[archgrove]

The Monty Hall problem continued to short circuit the “intuitive reasoning” part of my maths brain until I saw it explained as “Imagine there are a million goats and one car. You pick a door, Monty opens 999,998 doors. Should you switch?”, to which the answer is clearly “Of course” (you had a one in a million chance the first time). Then reduce from there.

[marcosdumay]

The problem with the Monty Hall is that there is a hidden rule that people never explain. If you put there rule there explicitly, it's reasonably easy to see that the presenter is messing with your up-front probabilities when he opens a door.

[toast0]

When the problem was originally going around, this wasn't so hidden. There was a cultural awareness of who Monthy Hall was and experience with the show Let's Make a Deal, and that he never showed the prize door (although, he was not bound to the presented rules). Although, personally, I remember being explicitly told that the host knew where the prize was and so knowingly opened a non-chosen, non-prize door.

My brain still wants to short circuit to two doors, one prize, random selection, 50% chance.

[ColinWright]

> My brain still wants to short circuit to two doors, one prize, random selection, 50% chance.

This seems to be a wide-spread belief/bias ... if there are two choices, two options, two outcomes, then it's 50:50.

OK, I buy a lottery ticket ... either I win or I lose, so it's 50:50, right?

Of course not. If you can start to see, by default, that two options are not usually equally likely, then you start to lose this misconception.

But really, you always have this situation: the prize is behind the door you chose, or it's behind one of the other two. That always remains true, so your probability of having the prize is always 1/3, even if you get to see behind one of the other doors. You're still sticking with your original choice, or switching to both the other doors (although one of them is open, so you ignore it).

[tantalor]

What rule is that

[mrob]

Monty will always reveal a goat regardless of which door you picked (which also requires him to know where the goats are).

[dmbaggett]

Exactly: what counterintuitively changes the overall probability of picking the correct door is Monty’s introduction of new information that biases the contestant to the right choice. Many years ago I was in such disbelief about this result that I wrote a little Monte Carlo simulation to “prove” I was right. In fact, it proved I was wrong!

[Biganon]

People always explain that. Unless they have no idea what they're talking about. It's an essential part of the paradox.

[Jensson]

The original problem formulation lacked it. Also this comic lacked it. Most people who describe the problem doesn't include the titbit. It is a very misunderstood point, you can even see it everywhere in this thread.

[Biganon]

You're entirely right, and in your other comment as well. I just recognized this was about Monty Hall and didn't even bother reading the details. I also assumed other people would do the same: mentally invoke the entire paradox. It is indeed true that someone discovering the paradox with this comic wouldn't have nearly enough info to understand it, and would be correct to say it's wrong.

[mcguire]

Second panel: "You pick one, then the host reveals that one of the other doors hides a goat. Should you switch?"

[mrob]

The "Math Goblins" comic does not explain it.

[jonplackett]

Based on him knowing where the prize is, it’s hard to escape the idea that he’s only showing you a new door because you picked the right one, therefore intuitively you feel like you should stick with your guns. At least if you’re wrong the way you chose, you won’t feel like you’ve been manipulated.

[klipt]

That's correct if Monty always opens the doors.

But suppose Monty only opens the other doors if you picked the prize, then obviously you shouldn't switch.

[Deestan]

That would be a different problem.

In Monty Hall, Monty always opens a door, and you're left to choose between your picked door and one other door.

[Jensson]

This isn't the Monty Hall problem though. The comic doesn't mention the show. It just has a similar setup so people just jump to the same solution even though it is wrong in this case. The correct answer is that it is impossible to know what the chances are if you switch since the problem lacks enough information to tell.

[Out_of_Characte]

How is it wrong in this case? your picked door had a 33% chance to have a price, since its one of three doors, now that a goblin is revealed the other door has a 66% chance of hiding the price. exactly like the monty hall problem

[Jensson]

No, you are wrong. The way the comic described it only describes a single instance of the host opening the door in this case. It could very well be that this host doesn't want you to win, and thus reveals another door and tries to goad you into switching only when you have picked the correct door. That is a very reasonable interpretation of a game like this, and in that case the correct answer is to not switch.

If you think that this description means that the host always opens a dud after you open the first, then you are wrong. It doesn't say that anywhere, and there is no reason to believe that this game show works that way. The only reason you think it does is that you connect it to the original Monty Hall problem which worked that way, which causes you to make the mistake thinking the same rules apply here.

[Out_of_Characte]

I thought about it some more, if the game is never played more than once then there's no strategy to develop for both the monty hall problem and the problem described in the comic, therefore that's a moot point. but if its played more than once then your evil host that sometimes opens a door doesn't make sense.

If the host always opens a dud then its a monty hall problem. but if the host sometimes opens a door, the host still can't do it every time you land on the price because then you wouldn't switch. so the host would have to sometimes open the door when you're on the price and sometimes not. But then when he opens a door it would indeed become a 50/50 so an evil host's best option is to never open a door in the first place.

[mcguire]

How exactly does the comic's version differ?

[mcguire]

Monty always opens a door concealing a goat.