Monty Hall Problem

Probability is based more on information than choice. The only situation in which you lose when always switching doors is if you initially select the car. This is a 1/3 chance, and so your chance of winning is 1-1/3, giving 2/3.

Imagine the same 3 door problem but with 100 doors. The goat is in one, the person opens all doors except the one with the goat and the one you chose. Do you switch? Yes, because you are not a retard.

I hear what you're saying, but if Monty opens one of the other doors and shows it's a goat, why doesn't the chance your first choice is a car now rise from 1/3 to 1/2, now there are only two doors left? Hasn't the question changed from what it originally was? Why is it different from if there were only two doors to start with? I'm serious about this, I really don't get it.

No it's 50/50
You either get the prize or you don't

this

found the STAT101 "F"-student

>t, Thomas Bayes

That's only one half of the information you have learned. You also know about the conditions of the trial with three doors. With this, you know if you initially select a goat (2/3 chance) the door you did not select is the car. The only lose condition when you always switch is initially picking the car first (1/3 chance).

So the logical fault is thinking that a three door minus one door problem is the same as a two door problem? Like I said at the start, I'm better at numerical analysis than probability. Let me try looking at it that way. Thanks.

Truly the GOAT among probability problems. R I P Monty.

Where's the well-deserved sticky?

The probability doesn't change because you learn nothing new when Monty opens a door. You know for certain before you make your choice that Monty will always open a door with a goat inside it, and so the probability that you have the car will remain 1/3.

...

You forgot Zbigniew Brzezinski

> Not wanting an animal that is the greatest of all time

And that's why I hated statistics in school. That shit just doesn't make any fucking sense.

If you pick a door and don't change it, you might as well have played the game without the host and just opened your door. You will have a 1/3 chance of winning.

Well, there are only two scenarios, one where you switch, and one where you stay.
Staying gives you a 1/3 probability of winning.
This means that there is a 2/3 chance that staying with your door is a losing strategy.

I thought about this and you do learn something new when Monty opens a door.

Before he opens the door there are three arrangements of car and goats: CGG, GCG, and GGC. After Monty opens one of the doors, you learn that only two of these arrangements was in fact possible, as he has eliminated the third by showing that there wasn't a car behind one of the doors. Since Monty knew he was going to do this before the game started the odds of getting the car were 50/50 the whole time. Monty's participation is crucial - it shows that the game was never a choice of three at the start, but was rigged to make it a choice of two, with the player's initial choice setup a distraction to cover it up. Since the rigging is in the player's favor, nobody gets hurt, and some drama is injected into the game.

I think if you discard the assumption that the initial choice was a fair one in three chance the problem becomes much simpler, if not trivial.

Only one of the arrangements is possible, as that is the true arrangement. The question is the probability with the knowledge of the player, for them there are three possible arrangements until a door is opened, unless I'm misunderstanding?

That is correct, only one arrangement actually happens. The player thinks there are three possible arrangements at the start, but once Monty opens a door he realizes there are in fact only two possibilities. Since Monty knew he would open a door there always were only two - its just that until he opens the door nobody knows which two of the three are the possible ones. Even Monty doesn't know - until the player picks a door, then he knows which door he has to open.

Its an argument of your frame of reference.

>I can't follow sequential procedures

Hi guys, thats true mathematicly speaking no doubt at that

>Steely Dan

Not Steely!

How's Mrs. Dan taking it?

This is so wrong it's almost accidentally right

They're not Schrödinger's goats, alright. Monty knew all along which two doors had goats behind them. He doesn't put them there after the contestant makes a choice.

Actually, they are. One is a goat-to-remain-hidden, and the other is a goat-to-be-revealed. At the start Monty knows which doors have the goats but he does not know which is which. If the player picks a door with a goat, it becomes the goat-to-remain-hidden automatically, and Monty opens the door to show the now identified goat-to-be-revealed. If the player chooses the door with the car, then Monty himself decides which goat is which by opening one of the doors.
How so? Look at the problem from Monty's viewpoint, not the player's.

>Before he opens the door there are three arrangements of car and goats: CGG, GCG, and GGC. After Monty opens one of the doors, you learn that only two of these arrangements was in fact possible, as he has eliminated the third by showing that there wasn't a car behind one of the doors.
>Monty's participation is crucial
These parts are entirely correct. However, it is not at any point a 50/50 chance. You have 1/3 chance of picking the car initially. Monty always reveals a goat. This means that if you pick a goat initially (2/3 chance), then Monty is forced to reveal the other remaining goat, leaving the car in the remaining door you didn't pick. In this scenario, switching will win you the car. Only if you pick the car initially (1/3 chance) will switching get you a goat. The point is that by eliminating one of the possibilities, Monty turns it from a 1/3 chance to a 2/3 chance - but only if you switch.

You are still thinking like a player. Monty's intervention ensures there were only two choices in the first place, as he HAS to reveal a goat in the mid-stage, and so that goat was never an option to be chosen.

>that goat was never an option to be chosen.
But it was. Every door was an option at first.

Perspective doesn't matter here. There's one correct answer and it's that switching will win the car 2/3 of the time.

The doors and the goats are not the same thing. Look at the game as a whole, not as a series of stages, like Monty does. The player does have a choice of three doors at the start, undeniably. But he only has the choice of the car and one of the goats because Monty will take the other goat away no matter what he does, and reduce the game to two doors. Because Monty is in control of the game (except for the final outcome) perspective matters very much.

I wrote that wrong, I meant to say "Look at the game as Monty does, not as a series of stages." Sorry.

I meant "game as a whole as Monty does"
I need a cup of tea

This problem is way less counter intuitive than most people think. Your first choice is 1 of 3 doors or 1/3. Monty, having knowledge of which is the right door, imparts some of that knowledge to your choice by limiting your options further.

It's largely counter-intuitive because it's usually presented from a first person perspective that doesn't give clear and compelling reason to believe that Mont is guaranteed to reveal a goat regardless of your choices.

It doesn't matter if he's guaranteed to reveal a goat. If he reveals a goat, you should switch. The question is never formulated in a way that asks you to switch before he opens a door.

>autism, the post

ok hot shot, show us that you understand the odds.
Let's change the parameters, say there are 11 doors to start with, the guest chooses two doors. Then Monty opens 3 doors with goats behind them. The guest now decides if he/she sticks with the original two doors or swaps them for another two of the yet unrevealed doors.
Is is good to swap, and what are the odds?

for clarity: 1 door has a car, 10 doors have goats, and you don't lose when/if your doors have 1car+1goat

Monty's knowledge is irrelevant. Revealing information about a door other your original choice doesn't say anything about the probability of your original choice, so it stays at 1/3. Therefore the final door has a probability of 1 - 1/3 = 2/3.

to anyone that's not convinced by the math, how about some empirical evidence? write a simple algorithm and test it yourself - switching door works.

I'd be more impressed if there were records of the results on the actual game show.

why though? If we can model the issue exactly it doesn't matter, not like there'll be interference from some weird factor in the game show.

It doesn't matter who you are, the odds remain the same. And just because there are two options that doesn't mean the odds are equally divided among them.

What is this list?

ded ppl

goats

I don't really know about the math involved in this, but I remember reading that many professional mathematicians were skeptical at first, including Paul Erdos who was later convinced after seeing a computer simulation.

Not that guy, but here's what I've got:

p(win on first pick)= 2/11
p(win on second pick, given incorrect first pick)= 1/3

So you go from 2/11 to 3/11 if you always switch. Good switch.

by "first pick", I mean your initial choice of two doors, "second pick" is if you swap them for a different set of two.

It doesn't. After the failed pick is shown, you have a fifty-fifty chance. The odds don't just magically favor the switched door. It's just a meme math professors make up to suggest that conventional math is incomplete.

we already had this bait post further up

Same here. Monty's doors have P=9/11.
After opening 3 of them, the 6 remaining share that P, each unopened door now having a (9/11)/6 probability. Two of them have 2((9/11)/6)=3/11

A great way to think about it is this:
When you pick a door, Monty offers you BOTH of the others, or the one you currently have. But he promises to keep one of the goats that you'll get to take it off your hands.
NOW, if you have picked a door, wouldn't you much rather swap to the other two?
Monty is essentially "giving" you 2 doors if you switch.

>If he reveals a goat, you should switch.

You're a brainlet if you don't understand why this matters and is crucial to the formulation of the problem. If Monty could choose to not reveal a goat then this degree of freedom allow him to manipulate the game, and you have to factor in his influence into the chain of possibilities if you are aware of this. Otherwise there's nothing to say that he didn't just specifically choose to reveal a goat in this instance because he knows you have the car, but wouldn't have otherwise, which would bias the results.

...

>Do you switch? Yes, because you are not a retard.

It's more about the rationale. A person feels like there's about a 1/100 chance of being right initially with a 100 doors, but with 2, there's maybe a 1/2, which is definitely a bigger chance, but that's not the right analysis.

The issue is that switching gets you the car 2/3 times.

But what if you pick a door, he opens a door, and then a second contestant rolls dice to see which door *they* pick. With 2 doors and 1 car, isn't this 50/50 odds of winning the car with a dice roll? But then you're also saying that *YOU* see one door with a 1/3rd chance of winning, and the other 2/3rds.

They can't be both right. Either the dice rolling 2nd contestant wins 2/3rds of the time which makes no sense, of you win 1/2 the time, which also makes no sense.

Wrong. Empirically wrong, in fact.

>Otherwise there's nothing to say that he didn't just specifically choose to reveal a goat in this instance because he knows you have the car, but wouldn't have otherwise, which would bias the results.

This would not change the fact that you have a 1/3rd chance of having the car if you stay. You should probably go back to whatever highschool you dropped out of and apologize to your math teacher if we're years into this Monty Hall discussion and you still haven't grasped this.

>This would not change the fact that you have a 1/3rd chance of having the car if you stay.

You don't seem to understand that probability is meaningfully based on the information you have and can only be specified purely mathematically if all the causal factors in play can themselves be described mathematically as random variables. If you choose a car, which is a 1/3 at outset, but then someone you trust comes up to you and says "hey, I actually got to check out what was going on backstage and know that the door you chose contains a car", then fine, you can continue to say that there was "1/3 chance that you chose the car initially", which is true, but are you still going to switch?

Likewise, if you know that Monty might only offer you the goat if you've chosen the car, making the scenario contingent on both your and his choices, then that's meaningful information that's you'd be stupid to not take into account.

I mean, just so you know, Monty Hall and Marilyn vos Savant have actually explicitly discussed this issue and agree with me on this. I really want you to think about what probability actually means.

>not change
bs

see

See Mythbusters

This is even more confusing

>Jan Kruis
It hurts. It still hurts.

This is still going on? Alright, guy here who showed the goats are Schrodinger - here's an analogy:

You are at an event with a door prize, and you have a ticket for it in your hands. You have to be present to win. The organizer says there are three tickets in the box, so you figure your chances to win are one in three. The guy next to you has another of the tickets, you don't see who has the third.

The organizer draws a ticket. It isn't yours, or that of the guy next to you. However, nobody still present claims the prize, and so the organizer says they're going to draw another ticket.
The guy next to you says "wanna swap tickets?"
Do you? Why?

I won't swap tickets because I'm afraid the guy is trying to nick my stuff.

I will swap tickets because it's the Monty Hall problem.

The ticket analogy above and the Monty Hall problem are mathematically identical. In both you initially have one chance in three, but one of the chances gets discarded. Why would you have one reaction in one case and the opposite reaction in the other?

Monty Hall was a very clever man. He used people's psychology to ramp up the drama on a simple 50-50 choice by dressing it up as a more sophisticated problem. In this he was no different than a huckster presenting the shell game, except that Monty was a honest man and was perfectly fine increasing the player's odds so they could win half the time.

I like this, because it forces people to actually think about the difference between this and what occurs in the game show instead of just being dumbasses and tricking themselves into believing they get it.

If after the first goat reveal a second contestant walks on the stage and, without further information, is asked to guess which door contains the car, he is indeed 50% likely to get it correct. This is because he does not benefit at all from the choice made in the first round. If, however, you tell him which door the first contestant picked, then he should be able to pick out the right door 2/3 times.

I know this isn't right but I can't quite put my finger on it.

But they don't win half the time.

I realize that the problems were identical and I just wanted to answer your question (but your reasoning about the solution is wrong).

I presented my reasoning to show how loss aversion could drive me to make an irrational choice.

The ticket analogy is not the same. In the Monty Hall problem, Monty Hall knows where the goats are, and can always open one of the other doors to reveal a goat. This is different from revealing a random door, and there happens to be a goat behind it, which would be analogous to the ticket problem.

Wait, I think I see it.

In the lottery, the winner is determined by drawing a ticket. This is entirely random. In this case, the first ticket is eliminated by virtue of being the only one that hadn't been selected. This is completely different from what happens in the Monty Hall problem, in which only one "ticket" is selected at first and then one of the other two is eliminated. Then, when it comes to the second drawing, the winning ticket is still to be determined - this is in fact what the drawing does. But in the Monty Hall problem, the winning door is determined from the start and known to the host. These two differences make the situation something else entirely from the actual Monty Hall problem. Both the manner of elimination and the selection of the winner are completely different. It would be as if there were two contestants on Monty Hall, they're both asked to pick a door, and a door is opened at random; if either of the contestants had picked it, a car is placed behind it. If neither had picked this door, then a car is placed at random behind one of the closed doors.

All that matters is that one door has been shown to be empty and then you get offered to change.

Knowledge doesn't matter. A door could have been blown open by the wind and you would a posteriori arrive at the same scenario.

It would be a different although not very interesting scenario if there was a goat or a car behind the door blown open (which is the prize again?).

If Monty doesn't know where the car is then he runs the risk of accidentally revealing it when he opens a door, ruining your chance of ever getting it.

And if that happened we wouldn't be here discussing if it's 2/3 or 1/2. We have to take that an empty door was opened and it doesn't matter which empty door (we also have to assume there's a prize even if we can't see it).

It's possible that an ignorant host could open the wrong door and ruin the whole thing, but we have to assume he didn't fuck it up since the premise is that of a door having been eliminated and also the alternative is trivial.

You could conditionalize the scenario by adding random door opening (chance of host fuck-up) but that's not what Monty Hall is about.

>And if that happened we wouldn't be here discussing if it's 2/3 or 1/2.

Exactly, that would 'select out' the occurrences where this happened, 0% of the time if you picked the car, but 50% of the time if you didn't. This means that seeing a goat would be more likely if you had picked the car, which would EXACTLY COUNTER the otherwise increased chance of getting the car when you swapped, given you see a 'goat' opened at random first, resulting in a 50:50 choice. If you don't understand this then you do not understand the problem.

This is only when the door is opened at random, yes? If Monty knows it's 2/3?

Its simply the fact that, by opening a door the host is giving you information because he will never open a door with prize, why autists cant understand it

>would exactly counter the otherwise increased chance of getting the car when you swapped

no, it wouldn't.

literally just think about this for one fucking second guys: you have a 2/3 chance of picking a goat on your first try. in these situations, you will win if you switch.

ergo, 2/3 chance of winning a car.

>muh 1/2

Look, brainlets, if you dont convince yourself you have to switch, do it the hard way, list every possible case and see it yourself, no, Monty revealing the prize is not a posibility

would the monty hall problem be harder if one of the goats had a bad attitude.

>Monty opens a door
>Nothing behind it
>"Where did it-"
>Sound of engine revving from behind door number 1

I understood this
thanks man, I'll sleep better than usual tonight

>the host is giving you information because he will never open a door with prize

This is exactly the point that most people, even those who claim to get it, don't understand at all, as you can see in this thread.

They don't get that probability is entirely based in information and that, if you had perfect information, then you could classify the probability of anything as either 1 or 0 given it happens or not.

This is the only good response in the thread. Your initial chance of having picked the car is 1/3. There is a 2/3 chance that one of the two doors you didn't pick hides the car. There must be at least one goat behind those doors, and it is removed, leaving the car an average of 2/3 times. Given enough plays, switching will produce twice as many cars as standing pat. This is not a 50/50 situation, and there is no information to be gained from the host opening the door - it will always show a goat.

C=car
G=goat

Here are the possible combinations

CGG, GCG, GGC

I pick a door, i choose the one on the right. There's only one(GGC) combination out of three that satisfies my pick meaning i have 1/3 chance of winning the prize and of course 2/3 chance of losing.
Monty Hall reveals a door with one of the goats inside it. Now i have these possible combinations, CG, CG, GC. But remember i picked the door on the right, therefore if i switch i have 2/3 chance of winning(CG, CG) or 1/3 chance if i stay(GC).

2/3 > 1/3 therefor it's better to switch

...

a weighted die still has 6 sides but the odds aren't all 1/6 anymore

what did he mean by this?

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