I still have trouble understanding why people struggle with the Monty Hall problem. It seems so obvious to me how the math works, yet very smart people have been fooled by it even after having it explained. I don’t get it.
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.
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.
(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.
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.
It is a bad combination of being more inclined to actually respond because of boredom and being less capable in doing so properly.
> 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 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.
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?
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?"
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.
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.
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.
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.
The information you get is which door to open if you are allowed to switch.
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.
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.
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.