I think part of it is that the answer depends on the fact that the host's strategy is to always open a door with a goat behind it that the player didn't choose. If the host instead opened one of the doors the player didn't choose at random, then one third of the time the prize will be revealed. The remaining two thirds of the time, switching and keeping your choice have an equal probability of getting you the prize.

Many descriptions of the problem don't tell you what the host's strategy actually is, so I'd guess that a lot of people who are "fooled" by the problem actually just have a different reasonable interpretation of what it's asking.

This is correct and very frustrating! For a deep dive into this issue check out:

https://statisticsblog.com/2011/11/23/monte-hall-revisited/

It's because the problem is usually stated incorrectly. The character in the comic gets it wrong.

The correct formulation is "Monty knows where all the goats are and is compelled to always reveal a goat." The alternative interpretations are "Monty opens a door at random that happens to reveal a goat" (in which case the 50:50 interpretation is correct), or "Monty reveals a goat only if you picked the car" (in which case you should never switch). All three interpretations are compatible with the version stated in the comic.

Hold on, going through the calculations for the second case you described (the, “picks one of the other 2 doors at random” one).

(1/3) chance you initially picked car. Then, (2/2) chance the door he reveals has a goat.

(2/3) chance you initially picked a goat. (1/2) chance he reveals the car (but we will condition on this not happening) and (1/2) chance he reveals goat.

Sanity check : these (1/3)(2/2) + (2/3)(1/2) + (2/3)(1/2) add up to 1, as they should .

For the cases where he doesn’t reveal a car, and we should switch: (2/3)(1/2)

For the cases where he doesn’t reveal a car, and we shouldn’t switch: (1/3)(2/2).

So, yes, (1/3)/((1/3)+(1/3)) = 1/2 , as you say.

What I personally dislike about the rephrasing is the ambiguity around the states "what happens when Monty selects the prize". I think that's the more important question.

If he reveals the prize you just lose, the chance to change your option only happens if he picks wrong.

It isn't a strange game. You pick one of 3 doors. The host then reveals the correct door and say you lost. Nobody would raise an eyebrow if that happened.

Lets take this scenario to show how changing would make you lose:

A conman runs a game where you pick one of 3 options and can win a prize. If you pick the correct option the conman will reveal one of the other and ask you if you want to switch. Switching then means you always lose. However if you pick the wrong one then he reveals the correct one and say you lost. So in this case following the rules of the original solutions means you are 100% likely to lose the game.

That people have such a hard time understanding this part of the problem is also due to not understanding statistics properly. If what I said above doesn't make sense to you then you don't understand the Monty Hall problem.

Thank you for your effort, unfortunately it turns out I am too sick currently to actually engage in it an demonstrate my point correctly.

It is a bad combination of being more inclined to actually respond because of boredom and being less capable in doing so properly.

But it has nothing to do with the Monty Hall problem. It would be a completely different game. The only common thing would be that they are both narrated as game shows.

It is the same game. The original problem reads:

> You are on a game show. There are three doors, one has a prize behind them. You pick a door. The host who knows whats behind the doors opens one of the doors and reveals that it was the wrong door. Should you switch?

This description doesn't contradict my scenario at all. it doesn't say "The host will always reveals a dud no matter what happens".

Now, your intuition tells you that my scenario feels wrong since the host doesn't always do the same thing every single time, but that is how math and statistics works. We only know that the host opened this door this one time, we aren't given information about how the host acts in general. My interpretation is therefore just as valid.

Of course, for the real game the host always opened a dud door. But the original problem description didn't say that, it just described a single instance. In a single instance there isn't enough information to tell. And no, if you assume that the reader must know how the original game show worked then the problem isn't a properly defined problem, and the answer could be anything. That is how math works.

I think you're reading the description incorrectly. It is not meant to be a specific instance that may happen differently for other instances. "You" is an abstraction over all players of the game show, there's an implicit "forall you:" in front of the sentence.

I agree that this is not as precisely stated as it could be. And unfortunately there's a type of logic puzzle that relies on people understanding these implicit quantifiers and being prepared to ignore them, which sometimes makes it hard to determine if they're supposed to be assumed present or not. But the most common interpretation is to include it.

> I think you're reading the description incorrectly. It is not meant to be a specific instance that may happen differently for other instances. "You" is an abstraction over all players of the game show, there's an implicit "forall you:" in front of the sentence.

But the problem described a specific instance. You are reading it wrong. Right now you are just digging deeper into a hole.

If the problem actually was told as you describe then you would be right, but it isn't. It just describes a single instance, you are at the show, the host opens a door, what do you do?

You completely missed the point of my comment. The wording describes a specific instance. And additionally, the rhetorical convention implies that there is no instance of the problem for which the description doesn't hold. This part is unwritten; the reader is meant to apply it from context. It seems like this is why there's a big disagreement about this particular write-up. Some people are applying this additional constraint, and others (yourself) aren't. If you counter that this constraint isn't written, and therefore it shouldn't apply, I'm sorry but you're wrong. The author clearly intended it.

Actually, at least according to Wikipedia, the description is "the host will always reveal a dud no matter what happens".

According to Wikipedia, the original problem statement was like this:

"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?"

https://en.wikipedia.org/wiki/Monty_Hall_problem

It is annoying that so many people who don't understand math will downvote in these discussions. Always happens. The problem as described in this comic doesn't have the solution comic says it has. That is a fact. It doesn't say it is the monty hall show, it just says it happens once, that isn't enough to accurately identify the pattern which leads the solution to be 2/3 chance if you switch. If you don't see that then you don't understand this problem.

What exactly do you mean? You mean because it could be the casino scam you describe elsewhere, that they only open the other door if you picked the right door?

Then I think the confusion might stem from the concept of probability. I think what is meant here is probability from the point of view of the deciding person (given their knowledge). That does not have to be the real probability at all.

I don't have a good example and I am not currently well read on it, but I remember from some Bayesian book the example of the lawn being wet. So either it has rained, or the lawn sprinkler had been on. And you can reason about the probabilities. But of course either the one or the other happened (in a simplified logic world). There are no probabilities about it. The probabilities are only the estimates of the person trying to infer if it has rained or not.

OK so strictly speaking, you could create zillions of variants of the Monty Hall Problem from the scarce description. Maybe the host looked at the player funny before the show, so the player raises his estimates for the likelihood of the host being a crook (as in your casino game). Or any number of other things could affect their reasoning.

So it would be merely convention of the mathematical puzzle to assume the simplest case of everything else being random. And then of course there will be puzzles that trick you exactly with that assumption, to teach you that you shouldn't make too many assumptions. OK.

What would he reveal in case 3) if you haven't picked the car, if not a goat? That variant doesn't seem to make sense.

Likewise for 2). If Monty reveals the car, you just pick the car. I also don't think his knowledge really matters. You have 1/3 chance of picking the car. One other door is revealed to not be the car. So odds for the other door to have the car is 2/3 - you don't even need a Monty in the story.

I can tell you why I struggle: my mind says that if I pick a door at random out of two I have the same probability of getting a goat.

So if I switch door and pick one at random it should be 50% right even if I pick the same one I had picked before.

I trust the smart people on this, but I'm going with the math goblins explanation from now on.

> my mind says that if I pick a door at random out of two I have the same probability of getting a goat.

In the proper setup of the problem, at no point do you pick a door at random out of two.

You start with three doors.

* You choose one, and you have a 1/3 chance that you have picked the car.

* There is a 2/3 chance that the door is behind one of the other doors.

Now, and this is a key point, Monty knows where the goat is.

* Of the other two doors, he can choose one that hides a goat. That doesn't affect your chance of being right, which remains at 1/3.

* Of the other two doors, he does choose one that hides a goat. That doesn't affect your chance of being right, which remains at 1/3.

* Of the other two doors, he opens one that hides a goat. That doesn't affect your chance of being right, which remains at 1/3.

Now there is one other unopened door.

* The chance that the door you chose holds the car remains at 1/3;

* The chance the car is behind one of the other two doors is 2/3;

* One of those two doors has been opened!

So the other closed door has a 2/3 chance of holding the car.

Yeah, my simplest explanation is the first guess you have a 1/3 chance. If you switch, you are actually just betting that you were wrong on the first guess… and there is a 2/3 chance you were wrong, so that is a good bet.

Indeed. Looking at this from an information point of view, the fact that the host can open a door tells you nothing as to whether your first choice was right. So the chance you picked the right door is always 1/3.

The information you get is which door to open if you are allowed to switch.

The confusing part is that him opening the door doesn't reveal any new information. In most cases knowing that an event happened changes other probabilities. Or in other words, probability of A is equal to probability of A given B in this case, where A is that your initial pick is correct and B is that he opened a door with a goat. For example, lets say that he opened two doors with goats instead, then what is the probability of A, ie your first guess being correct? It is 100% of course, since it is the only door left, him opening two doors with goats revealed new information about your door.

So the argument "your first pick is always 1/3, so you should always switch" is wrong. That doesn't normally apply. It is true in this case, but not true in the general case of problems like this since probabilities change as new information is revealed.

> The confusing part is that him opening the door doesn't reveal any new information.

It does reveal new information, just not about the door you chose.

> lets say that he opened two doors with goats instead, then what is the probability of A, ie your first guess being correct?

But you can't guarantee that he can do that. If you do guarantee that he can do that then you are giving me more information.

It's always the way, if you change the problem you change the analysis.

> So the argument "your first pick is always 1/3, so you should always switch" is wrong. That doesn't normally apply.

Just because it doesn't normally apply, that doesn't mean it's wrong in this case. It does apply in this case (as you say) and therefore it's right.

Knowing when it's right is hard, I agree.

If it helps any, you're supposed to be misled by the setup.

Monty could have asked the question as "Do you want to stick with the door you chose or do you think it's in one of the other two?" and skipped even opening one of them but that wouldn't have been as fun to watch.