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Comment by jml7c5 | Hacker News Reader

From reading the problem as posed in the article, the confusion seems similar to the "plane on a treadmill" thing, where people are interpreting the premise of the problem differently. (In plane on a treadmill: "the treadmill moves backwards @ -1 * plane airspeed" vs "the treadmill moves backwards to keep plane airspeed at 0". In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".)

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> vs "the treadmill moves backwards to keep plane airspeed at 0".

You might be correct in thinking people misunderstand it as that, but that's physically impossible, so I don't think you can argue that people have some sort of alternate understanding under which they are actually correct. That's just one type of wrong reasoning people might apply to the problem.

Add to that, in your hypothetical understanding of Monty Hall, that actually still doesn't change anything. Because even if the host picks a door completely at random, you still should always switch, it would just sometimes be the case that the host goes "ohh, that's to bad, I revealed the car and you can't win it now", but obviously that doesn't do anything to change your odds, because you're always better off or the same by switching, it's just in the unfortunate cases you'r switching from 0% change to 0% chance. But that doesn't do anything to change the fact that if he didn't reveal the car you're still going from 33% to 66%.

It's great that you're so sure of yourself, just a shame that you're wrong. If the host picks randomly, your chances are in fact 50-50.

Think of it this way: imagine every time the hosts picks the car the universe resets. Now imagine you see the host picking a goat. There's a 2/3rds chance in your universe you picked the car, and a 1/3rds chance you picked a goat, and so switching would seem like the bad option. This actually cancels out the effects of the normal Monty Hall problem, and we are left with a 50-50 chance.

> imagine every time the hosts picks the car the universe resets

Do you believe that we live in a universe that is reseting, every time Mr Monty hall opens a car?

As in literally Mr Monty Hall. From the game show. In real life.

Because the purpose of the question is to mimic the show.

You could also consider a case where they send in a new contestant every time the Random Monty Hall picks a car, and the result is the same.

> You could also consider a case where they send in a new contestant every time the Random Monty Hall picks a car

In the actual show, do you believe they are sending in a new contestant every time Monty hall picks are car?

Do you think that is what is happening in the actual game show?

No, obviously not. We're talking about expected values though, so it's a useful thought experiment.

> We're talking about expected values though

No, what people are talking about is the actual game show.

They are doing a statistical analysis of how the actual game show works, in real life.

So nobody is talking about a different, hypothetical situation, where the universe is reset, or the game show host brings out a new contestant every time.

Instead people are talking about how the actual show works. And your scenarios that you bring up, are not relevant, and therefore wrong.

> so it's a useful thought experiment.

No, it is a different thought experience that applies to a different situation.

It's funny to me that there always seem to be an over representation of rude people among those who get this problem wrong. There is absolutely no need to attack my person.

As for the problem, you are just plain wrong. We are talking about are regular old TV-quiz, there is not "universe resetting button" to save your logic.

In the regular situation, the host always remove a goat. because otherwise the quiz show is kinda boring. But in the modified version we are considering where he just picks a door at random, then in 2 of the 6 possible outcomes for the door the host picks, he'll be removing the car. Now since the problem has the host showing the doors content, that should be the end of the game show. Who want's to see someone pondering if they should switch between a goat and another goat right? But that doesn't change the logic at all, because the conclusion of "you should always switch" isn't impacted in any way. 0% to 0% is just no change. And in the rest of the cases your'll go from 33% to 66%.

Now of cause what would happen in your odd example where the host has a universe resetting button is that you don't have any choice at all, because no matter what the host will reset the universe until they maximize ratings, which might have you get the car or might have you not get the car, but there's no choice to be taken and the outcome that has maximum rating will always happen with 100%. That's why most stats problems avoid introducing "then the host resets the universe" in their problem description, it kind of ruins the whole point of calculating proabilities.

Think of it this way. In the original problem, there was a 1/3 chance of your original choice being the car, and a 2/3 chance of the other two. The host knowing which door to open gives you information without changing the odds, so there's still a 2/3 chance you picked wrong.

When the host does not know which door to pick, there's a 1/3 chance that you picked right, a 1/3 chance that the host opens the door with the car, and a 1/3 chance that you should switch. The host opening the door with the car ends that branch, so you're left with 2/3 of the original probability, and your odds for either branch remaining are 50:50.

I agree with you that "the universe resetting" is a dumb way to visualize or explain the problem. But if Mr. Hall opens the first door randomly (essential that it was a random choice), and it revealed a goat, then no, switching does not provide an advantage. You do not go from 33 to 67%. Instead, you are left with 2 options, both of which originally had a probability of 33% of containing a car, and which, now that the third option has been eliminated, are both currently 50%.

It does change your odds -- you're now living in a world in which the car didn't get revealed, and by Bayesian reasoning that means it's more likely that you live in a world where you picked the correct door.

I didn't state that it doesn't change your odds. Obviously going from 33% to 0% because the car isn't possible to win anymore will be a change of your odds. I stated it doesn't change anything, because in one situation you go from 33% to 66% by changing and in the other you go from 0% to 0% by changing. So there is no difference to the logic of always switching because it either improves your odds or keeps them the same.

If the host reveals a goat at random rather than by special knowledge, your odds don't go from 33% to 66% by switching. The case where the host reveals a goat is only two out of three, not three out of three because one third of the time, the host will reveal the car. In one of those two cases, you picked a goat and in the other you picked a car. So you're at 50% odds whether you switch or not, if the host doesn't know where the car is.

If the host does know, then there are three out of three cases where the host reveals a goat. In one out of three cases you picked the car but in the other two cases you picked the goat. So that's why your odds go up if you switch.

The problem is that there are two interpretations of the problem, and is is not obvious that the incorrect one is wrong unless you know that Monty knows what's behind all the doors, and chooses never to open a door with a car behind it. The question isn't "does Monty showing a goat change the odds that it's behind the door that neither you nor Monty picked", it's "does it change that probability from 33% to 50% or from 33% to 66%".

Scenario 1: The first, incorrect, interpretation of the problem is "You choose a door, which has either a goat or car behind it. Monty then chooses one of the other two doors, which will also have either a car or goat on it, and opens that door. You then have the choice of whether to switch doors or stay with your original choice".

Scenario 2: The second, correct interpretation of the problem is "You choose a door, which has either a goat or a car behind it. Monty then looks behind the other two doors, and chooses the one that has a goat behind it. If both have goats behind them, Monty chooses randomly. Monty opens his chosen door. You then have the choice of whether to switch doors or stay with your original choice".

    +-----+------+--------+---------+--------+--------+------------+------------+
    | Row | Car  |  Your  | Monty's | Result | Result | Frequency  | Frequency  |
    |     | Door | Choice | Choice  | Stay   | Switch | Scenario 1 | Scenario 2 |
    +-----+------+--------+---------+--------+--------+------------+------------+
    |   1 | #1   |     #1 |      #2 |    Car |   Goat |       1/18 |       1/18 |
    |   2 | #1   |     #1 |      #3 |    Car |   Goat |       1/18 |       1/18 |
    |   3 | #1   |     #2 |      #1 |   Goat |   Goat |       1/18 |       0/18 |
    |   4 | #1   |     #2 |      #3 |   Goat |    Car |       1/18 |       2/18 |
    |   5 | #1   |     #3 |      #1 |   Goat |   Goat |       1/18 |       0/18 |
    |   6 | #1   |     #3 |      #2 |   Goat |    Car |       1/18 |       2/18 |
    |   7 | #2   |     #1 |      #2 |   Goat |   Goat |       1/18 |       0/18 |
    |   8 | #2   |     #1 |      #3 |   Goat |    Car |       1/18 |       2/18 |
    |   9 | #2   |     #2 |      #1 |    Car |   Goat |       1/18 |       1/18 |
    |  10 | #2   |     #2 |      #3 |    Car |   Goat |       1/18 |       1/18 |
    |  11 | #2   |     #3 |      #1 |   Goat |    Car |       1/18 |       2/18 |
    |  12 | #2   |     #3 |      #2 |   Goat |   Goat |       1/18 |       0/18 |
    |  13 | #3   |     #1 |      #2 |   Goat |    Car |       1/18 |       2/18 |
    |  14 | #3   |     #1 |      #3 |   Goat |   Goat |       1/18 |       0/18 |
    |  15 | #3   |     #2 |      #1 |   Goat |    Car |       1/18 |       2/18 |
    |  16 | #3   |     #2 |      #3 |   Goat |   Goat |       1/18 |       0/18 |
    |  17 | #3   |     #3 |      #1 |    Car |   Goat |       1/18 |       1/18 |
    |  18 | #3   |     #3 |      #2 |    Car |   Goat |       1/18 |       1/18 |
    +-----+------+--------+---------+--------+--------+------------+------------+

In scenario 1, before any door is opened, you chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. Monty then opened a door which happened to have a goat behind it, which eliminates rows 3, 5, 7, 12, 14, and 16. Now you have a 6/12 chance of winning the car if you stay, and a 6/12 chance of winning the car if you switch, and this is how you come to the conclusion that there is no advantage in switching.

In scenario 2, you still chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. However, this time Monty's door-opening doesn't eliminate any rows with nonzero probability, since rows 3, 5, 7, 12, 14, and 16 have zero probability to start with. As such, you still have a 6/18 chance of winning the car if you stay, and a 12/18 chance of winning if you switch, so you should switch.

The presentation of the problem included the phrase "the host, who knows what's behind the doors, opens another door", so I think in that case the interpretation that Monty deliberately reveals a goat is more appropriate (in addition, obviously, to being what was intended), although I agree there is wiggle room enough to admit both possibilities and clarity about which case you are discussing is important as it changes the answer.

It does, but it doesn't emphasize it enough to allay confusion, particularly for those unfamiliar with the "gotcha" in the problem. It's easy to overlook in that description unless you know that it is the important clue to understanding.

Kind of like the classic riddle "John's mother has four children. They are named March, April, and May. What's the last child's name?" which uses underemphasis for almost comedic effect.

> as it changes the answer.

It doesn't. If the host randomly reveals a car, then you have 0% chance to win. If he doesn't you have 66% chance to win by switching.

I once thought that had to be the case. I wrote the simulation to demonstrate that I was correct. I learned that I wasn't. I encourage you to do likewise.

I think I now have an understanding of why the right answer is the right answer, but I thought I had that before; I am more confident in the answer than my reasoning.

I owe you an apology, you're correct, I was wrong. I can't pinpoint where my reasoning failed me exactly, but my confidence certainly was misplaced, and that's from someone who did do the simulation originally to understand the outcomes, but not to understand the distinction between the host deliberately or randomly revealing the door. I'm sorry, but thank you for you persistence in correcting me. It was a fruitful learning experience to dig into this problem again.

> In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".)

If the host opens a door with a goat, then it doesn't matter whether or not it was intentional.

Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter.

There's a difference here, that our language obscures, between procedure and hypothetical.

The intention of the host only matters if the contestant would have to choose the subsequent action (switching or not) before the hosts opens a door.

If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.

(Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.)

The intentions of the host do matter.

Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door.

I hope you will agree that this is equivalent to the problem as originally intended - Monty can be relied on to reveal a goat, and exactly why doesn't matter.

Breaking it down into equally likely cases, assuming the contestant picks door 3:

    A) The car is behind door 1, Monty picks door 1, Monty corrects.
    B) The car is behind door 1, Monty picks door 2
    C) The car is behind door 2, Monty picks door 1
    D) The car is behind door 2, Monty picks door 2, Monty corrects
    E) The car is behind door 3, Monty picks door 1
    F) The car is behind door 3, Monty picks door 2

When Monty reveals the goat behind (say) door 2, we know we're in case A, B, or F. All remain equally likely, and switching wins in A and B.

If Monty would not have corrected, then revealing the goat behind door 2 eliminates (the new) A as well, leaving us with only B and F, again equally likely.

If all of this remains unconvincing, I encourage you to write a simple simulation of the problem.

So, thinking about it really hard and reading about it online:

My comment was definitely wrong: If Monty could have opened a car door, but just didn't, then duh the probabilities for the car to be behind the doors are different than if Monty always opens a goat door. So in that way, the intentions of Monty, meaning how he chooses, definitely matter.

But I think your example here doesn't show that? Are you trying to illustrate the Monty Fall variation?

Sorry it took me so long to get back to this; real life intervenes sometimes.

I think what I was trying to do was frame the original Monty Hall problem as a variant of Monty Fall, in a way that (I hoped) makes it clear where Monty is doing work to convert some outcomes into other outcomes (and therefore producing different likelihoods).

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