Tricking Monty Hall | Hacker News Reader

Tricking Monty Hall

ignorethecode.net

50 pointszdw140 comments

The reasoning that made me get it was basically:

The key is that the host is quite constrained in which box he can show you: He must always show you a box with a goat and he can never show you the box you initially picked. So the choice the host makes is less random than your (initial) choice and can give you new information.

In particular, only two different things can happen:

a) You initially picked the car. In that case, the host is free to pick any of the two remaining boxes since both have goats in it and neither were picked by you. In this case, it would be obviously unwise to switch.

b) You initially picked a goat. In that case, the host has no choice at all: They must pick the one box with the second goat, which leaves the last remaining box as the one with the car. So you should absolutely switch here.

If you knew whether you're in situation a) or situation b), you'd already have won the game: You could switch as needed and would always get the car.

Obviously though, you don't know. However you know that a) happens when you initially picked the car and b) when you didn't. You also know, you initially picked the car with 1/3 probability. So you know that with 1/3 probability you're in situation a) and with 2/3 probability you're in situation b).

Therefore, if you're just always pretend you're in situation b) and switch, you'll win the car with 2/3 probability.

Yeah, this is exactly the easiest way to think about the problem. In summary, by switching you win when you initially picked a goat, which happens two thirds of the time.

This is the first thing to get, the second is that Monty won't pick the car. If you do not understand these two, then you will say 1/2.

I've heard the analogy with 1 000 doors and opening 998 of them before but that doesn't explain anything at all. If you think that makes you understand then you're mistaken; that made you understand something else.

The key is that you have 1/3 of being right, 2/3 of being wrong. So in 2/3 of the cases, you will be offered to pick the car. So 2/3 switching gives you the car and 1/3 loses you the car. So always switch, it has better odds.

The one which made it clear for me was: imagine you have 4 doors instead. You choose one at random. The host then chooses two of the remaining doors he knows has goats. Is it in your best interest to switch to the one he didn't open?

Suppose you have 100 doors. You choose one at random. The host then chooses 98 of the remaining doors he knows has goats. Is it in your best interest to switch?

In all of these situations, the basic question is: what is more likely - that you chose the one with the car, or that you didn't, and the door remaining is only remaining because the host knew it was the one with the car?

Yeah, simply imagining a much larger number of doors makes the logic obvious.

I used an entire deck of cards (52 of them) to demonstrate this to a friend, and he still wouldn't believe it. I gave up explaining it after that.

It's really easy to get people arguing about this one, even though the correct answer is so well documented.

I remember reading someone who couldn’t understand it - so he build a computer program (in python I seem to recall) to simulate it over millions of goes. And the numbers worked out as expected.

It's an easy python program to write, because there's nothing complex about the process. Pick a language here: https://rosettacode.org/wiki/Monty_Hall_problem

Every one of these prove the effect beyond a doubt.

To be precise they demonstrate it; a proof (mathematically) comes from axioms.

Any explanation that doesn't involve scaling up the number of doors to a arbitrarily large amount where the probability becomes obvious is really lacking if you ask me. The only reason this is so confusing in the first place is because it's only 3 doors.

With an infinite amount of doors, the chance of picking the right one at first is zero. When the host opens all doors except one other, it's almost certain that it's the door with the car.

It’s good for the aha intuition moment for some people. It’s the thing that made me get it as a kid. There’s something unsatisfying about it though. You can have 100 doors, and open 98 to leave the initial guess and 1 more. But this is scaling up the rule as “all but one”. I don’t see an obvious reason that the rule shouldn’t scale up as “open one more door”. Then you have 99 closed and 1 open and the result isn’t nearly as obvious, even though switching is the better move.

Yeah I thought about that, both +1 door and infinity-1 door options are both equally unfaithful to the original I suppose. Keeping the 50% door opening ratio would make more direct sense if you wanted to scale it up for other reasons, but isn't nearly as clear in intuitively showing the concept.

I agree that it feels weird at first, but thinking about it this way fixed that for me:

The whole reason for the apparent paradox is having the "50/50" choice at the end. You either keep the door you picked, or swap with the other door, so it feels like no advantage.

If they only revealed a single door, then you would have multiple doors to swap to, so you wouldn't have the "50/50" choice. The only way to keep the "50/50" choice at the end is to reveal all but one door and give you the option to switch to it.

You can distill it all to this:

If your initial guess is correct, then switching always loses.

If your initial guess is wrong, then switching always wins.

Everyone who understands the game can agree on those facts without controversy. If they don’t then there’s no point proceeding because they do not understand the rules of the game.

Therefore, the odds of winning when you switch are the inverse of the odds of winning from your initial guess.

If your initial guess wins 1/3 of the time, that means always switching loses 1/3 of the time and conversely always switching wins 2/3 of the time.

A slightly different reframing of this idea makes it even more clear to me:

When you make a choice, you divide the doors into two sets: the set containing your choice, and the set containing the other doors. Obviously your set has a 1/3 chance of containing the prize, and the other set has a 2/3 chance.

When the host reveals a door in the second set, it has no effect on those initial probabilities. The set as a whole still has a 2/3 chance of containing the door, and now you have the choice of selecting the only item in that set which you know isn't wrong.

This reasoning is just as obvious when you scale up the doors too. With 100 doors, your set has a 1% chance of containing the prize, and the set of unchosen doors a 99% chance. All but one door in that set are revealed to be wrong, then you get a "50/50" choice of swapping to that last remaining item... of course that's what you want to do!

Consider an alternative game

1. Pick a door

2. Monty asks if you want to switch to both of the other two doors

3. If you switched, Monty will remove a losing door from one of your two doors

I think it is easy to see that this game has the same odds as the original, and also that switching increases your odds from 1/3 to 2/3.

Like a lot of things in math, I think Monty Hall becomes easier to see when you look at some extreme cases. 3 doors is the lowest amount of doors where the paradox appears, ie no way to go smaller. So try a much bigger problem. I like this version:

After Sheherazade has told the calif 1001 stories, she tells him that (only) one of them is true, not fiction. She has him guess which one, and he picks one (say story #500). Then Sheherazade tells him that the true story is either the one he picked, or another one, say #312.

I think in this formulation, it is much easier to see that the calif would be well advised to switch his guess to #312, as his initial guess only had a 1/1001 chance of being correct. Monty Hall is the same problem with 3 instead of 1001, but the same principle holds.

Shameless plug, I am fascinated by Monty Hall and I had a hard time proving it to my friends so I made this: https://kr1stjans.github.io/monty-hall/ :D feedback appriciated :)

I haven't done the analysis, but one major assumption that the whole result is based on is whether the rules of the game are pre-established or just revealed after the first pick. With an adversarial game host who has the option to reveal or not, maybe the result changes, but more importantly, it explains the intuitionistic refusal for some to buy the argument.

In an adversarial setup you basically get 1/3 wins like you started with because the optimal host move is to provide no information whether they reveal or not

In an adversarial setup, the optimal host move is to permit you to switch only if you picked the car. You still get 1/3 wins with optimal play (never switching), but it's better for the host because it allows people who think it's the traditional game to get zero wins.

If the host strategy isn't explicitly specified, it's reasonable to assume an adversarial setup, because seeing the player lose after switching would be more entertaining for the audience.

I don’t think the premise is that the host is allowed to change your ability to switch at all or that the player is misinformed about the rules…

Obviously a game can be made unwinnable if you simply lie about the rules

> seeing the player lose after switching would be more entertaining for the audience

Have you ever watched game shows? Because it is far more satisfying and fun for the audience when players win, not lose. If you are out there rooting for people to lose, you're a miserable bastard.

Now that I think about it, I have watched a few episodes of game shows, and I think you're right. They are always set up so the audience identifies with the players and the hosts are the adversaries, so you want the players to win.

One thing to note is that 33% chance is still pretty good odds. Stats like that only really work with large numbers, and when you've got a single chance, neither of the outcomes is really surprising.

IOW, if every participant switched, the show would be handing out prizes 2/3rds of the time, but that 1/3rd of participants would still be going home empty handed. Do you want to be that one? If participants randomly chose whether to switch or stick, how often would the show hand prizes out?

Now we've got multiple conditional probabilities.

And that choice is what makes casinos, lotteries and games like these work: we want to succeed despite the odds.

My favourite casino game is craps. I find it amazing they created a game that has two basic strategies you can play and the one that has better odds is the "dark side" and is frowned upon to play that strategy.

Sometimes I'll lose a ton of money and a superstitious pass line bettor will make a quip about how I'm playing the dark side and deserve it for betting against the table. My usual reply is along the lines of: we're all playing against the casino and the dark side beats the casino more often than the pass line ;)

Craps is a social game, probably more so than any other table game in a casino. I find the minuscule odds advantage of playing Don’t Pass Line bets to be not worth “being the bad guy” at a table full of superstitious gamblers out having fun, but that’s just me. I’m sure the casinos love this artificial stigma.

I always wondered: how juicy would they have to make Dont Pass to overcome the social stigma at the table? Would it be enough to pay instead of push on 12? How much house edge would cause everyone to switch to the dark side? I have no doubt casinos have spent millions of man-hours of research to understand this.

EDIT: In addition, most craps games offer "odds" bets once a point is established which have zero house edge, so by taking the maximum odds, you reduce the house edge on Pass and Don't Pass to the point where it's not really probabilistically significant which side you play.

It reminds me of S. Korean "fan death."

The common belief that sleeping with a fan on can kill you by suffocation (among other "explanations") was and is a way to convince people to conserve electricity, and nothing more.

The artificial stigma is a way to convince people to not use the more successful strategy against the house.

> is a way to convince people to conserve electricity

According to Slate, the belief is older than that:

> Internet conspiracy lore sometimes blames the legend on a 1970s-era government campaign to conserve electricity, but in fact these warnings are generations older, dating almost back to the introduction of electric fans to Korea.

https://slate.com/human-interest/2013/01/fan-death-korean-mo...

Apparently we don’t know how the myth began.

https://en.wikipedia.org/wiki/Fan_death#Origins_of_the_belie...

Nice. Thanks for the links.

We had a Korean ESL student live with us for a few months. I was shocked when he was shocked that we run the fan blowing down the bedroom halls. I had never heard of this Fan Death, and he was having a quandary because he grew up believing that it kills people, and obviously we did it a lot.

> casinos have spent millions of man-hours of research

You'd be surprised, casinos, in fact, spend almost zero time actually working through the math of their games. All the established table games are taken for granted, and for anything new -- entirely new games or side/extra bets -- the responsibility lies on the game inventor to have a trusted 3rd party validate the math (this is usually enforced by the gaming commission) and the casino simply refers to the validated results when considering implementing the new game/wager.

Any consideration of "social stigma" about a wager is nil. As an example, baccarat is fundamentally a boring game with zero decisions, however the actual gameplay, where players have all kinds of superstitions and are allowed to touch, tear, fold the cards, can be extremely social and often quite exciting. Casinos could deal blackjack in a similar, social fashion where all players share a single hand, but none do. And certainly, no casino would shift the odds in the players' favor (pay on 12 on Don't) for some kind of social impact, as any positive benefit would be 10x, 100x subsumed by professional bettors exploiting the positive EV.

If participation randomly chose whether or not to switch they would walk away with a prize 50% of the time. Assuming randomly == equally likely

I like the variant: You pick a door. An earthquake hits and one door opens that is a goat. The earthquake of course had know knowledge of that it's just what happened.

Should you switch?

I really thought I understood the problem until I got that variant wrong.

There's a good video with a rant about the problem I will link to it when I find it.

Yep, this is the key. The host knows where the prize is, and the rules of the game force him to use and reveal some of that knowledge when he chooses the door to open.

That rule is fundamentally what changes the odds, but it's worth noting that Monty only revealing goats was not one of the rules of the actual Monty Hall gameshow, nor was it explicitly stated in the original framing of this problem (though arguably it was implied). The somewhat ambiguous description of the problem in the newspaper column that popularized it may have contributed to the confusion.

Yes, definitely. However, it's a pretty reasonable inference, because Monty would not open a door without being sure that it would not contain the prize because that would just destroy the dramatic tension. "Oops, sorry, guess you lost. Didn't see that coming. Better luck next time."

> Should you switch?

Why not? There is no harm in doing so if the probabilities are the same anyway...

What helped me visualize is this:

Say the game has a million boxes. You pick one at random, the host opens the other 999,998 boxes (which all contain goats), and asks if you should switch.

At that point, it's very obvious you really should switch, as the original box has a one in a million chance of being correct, but the second box that is now unopened is wink wink nudge nudge maybe the one you want.

That's really a good way to visualise it. The host opens all the other boxes except your pick and one specific other box somewhere in the middle. It's clear that something special must be going on with that box.

I see this posted a lot but it doesn't click with me. It only feels obvious once I know how the probabilities work - meaning I have to get the base case first.

Picture, in your mind or maybe draw on paper, 20 brightly lit doors in a row where 19 are labeled L and one is labeled W. Out of those 20 doors, one door gets randomly selected. It's almost certainly going to be an L. After choosing it, the lights go out above 18 of the other L doors so only the W door, and the L door you probably chose, remained lit.

If you stayed with your existing selection— one of two doors that remains lit— you've still almost certainly got an L door. If you switch to the one other door, with all of the other L doors eliminated, the only way it's not the W door is if you chose the W to begin with.

Two things about this

1. It requires that I already get the point about the probabilities.

2. You don't get a benefit in Monty hall if the host picks randomly - the description requires that they deliberately leave the winning door available. Any situation that gives you the same intuition but you just picture as happening without intent has given you the wrong feeling.

I don't find that increasing the number of doors makes it feel any different.

> You don't get a benefit in Monty hall if the host picks randomly

But the host doesn’t pick randomly, they always pick a goat. That is an integral part of the Monty Hall problem.

Picking randomly wouldn’t make sense. You’d pick a goat, then the host could randomly reveal the car and the game would end anticlimactically.

Exactly— the entire point is that the host always eliminates every losing answer that you didn't choose unless you chose the right answer initially. If you keep your pick, you have a 1 in 3 chance of winning, and if you switch, you have a 1 in 3 chance of losing.

Yes, it is this point (and reinforcing it) that explains the problem to me. Simply increasing the number of doors doesn't help - and it shouldn't.

I didn't simply increase the number of doors— I created a visual of a more extreme version of the problem without making it so extreme that it's abstract. That sort of thing helps a lot of visual thinkers and I'm not sure why you're being so pissy about it not helping you. Frankly, I'm sorry I tried.

I'm not being pissy about it, I'm explaining why this approach doesn't click for me. I'm not sure why, after being told that this type of explanation doesn't work for me and being told exactly why, you are surprised when I say that again. I'll try and help explain why in more detail.

The point is that for me the key bit of information is yet again missing from your explanation.

> Picture, in your mind or maybe draw on paper, 20 brightly lit doors in a row where 19 are labeled L and one is labeled W. Out of those 20 doors, one door gets randomly selected. It's almost certainly going to be an L. After choosing it, the lights go out above 18 of the other L doors so only the W door, and the L door you probably chose, remained lit.

It's extremely important the reasoning behind why the 18 other doors are picked is highlighted.

The game show deal or no deal has basically this, but with 20 or 25 or something boxes. At the end they get to switch - and it doesn't make a difference. It wouldn't make a difference if the host opened the boxes or the contestants. It wouldn't make a difference if the host opened them and knew what was inside.

What would make a difference is if the host opens the boxes while deliberately avoiding opening the jackpot.

Lets reframe deal or no deal in a way that highlights the key part for me:

There are 25 boxes with sums from 1p to £1M in them. Nobody knows what's in them. You pick one randomly, and the others are placed in a pile. You now get to choose one of two things:

1. Keep the box you chose randomly

2. Take the highest value box out of the pile of 24

What if there were 24 boxes worth nothing and one with £1M in, does your strategy change?

What if there were 23 boxes with nothing and one with £1M in, does your strategy change?

What if there were just three boxes?

Yes. The original phrasing is ambiguous though and the one I replied to doesn't say this. Without this very specific point, the other intuitive answer about switching is actually correct.

> Picking randomly wouldn’t make sense. You’d pick a goat, then the host could randomly reveal the car and the game would end anticlimactically

If you want to try and take the riddle that way, it also wouldn't make sense to always offer the switch. They offered the switch this time, but is that a bluff? Would make for a lot better tv show.

But just like prison guards don't set up elaborate schemes for 100 prisoners to be let go, we constrain ourselves to the invented universe of the riddle.

If you read a framing of lights with l and w and no setup explaining how the lights are logically changed the answer is not clear. If it is - your intuition is wrong.

The original phrasing of the problem is not ambiguous at all:

> 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

You can change the rules however you like but then it’s no longer the Monty Hall problem, and that’s what we’re discussing: a brain teaser with well defined rules and probabilistic interpretation.

That's not the original phrasing because that's from 1990, have a look at the wiki page you linked. The original problem does not specify that Monty will never open a box with the keys in. The game would still work if this was the case as Monty offers to buy the box before doing so. In the original problem, Monty says the chance is 1/2, which would actually be true if they had opened them randomly. (edit - a game where the host offers to buy the box you have before another one is opened is essentially the end part of deal or no deal).

Even the 1990 version however is ambiguous when it comes to puzzles. It just says they open a door which has a goat.

The solution originally posted explicitly assumes it to be the case, which is fine.

The million box analogy isn't very clear IMO, because it assumes the host always opens all boxes except one. In the original scenario, "all except one" is identical to "exactly one", so "exactly one" might be the correct host behavior.

There is no "correct" host behavior. One behavior helps explain why the probability is 66%, the other doesn't.

No, the host will only open 1 box, not 999,999. Otherwise it's not the same problem at all.

How is it different?

Here's a way to think of it that some people fine helpful. Assume N boxes, the host knows where the prize is, always reveals N-2 goats after the content picks leaving just the contestant's box and one other box unopened, and always gives the contestant a chance to switch after revealing the N-2 goats.

A person who does not switch wins if and only if their initial pick was correct. The probability of that is 1/N where N is the number of boxes.

A person who switches wins if and only if their initial pick was not correct. The probability of that is 1-1/N.

Switching is (1-1/N)/(1/N) = N-1 times as likely to win.

What this problem always fails for me is that the show host doesn't pick randomly. What if there was option of opening the car?

Also, does the goat come butchered or do you need to do it on stage?

If the host picked randomly then there would be a 1/6 chance that the show would end rather anticlimactically and you would be indifferent in switching. Now this 1/6 becomes your edge turning your 1/3 to 1/2 after the goat is revealed.

EDIT: 1/3 not 1/6

1/3 not 1/6. There are six outcomes of you picking a door and the host opening one of the remaining doors. In two of them you pick a goat the first time and then the host reveals the car.

I wonder if a decision tree would help more people since it’s possible to visualize every possible outcome of every possible game.

That's the thing that makes it click for me, and why there's a difference if the host is random or not.

If they pick randomly, there's no benefit to switching. No harm either.

If it's random, 1/3 of the time you're right on your first guess and switching doesn't help. 2/3rds of the time you're wrong - if the host knows and deliberately doesn't open the door showing the car switching means you win. If however he opens randomly, half of those cases show the car and you can't win - so the breakdown is 1/3 switching wins, 1/3 switching fails and 1/3 the car is shown and you lose regardless.

This is an extremely important part of the problem that is always glossed over in explanations of the Monty hall problem, and it really bothers me that it's ignored because it is crucial to determine what is the "correct" decision.

If the game show host is opening a door at random (i.e. it's possible that he opens the door with the car behind it OR the door you already picked) then the outcome is 50/50 whether you switch or not. But if the host knows what's behind the doors, and purposefully opens one of the remaining two with the goat, then you should switch doors to increase your odds to 66 percent.

On the above article this is only briefly addressed and not explained. In many tellings it is not addressed at all.

Not "rigging" the decision, would be pointless in the game. Yes, it is failed to be mentioned, but if the host revealed the car, then what's the point?

In this article it's addresses with this sentence

> The show's host then reveals the contents of one of the two remaining boxes, but, importantly, always a "goat box."

but it could be made slightly more explicit as:

The show's host, who knows the location of the car, is then compelled to always reveal one of the two unselected boxes as containing a goat.

As far as solving the problem in an academic sense, your explanation is clearer. However, one of the things that makes the whole thing fascinating is that the actual players are not told this specific information, that the host knows. If you watch the show religiously it might dawn on you, hey, he must not be picking the preview door randomly because he has never accidentally revealed the car -- but that's beyond most people's observation.

In practice, the revelation of one door is likely to make the player even more committed to the original choice -- the odds (seemingly, not actually) just went from 33% to 50%, momentum is good, I'm feeling lucky, no way am I going to switch now!

> If you watch the show religiously it might dawn on you, hey, [...]

You can't watch a hypothetical game show. Let's make a deal contains a myriad of different gameplay formats.

The "Big Deal" -- the game where the player chooses one of three doors -- was the finale of every episode of Let's Make a Deal under Monty Hall. Still is, in the Wayne Brady revival. Not hypothetical.

> If the game show host is opening a door at random (i.e. it's possible that he opens the door with the car behind it OR the door you already picked) then the outcome is 50/50 whether you switch or not.

I don't think this is true. Assuming you mean that if the host opens my box, I switch to one if the other two at random with equal probability (even if he opened my box and it was the car), and assuming if the host doesn't open my box that I stay or switch to the other unopened box with equal probability (even if the host opened the box with the car), I think the odds are 33% that I win the car, and 66% that I don't.

Consider it framed this way:

You pick door #3. The host says they're feeling generous so he is going to give you a free door. He blindly reaches into a bag with three marbles labeled 1,2, and 3. He pulls out the marble labeled 2. He opens door 2 and there's a goat behind it. Do you now switch to door 1? The answer is that it doesn't matter, it's 50/50.

Right, in the world where the door he already opened was done randomly (i.e. it could have been a goat or it could have been a car, and it could have been the door you also chose), then if he happened to both (a) open one of the doors you didn't pick, and (b) he opened a goat door, then switching doesn't matter.

But that's only one branch of the probability tree, and if you investigate all of them (assuming he chooses a door at random equally, might open yours, and might open a car), then it's not 50/50, it's 33/66 I think.

Perhaps you just weren't clear enough that you think the 50/50 is limited to only a subset of the possible outcomes in your scenario.

What would "switching" even mean if the door that was opened was the door that you had picked and the car was there?

What would be the point of switching to the remaining door if the door that was opened was not the one you picked and the car was there? (The probability of getting the car would be zero either way.)

You're right that the subspace of scenarios where (a) and (b) happen is just part of the full space of things that could have happened - but it's the only part relevant for the problem. (We're explicitely told that (a) and (b) are true!)

That's the point though, no? The only way to justify the 50/50 claim made above that I can see is that the host must choose the door to open at random from among all 3 doors. So scenarios do arise where the host opens your door and it is the car, or that the host opens some other unchosen door and it is the car.

If you don't allow for those possibilities, then the host really isn't making their choice with equal probability.

The 50/50 scenario doesn't make sense.

> If you don't allow for those possibilities, then the host really isn't making their choice with equal probability.

I don't understand your point. The host can make the choice with equal probability and all those possibilities were initially allowed. Once that a choice was made some possibilities are obviously no longer allowed.

> The 50/50 scenario doesn't make sense.

This is the original problem: https://web.archive.org/web/20130121183432/http://marilynvos...

  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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. [It's implicit that his knowledge is used to pick a goat with certainty.] He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

This is one possible 50/50 scenario:

  You pick a door, say #1, and the host picks a number between 1 and 3 from a hat, say #3, and opens that door, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

The answer in the first variant is "yes, switching from #1 to #2 doubles my probability of winning". The answer in the second variant is "it doesn't matter, the probability of winning is the same".

The scenarios that didn't happen are irrelevant. We are not been asked to commit to a strategy in advance. We are asked if we want to switch from door #1 to door #2 after we learn that door #3 was not the good one. That depends only the scenarios that still remain possible and their probabilities.

A much simpler problem to stress that the question is whether we want to change conditional on what we know - not all the things that were possible a priory:

  You're given the choice between two evelopes. One contains $1 and the other $10. You open an envelope and find $1. Is it to your advantage to switch to the other envelope?

The answer is obviously yes. If you found $10 the answer would obviously be no. But if you have to commit to switching or not before you see what was your first choice the answer would be "it doesn't change anything".

Your answer to the question is conditional on what may have happened. You don't have to "allow for other possibilities" that could have happened but you know already that didn’t.

The original situation covers all scenarios. Any choice of door you make and all possible doors the host might open. The odds are the same every time.

The proposed 50/50 scenario only makes sense for some possible situations - those where the host draws a door at random AND the door isn't the same one I chose AND the door opened isn't the car... once you put that situation on equal footing with the original one (all possible doors chosen and all possible doors opened) then it isn't 50/50 any more. Some of those situations are 100% wins (I chose door 3, host opens door 3, car is behind door 3) and some are 100% losses (I chose door 2, host opens door 1, car is behind door 1). Once you look at all outcomes, 50/50 doesn't apply. It becomes 33/66.

If your point is that you can cherry-pick a subset of outcomes and get 50/50 odds then I can't really argue, but I'm also afraid I don't how that's useful.

> If your point is that you can cherry-pick a subset of outcomes and get 50/50 odds then I can't really argue,

Why do you say “cherry-pick a subset of outcomes”? It’s the subset of outcomes relevant for the question.

Consider the simple problem that I proposed: “You're given the choice between two envelopes. One contains $1 and the other $10. You open an envelope and find $1. Is it to your advantage to switch to the other envelope?”

Do you agree that the answer is “yes” or would you argue that “once you look at all outcomes” the 100% probability of getting $10 by switching doesn’t apply?

> but I'm also afraid I don't how that's useful.

It’s useful to answer the following question:

“You pick door #1. The host picks a number between 1 and 3 from a hat and opens the corresponding door #3. There is a goat behind the door. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?”

What’s your answer?

That's about as relevant as this:

"I claim the odds of flipping a fair coin 100 times in a row and getting heads every time is 50/50, but only in the situation where you happened to flip 99 heads in a row already".

Of course it is true, but again, not useful in my opinion.

What about the envelopes problem? Do you think that claiming that you should switch the envelopes “but only in the situation where you happened to” find $1 in your first pick is not relevant? The question asked is precisely what to do in that particular situation! Knowing that if you switch you get more that if you don’t seems useful…

I’m also curious about your answer to the following question, really.

What do you think that the probabilities would be if you you were to find yourself in that situation? A hint: the probability that the car is behind the open door is zero. What about the others?

Don’t you have any answer at all?

I'm not sure why you keep asking the same question. The answer to the second one is here: https://news.ycombinator.com/item?id=36752870 and the answer to the first one is basically alluded to here: https://news.ycombinator.com/item?id=36758795

In your hypothetical situation, the rules are something like:

"Pick a door, then the host will randomly select one of the other doors. If the host doesn't select your door (1/3 odds) AND the host doesn't open a door with a car behind it (2/3 odds) then you can choose to switch or keep your door."

The odds of winning _after_ the host already beats 1/3 and 2/3 odds is indeed 50/50, but the odds of playing the game _overall_ is 33/66.

The only real way I can see of making this situation analogous to the Monty Hall problem is to consider the odds of the whole game (like Monty Hall does, which is always 66% chance of winning by switching, every time, no matter the path the game takes from start to end). If you want to cherry-pick just the situation where the host ALREADY beat some odds, then you do get 50/50, but it really is a non-sequitur in this thread (or in any thread about Monty Hall).

> The odds of winning _after_ the host already beats 1/3 and 2/3 odds is indeed 50/50, but the odds of playing the game _overall_ is 33/66.

The problem is not about the game overall anymore than the question about whether you'd like to switch the envelope after you got $1 is about the game overall. The question is whether you want to switch conditional on the situation where you got $1!. That's not cherry-picking - it's considering the exact situation described in the problem.

> The only real way I can see of making this situation analogous to the Monty Hall problem is to consider the odds of the whole game

The Monty Hall problem is not about "the whole game". The Monty Hall problem is quite explicitely about what would you do in the following situation:

1) you were given the choice of three doors: one car and two goats

2) you picked a door

3) the host opened another door

4) there was a goat behind that door

After all those things have happened you're being offered to switch from your initial pick to the remaining door.

What is the probability - after all those things have happened - that the car is behind each of the doors?

If you don't agree that this describes the problem there is no point in reading further. We can just agree that we have very different understandings of what the problem is about.

-----

If you agree with the description above the answer to the question depends your assumptions about how the host's choice in step 3) was done.

There are many different assumptions that could be made.

If he was avoiding the door you picked and the door with the car - as the original formulation of the problem implies - the probababilities are 1/3 and 2/3 (for the door you picked and the remaining one respectively).

If he was completely ignoring the location of the car when he made the choice the probabilities are 1/2 and 1/2 (it doesn't matter if he was avoiding your door or not).

> The problem is not about the game overall > The Monty Hall problem is not about "the whole game".

The Monty Hall problem is about the whole game. That's the difference, and that's why the 50/50 claim doesn't make much sense here.

> The Monty Hall problem is quite explicitely about what would you do in the following situation: [...] After all those things have happened you're being offered to switch from your initial pick to the remaining door.

Steps (3) and (4) in the Monty Hall problem are a bit more nuanced - the host opens a door guaranteed to be a goat, because the host knows where the goat is.

At this point, when you choose to switch or stay with your door, you have a 1/3 chance of winning if you stay, and a 2/3 chance of winning if you switch.

But the key point that you are missing is that this is the same for every single time you play the entire game. It doesn't matter where the car is in step (1), or which door you choose in step (2), or which door the host opens is step (3) - if you always switch doors, no matter what, every single game you double your odds.

And in the 50/50 scenario posed, that just doesn't hold. It isn't analogous.

> But the key point that you are missing is that this is the same for every single time you play the entire game. It doesn't matter where the car is in step (1), or which door you choose in step (2), or which door the host opens is step (3) - if you always switch doors, no matter what, every single game you double your odds.

The key point that you are missing is that it doesn't need to be "the same for every single time you play the entire game" to have a well-defined solution for the particular realization of the game described in the problem statement.

And the solution for the problem where (1),(2),(3),(4) happen with the host picking a door at random is 50/50. That's what cman1444 said.

You're asked a question conditional on (1),(2),(3),(4) happening and the things that didn't happen are irrelevant.

Just like if you're asked if you want to switch the envelopes when you got $1 you will say "yes" and it's completely irrelevant that you could have got $10 and you would have said "no" in that case - because you didn't.

In conclusion, some people find the problem where (1),(2),(3),(4) have happened and the host was picking a door at random (and its 50/50 solution) interesting and you don't. De gustibus non est disputandum.

I addressed this above. See the coin flipping example.

> I addressed this above.

I'm not sure what is "this".

> See the coin flipping example.

"I claim the odds of flipping a fair coin 100 times in a row and getting heads every time is 50/50, but only in the situation where you happened to flip 99 heads in a row already".

If I'm told that I'm in a coin flipping contest and (1) in my first flip I got heads, (2) in my second flip I got heads, ..., (n) in my n-th flip I got heads, ..., (99) in my 99-th flip I got heads and then I'm asked what's the probability that I get to 100 heads in a row there are different answers that I could give.

I could assume that the probability of every single flip is 50/50 and then my answer will by 50%. Or I could give an answer higher than 50% if I'm able to control the outcome - or if by now I suspect that the coin has two heads. Or I could give an answer lower than 50% if I suspect the game is rigged and they are going to make me lose now.

Whetever my assumptions about the upcoming flip, my answer will be conditional on (1),(2),...,(99) having happened already. There is no point in thinking about the "whole game" at this point in time. I already got heads 99 times - just like in the Monty Hall problem I already picked a door and he already picked another door with a goat.

Sure it's only one branch of the probability tree of the game as a whole.

But the way the problem is frequently framed is this (I acknowledge that the OP does briefly make the important distinction though):

"The game has three doors, two with goats and one with a car. You pick one door. The host selects another for free and shows you a goat behind it. Do you want to switch doors?"

My point is that, given this scenario, to make an informed decision you need to ask the host: "Did you know where the goats and cars were? And did you open the goat with certainty, or did you do so at random?"

The hosts answer to this question then determines whether you should switch to gain a 33 percent advantage or that it does not matter.

It’s also helpful (but not essential) to realize that the choice of which door to open has two parts, a rule and a decision. If the participant is doing a switch or a stay strategy, the decision aspect has no impact on winning.

The rule part is Monty can’t reveal the car. The decision comes up if the participant chooses the car first. Monty will have two doors to choose from to open. But if you have a stay strategy you’ll win for either choice, and if you have a switch strategy you’ll lose no matter what. So his decision doesn’t matter.

This makes it very easy to simulate. It also might lead to people to understanding the solution.

If you pick wrong first, you’ll see staying always loses and switching always wins. The probability of picking right first is 1/3, of picking wrong first 2/3. Since switch always works when you pick wrong and you pick wrong 2/3 times, go with that.

Where I live most cars have 4 doors, but since we are expected to pretend this is a well-posed problem we must assume there are literally 3 doors at the game host location.

Evidently the participant entered through one of them and already knows whats behind one door. If he saw the car during entry, the correct choice would be the door of entry. The host can open (and close, and open, ...) doors with goats all day long, the participant would know the car is where he just saw it.

If the participant did not see the car during entry it must be behind one of the remaining doors: either behind the back door, or behind the emergency exit door.

Clearly it would be highly illegal to obstruct the emergency exit door with a car, so under Standard Assumptions (see assumption 42, the last one in the code) we can assume the host did not park the car behind this emergency exit door. Especially since emergency exit doors open outwards.

Hence the car was parked behind the remaining backdoor, and since this follows from the Standard Assumptions we can conclude the participant made the same (by definition valid assumption) as described here, since its a Standard Assumption, and assumption 1 of the Standard Assumptions is that in the absence of necessary knowledge, everyone makes the Standard Assumptions.

Given that the participant (using the standard assumptions) knew immediately which door hides the car and given that this door was his initial "guess", the correct solution (under Standard Assumptions) is to stay with the initial choice.

This is a good way of putting it and I think would resonate with some people!

I've not been able to get my dad to understand why you should switch with 1,000,000 boxes because he refuses to believe the 3 box and 1,000,000 box case is the same. I can't even convince him a 3 and 5 box problem are the same.

I actually think I could convince him switching is correct using this strategy because I suspect if he thinks he's tricking or cheating at the game he'll be more open to the explanation.

Are we siblings

My problem with Monty Hall is that it’s not a repeated game. Average probabilities don’t help me.

This is the point Taleb makes in a lot of his books. You are not the average probability. You only get one play through [life] and the only thing that matters is how the dice fall for you. Even a 90% average chance of getting $prize, is still a 10% chance of walking away with nothing and you better be ready to get that one because you very well might.

The strategy to increase your average probabilities works in poker. Because you play multiple hands and even multiple games. In Monty Hall, you’re just making a guess and no matter how clever you are about it, it’s still just a guess. There is no strategy because the game doesn’t last long enough for strategy to matter.

Now if you had an iterated monty hall, then yes, bring out the maths.

> There is no strategy because the game doesn’t last long enough for strategy to matter.

It makes it much more likely that you win a car. You don't see any value in doubling your chances of a big win?

> You don't see any value in doubling your chances of a big win?

You’re improving your average chances. But you’re not improving this game’s chances. If you play 10 games, your average chances matter. If you’re playing 1 game, reality was already set before you started the game. The car doesn’t move because you improved your average probability.

I don't get what you're trying to say.

The car doesn't move, but your pick moves, even in this single game.

What do you think “this game’s “ chance of winning is by following the strategy vs not following the strategy?

> But you’re not improving this game’s chances.

You’re quite wrong. That is precisely what you are doing.

Unless you are arguing that you don't really have a choice because your actions are predetermined you are absolutely increasing your chances of winning this game.

Yes, the whole point is that the car doesn't move. When you initially chose, you had a 1-in-3 chance of correctly choosing the door with the car. The host is now offering you to switch over to, essentially, "all the doors you didn't choose", which means 1-(1/3). As you state, you've only got one opportunity to play, and the car itself hasn't moved, but you can switch your choice. Stick with the initial 1-in-3 chance, or go to the other side, i.e. 2-in-3 chance?

> You’re improving your average chances. But you’re not improving this game’s chances.

If the strategy won't improve your chances in any single game in a series of games - how can it improve the average chances?

Sounds like taleb has undermined your understanding of math then.

A 90% chance is worse than a 99% chance. The possibility that you get nothing is not an argument to ignore the context.

Consider Russian roulette. Are you seriously going to argue that the number of bullets in the gun doesn’t matter if you only play one time?

> The possibility that you get nothing is not an argument to ignore the context.

Ah but I’m not ignoring the context. I’m expanding the context beyond “fun math puzzle” to “wow this is a game not worth playing because there’s no edge”.

Again, only applies to the non-iterated version. If you find an analog of monty hall that gives you many at-bats, play away.

No it doesn’t! This is an astoundingly stupid argument. Doubling your odds of winning is an edge.

  "wow this is a game not worth playing because there’s no edge"

The cost of entry is $0.

The prize is a brand new car.

The optimal strategy gives you a 2/3 chance of winning.

Well, even if I try to embrace your argument — life itself is an iterated version of a more general decision game. It doesn't have to be Monty Hall specifically.

Over the course of your life, if whenever you have a choice to make you choose the lower probability strategy (here with the Monty Hall example it's not even an expected-value vs. variance tradeoff, switching is just all-around more optimal) you will most definitely end up worse off in the end.

I made the same argument the last(?) time a Monty Hall article made the front page.

This does not work as an argument against the Monty Hall problem.

It does work against simplistic arguments about which of two or more probabilistic games one should prefer, because it's basically an argument that linearity if expectations doesn't necessarily apply in non-repeated games.

It sounds like you embrace the frequentist interpretation of probability, but the bayesian aka subjectivist interpretation of probability says you can make predictions on the likelihood of single events based on your knowledge.

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

No. Wrong. That is not what frequentists would say.

If I tell you that I'm rolling a die, and ask you whether I rolled a six or not a six, offering you 100$ if you guess right, you very obviously should say "not a six", even if we only play the game once.

No, reading Taleb I realized that if it's not repeated it’s just a guess and winning or losing is a 50/50 thing. There is no strategy because the game doesn’t last long enough for strategy to matter. [/s]

It depends on if I look at the die before I decide on the "N or not N" options.

That's not correct, because if you agree that one kind of result tends to happen more than the other, then it is easier that the single game that you are currently playing belongs to the larger group. When you play once, it is like if you were randomly choosing a game of all the possible ones that you could play, in which case you are more likely to choose one of those that are more numerous.

It's the same reasoning that you would apply in other aspects of life. Imagine you have to travel to some town and there are only two possible roads: A and B. The stats tell that on road A there tend to occur 100 fatal accidents per year, while on road B only one accident has been registered in the last 10 years.

The causes of that disparity may be multiple, like, for example, that road A has many curves with precipices while B does not. But you don't even need to know those causes; the starts themselves tell you that as result it is easier to get a falal accident on road A than on road B. But, of course, it always exists the possibility that you survive in any of them, and also that you have an accident in any of them.

So, if you were to travel through one of them just once and had to choose, I don't think you would say that those stats don't matter only because you will travel once. You would prefer to take the safer road B.

In Monty Hall game, deciding to stay would be like taking road A, and deciding to switch would be like taking road B.

Moreover, if the game consisted in that instead of revealing a goat, the host gave you the opportunity to reject your first choice and instead check inside the other two doors and take which you prefer from them, it is obvious that it would be better to switch to the other two (unless you think 2 is not greater than 1), and that's true regardless of if you are playing just once or multiple times. Anyway, by checking those two doors you would necessarily find at least an incorrect one, as there is only one prize in total.

If you notice, as the host knows the locations of the contents and always removes a losing door from those that you did not pick, then it is like if he was doing that work for you of checking inside those other two doors, eliminating from them the incorrect one that you would have found anyway, and leaving closed exactly which you would have picked if you were who had checked inside those two doors.

So, if you agree that it would be better to switch to the other two doors, you must also agree that it is better to switch to the other single one that the host is offering, as both ways win in exactly the same cases.

I think the only way to demonstrate it to lay people is to rephrase it as the odds of picking the wrong one right off the bat. If it's 2/3 odds that you chose wrong, and I remove one card, did that change the odds that you already chose wrong?

The trick is focusing on the 1/3 odds of winning, and not mentioning there is an alternate (and correct) way of looking at the question - the initial odds of losing.

Bifurcate the world into the 3x3=9 chances of which door the prize is behind and which door the contestant guessed and write down the permutations. These are two independent random events and the permutations are equally likely with a 1/9th chance each.

Now for the cases where the contestant guessed the door correctly -- "11", "22", and "33" the right move is to stay and it doesn't matter which door Monty Hall reveals.

For the cases where the contestant didn't guess the door correctly -- "12", "21", "13", "31", "23", "32" the right move is always to switch (Monty Hall will have always revealed the third door in each permutation).

The probabilities of each are 1/9th so they add up to 1/3 chance of failing if you switch and 2/3 chance of succeeding if you switch. So always switch.

If you want to complicate things you can add the door that Monty Hall shows as a third option, but you need to be careful about the probabilities:

"112", "113" - these have to add to 1/9th so each has 1/18th chance (50-50 chance MH picked one door or the other) and both fail if you stay, similarly for "221", "223", "331", "332", 6*1/18 = 1/3rd.

"123" -- MH is forced to pick "3" so this is just 1/9th chance of success, same for "213", "132", "312", "231" and "321". So this is 6*1/9 = 2/3rd.

One error would be writing down all the permutations and assuming they were equally likely, but "112" and "113" must sum to the odds of "11" which is 1/9th, while "123" is the same odds as "12" which is 1/9th.

I think this approach shown in the article is not great. It adds additional layers of thinking where none are needed. You don’t lie. It also doesn’t feel like it would help explain variants like 5 doors, and only one goat is revealed.

I would argue that if you start out knowing that you will switch, you essentially do lie, because you initially pretend to pick a box (or door) that you know you will not end up opening.

It is true that this approach adds additional layers of thinking, but the problem is that without those layers, the solution to the problem is simply not intuitive for the vast majority of people. Adding these additional layers helps at least some people gain a more intuitive understanding of why switching helps.

Ehhh. Idk. I think it is not great.

A much easier framing on a similar level is simply saying you pick a door, and then the host says you can either open the chosen door, or both other doors one at a time.

It’s the same thing. But if you don’t get why the state is unchanged even after opening one of the doors I don’t think you get it with this whole lying thing

To the best of my knowledge, this "question" was first presented to the general public around 1990 via a popular syndicated newspaper column, and quickly "went viral" the way things did back then, by being repeatedly discussed in newspaper and magazine columns all across America.

It's fascinating that 33 years later, there is still no simple explanation of the optimal strategy, such that most people would say "a-ha" when they hear it. In fact, most people actually express skepticism or disbelief when presented with the "correct" strategy.

When people pose this "puzzle" verbally to another, there usually is a back and forth questions and answers about the exact problem statement. This is where most "puzzle" posers mess it up. In fact vos Savant messed this up herself.

A proof is a sequence of steps, which are to be justified individually. Justifications may be general theorems from the field of mathematics, its axioms, or otherwise givens from the problem statement.

Hiding withheld "given(s)" in the "solution" instead of providing them in the problem statement means the literal problem statement is ill-posed.

Check the section "standard assumptions" on Wikipedia:

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

"Under the standard assumptions, the probability of winning the car after switching is 2 / 3. The key to this solution is the behavior of the host. Ambiguities in the Parade version do not explicitly define the protocol of the host. However, Marilyn vos Savant's solution[3] printed alongside Whitaker's question implies, and both Selvin[1] and Savant[5] explicitly define, the role of the host as follows: [...]"

When I was first posed the problem, I told the person there was not enough information to solve the puzzle, I saw the Bayesian catapult under a specific assumption, but I realized the problem statement was underspecified and unposed. So while inquiring the acquaintance to explain "how it happened that there was a goat behind the door" the person just reiterated the already communicated sequence of events as a fairy tale, and this went on until he provided the solution. Of course his explanation was the Bayesian catapult, and I responded that the problem statement did not contain the requisite facts and he must have mistold it. So I looked up the problem on Wikipedia only to discover that vos Savant herself ill-posed her problem.

Its perfectly fine for a step in a proof of solution to refer to a given. But that given must either be directly present in the problem statement or be rigorously derivable from it. This is simply not the case.

If it isn't given it's not a given.

> In fact vos Savant messed this up herself.

I don’t think it’s _that_ bad.

She writes that “the host, who knows what's behind the doors, opens another door which has a goat.”

Why would the problem statement say that the host knows what's behind the doors (devoting 10% of the description to making this clarification) if not to indicate that he knowingly opens another door which has a goat?

The host knowing what's behind the doors doesn't specify the host's behavioural protocol.

There are multiple host protocols that are compatible with the stated run of events. They result in different probabilities.

For example protocol 1: if the player chose the door with a car behind it, open one of the doors with a goat, else don't open any doors and immediately reveal the player got a goat and "lost". In this scenario it's better to stay with your pick, i.e. the opposite advice of the "solution" since under this protocol the host opening a second door indicates you guessed the right door. Your selected fragment "the host, who knows what's behind the doors, opens another door which has a goat.” does not exclude this scenario.

Example protocol 2: after the player has chosen a door, the host throws a coin for the remaining 2 doors, but only opens that door if it has a goat.

You can make countless variations that are perfectly compatible with the givens in the problem statement yet have different probabilities associated with them. The problem really is ill-posed. If more than a 1000 PhD's wrote back in complaint, then perhaps a more probable explanation is that while reading the problem statement they assumed distinct host protocols, and if that happens, its worth checking if the givens are actually part of the problem statement, and not extraneously appended in the "resolution". Otherwise the puzzle just becomes "guess the probability of what I supposedly convey".

> The host knowing what's behind the doors doesn't specify the host's behavioural protocol.

I agree. I just said that it was not _that_ bad.

> If more than a 1000 PhD's wrote back in complaint, then perhaps a more probable explanation is that while reading the problem statement they assumed distinct host protocols,

I wonder how many of them pointed out that the problem was ill-posed - instead of assuming distinct protocols or just failing to solve properly the problem.

At least one those PhDs went from

"Since you seem to enjoy coming straight to the point, I’ll do the same. You blew it! Let me explain. If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2. As a professional mathematician, I’m very concerned with the general public’s lack of mathematical skills. Please help by confessing your error and in the future being more careful."

"I wrote her another letter, telling her that after removing my foot from my mouth I'm now eating humble pie. I vowed as penance to answer all the people who wrote to castigate me. It's been an intense professional embarrassment."

>> The host knowing what's behind the doors doesn't specify the host's behavioural protocol.

> I agree. I just said that it was not _that_ bad.

It's great we can agree the problem is ill-posed. But that is a binary thing, either it is ill-posed or it is not. It can't be a little bit ill-posed.

I don't believe it's a coincidence that the ill-posed problem gets the high profile attention. It's not easy to design a proper puzzle, so it's tempting to simply hide a non-given "given" in the "solution".

The best way to not get caught is to have separate identities ill-pose-without-the-given and "solve"-with-the-given. Flame wars guaranteed...

EDIT: now that I think of it, I even fail to verify that the poser "Craig F. Whitaker" didn't collude, or even that he existed.

Looking at the archive.org copy of vos Savant's own website:

https://web.archive.org/web/20130121183432/http://marilynvos...

I see something very suspicious about the first 2 letters quoted (including the one you requoted): They are ostensibly written by different people, yet use very similar wording, and essentially contain the same information sentence by sentence.

First I will repost both integrally:

Since you seem to enjoy coming straight to the point, I’ll do the same. You blew it! Let me explain. If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2. As a professional mathematician, I’m very concerned with the general public’s lack of mathematical skills. Please help by confessing your error and in the future being more careful.

Robert Sachs, Ph.D. George Mason University

You blew it, and you blew it big! Since you seem to have difficulty grasping the basic principle at work here, I’ll explain. After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the odds are the same. There is enough mathematical illiteracy in this country, and we don’t need the world’s highest IQ propagating more. Shame!

Scott Smith, Ph.D. University of Florida

Both use "blew it", where there are many other alternatives like "mess it up" (which I used).

Robert: If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2.

Scott: After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the odds are the same.

Both also somehow connect this with public literacy and public shaming.

Robert: As a professional mathematician, I’m very concerned with the general public’s lack of mathematical skills. Please help by confessing your error and in the future being more careful.

Scott: There is enough mathematical illiteracy in this country, and we don’t need the world’s highest IQ propagating more. Shame!

These people at least seem to exist, but something strange is going on.

It's also regrettable I can't find the original Parade articles on the internet archive...

> I don't believe it's a coincidence that the ill-posed problem gets the high profile attention.

In this case even the well-posed problem(s) can be highly controversial as it can be seen from other comments - including someone who claims that probabilities are meaningless here because "it’s not a repeated game."

> something strange is going on.

The New York Times may be part of it then, I got the second quote from https://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-do...

Sadly I can't read the NYT.

> In this case even the well-posed problem(s) can be highly controversial as it can be seen from other comments - including someone who claims that probabilities are meaningless here because "it’s not a repeated game."

I will take a look at such a comment if you link it, but I'm not going to actively search for such comments as I don't plan to play Monty Hall police.

The most intelligent probably just responded on the internet when they first got acquainted with it, and upon realizing the problem is ill-posed just ignore it. So the visible flamewar tends to be between people who don't realize the problem is ill-posed...

I was referring to https://news.ycombinator.com/item?id=36736947 but I didn’t expect you to read all the exchanges or comment on them.

“In Monty Hall, you’re just making a guess and no matter how clever you are about it, it’s still just a guess. There is no strategy because the game doesn’t last long enough for strategy to matter.”

The point is that the discussions are not always related to a misunderstanding about the protocol followed by the host to open a door.

Someone else doesn’t accept that (after the host’s protocol is fully specified) the problem is conditional on the stated run on events - insisting that it only makes sense to consider “the whole game”: https://news.ycombinator.com/item?id=36737326

> It's great we can agree the problem is ill-posed. But that is a binary thing, either it is ill-posed or it is not. It can't be a little bit ill-posed.

Would the problem be well-posed if the original problem statement was amended to fully specify the host’s behavioral protocol?

Yes if the host's behavioral protocol and the rules of the game were specified (deterministically or even probabilistically) the problem would be well-posed. I assume you mean the problem statement proper were to be amended, and not amending just the "solution", i.e. add and move the requisite givens from the solution to the actual problem statement.

Of course, it would then just reduce to a simple rote calculation, instead of bait and switching.

Another issue is that it is impossible to objectively agree on the "right" problem statement, given the ill-posed problem statement. An ill-posed problem statement is really a set of different problems each with their own answers.

If you decide to do this, I highly recommend providing multiple fixed problem statements, and explicitly indicating the distinction from "the" Monty Hall problem, yet the connection to it: make sure each fixed problem statement is compatible with Whitaker's original sequence of described events.

Make sure the reader understands that "the" Monty Hall problem does not in fact exist. Double check you don't use "the Monty Hall problem" anywhere, as even the name reinforces the idea that we are talking about a specific and thus well-posed problem and not a set of distinct and thus ill-posed problems. (I assume I probably did somewhere use the phrase in the thread above... sigh).

If your medium allows comments etc. make sure the reader is immediately aware of this, otherwise the comments will probably fill with readers unintentionally re-mansplaining the "proper" solution of "the" Monty Hall problem "under standard assumptions" set by the Ministry for Standardizing the Interpretation of Certainly Not Ill-posed But Just A Little Understandardized Problems.

The question is whether “Is it to your advantage to switch your choice of doors?” but the objective is not specified. Do I want to get the car? Do I want to get a goat? The latter objective is compatible with the problem statement.

Let’s say that it is amended again to fully specify the objective. Is it well posed now?

We are not told how the contestant picks a door and how the person responsible for putting the car and the goats behind the doors makes the choice.

There are multiple ways that the choices may have been done which are compatible with the stated run of events. It’s not a given than the location of the car and the door selection are independent.

The contestant may have seen the car, or may have heard the goats, and may have used this knowledge to pick a door. Or maybe the contestant doesn’t knew anything and it was the producer who knew that the contestant would pick door #1 because that’s his lucky number or whatever and intentionally put the car there or avoided doing so.

Game theory disregards the true ulterior objective of a participant, the game rules decide if there's a winner or a draw etc...

For example a king may force an unwilling philosopher to play chess together. The philosopher indeed intentionally loses the game. But according to the rules of chess the king still won.

A necessary condition to compute any probabilities is that the game rules are specified (and the ill-posed Monty Hall problem can be viewed as a single player game with ill-posed rules, or as a 2 player game of host vs contestant with ill-posed host protocol).

That doesn't mean it's a sufficient condition. Don't ask me to specify an explicit checklist to verify a natural language problem is well-posed: the burden of proof lies with the poser. That said, even though the burden of proof shouldn't lay with the challenged, those who are challenged are free to illustrate counterexamples exposing the ill-posed nature of the problem, such as example protocol 1 listed above.

If you truly like formally verifiable puzzles you could consider the formalization of Fermat's Last Theorem. Observe the huge contrast with the ill-posed Monty Hall problem:

X ^ N + Y ^ N = Z ^ N

has no solutions in the strictly positive integers for 3 <= N.

Every high school kid knows what exponentiation is: the power X ^ N is a repeated product of the same base X, where the base appears an exponent N number of times.

Its trivial to understand the question. Its trivially provable that the formula is well formed, and thus has a meaning. But yeah, with this type of puzzle it's less obvious how to rile up the less educated against those who practice careful rigorous precision. And then collectively repeat pretend those who protest "don't get it". Every time it passes I see the same things: a bunch of people pretending they nerdsniped whoever questions the ill-posed nature of the problem. Except they didn't even hit. We are expected to collectively pretend what a bunch of dummies the more rigorous and careful are, for not jumping to conclusions or for not playing along with pretending the hidden given was supposedly really given. It's just organised gaslighting intelligent people, who already know bayesian statistics, and lambasting them for not understanding it when rather often they do, but moreover they understand the problem is ill-posed, even if they don't use the mathematical jargon (they might request some more alternative runs of the game in order to make sure they understand the rules of the game).

To pull a Monty Hall is rather easy (please don't actually go out and do this):

1. just take any textbook problem with solution

2. make sure you understand both problem and solution.

3. eliminate a given from the problem statement.

4. fun and attention (in this world probably ad-profit)

Make sure to select an easy enough problem: you don't need to defend your practices if the fanboys will mansplain your solution over and over for you. The easier the problem the larger the group of fanboys in your defense.

Make sure to select a statistical one: whenever people object ambiguity of the problem statement and describe other possibilities they can be derailed as if they are not talking about the existence of other valid interpretations of the ill-posed problem statement, but as if they are talking and incorrectly analyzing probabilities and possibilities within the Holy interpretation of the ill-posed problem (pretending the hidden given was present in the original problem statement).

Make sure to separate authorship of the problem statement and your "solution", this gives plausible deniability when people point out the problem is ill-posed.

I feel like most explanations are overcomplicating it.

By the rules of the game, the option of switching is asking you the question "Do you think you chose the correct door out of the three?". If you think you did, you don't switch. If you think you didn't, by rules of the game, switching will get you to the right door. Since you had a 33.3% chance to choose correctly and 66.6% to choose incorrectly, betting against yourself and switching is probably the correct choice.

I did a presentation on this for a discrete math course in undergrad.

IIRC I did the N box case and took the limit to infinity and, unsurprisingly, got the result that you should always switch.

I like this explanation. Another equivalent one that I wrote about a while back[0]:

You can choose to have:

- The most valuable prize that’s behind door A, or

- The most valuable prize that’s behind doors B or C

When you look at it this way, it’s obvious that you would rather have the most valuable prize from the ‘other’ doors, and the way to do that is to switch.

[0] https://www.encona.com/posts/monty-hall-problem

I've made this "intuition" for the Monty Hall into a game [1] so I can test it for myself.

I remember being intrigued by the notion that you're better off switching.

1 - https://victorribeiro.com/montyhall/

Keep in my that this was made a long time ago, I wasn't even a programmer back then, I was more a webmaster.

Here are a couple of different explanations using Bayes' Theorem as well as a more informal one. https://dev.to/santisbon/the-monty-hall-problem-28om

The reasoning I've always used to understand this problem is to imagine that you and the host both choose at exactly the same time. Then 2/3 of the time you've both simultaneously selected both boxes with goats, and therefore 2/3 of the time the unselected box is the car.

And here’s an explanation from a theater show by Derren Brown @ 5:43 - https://youtu.be/GPoPSNxV1D4?t=342

I wonder: has anyone went through the old tapes and confirmed that the people who stayed won about a third of the time, and the switchers 2/3 of the time?

I agree with the math, but wonder how unscrupulous the show was.

if you play this game repeatedly, 2/3 of the time your first guess will be a goat, and Monty will then always show you the other goat, and if you switch you will see the car behind the remaining door.

if you play the game once, you should follow the same strategy.

don't explain more than that, the person just needs to grok that. a restatement of the main idea might help:

2/3 of the time your first choice is a goat therefore 2/3 of the time Monty is showing you the other goat therefore 2/3 of the time if you switch you will be switching to the car.

Here's my attempt to explain the Monty Hall with code: https://github.com/DeegC/monty_hall_paradox

I honestly wouldn’t mind a goat.

This is incorrect, the goats and car are behind doors. They are not inside cardboard boxes.

Reply on news.ycombinator.com