Monty Hall is psychological bullshit

But that's not true, if you do just alternate between the two as you say, the results will be 2/3rds and 1/3rd, these suppositions are demonstrated with an equal sample of switches and stays, which is what you're after. Also monty never opens the door you "picked" which you seem to be forgetting. That would also lead to a 50/50 case

When you "choose" your first door, you're saying "Monty, you cannot open this door, you must open another door with a goat." That is a meaningful distinction. The first pick is far from meaningless.

> That's still not what I'm talking about
Then maybe you should try making a coherent post rather than just ranting.

>That is a meaningful distinction.
No, it's not. Which physical door is selected is meaningless since probability only cares about the quantities and not positions.
The end result of the first decision is ALWAYS the same, regardless of which first door you pick.

This argument isn't relevant because I'm arguing the semantics of framing it as "stay or switch" rather than "1 or 2" when the first decision will always have the same outcome no matter what you do. When framed as "stay or switch", the math checks out. I'm not arguing that. I'm saying it shouldn't be framed like that.

My first choice had NO statistical influence in the outcome of the 2 remaining doors (there will ALWAYS be 1 car and 1 goat in play. Their positions are irrelevant). Maintaining any attachment to my first choice is illogical and statistically irrelevant. I don't care what my first choice was. I now have two doors in front of me. I can pick either door one or door two.

Just extend the problem to 1000 doors

Initially, you have 1/100 chance to pick the car on your first guess

If you pick a door in the other set, Monty opens every other door besides that one (and they all have goats)

The set has 99/100 chance to have the car in it, so 99/100 chance the car is in the second door you picked. Remember, monty knows where it is and can only open doors with goats.

The first door is put into "reserve". That door had a 1/3rd chance of having a car. Monty then opens another door with a goat. There remains a 1/3rd chance that the car is behind your reserve door, but a 2/3rd chance that one of the two other doors had the car, and one has already been eliminated for you.

After your chosen door is set aside, what you have is "there is a 1/3rd chance door 2 had a car, and 1/3rd that door 3 had a car." He then opens one of these two, leaving the other that can still be opened. You're opening two doors, monty opens one, you open one, the probability that the car was behind one of these doors is 2/3rds. Suppose monty doesn't open one of the two doors, if you were instead told "You can open both doors you did not select" (remember one must have a goat regardless) would you argue that that isn't giving you a 2/3rds chance?

>when the first decision will always have the same outcome no matter what you do
Whose first decision? yours? If so, no, because you will pick the car first 1/3 of the time.

>My first choice had NO statistical influence in the outcome of the 2 remaining doors
Yes, so why are you arguing that your first choice reduces the probability of the car being in the second set to 1/2 when it starts off as 2/3??

Honestly let's not make this more complicated than it has to be, the omniscient monty is just a roundabout way of saying "Go ahead and open both doors" (by process of elimination one of those two will always have a goat, so whether you or monty opens it is irrelevant)

>The first door is put into "reserve"
This is a framing semantic and not statistically relevant. It does not matter which door is put into "reserve". The result of your first decision will always be one car and one goat in play. There is no reason to maintain an attachment to the reserve door, as it does not contribute to what the options will be in the second round.
Remember that the physical position/order of the doors are meaningless, only the numerical quantities of car and goat are relevant.

By framing the problem as something that is dependent on your first choice (when objectively it isn't), you get the 1/3 2/3 probability

It doesn't reduce the probability, it reframes the entire problem. "Switch or Stay" is a subjective semantic that relies on your first decision. In this lens, the probability works out to be 1/3 to 2/3. But if I look at the EXACT SAME DECISION with a different lens that treats it as an isolated decision with no previously picked door (which is completely valid since your initial decision did not contribute to the current state of the 2 remaining doors), it's 50/50. Exact same problem, different probabilities purely because of how it's framed.

Monty opening the door is the same as you opening the door, therefore if you switch you're opening 2 doors, instead of just the one you set aside. Suppose you pick door 1. And they tell you "Okay, you can pick door 1, or you can pick both door 2 and door 3, if either has the car you win." What would you do? I know i would open both.

>arbitrarily redefining the problem and removing the context changes the odds!!! pseudoscience!!!

It's not an arbitrary redefinition, it's a logical redefinition eliminating irrelevant data
"pseudoscience" was just bait to make people respond (it worked)

Another way I think of this is that you have your original 1/3rd chance, and you also have a 2/3rds chance that the car was behind either, in the set of the two other doors (remember monty can never open your first choice, and can never open the car) So you have 1/3rd chance, and you have 2 "1/3rd" chances, but he eliminates one of those two, so the odds stack behind that last door, leaving 1/3rd and 2/3rd. The "irrelevant data" of monty not being able to open your door, and monty having to open a goat door only, allows you to take advantage of the probability that the car will be contained within that set.

The context of the problem isn't irrelevant. You can't just pretend the other doors didn't exist, otherwise the problem would've just been two doors from the beginning.

>You can't just pretend that the one door you objectively can never select is irrelevant

math.ucsd.edu/~crypto/Monty/montydoesnotknow.html
And to reiterate, you are absolutely correct that the probability is 50/50, if and only if monty does not have to open the door with the car, as is demonstrated in this example.
Sure, but which door you "objectively can't select" is decided by the first door you did select, suppose the goats are behind 1 and 3, if you pick 1, he has to open 3, if you pick 3, he has to open 1. If you pick 2 he can open either, but remember, there's only a 1/3rd chance the car is behind 2 (for the sake of argument i've already told you where it is.) He cannot and will not open your door or a car door.

>the problem would've just been two doors from the beginning.
The first decision only exists as a mind game that - while bearing no relevance to the ultimate decision as it does not influence the second decision's setting - causes the player to consider additional irrelevant data regarding their ultimate choice as they have developed an arbitrary attachment to a certain door. It's just a psychological trick to make the situation more tense.

*monty does not have to open the door with the goat

Monty is giving you the ability to open the two doors you didn't select, giving you the probability that your door is in that set of 2 doors (2/3rds of the doors available) but if monty can also open the car door, the it is 50/50 instead.

You can't because it's equally likely within the set of the other doors that it could be behind one of those "non-selectable" doors and all the other doors become non-selectable.

>Sure, but which door you "objectively can't select" is decided by the first door you did select, suppose the goats are behind 1 and 3, if you pick 1, he has to open 3, if you pick 3, he has to open 1. If you pick 2 he can open either, but remember, there's only a 1/3rd chance the car is behind 2 (for the sake of argument i've already told you where it is.) He cannot and will not open your door or a car door.

The order of the doors is still irrelevant to probability
No matter what, there is always one goat door you can NEVER pick. It doesn't matter which of the two physical doors that is. The objective outcome of the first decision is eliminating one goat door from play. The subjective outcome is causing an attachment to a certain door which then acts as a mind trick for the player when deciding what their second decision should be.

There is a 1/3rd chance the car is behind the door for each of the three doors. You can reserve whichever you want, and monty will always select (100% instead of 50/50) the door with the goat that you didn't select. That 100% chance is crucial to the experiment. If monty has a 50/50 chance of opening the goat door out of the two you didn't select, then experimentally you will get a 50/50 probability between switch and stay or 1 and 2 or whatever you want to call it. But that is not the case in the original problem.

The order of the doors is irrelevant, but when you begin the doors will always hold whatever is behind them. When you start, suppose door 1 has goat, door 2 has goat, door 3 has car. This setup will stay constant throughout that round, they are no longer variables, and therefore they are influenced by these rules ("monty knows") that we have set in place. You seem to be treating these as independent variables that can change arbitrarily throughout the round.

...

>No matter what, there is always one goat door you can NEVER pick
No matter what, if all other things are equal, the subset of two doors will have car behind one of them more often than the subset of one.

His problem is that he's treating the probability as a 50/50 roll of the dice each time you open a door instead of 3 predetermined constants.

>You seem to be treating these as independent variables that can change arbitrarily throughout the round.
Not quite. I'm treating the two decisions as independent events and eliminating the framing of 'stay and switch'. My argument is that this is a valid view because your first decision has no impact on its outcome (one goat is always eliminated, one goat and one car are left in play). The first decision allows one door to be defined as "yours" which then allows for the question of 'stay or switch' which is dependent on the first choice and results in the 1/3 to 2/3. But if your decision had no bearing on the outcome, it should be discarded and isolated. Maintaining an attachment to a door serves no relevance.

This isn't what I'm talking about. Think higher level abstraction and lose the attachment to 'stay or switch', which relies on the first decision. My argument is that the first decision can be discarded completely since you are unable to alter its outcome (there will always be a goat and a car left in play)

So basically, you're positing a different problem. You are correct, in the context of stay or switch and monty knowing from the start, it is 1/3rd and 2/3rd. In a purely random context it's 50/50. But the one you're talking about is no longer the monty hall problem. Can you at least accept that under the rules of the monty hall problem the 2/3rds chance holds?

>But if your decision had no bearing on the outcome
Your decision does have a bearing on the outcome because Monty will never open your door, it will always be one of the two in the other set.

This.

OP it makes more sense if you assume that the car is in a specific door and you pick which one to open at random. If you work through all the choices you see that switching wins 2/3rds of the time.

"This question seems to have a non-intuitive answer. Why were so many convinced that Marilyn Vos Savant was wrong? They had all decided that it did not matter if the contestant switched or did not switch. There may be a reason so many disagreed with her. Omitting one phrase in the statement of this problem changes the answer completely and this might explain why many people have the wrong intuition about the solution. If the host (Monty Hall) does not know where the car is behind the other two doors, then the answer to the question is "IT DOESN'T MATTER IF THE CONTESTANT SWITCHES." The change in the statement of the problem is so slight that this might be the reason this problem is such a "paradox."" from math.ucsd.edu/~crypto/Monty/montybg.html

You're in good company, many before you have asked the same questions, but you're still wrong.

>In a purely random context it's 50/50
It's not purely random, it's eliminating data that has no impact or relevance to the outcome.
>Can you at least accept that under the rules of the monty hall problem the 2/3rds chance holds?
Of course, I even say so in the OP.

>it will always be one of the two in the other set.
It will always be a GOAT in the other doors. Always. No matter what you pick. The fact it's a "set" is irrelevant framing from this perspective, as is the entire concept of 'staying' or 'switching' based on a previous decision that had no relevance to its outcome. If you reject the notion that you are making a decision relative to another door, it reframes the problem

I get you're trying to be smug at all but this is not at all what I'm talking about

...

What you're talking about is not the monty hall problem. You are right, but it's not the same problem.

This shit isn't about a fucking car
It's to see if people are going to stay with their choice or change it for small chance of improvement

If your argument is that the MHP is fundamentally about the question of staying or switching (rather than just being the methodology used for door opening and the fact two door choices are made) then I can agree

You're acting as if your intial choice has no impact, but that's wrong because you could pick the car initially and lose by switching. You're ignoring the fact that choice of picking the first door absolutely matters; you have a 1/3 chance of picking the car and a 2/3 chance of not picking it on the first door.

Does Monty opening the goat door retroactively change the odds you took at the time? No, because there's still a 2/3 chance you picked a goat, the existence of the other goat doesn't change the odds that your door has a goat.

>It will always be a GOAT in the other doors. Always. No matter what you pick. The fact it's a "set" is irrelevant framing from this perspective, as is the entire concept of 'staying' or 'switching' based on a previous decision that had no relevance to its outcome. If you reject the notion that you are making a decision relative to another door, it reframes the problem
You don't get it OP.

You know that the chance of the car being in one of the two doors you don't pick is 2/3rds, right?

>After your first decision, the host will ALWAYS open a goat door. There will ALWAYS be one goat and one car remaining. You have no impact on this outcome no matter which door you pick. It will always end with the same result. The first decision has zero relevance.
No it has relevance. The only time the host has a choice in which door to open is when you pick the car first go, which happens only 1 in 3 times. The other two times out of three the host is telling you which door he had to pick, and by extension which door he wasn't allowed to pick - the door with the car in it.

So a third of the time he tells you nothing useful, two thirds of the time he tells you exactly which door has a car in it.

nice proof brah

If you initially picked the car, the second question will not be about a goat and a car. it will be about two goats. So your whole argument falls apart there.

since monty always opens the goat door, switching will result in the car 100% of the time if you picked a goat door first. And you picked a goat door first 2/3 of the time. Its really not so hard to understand.

the only reasons real mathematicians and people that have actually passed a probability course don't debunk this shit is because, it is like talking to a fucking wall.

you are thinking about it the wrong way. its really a one-dimensional game, since montys pick is not random. he will always eliminate a goat for you. so the real probability event here is the decision if you remain or switch after monty eliminated a goat for you. and the simple answer is switching is always better, because the probability you were wrong the first time is 2/3 and with switching you can turn that into right.

Your first pick "pins" a door. It becomes unavailable for Monty to reveal. He will still always have a goat left to reveal, leading to the choice between a goat and the car, but you have a chance of 2/3 of having pinned a goat door yourself.

Event 1: The probability of choosing a goat if you stay - 2/3 (your initial choice when you knew there were 2 goats behind 3 doors)
Event 2: Probability of choosing a goat if you switch: 2/3 x 1/2 = 1/3 (second event MUST rely on first event in this context)

So there is a higher likelihood of getting a goat if you don't switch.

...

>math.ucsd.edu/~crypto/Monty/montydoesnotknow.html
Nobody has a right to post ITT until they play the game at least once here.

You can still pick the door that has been revealed

You can't pick a door that Monty revealed, that's not how the problem works.
Why would you want to pick a goat anyway?

thank you, this is the post is was looking for

See for yourself

A core assumption of the problem is that you don't want a goat so you won't pick the door that has been opened to reveal a goat.

No, if you switch after he opens the door, you have a 67% chance of winning.

Yes at the end there is only a choice between a goat and a car. But the chance off getting a car is not 50% because there is a 2/3 chance you started with a goat.

You have a 67% chance to pick a goat the first time. It's more intuitive if you think of it like that.

It doesn't matter if he knows or not.

If he removes one door and gives you another chance you're still likely sitting on a goat so it's in your best interest to switch.

ie the chance that the other door is a goat is still only 33% vs 66% that you're already sitting on the goat.

for me it never made sense till I framed it this way.

Came here to post something like this.

If there are n doors, and you guess at random which one has the car behind it, then the probability of you being correct is 1/n.

The odds of it being in an unchosen door is (n-1)/n. If all but one of the unchosen doors are opened and the car is not revealed, then the probability of the car being behind a door you didn't choose is still (n-1)/n.

If you pick one out of 1000 doors, then some guy says it's in the one you chose or this other one, would you switch?

If someone came into the experiment when there were only 2 doors left, then the odds of them getting the correct door would be 1/2.

>If someone came into the experiment when there were only 2 doors left, then the odds of them getting the correct door would be 1/2.
Wait what if they picked the door the host didn't open that you didn't pick?
Shouldn't it be like 99%+?

They just wouldn't know unless you told them, how likely they'd be to win.

Other people have said this, but what you're describing is not the Monty Hall problem.

If Monty opens a goat door ahead of time and *then* lets you make your decision, then you're correct that it's 50/50.

So what's different if he lets you make your decision and then opens a door?

The goats and car are in one of three equiprobable orders: CGG, GCG, or GGC. The key point is that this ordering can never change from the time one of the doors is opened or chosen.

So as soon as you choose a door, one of the three orderings has been determined once and for all, and you need to stick with this ordering throughout the rest of problem. The ordering of car/goats is still unknown to you, but it is something definite and immutable, and no longer "random" like it was before you picked the door. So when Monty opens the goat door, the remaining car and goat do not get "reshuffled" between the two remaining doors. Thus it is misleading to talk about a "50/50 shot" at this point, because the positions of the car and the remaining goat have already been determined. At this point, you have to analyze the three separate cases, (i.e., CGG GCG, GGC), breaking each one up into three subcases depending on which door you pick (of course, if you just want the answer of 2/3, there are quicker ways to do it). Note that the information "you picked a door and Monty opened another door with a goat behind it" is consistent with all 3 of the original possibilities.

What happens if instead monty opens a goat door and then lets you choose? From the moment he opens it, one of the three possibilities CGG, GCG, GGC is fixed. But obviously you can logically rule one of these out, leaving you with only two possibilities.

tl;dr the ordering CGG, GCG or GGC is random, but it is fixed once and for all as soon as a door is either opened or chosen. choosing a door first, like in the original problem,does not eliminate a priori any of the 3 possibilities, but opening a door first does.

Are you the Canadian from that /pol/ thread, who just couldn't wrap his head around probability theory?

>But when you frame the exact same decision as an isolated event
But it's not an isolated event, which is why it works

You got it all wrong, and somehow right to.

Most "geniuses" trying to support the result of the Monty Hall problem use the bullshit "well, initially it's one third, and atfer a door is open, when you chose the other door your choices are then 1. ISN'T IT OBVIOUS!1!1 (even though it ends up being 1/ and 2/3)".
It's a shitty way to explain it, but the retards think it makes them smart.

But you pointed out something important. the HOST is the one fucking with the probability, not you. Opening one of the door is not the slightly convenient event that helps you change the probability, it is what changes the probability. It's not that the chances were 1/3 initially, and then 1 half. As you said, it's because the host is specifically selecting that you didn't chose initially, and would of been harmful if you did.
Even if you don't believe the theory, experimental results show that not switching
is 1/3 chance and switching is 2/3, doesn't matter how you do it, may it be using a program or by hand.