching thus always yields the winning door since it is the only other door remaining.

Thus, switching carries the 2/3 chance of having picked the wrong door into a 2/3 chance of switching to the correct one.

>> No.9385971

does this question depend upon the knowledge the person has of what had happened?

Say there's a 100 doors, 1 has been picked by the player and the other 98 shown to have goats by the host. But now a third person who has no idea of what went on comes into the scene. Would this person be playing by the first person's possibilities or would they reset making it a 50/50?

Also, if the guy chose to get rid of a door but in completely random way, would this change the probabilities or would it still be a 2/3 chance by switching?

>> No.9385988

Even if the first person decides to choose between the two doors randomly the probability will be 50%. If he decides to ALWAYS switch it will be ~66,6*%.

If the third person learns or knows that the guy is in such a situation and gets to choose doors and decides to ALWAYS switch it will be again ~66,6*%. If he chooses randomly 50%, wheter he knows about the situation or just chooses between two doors (even though the probability calculation is different).

>> No.9386138
File: 5 KB, 250x174, q5OL30E.jpg [View same] [iqdb] [saucenao] [google] [report]

>posting raw code

>> No.9386513

Fuck are you on about m8

In a given instance of the game being played, switching is twice as likely to get you the car than staying. That's all there is to it. What's this nonsense about choosing strategies? You don't have to play more than once, multiple trials just demonstrate the odds.

>> No.9386522

I mean you're on the game show once. You don't decide to "always" switch. You switch once. Which, I guess, is always, for as many times as you get to.

>> No.9386551

Twice as likely to win but still barely better than 50/50 is not an argument.

>> No.9386565

It's as good as it's going to get for this problem, which is the point. It also gets better with more doors. What do you actually want?

>> No.9386879

It matters.

Choosing randomly at the end will get you odds of 50%.

1/3 * 1/2 + 2/3 * 1/2 = 1/2 = 50%

>> No.9387248

If he decides whether or not he switches, the probability will be 66.6% for the times he switches and 33.3%for the times he does not switch. Assuming he survives being beaten by his wife's purse for being a dumbass for not switching.

>> No.9387306
File: 41 KB, 410x250, monty_diagram.png [View same] [iqdb] [saucenao] [google] [report]

you change = 2/3 car

you stick = 2/3 goat

>> No.9387851

>You either DO or DON'T get hit by lightning. It's a 50% chance.

>> No.9387857
File: 30 KB, 1600x1200, mh.png [View same] [iqdb] [saucenao] [google] [report]

>> No.9387920

I'll just leave this here.


>> No.9387954


I just played this for a lil bit and got 6 cars out of 11 attempts without switching... so..

>> No.9388183

>I was only pretending to be retarded.

>> No.9388189


>> No.9388191

This is an altered version of the game where you can immediately lose just by picking a door since the option exists for the car door to be revealed.

>> No.9388452

I'm well aware but that's not the point at all. Please explain to me the functional difference between "choosing the strategy to always switch" and just switching the one time it comes up because after all you're only on the game show once. Tell me why that is a relevant distinction. I don't understand the whole point of that massive post which seems to be very careful and deliberate in its wording but makes no sense. I don't see the problem with the original question.

>> No.9388455

"Switching" implies an awareness of the original choice, or at least an awareness that a choice was previously made. It's entirely different from just having to choose between two doors.

>> No.9389111

Choosing to switch is not a random choice. You don't flip a coin. Think of it like ... he learns of the game and chooses to switch before the game has even started. Doesn't even really matter if he does it on the spot.

You take a decision and stick to it, no matter what. You can calculate the odds of winning based on THAT choice. Choosing between the two options randomly has different odds.

That you only appear on that game show once does not matter for the odds.

Mind that the value of the odds will approached exactly in infinitely many games, but you could still suffer an improbable loosing streak of n loosing games where you decided to switch. It's unlikely but that's the way it is.

Switching randomly (with or without knowledge of the previous step) or choosing randomly between two doors have different probability calculations but the same odds of 50%.

>> No.9389127

You told me literally nothing I didn't already know and I'm still confused as hell as to what the fuck >>9385853 is actually getting at. As far as I'm concerned the original question covers the problem in its entirety and the answer is a straightforward "yes".

Also, with regards to switching, the point is that switching means you're aware that you're switching from something to something else. Even if you don't know which door was picked, switching will work just as well for you as if you had picked it yourself.

>> No.9389147

Basically the way the question is most of the time asked turns it into a trick question, probably by conflating some views.

It does not help that the probability is 50% if you choose randomly.

>switching means you're aware
Well choosing randomly would probably be the choice choose wehter you switch randomly.

So we have with different odds each.
* Switch
* Stay
* Choose randomly whether to stay or switch

>> No.9389201

If you play the game once you will either win or lose.

If you play multiple times, you will win more often by switching, but still have chances to lose by switching. There is no 100% win method and there is little to console a contestant who switched yet lost.

>> No.9389232

>christmas tree ascii on /sci/
reported for religion

>> No.9389235

no. one of the three blue goat boxes disappears after the GM opening the first door

>> No.9389250
File: 12 KB, 220x229, 1512533156481.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.9389255

If you don't understand why something is happening then the simulation is probably wrong.

>> No.9389263
File: 30 KB, 481x425, udenseretard.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.9389376

Isn't there a proof that switching gives a higher probability of winning the prize? If so, why the fuck are these threads still a thing?

>> No.9389381

It's an extremely simplified form of superposition, some people can only think in terms of absolute states.

>> No.9390189

>Basically the way the question is most of the time asked turns it into a trick question, probably by conflating some views.



>> No.9390766

>If so, why the fuck are these threads still a thing?
because the ride never ends

>> No.9390949



>> No.9391144

>choosing randomly whether you want a 1/3rd or 2/3rd chance of getting the car

>> No.9391378

Which coincidentally gives you 50% odds!

Even if it is more complicated to calculate than simply choosing randomly between two doors.

(Not) taking your audience's previous knowledge into account. And then complaining or boasting how stupid they are.

>> No.9391937

Is the Monty door problem mathematically similar to the question of whether light is a particle or a wave?

>> No.9391954

only if the goats are autistic

>> No.9391990

mathfags will try their damndest to tell me to switch but either door originally had a 1/3 chance of having a car. switching doesnt increase your initial odds lol

>> No.9392019

No. I guess?

Try imagining it with 9999 doors.

If you choose to switch the probability is:
(Chose correctly) 1/9999 * 0 + (Chose incorrectly) 9998/9999 = ~99,99989%
(Because the GM removes 9997 doors that are incorrect, after the first option has been chosen)

>> No.9392127

>(Not) taking your audience's previous knowledge into account. And then complaining or boasting how stupid they are.
I feel like you're poorly arguing a very trivial point.

>> No.9392274

No matter how many doors there are it's always 50/50 either your wrong or your right

>> No.9392286

>Getting more information doesnt change the odds

If you dont understand Monty you are retarded

>> No.9392295

>OP either is or isn't a faggot
>Despite gay people making up less than 10% of the population, the chance that OP is a faggot is 50%

>> No.9392673


>> No.9393025

>the chance that OP is a faggot is 100%

>> No.9393045

only in the same way if i think i bought a winning lottery ticket, it's always 50/50 either i'm wrong or right

>> No.9393061

this board is worse than facebook, brainlets who don't understand simple probability and trolls everywhere

>> No.9393068

Imagine a game where the host just straight-up asked you if you want to open one door or two doors. What's your best strategy?
Obviously you choose to open two, because it gives you two chances to win, not one.

Now imagine instead he asked you to pick a door first, and then gave you the option to switch and open the other two. Does that change your strategy?
No, because you know the question is coming, you can just pick the door you don't want to open and then "switch" to the two you do.

If you choose to open two doors, the host tells you what order to open them. He always makes sure that the first door is a goat. Does that change your strategy?
No. You're still opening two doors. The order you open them doesn't matter.

Instead of telling you to open the first door, the host just opens it himself. Does that change your strategy?
No. Two doors are still being opened. Who opens them doesn't matter.

Finally, instead of asking if you want to switch first and then opening a door, the host opens the door and then asks if you want to switch. Does that change your strategy?
No. The door is going to be opened immediately anyway if you switch. Doing it first changes nothing.

>> No.9393077
File: 4 KB, 211x239, f2f4f2f24g.png [View same] [iqdb] [saucenao] [google] [report]

You fucking idiot. The probability changes because YOU HAVE GAINED INFORMATION.

>> No.9393086

you now know that one of the doors you did not choose had a goat.
which is something you already knew before.
no new info
There was originally a 2/3rd chance the car was behind a door you didn't choose, and now that entire 2/3rds chance has been consolidated behind the only remaining unopened door you didn't choose.

If at the beginning you had the choice beween choosing door 1 vs choosing door 2 AND door 3, which choice would you make?

>> No.9393315

>trivial point
Unless when talking about WHY it fucks up peoples reasoning.

>> No.9393687
File: 135 KB, 338x220, deepfried_1514270854817.png [View same] [iqdb] [saucenao] [google] [report]


>> No.9393879

I understand it mathematically but I think the question is still stupid because it changes the scenario partway through. That's the part that screws with people. They're thinking about the final choice (two doors only) as an isolated case, and of course if it were always just the two doors with 1 prize then the probability is 50/50. In the true scenario, at that point, the host IS still asking you to choose between two doors, one of which hides the prize. It's a trick. The third door that originally made it a two-thirds chance gets removed from the scenario, but the probability still behaves as if there are three doors in play? It's not unreasonable for people to think the chance becomes 50/50.

>> No.9393904

I think I understand you, OP.

Probabilities is just a hack. It's a glitch in reality, and the fact that it CAN be used in real situations is a mind blower.

The thing with probabilities is that they don't say anything about the event you're looking for. It tells you what to do.

Take the chance of raining, for example. If the chance of raining is at 30% or if it is at 80%, your knowledge about the actual raining event is zero. You don't know if it is going to rain or not, and never will. But if the chance of raining is 80%, you might want to carry an umbrella.

The same goes for the Monty Hall problem. The knowledge about which door contains the prize is zero at every instance. But you can narrow down what you should do in order to obtain the prize with probabilities.

>> No.9393910

The reason for that is that people just don't get how the second choice isn't simply a choice between two doors. That part is counter-intuitive but it doesn't make it a trick question. I don't why you're bringing all this other shit into it.

>> No.9393915


This. There's no reasoning as to why probabilities work in a deterministic universe.

>> No.9393917

That's basically it, probability gives you the opportunity to make an educated guess based on incomplete information.
In reality, in the Monty Hall problem, there's no 2/3 or 50%, one of the doors has a 100% chance of having the price behind it, and the two others have 0%. Probabilities aren't a physical property of the universe as far as we understand it.

>> No.9393919


>> No.9393969

>what is QM

>> No.9393973
File: 17 KB, 240x297, Screenshot_20171223-184103.jpg [View same] [iqdb] [saucenao] [google] [report]

LOL at your faulty logic

>> No.9394249

P(Your original choice is correct) = 1/3. This fact isn't negated by opening another door.

>> No.9394250

Additionally: IF you choose randomly between the two possibilities (Stay, Switch) the probability is actually 50%!

>> No.9394252

Probabilities are limits that converge to a specified probability.

>> No.9394255

Because it matters IMO.

>> No.9394266

That's a neat little fact but still entirely irrelevant to the question.

>> No.9394340

My compeng professor would have discarded identical results as poor optimization or "don't cares". There's better ways of weighting a result than duplication. There's 2 results, which make unweighted results 50/50. You can certainly get to 1/3 and 2/3 by giving more weight to one result, but pretending that there's more results that cause that probability is just deception.

>> No.9394361

Your professor is an idiot.

>> No.9394468


the will always be one goat behind one of the 3 doors, thus: it can be (1 being the goat:)

1 0 0
0 1 0
0 0 1

>> No.9394493

saving literal billions of transistors on this sort of thing isn't idiocy.

>> No.9394561


>> No.9394564
File: 15 KB, 679x192, firefox_2017-12-26_13-01-29.png [View same] [iqdb] [saucenao] [google] [report]

if you ignore everything that happened and just look at the last 2 doors, you are correct, it's 50/50. but what kind of fucking retard are you if you ignore the whole point of the problem?

>> No.9394590

no, he's an idiot for misunderstanding the professor

>> No.9394599

no you fucking retard. if you had two cards, i select the wrong one and you tell me i selected the wrong one and i get to choose again, the probability would be 100% every time.

>> No.9395241

because you're randomly choosing whether you want 2/3rd or 1/3rd chance of winning, and that averages to your 50%

>> No.9395392

Wow...this discussion should not take more than 5 minutes. If you actually believe the OP you do NOT belong on this board.

>> No.9395473

No shit.

>> No.9395941

This is the best explanation out of all the other examples.

/thread everyone, no need to argue anymore.

>> No.9395954

The first person doesn't choose between the two doors randomly, the question said he has ALREADY chosen a door.
After this choice one of the losing doors is revealed. Because the person is more likely to choose a losing door (2/3 chance of choosing a loser versus 1/3 for a winner), it is much more likely that the other door is the winner, so it is best for the person to switch their choice.

>> No.9395974

>try imagining the problem as a completely different problem
No. These more-than-3 door examples do not help anyone. Or maybe they do. Probably not. If they can't understand it's not 50/50 when the problem boils down to only two doors, whether they start with 3 or 100 doesn't change anything, moreover you have a 99% chance of winning by switching in the 100 doors version, which also skirts past a logic problem presented by the original question of even if it might be 66% win by switch, there is still a big chance to lose by switching. The issue isn't about finding a method that guarantees wins, yet virtually guaranteeing wins is what happens when switching in 100 doors.

>> No.9395977

I disagree. More doors makes the explanation more intuitive for many people.

>> No.9395979

I do agree that some particularly stupid people (see this thread) are beyond help, though.

>> No.9395984

Switch to Monty knows.

>> No.9396059

it reveals that Monty is opening (n-2) doors, not just 1

>> No.9396063

Man I hate probability

>> No.9396064


This is literally Probability 101, here's a lecture about this very problem if brainlets still don't understand

>> No.9396389

Low quality bait

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