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I think the problem statement is clear, and is independent of how the TV show actually operated.

The problem posed is that you have three closed doors, behind one of which is a car, and behind the other two are goats. You get to pick one of the closed doors, and will win whatever is behind it (you want the car).

One of the doors you did not pick is now opened, revealing a goat. You therefore now know the car is either behind the door you picked or behind the other closed door.

In this setup, the question is: if offered the chance to switch your selection from the door you initially picked to the other closed door, should you switch? Does it make a difference?

Mariyln Vos Savant said switch. The peanut gallery (some statisticians who should have known better) said it makes no difference - with two doors left you've got a 50/50 chance of winning the car whether you switch or not.

First off, you can run multiple simulations of the game (with random positions of the goats and car), and prove for yourself that always switching is a better strategy, so it's sad that anyone would insist that it's not, even if that seems logical.

The simplest way to explain why switching is the correct strategy is this:

When you make your initial selection, you have:

a) A 1/3 chance of selecting the car, and

b) A 2/3 chance of selecting one of the two goats

So, 1/3 of the time you selected the car and will therefore lose if you switch, and 2/3 of the time you selected a goat and will therefore win if you switch. Switching will therefore be a winning strategy 2/3 of the time.

It absolutely is not "independent of how the show operated." The host knowing in advance where the car is, (thereby ensuring the revealed door is always a zonk) is essential. If the host just picked a random door every day, and sometimes did reveal the car prematurely, then it would in fact be a 50-50 decision to switch.

The math here, for those wondering, is that “informed host has a 0 chance to pick the car” becomes “ignorant host has a 1/3 chance to pick the car”. Contestant’s original pick is still 1/3 and the unpicked door goes from 2/3 (informed) to 1/3 (ignorant).

That doesn't change anything.

Obviously if the host reveals the car then you've lost, but we're not talking about such a case. Regardless of whether that is happening some of the time, there is always a 2/3 chance your initial pick was wrong and you'd be better off switching.

If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this point, is a 50-50 proposition. Or 1/3 to 1/3 if that makes more sense. Switching is the advantageous move only when the host always reveals a zonk first, and that's why it does in fact matter "how the TV show operates."

Another show, Deal or No Deal, often came down to 2 suitcases (from an initial 26), one of which was worth a six-figure amount and the other a tiny amount. Something like $200,000 vs $500. The logic of the Monty Hall problem would suggest that since the player had 1/26 chance of initially choosing $200,000, they should switch suitcases at the end. But because all 24 previous reveals were random, there is actually no advantage to doing so.

> If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car.

No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of there being a car anywhere, when we know for sure (3/3 chance) that it is somewhere!

The reality is that the contestant's original pick has a 1/3 chance of being right, and the other other two doors have a combined 2/3 chance of the car being behind one of them. Once one of those doors has been opened and the car shown not to be there, then this now means this combined 2/3 chance lies with the other door (which you should therefore switch to). Note that the probabilities now add up as expected: 1/3 chance original pick + 2/3 remaining choice = 3/3.

The host’s knowledge does contribute to the 2/3, because that knowledge prevents an outcome from occurring. Let us consider the switching strategy and instead say selecting the correct door awards 6 points. Our question: what is the expected value of this strategy when the host knows, vs when the host doesn’t know.

Consider the scenario where the player’s first choice is the winner. This happens 1/3 of the time, and the host’s knowledge doesn’t change their behavior. With the switching strategy, the player always loses in this case, so the contribution of this scenario towards the expected value for either case(host knowing vs not) is 0.

Consider the other scenario where the player’s first selection does not contain the car, and note that this situation happens 2/3 of the time. For convenience, label the player’s selected door 1, and the car containing door 2.

If we assume the host has knowledge, then the host will select door 3. If the player decides to stay, they lose. In this scenario, the switch strategy always wins, so its contribution towards the expected value when the host knows is (2/3 * 6) = 4.

If we assume the host doesn’t have knowledge, he will pick either door 2 or 3 with equal probability. In the event he selects door 2, the winning door, so half the time, the player loses because they can’t switch. This occurs half of the time within this scenario, so this sub -scenario contributes 0 points to our expected value. If the host selects door three, then the switch strategy wins, meaning this sub-scenario contributes (2/3 * 1/2 * 6) = 3 points to our overall expected value.

When the host has knowledge, the scenario where the player loses because of the host selection will not occur, leading to the 2/3-probability. Without the host having this knowledge, our event space changes and the player switch strategy only has a 1/2 chance of winning.

What do you think is incorrect about the code here which produces 1/2 if Monty randomly chooses? https://news.ycombinator.com/item?id=39515082

Nothing .. I was wrong.

Switching only wins when Monty avoids opening the car door. If Monty opens a random door, then switching makes no difference.

The flaw in my logic above for the random case is that it's 1/3 chance orig pick was right, 1/3 chance unopened door, plus 1/3 chance that Monty opened the door with the car behind meaning you never even got a choice to switch.

> The simplest way to explain why switching is the correct strategy

In my experience the most _intuitive_ explanation is to simply ramp it up to 100 doors, with Monty opening 98 of them, to make it clear that switching offers you the benefit of all unopened doors.

Maybe different explanations resonate with different people.

Another explanation:

After initial selection there is 1/3 chance car is behind your door, and 2/3 chance it is behind one of the other two doors.

When one of the other doors is opened revealing a goat, we NOW know (new information to be taken advantage of!) the 2/3 chance represented by those two other doors lies with the unopened one.

So, your choice is stick with original choice which has 1/3 chance of being correct, or switching to the other door which you now know has a 2/3 chance of being correct! Obviously you switch.

The big lie to "with two doors left it's 50/50" is that opening doors makes no difference. You're original choice had 1/3 chance of being correct, and you can't go back in time and change that.

“People” seem to think whenever you have a number of options, each option is equally likely.

I get into the car today - either I die in a crash or I don’t.

Hmm, I didn’t die, I survived the 50:50.

Their model doesn’t allow accounting for varying probabilities for different outcomes.

It's a lot easier to see the 2/3 nature if you imagine that instead of revealing the goat and then asking if you want to switch, Monty said "you've chosen this door that has a 1/3 chance. I'm offering you to switch to choosing both of these two remaining doors, at least one of which definitely has a goat by definition." This is functionally the same offer. It's 2/3 to switch because you're effectively choosing two doors and having Monty automatically grant you the car if it's behind one of them.

The initial door reveal is misdirection by an informed host, such that it has no effect on the odds, which are always, players choice of door is 1/3. Imagine instead:

After the player picks a door, the host states out loud, "we started with 2 zonks, you picked one door, so there must be at least one unchosen zonk... and I'm going to show you one behind door #X (door opens with zonk)... now, would you like to switch?"

Wouldn't make for good TV but it describes what is actually happening. Knowing that the host will always show an unchosen zonk is the key to realizing that opening that door has no impact on the player's odds (1/3 originally, 1/3 after the reveal, thus switching is 2/3).

> The initial door reveal is misdirection by an informed host, such that it has no effect on the odds, which are always, players choice of door is 1/3

Yes, exactly. Which means when you switch, the host is effectively allowing you to choose both the other doors, which gives you 2/3 odds.

Also regardless of whether she's a woman or she's known for intelligence, a response in which you say "You're wrong" needs an extra second's thought to consider whether, in fact, they are wrong, or maybe just their understanding is different from yours and you need to reconsider with their context.

Some time back in an HN thread about programming languages I read a response in a thread about inference which said that a particular feature doesn't really save on typing. And it seemed ludicrous. Until I realised the author of the comment has seen the word "typing" (which everybody else in the thread is reading as "The thing you do with a keyboard") and they're interpreting it as "The thing the compiler does". And yeah, with that perspective of course inference isn't saving typing, all your variables still have types, we were just happy because we spent less time operating the keyboard to write a program. Their comment was not wrong and we weren't correcting them, we'd just understood things differently!

Edited to add, link: https://news.ycombinator.com/item?id=36593110

I almost never say "you're wrong", no matter how confident I am. Because I can be, and often am, wrong myself. If there is a disagreement, a miscommunication, etc, why not instead work with the person to find where you two differ and look for common ground? If the other person really is wrong, it's almost always naturally revealed that way.

Also it’s a lot more fun to simply shout “FALSE”

>> The game would be reduced to "does Monty feel like gifting you a car or a goat today".

But, in fact, that is the way the real show worked. He was more likely to give you a car when he hadn't given away any cars for a while. And Marilyn's description of the problem included no information about whether or not Monty offered to swap doors rarely, frequently, or always, or about the previously observed outcomes depending on whether or not the contestant took the bait. And she surely did not say that Monty was a robot and had no interest in what happened on his show.

The way the game actually worked is like how poker and life work until proven otherwise. Every high-stakes game has a sucker, and if you do not know who is the sucker, it is you. Read the story in which a NY Times reporter interviewed Monty about the controversy and you will learn how it really worked.

If you're playing the game as stated, and if the host opens a door revealing a goat, you must switch for the best odds.

The offer to switch doors is part of the problem statement. You can't justify your decision based on the possibility in the real game that you might not get the offer to switch (which would mean that when you do it might be an intentional red herring).

Even if there's a 50% chance there are all goats, or a 75% chance, you still switch. That reduces the overall probability of a win, but it doesn't affect the relative probability from switching. The only reason you might not want to switch is if the first choice had been forced, such that there was an above-random chance that your first pick was the car.

The person you are responding to is pointing out that if the host doesn't offer you to switch all the time, then it isn't the best strategy to always switch when you are given the option. For example, knowing that everyone switches when given the option, the host might only offer to switch when you've picked the correct door. In that case, not switching would be better. Hence the poker aspect.

I'll admit that it might be a little more complicated than that. Does life require that we make default assumptions? Is the default assumption that when we are seeing and playing a game for the first time, we should assume that the other players are following the simplest and most impartial strategy that can explain their behavior a good default assumption? Implicit default assumptions are rampant in human efforts to resolve the hard problems of life. Can anyone recommend ways to identify them, to find alternative default assumptions, choose good ones, and to be aware of their impact on conclusions.

> it just wouldn't make sense in the context of the game show for the presenter (who knows where the goats are) to ever open a door to reveal the car and give you the option to switch

No, but it would make perfect sense for him to open a door at random and - if doing so reveals the car - tell you sadly that you've lost as a disappointed klaxon plays. Indeed this is almost the only possible way that the initial door opening could realistically make sense in a game show; if it works in any other way, there's no tension when the first door is opened since there's no risk!

> there's no tension

There certainly is. It's a high-stakes scenario and the door-opening presents new information. It makes the conclusion feel less certain.

Revealing the car would conclude the game, of course. Unless the contestant were holding out for the better goat.

In reality, Monty Hall often did not behave in the way the problem describes, that might be the source of confusion surrounding this. (e.g. sometimes he would reveal a prize and imply there might be a better prize behind one of the doors. sometimes there were only prizes and no goats, etc.)

> Not only is Marilyn's interpretation grammatically valid

I would consider both interpretations equally valid. If anything, in law we have the "rule of the last antecedent" (phrase/clause modifies nearest antecedent noun) so that the hater's interpretation could be more correct if based only on this rule.

This language is ambiguous: "host ... opens another door, say #3, which has a goat."

The two options are Marilyn's interpretation of the door always containing a goat (invariant version), or that one door he opens (say #3) just happens to contain a goat (hater's version) but it could have contained a car.

As suggested by someone below, a much clearer formation could have said "host opens a door with a goat, say door #3" to avoid this particular ambiguity.

There's also the issue of usage of "which" signifying a non-restrictive non-essential clause.

If the the door contains a car, then it doesn't matter whether you switch or not. If it is a goat you get the problem as intended.

You still have the intended problem as a subproblem, so this is just a distraction and not a serious objection.

If a goat was revealed randomly instead of intentionally, there is no advantage to switching. See the "Monty Fall Problem" in this paper already linked elsewhere in the discussion: https://web.archive.org/web/20230706235720/https://probabili...

To use the 100 door extension some people find helpful: If Monty always reveals the 98 doors that don't contain goats, then 99% of the time (every time you picked a non-car door), the other door will have the car. If Monty is opening a random set of 98 doors, then 98% of the time Monty will reveal the car, 1% of the time you'll have picked the car on the first guess, and 1% of the time the car is in the other door. When you're in those 2% of cases where no car is revealed, you have a 50/50 shot of being in either of the 1% states where that happens.

On this topic, please watch what I think is one of the greatest scenes in movies: Kaspar Hauser directed by Warner Herzog [1].

[1] https://youtu.be/C9uqPeIYMik?si=ZjJIH1RhQFTN_ARR

Hey that riddle was in Labyrinth (with David Bowie as the goblin king!)

https://www.youtube.com/watch?v=grOpvXBmTx8

> it just wouldn't make sense in the context of the game show for the presenter (who knows where the goats are) to ever open a door to reveal the car and give you the option to switch.

None of it makes sense because it is not a real game show, it is a thought experiment.

Apropos the topic of being confidently incorrect, it was indeed a real show: https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal

Hosted by a real Monty Hall: https://en.wikipedia.org/wiki/Monty_Hall

It even had the gag prizes like goats (or, as in the photo from the wiki page, a llama)

The Monty Hall problem is a thought experiment (based on a real game show). As evidence, see this letter from Monty Hall himself ("no trading boxes after the selection"):

Dear Steve:

Thank you for sending me the problem from "The American Statistician."

Although I am not a student of a statistics problems, I do know that these figures can always be used to one's advantage, if I wished to manipulate same. The big hole in your argument of problems is that once the first box is seen to be empty, the contestant cannot exchange his box. So the problems still remain the same, don't they. . . one out of three. Oh, and incidentally, after one is seen to be empty, his chances are no longer 50/50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so. It was always two to one against him. And if you ever get on my show, the rules hold fast for you -- no trading boxes after the selection.

Next time let's play on my home grounds. I graduated in chemistry and zoology. You want to know your chances of surviving with our polluted air and water?

Sincerely,

Monty

https://web.archive.org/web/20100408200824/http://www.letsma...

I think importantly, the "Monty Hall Problem" as described is different from how any iteration of the Big Deal at the end of Let's Make a Deal is actually run. Hall himself recognized this and mentioned it in a NYT interview in the 90s. It's paywalled, but the excerpt is here: https://en.wikipedia.org/wiki/Monty_Hall_problem

Hall understood the problem, giving the reporter a demonstration with car keys and explaining how actual game play on Let's Make a Deal differed from the rules of the puzzle.

When I first heard about the puzzle, I struggled with the solution precisely because I kept making assumptions about the puzzle based on my knowledge of the show. It wasn't until I thoroughly read the question, setting aside those assumptions, that I realized it is materially different from the show and I was able to really understand the solution.

The name of the problem is obviously taken from the real game show, but the real show did not work like in the thought experiment. So you cant just apply game show logic.

The “monty hall problem” (in the form stated in the article) is really a trick question, since it hinges on some assumptions which is never stated.

Even when stated correctly, the problem is a fun and counter-intuitive problem. But when leaving out critical information, you create a very differet problem of guessing what the unstated assumptions are.

Yes, the actual show was mentioned in the article.

Also, Let's Make a Deal has been revived and canceled many times, and has been on the air since 2009.