[bemusedthrow75]

You've restated the problem (incorrectly) -- changed it.

There's still a goat you could show me. And it is a fact that you show me a goat. Nowhere in the problem does it suggest there is a chance you show me a goat.

I do honestly admire your dogged commitment, and I think the way you are committed shows up an important point about the article and the history of the problem.

Which is that one can quite clearly fairly argue the point, as you are doing, without resorting to misogynistic or patronising rudeness as so many did at the time!

But you're still wrong. :-)

And I'm going to leave it here.

[rspeele]

I am curious how you feel about the "Monty Fall"/"Monty Crawl" problems linked elsewhere in the thread.[0]

For my part I am somewhat sympathetic to jncfhnb's point. The exact phrasing that Vos Savant was asked did not specify the rules that the host was required to open a door, nor that it always must be a goat door. It simply says that the host has knowledge of what's behind the doors, and in this particular iteration of the game, he showed you a goat and asked you about switching.

That does not exclude a scenario where the host is a manipulative fellow, who chose to show you the goat only because he knows you are about to win, trying to convince you to lose. A contestant on the real show would surely worry about this possibility.

Of course, the people who wrote to disagree with Vos Savant almost never said "the problem is not fully stated", they said "it's 50/50 you fool", which isn't right. Additionally, since it wouldn't be a math problem at all if we let the host have agency, it is reasonable to assume the unspoken rule that he does not, leading to Marilyn's correct 2/3rds answer.

[0]https://web.archive.org/web/20230706235720/https://probabili...

[bemusedthrow75]

It doesn't make a difference what causes Monty to reveal a goat.

A goat behind a door the contestant did not choose is eliminated. That is all that matters. It could have been done by a Heath-Robinson machine, or a passing lonely shrubber, or elves.

Monty isn't a floating variable in the puzzle who makes choices. His choice is fixed. Which I imagine is why vos Savant adds the actually extraneous information that Monty is fully aware what is going on behind the scenes -- to underscore the concept that Monty isn't a variable.

The fact that a goat behind an unchosen door was revealed is what is crucial to the setup of the entire puzzle.

And that -- despite jncfhnb's protestations -- is information that makes the puzzle determinate.

[rspeele]

> It doesn't make a difference what causes Monty to reveal a goat.

Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door.

The information gain only happens in the original game because we know that Monty was forced to avoid the winning door in the (66% likely) case where we didn't already pick it.

[bemusedthrow75]

Nope.

But I am going to leave this to someone else to explain. I'm tired out now.

[jncfhnb]

Suppose Car is C.

The only possible options are (equally likely)

You choose A and get shown B You choose B and get shown A You choose C and get shown A You choose C and get shown B

4 options. You lose 50% of the time.

The 2 options where you would normally get the 2/3 odds are explicitly ignored when you are told Monty chosen randomly and randomly got a goat.

You choose A and get shown C or you choose B and get shown C are forbidden.

50/50 probability on the dot.

If Monty chooses with intention then the options where you choose goat and are shown goat double in probability so you get back to 2/3

[rspeele]

They are two different problems. Map out all the scenarios exhaustively and you'll find the difference.

In both cases we were originally 1/3 chance of being right. That is not in dispute.

In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car.

In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that:

1. We picked the car and saw Alice

2. We picked the car and saw Bob

3. We picked Alice and saw Bob

4. We picked Bob and saw Alice

Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others.

This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box.

[0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

[rspeele]

Since this bugged me all day, and I suspect you are the kind of person where it bugged you all day, too, here is a better description of the "fall"/"hall" distinction.

I think we can agree that these are the six possible, equally likely, configurations of the problem starting from me having chosen door 1. G1 here is "goat 1" and G2 is "goat 2". For each possible prize behind my chosen door, there are two possible configurations of the remaining prizes.

    My Door | Door 2 | Door 3
    Car     | G1     | G2
    Car     | G2     | G1
    G1      | Car    | G2
    G1      | G2     | Car
    G2      | Car    | G1
    G2      | G1     | Car

With the "Monty Hall" problem, Monty uses his knowledge to always open a goat door. Thus we see the following revealed options and resulting 2/3s probability of switching succeeding. This is the classic version of the problem.

    My Door | Door 2 | Door 3 | Monty Reveals | Switch Result
    Car     | G1     | G2     | Either        | Lose
    Car     | G2     | G1     | Either        | Lose
    G1      | Car    | G2     | G2            | Win
    G1      | G2     | Car    | G2            | Win
    G2      | Car    | G1     | G1            | Win
    G2      | G1     | Car    | G1            | Win

With "Monty Fall", the first thing that happens is a randomly chosen door, that isn't our own, reveals a goat. This is interesting. In the classic problem we were always going to see a goat next, because those are the rules Monty plays by. But in this case, the fact that we randomly found one wasn't guaranteed.

Essentially, you are blindfolded and throw a dart at the 2x6 grid of cells under the headers "door 2" and "door 3", and I tell you that the cell you've hit is a goat. What do you know about the row you hit being a switch-or-stay row? Well, half the possible goats you might've hit are in the first 2 scenarios where you should stay, and half the possible goats are in the last 4 scenarios where you should switch. So you're at 50/50. You don't have any new information to switch on.

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car    | G2(e)
    G1      | G2(f)  | Car
    G2      | Car    | G1(g)
    G2      | G1(h)  | Car

You are just as likely to be looking at (a), (b), (c), or (d) (so you should stay) as you are to be looking at (e), (f), (g), or (h) (so you should switch). It is 50/50 in this version of the problem.[footnote]

This may make it confusing going back to the original. I seem to have shown that both ways make sense but still, how is it different? Imagine it like Monty is doing a random dice roll for which door to open, and he simply juices the outcome by correcting it to the goat door when a car door is selected, since he can't reveal a car and spoil the game. Now we have these equally possible scenarios (a) through (l) for his fair dice roll...

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car(e) | G2(f)
    G1      | G2(g)  | Car(h)
    G2      | Car(i) | G1(j)
    G2      | G1(k)  | Car(l)

Which he corrects, avoiding cars, to:

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car    | G2(f,e)
    G1      | G2(g,h)| Car
    G2      | Car    | G1(i,j)
    G2      | G1(k,l)| Car

Now we are back to the original game scenario where we see a goat no matter what. And we can see that 8 of the possible ways we might have arrived at seeing this goat come from "switch" rows while 4 come from "stay" rows.

[rspeele]

[footnote] If this part of the explanation is bugging you, consider these related problems:

Problem 1. There are two opaque, externally identical bags, each containing 2 marbles. One bag contains 2 black marbles. The other bag contains 1 black marble and 1 white marble.

You choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in the chosen bag is black?

Answer: .elbram kcalb dnoces a dnif ot ylekil sdriht-owt era ew oS .gab etihw-dna-kcalb eht morf si eno ylno dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT

Problem 2. There are three opaque, externally identical bags. One bag contains 2 black marbles. The other two bags each contain 1 black marble and 1 white marble.

Again you choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in your chosen bag is black?

Answer: .tnecrep ytfif era elbram kcalb dnoces a gniward fo sddo ruO .gab kcalb-lla eht nesohc gnivah fo sddo ytfif-ytfif ta won era ew oS .sgab etihw-dna-kcalb tnereffid owt eht morf era owt dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT

You can connect Problem 2 to our random door opening and a goat being revealed, in the Monty Fall (with an F) problem.

[jncfhnb]

Noooo

It is not the fact that a goat is behind the door. It is the fact that a goat door would always have been opened! These two facts are not the same!

[bemusedthrow75]

I'm sorry what now? Stop second guessing the puzzle which is clearly stated.

I'm so, so done with this now that I am actually going to render inoperable my only HN account so I cannot possibly come back to this, or any other thread.

[jncfhnb]

It is NOT STATED in THIS VERSION OF THE PROMPT that monte will always open a door and that it will always be a goat door.

It is simply stated that on one play of the game, Monte CHOSE to open a door and it was a goat. If this is truly all you know, you have learned nothing.

It is only useful information if monte explicitly opens a goat door for you. If he opens a door at random and gets goat your information gain goes away even if it’s the same door!

I’ve literally mapped it out for you else where. There’s only 4 outcomes. You can check yourself.

2:2

[jncfhnb]

I have not restated the problem.

You’ve incorrectly assumed that I would be showing you a goat in every case. But that is not included in the prompt. All you know is that you were shown a goat on one play. This is possible in the setup I’ve listed. You landed in that 33% case where you chose a car.

There is nothing in the prompt that says we are not playing my variant of the game.

Being told that you SAW a goat does not mean you would ALWAYS SEE a goat if the previous conditions had gone differently. And that is why you’re wrong :)

Again > all you know is that he opened a goat door

[bemusedthrow75]

> You’ve incorrectly assumed that I would be showing you a goat in every case. But that is not included in the prompt.

Yes. It is. It is a fixed part of the scenario. Monty opens a door and shows you a goat. He knows it is going to be a goat (he is "well-aware of what is going on behind the scenes"). He's showing you a goat as part of the problem which is: should you switch?

Again: think through what the problem actually SAYS:

> Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.

> “Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?”

No matter what: the goat Monty shows you is not a matter of chance. It is a fixed part of the problem. It's there in writing: he shows you a goat. Full stop.

[jncfhnb]

No! Read the prompt very carefully.

> Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.

You know nothing except you picked door 1 and then he decided to show door 3.

You do not know if he would have shown door 3 had you picked door 2! This is NOT in the prompt. You cannot assume that.

The experience of this contestant is fully and equally possible within my version of the game too!

[bemusedthrow75]

> You do not know if he would have shown door 3 had you picked door 2! This is NOT in the prompt. You cannot assume that.

Monty isn't a contestant. Monty's action and outcome is part of the fixed description of the problem. He opens one of the doors and reveals a goat. Full stop.

If you think revealing one of the two goats (the other of which might or might not be behind your current selection) isn't information of value in assessing whether your chances in that scenario are improved by switching, I'd encourage you to consider why.

And now I really must leave it to someone else to help you, if they will. But I very much appreciate your politeness.

[jncfhnb]

> Monty isn't a contestant. Monty's action and outcome is part of the fixed description of the problem. He opens one of the doors and reveals a goat. Full stop.

Incorrect and I invite you to cite the line of the prompt that says otherwise. This is the critical missing piece of information that makes it the commonly understood “Monty hall problem” and not the improperly stated prompt that we actually have.

“Monty did this” != “Monty would have done this regardless of your previous choices”.

Unless this is stated, you just have an incomplete problem. If you don’t know the mechanics by which Monty decides to share information, it does not give you probabilistic info.

Probability questions frankly often require you to make judgements about the probable starting conditions based on information given so far. You frequently need to recognize that your current situation is the result of prior processes that are or are not defined.

[habinero]

No, you're solving the wrong thing. This is a thought experiment, not a real life scenario with ambiguity.

In this thought experiment, MH will always open a door to a goat on purpose and always ask if you want to switch, because the entire point of the thought experiment is that people find the 2/3 odds if you switch unintuitive and baffling, leading to extremely long thread chains.

You're doing the equivalent of saying "why can't we wait and see if the cat meows" when talking about the Schrödinger's thought experiment. It can't because that's not the point of the exercise.