Tricking Monty Hall

https://ignorethecode.net/blog/2023/07/15/tricking_monty_hall/

by zdw • 3 years ago

50 140 3 years ago

necovek• 3 years ago

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.

evandale2• 3 years ago

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 ;)

ryandrake• 3 years ago

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.

jncfhnb9• 3 years ago

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

stavros3• 3 years ago

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.

xg152• 3 years ago

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.

IanCal11• 3 years ago

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.

chefandy3• 3 years ago

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.

mrob3• 3 years ago

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.

stavros3• 3 years ago

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

oefnak• 3 years ago

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

stavros3• 3 years ago

How is it different?

deeg• 3 years ago

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

evandale2• 3 years ago

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.

arithma2• 3 years ago

Are we siblings

Swizec3• 3 years ago

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.

IanCal11• 3 years ago

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?

jncfhnb9• 3 years ago

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?

jdechko2• 3 years ago

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

atq2119• 3 years ago

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.

Null-Set• 3 years ago

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

LukasMathis2• 3 years ago

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.

EGPRC• 3 years ago

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.

ksaj5• 3 years ago

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.

Ekaros• 3 years ago

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?

drexlspivey• 3 years ago

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

IanCal11• 3 years ago

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.

IanCal11• 3 years ago

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.

cman14443• 3 years ago

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.

dylan604• 3 years ago

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?

tromp• 3 years ago

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.

jjnoakes8• 3 years ago

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.

travisjungroth2• 3 years ago

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.

DoctorOetker8• 3 years ago

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.

rain1• 3 years ago

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

arithma2• 3 years ago

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.

jncfhnb9• 3 years ago

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

mrob3• 3 years ago

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.

mikrl• 3 years ago

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.

xg152• 3 years ago

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.

FartyMcFarter• 3 years ago

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.

hurril• 3 years ago

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.

ComplexSystems• 3 years ago

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?

TulliusCicero• 3 years ago

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

ksaj5• 3 years ago

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.

moffkalast2• 3 years ago

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.

travisjungroth2• 3 years ago

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.

nkrisc• 3 years ago

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.

diputsmonro2• 3 years ago

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!

jncfhnb9• 3 years ago

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.

LukasMathis2• 3 years ago

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.

jncfhnb9• 3 years ago

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

rahimnathwani2• 3 years ago

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

andy8005• 3 years ago

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.

tzs• 3 years ago

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.

patrickthebold• 3 years ago

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.

lisper2• 3 years ago

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.

slavik81• 3 years ago

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.

kgwgk16• 3 years ago

Should you switch?

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

johnfn• 3 years ago

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.

voyager1• 3 years ago

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 :)

atum47• 3 years ago

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.

silisili• 3 years ago

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.

activitypea• 3 years ago

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.

jamjamjamjamjam• 3 years ago

I honestly wouldn’t mind a goat.

fsckboy• 3 years ago

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.

c7b• 3 years ago

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.

Orlan• 3 years ago

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

lamontcg• 3 years ago

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.

DoctorOetker8• 3 years ago

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.

kgwgk16• 3 years ago

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?

DoctorOetker8• 3 years ago

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".

11235• 3 years ago

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

Tricking Monty Hall

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