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Math Goblins

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72 pointsColinWright74 comments

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.

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.

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.

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.

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

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.

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.

How exactly does the comic's version differ?

Monty always opens a door concealing a goat.

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.

What rule is that

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

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.

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

The "Math Goblins" comic does not explain it.

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.

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

Right, the host revealed a goat this time. It didn't say the host always revealed a goat. The sentence is very clear on that point.

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.

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!

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.

> 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).

The votey (red button to the bottom right of comic) says i^i is a real number. I didn't understand how at first, but the proof seems straightforward: https://math.stackexchange.com/questions/216871/why-is-ii-re...

That proof uses Euler's formula, which is the real brain melter.

I vaguely remember a 3blue1brown video that explains this really well too. Something to do with thinking about it in terms of rotations on the complex number plane which can land you back on the axis of real numbers. That was the most intuitive thing that I remember seeing

https://youtu.be/pq9LcwC7CoY

Overly pedantic but I think i^i is actually not well-defined (and in general exponentiation of complex numbers is not well defined) since e^(2 * pi * i) = 1 so for instance we would have i = e^(0.5 * pi * i) = e^(2.5 * pi * i) and then i^i = e ^(-0.5 * pi) = e^(-2.5 * i) but those last two numbers are both real and not equal so complex exponentiation only works when you pick a branch of the complex logarithm. Anyway...

Yes, though regardless of which branch you pick to define i^i, it's value is real.

By this point you're probably used to the idea that most of the real numbers are irrational numbers. That is, even though there are countably many rational numbers, there are uncountably many real numbers.

Numbers like pi and e, even though they are irrational, can still be described. There is still an algorithm that computes their digits. The same goes for 0.112358132134... or Apery's constant. But the vast majority of real numbers can never even be described.

Proof: Any description of a number (any algorithm that computes its digits) is a finite string of symbols. There are countably many finite strings of symbols, but uncountably many reals. Math goblins.

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.

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.

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.

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.

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.

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

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.

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 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 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/

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.

I wonder if you could trick people who mistake Monty hall like this.

Create a Casino, let users pick one of 3 doors, and give them 4x the money back if they pick the right one. Seems like bad odds for the casino, right? But the trick is, if they pick the right door you open another and give them the option to choose. Naïve as they are they will think that now they have 2/3 chance to win if they switch. But overall with these rules they have 0% chance to win. It will work as long as you only play the game once per person.

The Monty hall problem is a very nice example of why reasoning with conditional probabilities is hard. If you have studied conditional probabilities and encounter the Monty Hall problem it is easy to think you are in the situation to apply your knowledge and reason like that:

I know that the prize is not behind door A (because the host showed me the goat there). Given that the prize is not behind door A calculate the probability that it is behind door B. Well that's obviously 1/2. Why? Just use conditional probabilities that's what they are for. (The calculation is trivial: Event X="prize behind B", event Y="prize not behind A", P(X)=1/3, P(Y)=2/3. P(X and Y)=1/3. P(X|Y)=P(X and Y)/P(Y) = 1/2.)

Exactly why this argument is wrong is the subtle part. Of course the argument doesn't put to use part of the information given - which door I chose originally and what contract I had with the host, so this is normally enough to make one uncomfortable, but still we know that, e.g.

"In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (...) has already occurred." (source: wikipedia)

Having thought about it I find that adding some more precision to the intuitive motivation of conditional probabilities might be helpful. I certainly think twice now before using conditional probabilities to model the real world, but realizing that I should has cost me quite a bit of head scratching.

I think most of the Monty Hall confusion comes from focusing too much on the question and the math and missing the important part: The confusing setup is part of the game! Essentially it works like this:

Host: Contestant, please choose one of the 3 doors

Contestant: I choose door #1

Host: Confusing jibber-jabber to raise tension and keep this fun to watch

Contestant: I don't know what to do!

Host: Do you think the prize is behind your selection, door #1, or behind any of the others?

Expressed this way there's no confusion but it's the exact same problem. The contestant first selected 1/3 doors and the host is offering to let Contestant keep that 1/3 or choose the other 2/3. Any obfuscation in between is intentional for the sake of the game.

Let’s say 1 in 50 people have a genius-level IQ. What are the odds that a random group of 50 people doesn’t have one such person? The answer is 37%, or 1/e. Why? Because the chances a given person isn’t a genius is 49/50. For 50 people, the odds that none are geniuses is (49/50)^50, which is about 37%. And limit as n approaches infinity of ((n-1)/n)^n is 1/e. Math goblins.

Surely there must also be Math Gremlins, who are chipping away at its foundations. CF Gödel and Chaitin’s Incompleteness Theorems.

Actually gremlins are a bit different creatures: https://en.wikipedia.org/wiki/Foo_fighter

The way he presented it there's a 50% chance that opening another door reveals a goat. I've of the other doors will always contain a goat The key piece of information that is merely implied by the Monty Hall problem but not explicitly defined is that the host knows which door is the winning door and deliberately opened a losing door. If the host does this purposefully then you should switch your answer because two thirds of the time you will have chosen a goat and the host will be forced to reveal the other losing door. However if the host has opened a door at random and it happens to reveal a goat then changing your answer will not improve your odds. There is a 50% chance that the remaining door is a goat rather than a two thirds chance. He could have just as easily revealed the prize. Explained in this way the Monty Hall problem is not confusing. The wording is what was confusing.

Also, there are no electrons. (https://www.amazon.com/There-Are-Electrons-Electronics-Earth...)

Right, good luck with that.

2 doors to choose from, 1 price, 50% chance of winning, period.

And I don't care if there were initially a million doors to choose from.

The missing piece is: have the doors been allocated before the reveal and allowing the switch? If yes, then the chance is 2/3 for switching.

If there are two doors, and one prize, and you can choose one door, then you're right, it's a 50% chance.

That's not the situation, and the chance is not 50%. I don't care if you agree or not, you are mistaken if you think the chances are 50:50.

But that is the situation, there are exactly two doors and one price left.

We are either talking past each other, or you are missing the point. In the Monty Hall problem it's not 50:50. Really. It's not.

* There are three door, one conceals a prize;

* You choose one door;

* The host, knowing where the prize is, opens one of the other doors to reveal a non-prize;

* You have the option to stick with your choice, or to switch;

* If you stick, your probability of having the prize is 1/3;

* If you switch, your probability of having the prize is 2/3.

There's enough discussion elsewhere, or you can write your own simulation (but make sure you simulate the right thing, and not something different).

As others have pointed out, you need to know in advance that (a) the host will always open another door, (b) the host knows where the prize is located, and (c) the host will never reveal the prize.

But given that, the odds really are not 50:50.

I am aware, and at one point I accepted the whole thing as is; but the older I get the more I feel like it doesn't make any sense to me, theories and proofs be damned.

I just can't see the point in switching; opening the door provides new information, but that information applies equally to both remaining doors.

Sorry if that makes people upset, but what am I supposed to do?

> opening the door provides new information, but that information applies equally to both remaining doors.

This is intuitively obvious, but where it goes wrong.

You've chosen door A. Doors B and C remain, at least one hides a goat, and the host knows where everything is.

They can always open one of B and C, so the fact that they do so tells you nothing about A.

But if they open B it tells you they didn't open C, and vice versa. So when they open a door they are telling you nothing about the door you chose, and something about the other unopened door.

> ...what am I supposed to do?

You can try writing a simulation and watch it work, but another way to think about it is this.

Suppose you choose one door, put your hand on it, then close your eyes. Now you're given the choice of sticking, or switching to both the other doors. Clearly it's to your advantage to switch. Now with your eyes still closed the host opens one of the other doors. You can still stick, or switch. In effect, switching lets you choose both of the other doors, but you can ignore the one the host opened.

Does that help?

How is this the top comment? Someone with the hubris to think that their intuition is better than dozens of proof of the Monty Hall paradox?

Or maybe I missed the joke

The comic didn't fully describe the problem so there is no correct answer. Like 99% of people who think they understand Monty Hall don't really understand it. If you can't spot when the problem isn't accurately described then you don't understand it.

It was likely because HN often puts new comments at the top of the thread for a few minutes to give them exposure. (I suspect it probably puts different new comments at the top for different visitors.)

The correctly described monty hall problem has 2/3 chance if you switch. This isn't the correctly described monty hall problem, so there is no way to know what the chance is if you switch. It doesn't even say it is the monty hall show, so you can't even patch it up using external knowledge. Instead these are just some loose meme bits that perpetuates a lie because the author didn't fully understand the Monty Hall problem.

You are being repeatedly downvoted because you are repeatedly saying that the description is not of the classical Monty Hall problem. Here's the description in the comic:

* Suppose you can pick door A, door B, or door C.

* Two doors hide a goat. One hides a prize.

* You pick one, then the host reveals that one of the other doors hides a goat.

* Should you switch?

Let me say that I think that in this description it is fairly clearly implied that:

a) The host reveals the presence of a goat by opening one of the other doors;

b) The host will open a door, regardless of whether you picked the prize or not;

c) The host knows where the prize is, and specifically opens a door to reveal a goat.

I believe (a) because we know by counting that one of the other doors hides a goat. If saying "the host reveals that one of the other doors hides a goat" is not content free, then it must imply that the host actually does some revealing.

I believe (b) because we are told that the host does reveal a goat in (a). With no conditioning mentioned I believe it is fair to assume that the host will always open a door.

I believe (c) because if the host does not know where the prize is, then they must open a door at random, and would not be guaranteed to reveal a goat. I believe the statement as it stands precludes that.

So, can you tell us exactly why you think this is an incorrect description? Can you tell us exactly why the probability of winning the prize upon switching are not improved to 2/3?

The implications are not clear. I suspect you only think they're clear because you already know the classical problem, or perhaps because you have more faith in the good intentions of strangers than I do. My personal reaction to the scenario described in the comic would be that the host was trying to trick me because it would make for more dramatic television (only revealing the goat because I'd picked the car), and the correct decision would be to refuse the offer.

Your belief (a) is compatible with my common sense assumption that the host's goal is to maximize drama by making me lose the car, not to help me. Your belief (b) is incompatible with my assumption, but it is not required by the wording of the Math Goblins problem. Your belief (c) is compatible with my assumption.

All of Jensson's comments in this discussion are correct, and they are being unfairly downvoted.

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