Monte Hall Problem

Have you heard of the famous Monte Hall problem in statistics? It’s freaky. And I believe it offers the best evidence that our reality is subjective.

The set up is this. Game show host Monte Hall offers you three doors. One has a car behind it, which will be your prize if you guess that door. The other two doors have goats. In other words, you have a 1/3 chance of getting the car.

You pick a door, but before it is opened to reveal what is behind it, Monte opens one of the doors you did NOT choose, which he knows has a goat behind it. And he asks if you want to stick with your first choice or move to the other closed door. One of those two doors has a car behind it. Monte knows which one but you don’t.

The trick here is that most people assume it makes no difference if they stick with their original choice or move to the other door. They believe the odds are 50% either way, since there are only two choices and you don’t know anything about either choice. But mathematicians say that is wrong. You substantially increase your odds by switching doors.

That is interesting enough on its own. (I’ll give a link later that explains the math of it.) But here is the freaky part. You only improve your odds by switching doors if Monte Hall knows what is behind each door. If he simply got lucky and opened a door with a goat behind it, your odds are unchanged. In other words, your odds are changed by Monte’s knowledge, and your knowledge that Monte has that knowledge.

If reality were objective, statistics wouldn’t be influenced by knowledge. That means your world is either partly created by your mind, or you are a hologram created by some other mind, and there are a few bugs in the software.

Here’s a link to more than you want to know about the Monte Hall problem.

http://en.wikipedia.org/wiki/Monty_Hall_problem

You see the same sort of thing happen in the classic double-slit experiment in physics. The result of the experiment changes if the observer has additional information about what slit a photon passes through. Again, knowledge changes the real world. That can’t happen in the world you imagine you are living in. It has to be a bug in the hologram program. At the very least it shows that your reality is subjective.

Here’s more on the double-slit experiment.

http://en.wikipedia.org/wiki/Double-slit_experiment

DMD

Comments

[Sarah]

This is really great, is this column still in print in one newspaper?

Sarah
Note Promote Editor
Http://www.NotePromote.com

[Stephen Young]

I have written a short JavaScript program that simulates the Monty Hall game. Click on a door to select it and then a goat will be revealed behind another door. You should then click on Swap or No Swap. The result is then counted according to whether you won or lost and whether you swapped or not. You can save yourself a lot of clicking by playing 1000 games at a go by selecting "1000 Plays Swap" or "1000 Plays No-Swap"
The page is at http://members.chello.at/stephen.joung/monte.html

[Chris L]

I know I'm getting in late on this, but I feel the need to clarify something. (I was going to weigh in on the Monte Hall problem, but that has been discussed to death, and I have no idea where your mind is on it now, ditto for the double slit, but I'll be brief.).

The reason given by modern physics for the phenomenon in the double-slit experiment that you discuss is the following.

As the electron is travelling through the vacuum towards and after the slit, it does not exist in a particular place. If you observe it as it goes through the slit, then you interact with it and fix it in a position in space and time. It's not the knowledge of which slit it goes through that changes things, it's the interaction that is caused by the act of observing it. If nothing interacts with the particle, then it truly does go through both slits. Whereas if there happens to be some dust in the way of the slits, then even though we still don't know which slit it went through, it will have interacted with the dust and "picked" a slit.....ok this is turning out to be not so brief, so I'll just end it there. I'd love to continue this discussion, but not with posts in a forum.

[HGB]

Most people have said that your statement

"You only improve your odds by switching doors if Monte Hall knows what is behind each door. If he simply got lucky and opened a door with a goat behind it, your odds are unchanged."

is false, and that it doesn't matter whether Monty knows what's behind the doors or just gets lucky and happens to open a door with a goat behind it.

In fact, you are correct, and it illustrates the point I was trying to bring up with my earlier example.

What people consistently miss is that, although there is a 1/3 chance that Monty will open the door with the car, this probability is -dependent- on whether you initially choose the correct door or not. Clearly, if you initially choose the door with the car, then regardless of what remaining door Monty chooses, the game will continue.

However, if you initially choose the wrong door, then there's a 1/2 chance that the game will end because Monty revealed the car.

The critical thing to realize is that, once you know that Monty randomly choose a door and picked one with a goat, this provides evidence to decrease the probability that you initially picked the wrong door. In fact, there's a 1/3 chance you picked the correct door, and there's a 2/3 chance that you picked the incorrect door -multiplied- by a 1/2 chance that Monty also picked the incorrect door, which gives you 1/3 as well. (The other 1/3 is when Monty picks the correct door and the game ends.)

Thus, you have a 1/2 chance either way.

This is related to what I was driving at with my earlier example:

Suppose there's a disease that affects 1/1000 of the population. A test has been developed to determine if you have the disease, that is 99% accurate, that is to say: if you have the disease, then the test will come back positive 99% of the time. If you don't have the disease, then the test will come back negative 99% of the time.

You want to know if you have the disease, so you go get the test and it comes back positive. What's the probability that you have the disease?

Most people will say, given that the disease is 99% accuracy, and it came back positive, you have a 99% chance of having the disease.

However, this is false. Formally, you need to use Bayes Rule to do the computation. Intuitively, the explanation is that, since the disease only affects 1/1000 people, there's a much higher probability that you were one of the 999/1000 people who are not affected and got a false positive with probability 1/100, then you were one of the 1/1000 people with the disease and got a true positive with probability 99/100. Do the math, and you'll find that, given that the test came back positive, there's about a 10% chance that you have the disease, not 99%.

[Piache]

There is one thing wrong on your description.

"You only improve your odds by switching doors if Monte Hall knows what is behind each door. If he simply got lucky and opened a door with a goat behind it, your odds are unchanged."

This is not true.

If he simply got lucky and opened a door with a goat behind it, your odds changes in the same way.

Your odds changes not because Monte Halls knows what is behind each door, they change because YOU know one of the doors does not have the prize.

It doesn't even matter if there is someone opening doors.

Suppose that you pick a door, then a computer randomly picks one of the remaining doors and shows you the content.

If the door has a goat, then if you change doors you increase your odds.

If the door has a prize, then your odds also changes: the odds of winning are 0.

[Amun]

Using the the uncertainty principle to claim that the universe changes depending on what you know of it and is therefore subjective has always bothered me. In our existance, we cannot know something without observation. That is a limitation of what we are. If we were some kind of creature that could view the path of photons without disturbing them then this argument would never have existed.

[Harry Johnston]

Dave: quantum entanglement definitely does *not* allow faster than light communication. That is, you can't actually send your "morse code" faster than light. This can be demonstrated in the general case by choosing an appropriate formulation of quantum mechanics, but you can also take any specific proposed design and if you do the maths properly you find that there is no way for an experimenter at the "receiving end" to determine the setting at the "transmitting end". You can't even make a better than random guess.

In the specific design you describe the "receiving end" will always be uniformly lit (an equal probability of a photon arriving at any point) regardless of the layout of slits at the "transmitting end". The only correlation you'd get between the two ends is that when a photon hits the screen at the transmitting end (as opposed to missing the slits) it will have landed in one of two specific areas of the screen at the receiving end - note that I'm assuming that the initial direction of travel is the entangled property.

People often say that quantum entanglement is weird, but once you've really got your head around quantum mechanics, quantum entanglement is no stranger than classical correlations. (It's only the presence of non-commuting observables that makes the Bell experiment so counter-intuitive.)

Imagine having one red and one white marble in a bag, taking one out at random, and sending it to Alpha Centauri by special delivery without ever looking at it. Then look in the bag. If you see a red marble, you instantly know that the white marble is at Alpha Centauri - spooky action at a distance?

(For what it's worth, I'm not a professional physicist, but I did complete a PhD in quantum optics.)

[Kristina L.]

I think this proves that statistics are subjective more than that reality is. I've been a little suspicious of both statistics and proof by induction for quite a while.

[Douglas]

An example of how knowledge changes probability:

Say you have two cards, one with "this is the winning card" written on it, and one with "this is the loosing card" on it. To win, you have to pick the winning card. People who can read will win every time, they have a 100% chance of winning. A blind person, who cannot read the cards, will have to pick at random. They will win 50% of the time.

This is the same as in the Monty Hall problem, Monty knows which is the winning door and will always leave it for you to pick.

[Matt McNamara]

The easy way to rationalize "The Monte Hall Problem" is to think of it the other way...

One door has a prize. Two do not. You have a 1 in 3 chance of picking the correct door. However, you have a 2 in 3 chance of picking the WRONG door. By switching your answer after Monte's reveal, you swap your odds.

[Joachim Dyndale]

I believe this is an example of a purely theoretical problem. In reality, one might as well ask Monty to remove a goat before the game begins - giving a 50% chance of choosing the car.

Practically, there is no difference between the two scenarios for the contestant.

[E]

Gordon goosemonster wrote:

[Not really anything to complex here.

You initialy have a 1/3 chance of wining.
You are then shown a goat.
Your odds change to 1/2 as you now have the option to choose again. You don't actually have to change your door, choosing to keep your original selection still gives you a 1/2 chance.
Any other way of looking at it is just typical American stupidness.]

You'll have more credibility about "typical American stupidness" if you're less of a dumbass yourself.

It is counterintuitive.

But my God, there are links with pictures to explain it. Even someone like me, a typical stupid American... who went to public school... can get it.

Initially you have a 2/3 shot at picking a goat.

On pick two, if you picked the car first and switch, you'll lose 1/3 of the time.

THEREFORE, if you did Not pick the car first and switch, you'll Not lose 2/3 of the time.

That is not even money.

If you miss the car on your first guess, the reveal combined with the switch only leaves winning as a possible outcome. So your chance to win becomes the same as your initial chance to lose.

And if you look carefully, you'll see Monte Hall, with the score so far...

Typical Stupid American: 1
Arrogant Foreign Goat-Fucker: 0

Would you like to switch doors?

[stefan]

It is unlikely that the one door you pick has the car. That's the 1/3 thing. And yet the car could in reality be there the one time that you play the game.

The probability is worked out by considering multiple plays. But if you only play once, as a contestant, as implied by the game show scenario, then it makes no difference switching.

I think this is why many people get the answer "wrong" from a mathematical point of view, but are right in practice.

It's not a difference between object and subject, it's a difference between multiple and single plays.

The odds being against you doesn't prevent you from winning in reality. Every week some guy wins the lottery.

[Matt]

The perspective of the contestant is subjective, but the position of the car is objective.

[James F. R. Wright]

The idea that "your odds are changed depending on Monte's knowledge" is freaky and blows your mind shouldn't do and isn't since it's a misnomer. The set up where the presenter knows which curtains the goats are behind is a completely different set up to the one where he doesn't know. So each set up has its own probability.

[marco polo]

I used to think you were crazy-smart and eloquent.

Now I'm not so sure, maybe your spectacular presentation and humor bamboozle me most of the time, and you're secretly diluting my IQ.

Please post again on this topic, to acknowledge you didn't have your coffee today, or at least try making your point again? I mean jeez

[John]

Scott: "In other words, your odds are changed by Monte’s knowledge, and your knowledge that Monte has that knowledge."

The odds are set by what the question states that Monty will do. Your speculation about his knowledge is specious; he could be a robot and the odds would be the same.

The knowledge of the experimentor ("you") affects your assessement of the odds. But not the odds themselves - they don't depend on what the experimentor knows.

Quantum indeterminacy is interesting and wierd precisely because it is fundamentally different from conventional causal probability. If you can't see the difference that doesn't mean you've discovered something. It means you've failed to "get" what Bohr, Heisenberg and others did discover.

[John]

Scott: "In other words, your odds are changed by Monte’s knowledge, and your knowledge that Monte has that knowledge."

The odds are set by what the question states that Monty will do. Your speculation about his knowledge is specious; he could be a robot and the odds would be the same.

The knowledge of the experimentor ("you") affects your assessement of the odds. But not the odds themselves - they don't depend on what the experimentor knows.

Quantum indeterminacy is interesting and wierd precisely because it is fundamentally different from conventional causal probability. If you can't see the difference that doesn't mean you've discovered something. It means you've failed to "get" what Bohr, Heisenberg and others did discover.

[Greg]

You show promise in acknowledging that reality is a subjective experience, but show cartoon logic in some of your other conclusions.

Since I've argued with other people about this subject in the past I've put my response on my blog @ http://gregbecerra.blogspot.com/2008/04/mysticism-of-monty-hall.html

Times like this makes me wonder if you draw Dilbert or if Dilbert draws you.

[Chris]

All this talk about Monty having or not having any knowledge is silly and beside the point. Monty's knowledge is irrelevant. What is important is *your* knowledge. If your level of knowledge increases then your odds change; it doesn't matter if that knowledge comes from Monty himself or from knowing what is behind the door he picks.

As to the double slit experiment, anyone who's interested in understanding the double-slit experiment and modern quantum mechanics *needs* to read the essays here: http://www.overcomingbias.com/science/index.html. Some other posters have mentioned that link but it deserves to be reiterated. Scott, if you haven't read that yet then you definitely should.

[addendum to common sense]

Think of it like this: If a third party enters AFTER the host has already revealed a goat and doesn't know which door you originially picked, his odds are 1/2 to 1/2. You, though, have prior knowledge, so your odds are 2/3 to 1/3.

Scott, I think your problem is in your definition of probability. Probability DOES NOT describe reality; it quantifies the possibilities of random events. The more knowledge you have, the less unknown (which, some might argue, is behind our perception of randomness) the future is, and the higher probability you have of being right.

[common sense]

This is actually very simple.
Originally, you have a 1/3 chance of being correct. Once you choose a door, it doesnt matter what happens around you anymore - a nuclear explosion could destroy all life for all you care - your odds of winning are still 1/3. YOUR DOOR will only win 33% of the time.
So, when the host opens one of the other doors that contains a goat, that doesn't affect your original probability - the door you opened originally will still only win 33% of the time! Since there's only one door left, it HAS to win the other 66%!
My theory is an extension of the following scenario: Suppose you chose a door (we'll call this group A), and then (without opening a door) the host asks you if you want to switch. Well, your probability of winning is 33% now, so it should seem logical to switch doors, because 66% of the time the car will be behind one of the other doors (we'll call these doors group B), no? The thing is that group B contains two doors, so any INDIVIDUAL door you choose will always have a probability of 33%. You can't improve your chances by switching.
But if the host already revealed one losing door, switching works out for you, because he took a door out of group B! Now, group B, which has a 66% chance of winning, only contains one door!
The math works out even assuming that reality is real.

[Martin Bucknall]

My chances of winning aren't 1/3 but 2/3. You can get about in Edinburgh using a goat, but using a car it's impossible.

[yes]

lol "That is interesting enough on its own. (I’ll give a link later that explains the math of it.) But here is the freaky part. You only improve your odds by switching doors if Monte Hall knows what is behind each door. If he simply got lucky and opened a door with a goat behind it, your odds are unchanged. In other words, your odds are changed by Monte’s knowledge, and your knowledge that Monte has that knowledge.

If reality were objective, statistics wouldn’t be influenced by knowledge. That means your world is either partly created by your mind, or you are a hologram created by some other mind, and there are a few bugs in the software."

biggest load of crap i've ever seen. i know you're smart (you're in mensa for goodness sake) so if u do believe this it's just confirmation bias.

basic probability: Pr(A) is usually different to Pr(A given B). it's got nothing to do with holograms, and continuing to say something doesn't cause it to be true. stop trying to hypnotise people.

[Gordon goosemonster]

Not really anything to complex here.

You initialy have a 1/3 chance of wining.
You are then shown a goat.
Your odds change to 1/2 as you now have the option to choose again. You don't actually have to change your door, choosing to keep your original selection still gives you a 1/2 chance.
Any other way of looking at it is just typical American stupidness.

[ChrisB]

I think almost nobody on here has appreciated what Scott is getting at on this one. Yes, if you know your math, it is no mystery that if the game show host effectively says to you, “you can stick with your choice (2/3 prob of being wrong) or you can swap to these two other doors and if the car is there (2/3 prob of being the case) I'll guarantee you a win by opening a goat door”, then swapping would be in your interest.

It is also understandable that if the game show host tells you honestly that he also has no knowledge of where the car is and he says to you, “you can stick with your choice (2/3 prob of being wrong) or you can swap to these other two doors and I’ll randomly open one and what’s behind the remaining door is yours”. If you were wrong on your first choice (2/3 prob) there’s a 1/2 chance that he’ll open a goat door and you win. Thus there is a 1/3 chance of winning if you swap and a 1/3 chance if you stick so deciding to swap before he opens his door gives no advantage.

This last scenario however is not the game. You don’t have to decide on swapping until after he has opened his door. This is where my brain just doesn’t know any more. I’m thinking, if he opens his door and there’s a goat, regardless whether he opened the door randomly with no knowledge or intentionally with knowledge my choice is now stick with my original choice (2/3 prob of being wrong) or go for the last door which if my first choice was wrong must now contain the car with certainty. I’m still thinking therefore that if I can make my choice after seeing the goat I would now be wiser to swap. His prior knowledge is irrelevant surely? We both have now gained some knowledge. The math experts however tell me that this is not the case. Although the situation, after the goat is revealed, is physically the same it is different because the game show host made a random choice.

Before you all rush to tell me, I know that before the game started there was a 1/3 chance that he would reveal the car but psychologically I don’t care about this case as then swapping or sticking makes no difference, as I loose every time presumably.

To illustrate the dilemma better what do you do if you don’t know whether he opened the goat door with knowledge or randomly. Or what would you feel in the famous 52 deck of cards scenario if you select a card at random and he turns over 50 cards at random and the target car isn’t there. I know it’s mathematically wrong but I’d be thinking if the target card was still in the pack it would be extremely unlikely that he wouldn’t find it in 50 trials, maybe that means I’m more likely to have it. Or I might be thinking I had a 51/52 chance of getting the wrong card and now there’s only one card left unturned from his pack so If I chose wrong (very likely) the unturned card is the right one with certainty so I should swap. The problem illustrates how our brains cannot assess risk logically but does that mean we are wrong?

[Michael]

"If reality were objective, statistics wouldn’t be influenced by knowledge."

Statistics AREN'T. What is influence is PROBABILITY. And it is obvious that probability is influence by knowledge. Isn't it?

[Robert]

Why did you consistently misspell Monty? It's spelt right in your link.

In neither case is it the actual knowledge that matters. The reasons the probabilities change is that when Monty knows which door the prize is behind, he never opens that door. If Monty doesn't know which door the prize is behind but just happened never to open the door with the prize behind it consistently in an experiment based on this, then the empirically measured probability would be the same as in the "Monty knows" case.

In the case of the double slit experiment, actual knowledge doesn't matter either. If you measure which slit the photon goes through, and then throw that information away or conceal it forever, or simply don't look at it, the interference pattern still disappears. It's the _measurement_ that's important, not the _observation_.

[William]

For some reason my post seems to have been lost, but to put it simply this example doesn't work since you're talking about a single instance, and statistics don't really work like that. You can talk about how there is only a 1 in 10 000 000 chance of getting hit by a falling 10 ton barrel of yak butter, and so you're safe.... until you're the poor git who's tombstone reads something pithy like, "Yak'd to death.".

Anyway, I only get to choose once, then statistics doesn't apply. If I get to choose an infinite number of times then I'll have nearly an infinite number of cars no matter whether I change or not, and leaving me with a serious parking problem.

[Tormod]

WHAT!? SERIOUSLY!?!

Do you mean to say that if someone with knowledge DIVULGES something to you, then you are in a position of making an informed decision improving your odds??

WOW!? That's crazy talk!!!!

(Sarcasm, yes)

[Anna]

I like the links posted by someone (sorry, couldn't find your entry again in the list to mention by name) because it helps prove a point.

I just used this simulator:

http://www.versificator.co.uk/misc/monty-hall-problem.html
and the first time I played, I kept my first choice of door and got the car.

Obviously, if I had many opportunities to play the same game, in the long run I would do better switching doors, but since the game is being played only once... I wouldn't use probability (long-run relative frequency) to make a decision in this case, and the simulator showed that there's nothing wrong with my choice:-).

[Chris B]

Too many Chrisses, I'd better get more specific.

@Other Chris / Dave, Monty random vs. Monty knowledge *would* affect the probability IF, and only if, it were possible for him to pick the car.

Scott already said Monty didn't get the car. Presumably, if you ran the experiment over and over, 1/3 of the time Monty would open the door and get the car, and you'd be right.

The times you run the experiment and Monty gets a goat, EVERY TIME that happens, the "switch" door still has 2/3 chance of being correct.

And that will ALWAYS happen, as long as Scott says "Monty opened a door and got a goat."

Which, by the way, is *EXACTLY* the same thing as Monty knowing beforehand that he would get a goat--Scott saying "he opened it and got one" is precisely the same thing, math-wise. There was 0% probability of getting the car, because Scott defined the problem by saying "he didn't get the car." How is that different from knowing where the car is?

The whole trick is to not make it more complicated than it is. There are three doors. You get one, with a 1/3 chance of being right. Monty gets 2, with a 2/3 chance of being right. Monty is offering you his 2, instead of your 1. That's the WHOLE math of it. All the rest is distraction to confuse you. As long as Scott never has Monty pick the car, the problem doesn't change.

In other words, it doesn't matter if Monty knows... as long as SCOTT knows. :)

[TallDave]

"There are 2 doors, so the odds are 1 in 2 that they already have the right door, and 1 in 2 that switching will get them the car."

For a person who walks in when there are only two doors, this is true. However, he is unaware of the prior knowledge of the person who chose a door; he has been indirectly informed by Monte that that door is less likely to be the correct one. An operation has been performed that indicated the other door is more likely to have the prize.

In the coin analogy, this is as though someone has weighted the coin to make it land heads more often.

Another way to consider this puzzle is to pick one card rnadomly from a deck. The odds of you picking the Ace of Spades is one in 52. Your friend helpfully removes 50 of the non-AOS cards leaving only your card and another. There's a 51/52 chance you did not pick the AOS and that the other card is the AOS.

Now, to another friend who walks in, there's a 50/50 chance either card is the AOS. But you know better.

It's a limited information problem. That may seem "subjective," but only the calculation of the odds is subjective. The cards will not come up AOS 50/50 for your friend; your card will be the wrong one 51/52 times.

[KD]

I've heard of this one before. I always figured the best way to do it is to either do a live experement and see how the odds change or try to do some computer model if you have a good random number generator.

[TallDave]

"That is interesting enough on its own. (I’ll give a link later that explains the math of it.) But here is the freaky part. You only improve your odds by switching doors if Monte Hall knows what is behind each door. If he simply got lucky and opened a door with a goat behind it, your odds are unchanged. In other words, your odds are changed by Monte’s knowledge, and your knowledge that Monte has that knowledge.

If reality were objective, statistics wouldn’t be influenced by knowledge. "

No, the freakiness is an illusion. Here's why: if Monte doesn't know which door the prize is behind, he could open the right one instead of a wrong one and oops! there's the prize. That's why knowledge matters here: it's the statistical difference between the likelihood of the scenarios in which Monte opens accidentally opens the right door. There's nothing subjective about it.

[Ben]

Chris-
You are implicitly assuming that one of Monty's doors contains the car and the other contains one of the two goats. It is also possible that your door has the car, and his two doors both have goats. There is a 1/3 probability that he has two goats, 2/3 that he has one goat and the car. So if he opens a door with a goat (choosing the door randomly), there is a 1/3*0 + 2/3*1 = 2/3 probability that switching will give you the car (where the 1/3*0 is the probability he has two goats * probability you will win if you switch, and 2/3*1 is the probability he has one goat * the probability you will win if you switch). You thus have a 2/3 probability of winning by switching if he shows you a goat, no matter what method he uses to pick the door he opens.

[Dave Oblad]

Ok, I couldn't resist still yet another comment.

I explained below how the double-slit experiment can be shown to form faster than light
communication via entangle light beams. To explain this I have to suspect that there
exists a common point that connects all other points in the universe. This common
point is the cause of quantum probability. It's pseudo random noise. I will even
suggest that when we get quantum computers working on analysis of this noise we
might find signals outside of Random Probability. Injected signals from ET's.

SETI, while a cool project, reminds me of a bunch of primitive secluded Islanders using
giant man-made ears to listen for remote drum beats. Lack of detecting such signals
forces them to conclude that they are the only humans on the planet earth.

Meanwhile, radio waves imprinted with "I Love Lucy" re-runs are passing through
their bodies. Perhaps, when we re-direct our focus to decoding Quantum Noise, we
might find re-runs of "I Love Fuzy" being broadcast this very moment from a galaxy
far, far away.. just a thought!

What?... I'm not supposed to reveal the Quantum Singularity Continuum yet?.. Ok!
Never mind folks ! (lol)

Best wishes from Dave :^)

[Dave]

Monty's knowledge DOES affect the probability outcome. When you picked your door originally, you had a 33% chance of being right, while the filed had a 67% chance of being right.

If Monty picked randomly, then he had a 33% chance of picking a car, which he would do 33% of the time. If he got lucky and picked the goat by accident, then the odds of both of the remaining two doors increase to 50-50.

If however Monty KNOWS that he is eliminating a booby door, then the odds on your door remain 33%, but the odds on Monty's singular unopened door increase to 67%, to reflect the original value of the field.

[Chris]

Scott, I have to echo what some of the others have said... something I missed by reading your post too fast.

Millions of people are wrong in thinking the Monty Hall problem isn't correct, but *you* are wrong in thinking that Monty *needs* to know what's behind the doors. He only has to know what's behind the doors to guarantee he never picks the car. If he tries one of his at random and *doesn't* pick the car, it's just as good as if he'd known.

In other words, when you pick a door, Monty has the other two. So your door has a 1/3 chance, and Monty's two have a 2/3 chance. Full stop.

Knowledge of the contents have nothing to do with it. Monty is telling you you can have your 1 door, or have the best of his 2 doors. All the rest is wording designed to make it counterintuitive, which works VERY well, judging from the constant outcry of the Disbelievers.

If Monty opens one of his and it's a goat, *whether he knew it or not beforehand*, Monty's two doors still have a 2/3 chance. Not knowing about it beforehand means that Monty *could* have opened the car, but you said he rolled the dice and got a goat, which is no different from having *known* he would get a goat. The two doors Monty had gave him a 2/3 chance of having the car and it's still true.

What you say *is* true if sometimes Monty can get the car, because he didn't know what was behind the doors. Makes for a pretty lame TV show and a pretty unhappy contestant. But if you only show the ones where it's a goat after all, then it's just as good as knowing which was which.

[Chris]

I noticed that a number of posters seem to believe that it's still in the player's interest to switch doors even it Monty picks a random door. I don't think anyone has corrected them yet directly so I'll pipe in.

In that case, the probability of winning has to include the probability that Monty will pick the car. In the original problem, there was no probability in Monty's decision; now that there is it changes the whole thing.

There's a 50% chance that monty will pick a goat and 50% that he'll pick the car. If he picks the goat then you switch and there is a %66 chance that you win the car. But if he picks the car then there is a 0% chance of you winning the car (but you still switch even though it doesn't matter because we've agreed on the 'always switch' rule). So your chance of winning is (50% * 66%) + (50% * 0%) = 33%. Not 66%.

[RandyR]

I am beginning to think that "JD" is a construct of Scott Adams. Notice it was the very first reply and it was by far the most vehemenantly against the correct Monte Hall analysis than anyone.

Perhaps someone else did write it, but I think Scott harvested from a different place and transcribed it in reponse to his own blog just to yank our chains.

[Paul]

This is right in your wheelhouse Scott: http://www.physorg.com/news127499715.html

A 13 year old kid correcting NASA's estimates of whether an asteroid would hit Earth.

[Mr. Wampus]

A more direct correlation to your post about knowledge changing the world is the "Schrodinger's Cat" experiment (read about it here: http://en.wikipedia.org/wiki/Schrodinger's_cat)

Schrodinger devised a thought experiment to illustrate the weirdness of quantum mechanics, one component of which states that a particle simultaneously assumes all of it's potential states and only locks onto a particular state when observed. His experiment envisions having a live cat sealed inside a box with a poison bomb triggered by radiation. A quantum particle is then introduced to the box. The particle of course assumes all of it's potential states including one that radiates. The cat then is both alive and dead inside the box and only assumes a form upon the event of someone opening the box and observing which state the particle ultimately assumed; one that radiates or one that doesn't.

Weird.

[jeff]

There seems to be a large number of people who are arguing this puzzle is wrong. Is it that you that are too lazy to read or too challenged to comprehend the problem?

The solution/explanation is all over the internet. Just because you don't get it, doesn't mean it ain't true.

Dance, dance, dance.......

[steve?]

Of course knowledge changes statistics. I might even venture to say the statistics is the mathematics of applied knowledge.

The most basic example I can think of are loaded dice. One assumes that each number on a six sided die has one in six chance of coming up. However, if the die is loaded, one can change one's prediction in light of the new knowledge.

[d-rob]

Well, you got it wrong that Monte's knowledge is required to improve your odds of winning. All that is needed is for Monte to reveal a goat, it doesn't matter if he did it intentionally or accidentally.

The point is that there's a 2/3 chance that your first guess was wrong. But as long as there's 2 other choices, you have no way of improving your odds. However after Monte shows you one of the wrong choices, there's still a 2/3 chance your initial guess was wrong, and only a 1/3 chance that the other door is wrong. So obviously its better to switch.

There's nothing mysterious about Monte's knowledge. The only reason the players odds wouldn't improve if Monte chose at random is because in that case Monte would have a 1/3 chance of picking the car and ending the game before the player could switch.

See what I mean?

[indo]

One of the first things u get taught in Statistics is the difference between random and non-random processes, and how the same set of equations (ergo logic) cannot apply to both.

If Monte knows, his selection of a door is non-random; i.e. u've altered the rules of the game: it's not the game that's altered the rules, so to speak.

To put it another way -> if Monte knows, you can also deduce certain things. If you deduce certain things, u've got more knowledge of the process. If you've got more knowledge of the process, u'd be an idiot not to make a better decision than if u didn't have that knowledge in the first place.

No offence ole' lad, but maybe u should treat statistics like u do voting ... it doesn't hurt to say u don't know! ;-)

(PS - Pls don't mind the post, tis not of my will.)

[David MacMillan]

Everyone has already corrected you in about a zillion different ways - might as well join the fray. Hope that my explanation of the double-slit will be a bit more readable.

There's a 1 in 3 chance you picked the car, and a 2 in 3 chance you picked a goat. Now if you picked a goat, then Monty is telling you where the car is. If you picked the car, he's just confusng you. So there's a 2 in 3 chance that he is telling you where the car is.

Basically you are betting that your original choice was wrong, which is a safe bet to make since there's a 2 in 3 chance that you were, indeed, wrong. But people don't like to think that they might have been wrong.

On to double-slit....

Light is an electromagnetic disturbance, not a particle...think about it like the droplets of water flung off of a wet dog when it shakes itself. The dog has little bits of water flying off of it...a vibrating electron has little bits of energy flying off of it. The flying bit of energy is sustained as an oscillation of electric and magnetic fields (just like the water droplet's individual molecules are bouncing around). However, since magnetic and electric fields have to align in some particular fashion, there are constraints on what path the bit of energy will take.

Left to itself, an individual bit of energy (a photon) will travel based on the overall electromagnetic vector scattering that an infinite number of photons would have. It's not a finite particle. However, if you observe it then you are constraining it to what you see it doing, not what it actually does.

It's really a bit more complicated than that, but it's not evidence that we are a hologram.

[Eric]

For the skeptics, I wrote a small Java program (a long time ago) that simulates the problem. Scott is right regarding the improvement of odds if you switch. He's incorrect about Monte knowing affecting the odds. The simulator doesn't "know" where the car is. Maybe he knew that was incorrect.

[AL]

I’ve read quite a few of the comments below and no one seems to agree with Scott about what he said about the Monty Hall problem. Scott is in fact correct about that if you know that the game show host doesn’t know behind which door the prize is and he only gets lucky by revealing the goat, you have a 1/2 chance of winning the prize either way. The reasoning is as follows:

If on your fist choice you happen to pick the door with the prize (probability 1/3) Monty will always reveal a goat. If on the other hand you pick a door with a goat on your first pick (prob. 2/3), Monty will have a prob. of 1/2 of revealing the prize (resulting on your loss). Provided he revealed the goat, you know that either you picked correctly or you picked incorrectly and Monty got lucky and revealed the goat. Both events are equally likely (1/3 prob. each where the remaining 1/3 prob. is that Monty revealed the prize), so provided Monty revealed a goat your chances are 1/2 either way.

Here is the reasoning in terms of conditional probability.

P(prize) = Probability of your first choice being the door with the prize = 1/3
P(goat) = Probability of Monty revealing a goat after your first choice = 1/3 + (2/3 * 1/2) = 2/3
P(prize | goat) = Probability of your first choice being the prize given that Monty has revealed a goat.

P(prize | goat) = P(prize)/P(goat) = (1/3)/(2/3) = 1/2

If Monty knows what is behind each door and deliberately chooses to reveal the goat no matter what, then as has been argued many times below you increase your chances to 2/3 by switching doors.

[Irishman]

Scott Adams said:
> If reality were objective, statistics wouldn’t be influenced by knowledge.

Nonsense. Statistics are a measurement, and like any measurement, they need a reference point. It's like asking, "How far is it to Albequerque?" From where? San Francisco? Dallas? Paris, France? Downtown Albequerque?

[Mark]

The other thing wrong with your interpretation of the Monte Hall problem is your assertion that the hosts knowledge affects anything. The "effects" of his knowledge are more logically seen as a limitation in the game.

[Mike Jackson]

You write "You only improve your odds by switching doors if Monte Hall knows what is behind each door.... your odds are changed by Monte’s knowledge." But Hall's knowledge doesn't affect what you do; his action does. So your conclusion that "If reality were objective, statistics wouldn’t be influenced by knowledge" doesn't follow. Are you really surprised that in a given situation, the chances of what happens next change based on what you do now? Your argument isn't based on subjectivity, really, just imprecise language.

[JEB]

what year are you in? i think this theory only applies to 1975

[Phil Goetz]

"If reality were objective, statistics wouldn’t be influenced by knowledge."

Really? So, say there are 2 doors, 1 with a prize. Just before you pick a door, Monty says, "By the way, the prize is behind door #1." That shouldn't increase your odds of winning?

[neopolitan]

I remember why I hated Stats at uni now. I reread what I posted and must admit that I screwed up.

It's 2/3. Damn that whole inability to retract posts thing.

cheers,

neopolitan

[JR]

Scott said: "If reality were objective, statistics wouldn’t be influenced by knowledge."

Probabilities/statistics are all based on knowledge. You can't give any probabilities of something happening if you don't know anything about it.

How can we say there is a 1 in 3 chance of picking the car at the start of the game? Cause we KNOW there is just one car behind three doors. If we didn't know how many cars and how many goats and how many doors there were in the game, we could not know the probabilities of picking the car. When Monte opens a door with a goat behind it later in the game, we simply gain some additional information that we can use to revise our probabilities.

It's just normal that probabilities would get influenced by knowledge. Probabilities would not exist without any knowledge.

[Bromond]

I'd rather think about Carol Merrill and the door SHE is standing in front of (RRRRRRRow!)...http://www.letsmakeadeal.com/carol.htm

[Robin]

It happens I knocked up a couple of simulations of Javascript simulations this a while ago, if anyone wants a try:

Knowing host - http://www.versificator.co.uk/misc/monty-hall-problem.html
Clueless host - http://www.versificator.co.uk/misc/monty-hall-problem-clueless-host.html

It amazed me and got me thinking all sorts of seemingly wacko things about consciousness too, but I think I have the boring old conventional maths right now: the cause of the change is that in the "clueless host" scenario you're eliminating all the cases where the host opened the 'car' door, so your sample of results are very biased. With the knowledgable host, those cases never existed.

[Mark]

Looks like someone went to the movies :-) This is on several blogs lately because they bring it up in the movie "21". Very good film btw, I recommend it.

I had never heard of this before watching the film, but found it interesting.

[dsg]

It makes more sense if you are wearing a chicken suit.

[neopolitan]

Damnit, I noticed some cut and paste errors in my last post.

Not doing it again, you can work out what I meant, I think.

[Craig Fratrik]

Photons aren't particles, so they don't "pass through" a slit. (They aren't really waves either.) For an interesting constructive definition see the series of posts about Quantum at:

http://www.overcomingbias.com/science/index.html

I think the first case might be described as a Bayesian problem. Which for many people, including doctors for whom the decision can mean life and death, does seem counterintuitive:

http://yudkowsky.net/bayes/bayes.html

[a]

Wrong.

Monte doesn't need any knowledge, being lucky is good enough.
Read the maths explanation properly.

[wernman]

Define the software developer of the hologram as God, and you have a Christian world view, although in terms that most of us don't think about it much. But it's true. CS Lewis talked about this world as The Shadowlands, in that everything here is only copies of what's in Heaven -- the Real World. Like a drawing is a 2-D "shadow" or "hologram" of something that's really 3-D. (Holograms are just hi-tech shadows -- CS Lewis wrote in the '40s; they didn't have holograms yet.)

[johnmac]

Funny I was reading about how this has screwed up 100's of experiments trying to prove cognitive dissonance.

It doesn't matter if Monty knows or doesn't know where the car is.

Three doors

A | B | C

Their initial probabilities

1/3 | 1/3 | 1/3

You pick "A" and the probabilities of where the car is are thus:

A = 1/3 | B or C = 2/3

So now B or C are revealed, Doesn't matter if it's random or not

Say it's C

Now the probabilities haven't changed A is still 1/3 and B or C is 2/3...only you know it's not C. Therefore there is a 2/3 probability that it is B. REGARDLESS of Monty's knowledge.

It works with the 100 doors as well. I think it's a great deal messier with deal or no deal. I'm not even sure they let you switch. But if you think you are going to win the Million you are an Idiot.

[liam]

Scott,

You are making an assumption here that is not true.Statistics only deal with the probability of a particular hypothesis being true, They don't actually change the truthfulness of this hypothesis. In other words, Monty's actions or the scientists observation don't actually change the location of the car or of the electron, they just change the probabilities associated with the belief that the electron or the car are in a particular location.

You seem like a fairly smart guy, I don't understand how you didn't get that. These are the types of arguments that the God-botherers use to discredit science. Are you testing us o mighty Scott?

[liam]

Scott,

You are making an assumption here that is not true. Statistics is a way of quantifying the probability of a particular hypothesis being true, it doesn't actually change the truthfulness of a hypothesis. In other words, Monty's actions or the scientists observation don't actually change the location of the car or of the electron, they just change the probabilities associated with the belief that the electron or the car are in a particular location.

So you start with a basic assumption and new information changes the parameters of your initial assumption, leading to a new set of beliefs. But this does not and cannot affect physical reality.

You seem like a fairly smart guy, I don't understand how you didn't get that, particularly as the argument you propose is typical of the types of arguments that the God-botherers use to discredit science.

[AC]

Ok, Scott, I read through the double slit article. This is not my area of expertise by any means. By saying that, I am sure my viewpoints will be dismissed.

"Knowledge" is not changing the experiment. Adding a "detector" for that single photon causes the experiment to change. Adding another variable to the experiment causes a change.

To use this experiment to make the jump of "Knowledge changes the real world" is not very responsible of you. (Ignores the DMD music in the background)

It seems obvious that knowledge about the real world changes human behavior. It's called free will. ;)

[neopolitan]

Scott,

You do realise that your argument is circular, don't you?

The only reason that the apparent probabilities appear to change is because Monty Hall knows something. If there was some Grand Programmer who knew all about us and fiddled around with our choices based on that knowledge, then yes, the fiddling around with our choices would be evidence that there is someone fiddling around with out choices.

If Monty has no privileged knowledge, then probabilities don't even seem to change. Basically this is the difference, if Monty Hall is the Grand Programmer, then probabilities appear to change. If Monty Hall is just some schmo, then probabilities don't change.

I agree with some here who say that the probability of picking a car is always 1/2 - from the beginning of the game. You really have to put a lot of effort into fiddling the figures to get the "right" answer. Here is some proof (sorry about the length, it's not rocket science, just a little tough to explain in fewer words):

----------------

Think about a slight variation of the game.

You have three doors to choose from, two have a goat behind and one has a car. Monty asks you to select a door, but not to tell him which door you have selected. Then Monty selects one of the two doors which has a goat.

These two selections are totally unrelated.

Now you have two possible scenarios.

1. Monty picks the door that you selected.
2. Monty picks a door that you didn't select.

You are left with two choices in both scenarios. You now know that one of the remaining doors has a goat and the other has a car. Maybe Monty picked your door, maybe he didn't, but he didn't know which door you had selected, so it didn't affect his selection process. So all you know is that you have a 50% chance of picking car from the remaining two doors. Mathematically, the selection you made before Monty made his selection is totally irrelevant, even if it may be important to you psychologically. Any decision you make now is between two doors, a 50-50 choice, although you may phrase it as "stick or switch" the outcome is the same.

Chances go like this:

(UC - you pick car, UA - you pick goat A, UB - you pick goat B, SW - you switch, ST - you stick. MA and MB is Monty picking goat A and B - under the rules he cannot select the car, he just picks a door in front of one of the goats entirely at random and we can force him to by simply not letting him know which goat is goat A and which is goat B)

UC - 1/3, MA - 1/2, ST - 1/2 ... total 1/12 (win car)
UC - 1/3, MB - 1/2, ST - 1/2 ... total 1/12 (win car)
total chance of winning the car if you select the car from the start 1/12 + 1/12 = 1/6

UC - 1/3, MA - 1/2, SW - 1/2 ... total 1/12 (win goat B)
UC - 1/3, MB - 1/2, SW - 1/2 ... total 1/12 (win goat A)

UA - 1/3, MA - 1/2, UC - 1/2 ... total 1/12 (win car)
UA - 1/3, MA - 1/2, UA - 1/2 ... total 1/12 (win goat B)

UA - 1/3, MB - 1/2, SW - 1/2 ... total 1/12 (win car)
UA - 1/3, MB - 1/2, ST - 1/2 ... total 1/12 (win goat B)

UB - 1/3, MA - 1/1, SW - 1/2 ... total 1/12 (win car)
UB - 1/3, MA - 1/1, ST - 1/2 ... total 1/12 (win goat A)

UB - 1/3, MB - 1/1, UC - 1/2 ... total 1/12 (win car)
UB - 1/3, MB - 1/1, SA - 1/2 ... total 1/12 (win goat A)

Your probability of winning a car is 1/6 + 1/12 + 1/12 + 1/12 + 1/12 = 1/2

Your probability of winning goat A is 1/12 + 1/12 + 1/12 = 1/4

Your probability of winning goat B is 1/12 + 1/12 + 1/12 = 1/4

So your probability of winning any goat is 1/4 + 1/4 = 1/2

---------------------

Now think about the standard version. You make a selection, then Monty makes a selection based on your selection.

Chances go like this:

(UC - you pick car, UA - you pick goat A, UB - you pick goat B, SW - you switch, ST - you stick. MA and MB is Monty picking goat A and B - under the rules he cannot select the car)

UC - 1/3, MA - 1/2, ST - 1/2 ... total 1/12 (win car)
UC - 1/3, MB - 1/2, ST - 1/2 ... total 1/12 (win car)
total chance of winning the car if you select the car from the start is 1/12 + 1/12 = 1/6

UC - 1/3, MA - 1/2, SW - 1/2 ... total 1/12 (win goat B)
UC - 1/3, MB - 1/2, SW - 1/2 ... total 1/12 (win goat A)

UA - 1/3, MB - 1/1, SW - 1/2 ... total 1/6 (win car)
UA - 1/3, MB - 1/1, ST - 1/2 ... total 1/6 (win goat B)

UB - 1/3, MA - 1/1, SW - 1/2 ... total 1/6 (win car)
UB - 1/3, MA - 1/1, ST - 1/2 ... total 1/6 (win goat A)

Your probability of winning a car is 1/6 + 1/6 + 1/6 = 1/2

Your probability of winning goat A is 1/12 + 1/6 = 1/4

Your probability of winning goat B is 1/12 + 1/6 = 1/4

So your probability of winning any goat is 1/4 + 1/4 = 1/2

------------------------

Now the statistics people haven't done this second analysis.

What they have done the first analysis and then cast out some of the options. Monty can't choose the car, so they do this:

(UC - you pick car, UA - you pick goat A, UB - you pick goat B, SW - you switch, ST - you stick. MA and MB is Monty picking goat A and B - under the rules he cannot select the car, where he picks the car, this is a null option. Note that you must have the option to select an open door if Monty is allowed to select the door with the car.)

UC - 1/3, MA - 1/3, ST - 1/3 ... total 1/27 (win car)
UC - 1/3, MB - 1/3, ST - 1/3 ... total 1/27 (win car)
UC - 1/3, MC - 1/3, ST - 1/3 ... total 1/27 (win car) - NULL OPTION

UC - 1/3, MA - 1/3, UA - 1/2 ... total 1/27 (win goat A) - STUPID DECISION
UC - 1/3, MB - 1/3, UA - 1/2 ... total 1/27 (win goat A)
UC - 1/3, MC - 1/3, UA - 1/2 ... total 1/27 (win goat A) - NULL OPTION

UC - 1/3, MA - 1/3, UB - 1/2 ... total 1/27 (win goat B)
UC - 1/3, MB - 1/3, UB - 1/2 ... total 1/27 (win goat B) - STUPID DECISION
UC - 1/3, MC - 1/3, UB - 1/2 ... total 1/27 (win goat B) - NULL OPTION

UA - 1/3, MA - 1/3, UC - 1/3 ... total 1/27 (win car)
UA - 1/3, MB - 1/3, UC - 1/3 ... total 1/27 (win car)
UA - 1/3, MC - 1/3, UC - 1/3 ... total 1/27 (win car) - NULL OPTION

UA - 1/3, MA - 1/3, ST - 1/2 ... total 1/27 (win goat A) - STUPID DECISION
UA - 1/3, MB - 1/3, ST - 1/2 ... total 1/27 (win goat A)
UA - 1/3, MC - 1/3, ST - 1/2 ... total 1/27 (win goat A) - NULL OPTION

UA - 1/3, MA - 1/3, UB - 1/2 ... total 1/27 (win goat B)
UA - 1/3, MB - 1/3, UB - 1/2 ... total 1/27 (win goat B) - STUPID DECISION
UA - 1/3, MC - 1/3, UB - 1/2 ... total 1/27 (win goat B) - NULL OPTION

UB - 1/3, MA - 1/3, UC - 1/3 ... total 1/27 (win car)
UB - 1/3, MB - 1/3, UC - 1/3 ... total 1/27 (win car)
UB - 1/3, MC - 1/3, UC - 1/3 ... total 1/27 (win car) - NULL OPTION

UB - 1/3, MA - 1/3, ST - 1/2 ... total 1/27 (win goat B)
UB - 1/3, MB - 1/3, ST - 1/2 ... total 1/27 (win goat B) - STUPID DECISION
UB - 1/3, MC - 1/3, ST - 1/2 ... total 1/27 (win goat B) - NULL OPTION

UB - 1/3, MA - 1/3, UA - 1/2 ... total 1/27 (win goat A) - STUPID DECISION
UB - 1/3, MB - 1/3, UA - 1/2 ... total 1/27 (win goat A)
UB - 1/3, MC - 1/3, UA - 1/2 ... total 1/27 (win goat A) - NULL OPTION

Initially there are 27 possible outcomes. Of those outcomes, 9 have you winning the car, 9 have you winning goat A and 9 have you winning goat B. But Monty is not allowed to select the car so the 9 outcomes where Monty does that are eliminated.

The remaining 18 outcomes have you winning the car 6 times, goat A six times and goat B six times.

However ... in six of the cases where you won a goat, you knew you were selecting a goat. That would be stupid decision, unless you really liked goats. We can eliminate those stupid decisions too.

This leaves us with 12 possible outcomes.

They don't just add up the remaining options (6 out of the 12 valid outcomes have you win the car, 50% again). What they do is assume that if you have two options and one is a stupid decision, then they assume that you have no choice! They also redistribute the chances of Monty picking a particular goat. They also must eliminate four of the remaining 12 options, where Monty would select your door (if you had selected a goat), and he is not allowed to do that.

This means it goes to this:

UC - 1/3, MA - 3/6, ST - 1/2 ... total 3/36 (win car)
UC - 1/3, MB - 3/6, ST - 1/2 ... total 3/36 (win car)
UC - 1/3, MC - 0/3 ... - NULL OPTION

UC - 1/3, MA - 3/6, UA - 0/2 ... total 0/36 (win goat A) - STUPID DECISION
UC - 1/3, MB - 3/6, SW - 1/2 ... total 3/36 (win goat A)
UC - 1/3, MC - 0/3 ... - NULL OPTION

UC - 1/3, MA - 3/6, UB - 1/2 ... total 3/36 (win goat B)
UC - 1/3, MB - 3/6, UB - 0/2 ... total 0/36 (win goat B) - STUPID DECISION
UC - 1/3, MC - 0/3 ... - NULL OPTION

UA - 1/3, MA - 0/6, SW - 1/2 ... total 0/36 (win car) - MONTY CHOOSES YOUR DOOR
UA - 1/3, MB - 6/6, SW - 1/2 ... total 6/36 (win car)
UA - 1/3, MC - 0/6 ... - NULL OPTION

UA - 1/3, MA - 0/6, ST - 0/2 ... total 0/36 (win goat A) - STUPID DECISION
UA - 1/3, MB - 6/6, ST - 1/2 ... total 6/36 (win goat A)
UA - 1/3, MC - 0/3 ... - NULL OPTION

UA - 1/3, MA - 0/6, UB - 2/2 ... total 0/36 (win goat B) - MONTY CHOOSES YOUR DOOR
UA - 1/3, MB - 6/6, UB - 0/2 ... total 0/36 (win goat B) - STUPID DECISION
UA - 1/3, MC - 0/3 ... - NULL OPTION

UB - 1/3, MA - 6/6, SW - 1/2 ... total 6/36 (win car)
UB - 1/3, MB - 0/6, SW - 1/2 ... total 0/36 (win car) - MONTY CHOOSES YOUR DOOR
UB - 1/3, MC - 0/3 ... - NULL OPTION

UB - 1/3, MA - 6/6, ST - 1/2 ... total 6/36 (win goat B)
UB - 1/3, MB - 0/6, ST - 0/2 ... total 0/36 (win goat B) - STUPID DECISION
UB - 1/3, MC - 0/3 ... - NULL OPTION

UB - 1/3, MA - 6/6, UA - 0/2 ... total 0/36 (win goat A) - STUPID DECISION
UB - 1/3, MB - 0/6, UA - 2/2 ... total 0/36 (win goat A) - MONTY CHOOSES YOUR DOOR
UB - 1/3, MC - 0/3 ... - NULL OPTION

The final 8 possible outcomes include four outcomes where you win a car, two in which you win goat A and two in which you win goat B. The redistributed probability is:

win a car = 3/36 + 3/36 + 6/36 + 6/36 = 18/36 = 1/2
win goat A = 3/36 + 6/36 = 9/36 = 1/4
win goat B = 3/36 + 6/36 = 9/36 = 1/4

Seem familiar?

But we can look at it another way, a way that I think is not valid, but heck, I am not a statistics guy.

You could look at the valid outcomes and force upon them the same probability, ie

UC, MA, ST ... (win car)
UC, MB, ST ... (win car)
UC, MB, SW ... (win goat A)
UC, MA, UB ... (win goat B)
UA, MB, SW ... (win car)
UA, MB, ST ... (win goat A)
UB, MA, SW ... (win car)
UB, MB, SW ... (win goat B)

This gives you the same probability as all other distributions I have looked at. 50% chance at the car, 25% chance at either goat.

To get the "right" answer, you have to do something else which is a little odd. You are obsessed by the car. If you win a goat, you don't care which goat it is, you only distinguish between whether you switched to the goat or stuck with a goat. If you win a car, however, you restrospectively acknowledge that you had a range of options - so, where MG is Monty picking a goat (any goat) and UG is you picking a goat (any goat), we have:

UC, MG, SW ... (win goat)
UG, MB, ST ... (win goat)
UC, MA, ST ... (win car)
UC, MB, ST ... (win car)
UA, MB, SW ... (win car)
UB, MA, SW ... (win car)

Then you force upon these equal probability (which we know they don't have) and finally you arrive at the "right" answer.

2/3 chance at the car, 1/3 chance at a goat.

It would seem that it is not Monty that changes the apparent probability, but our fixation on the car. In certain regions of certain countries (thinking Wales and New Zealand here), it is entirely possible that you would have a significantly higher chance of winning the better looking goat. (Admittedly, not everyone aspires to goats, so maybe the program directors would use sheep instead.)

cheers,

neopolitan

[Chris Orton]

The key to the analysis is that, if the host knows where the goat is, his choice is constrained, and this affects any shift in probability caused by the contestant changing his choice.

If A contains the car and B and C contain goats:-
If A is chosen the host can choose B or C …… case 1(2 choices)
If B is chosen the host can only choose C……. case 2(1 choice)
If C is chosen the host can only choose B……. case 3(1 choice)

In case 1:
B chosen – Holding wins
Switching loses

C chosen - Holding wins
Switching loses

In case 2 - Holding loses
Switching wins

In case 3 - Holding loses
Switching wins

Therefore summating all the instances:

Holding wins twice loses twice
Switching wins twice loses twice

The probability of success is not altered by switching

Q.E.D.

Probability needs careful thought – that’s how successful gamblers make their money-
By not gambling, but working it out!

[Chris Orton]

The key to the analysis is that, if the host knows where the goat is, his choice is constrained, and this affects any shift in probability caused by the contestant changing his choice.

If A contains the car and B and C contain goats:-
If A is chosen the host can choose B or C …… case 1(2 choices)
If B is chosen the host can only choose C……. case 2(1 choice)
If C is chosen the host can only choose B……. case 3(1 choice)

In case 1:
B chosen – Holding wins
Switching loses

C chosen - Holding wins
Switching loses

In case 2 - Holding loses
Switching wins

In case 3 - Holding loses
Switching wins

Therefore summating all the instances:

Holding wins twice loses twice
Switching wins twice loses twice

The probability of success is not altered by switching

Q.E.D.

Probability needs careful thought – that’s how successful gamblers make their money-
By not gambling, but working it out!

[Hacker Kitty]

I like puzzles like this... particularly the "Blue Eyes" one which is a little like your take on the Monty Hall problem. Mostly that is just about raw probability (you only have 1/3 chance of picking the right door first time, so 2/3 time, switching doors gives the right result). This problem really focuses on how common knowledge interacts with probability.

http://xkcd.com/blue_eyes.html
(btw, xkcd rules!)

It took me a long while to really understand this one

[kettu]

I think there is a clue in this sentence in the Wikipedia article:

"The reasoning above applies to all players on average without regard to which specific door the host opens and all individual players at the start of the game, but not to a specific player at the point the player is asked whether to switch given which door the host has opened (Morgan et al 1991)."

Since it doesn't apply to an individual player at the time of play, his odds remain 1/3. I don't think they would give away cars if there was a reliable way to stack the odds 2/3 in your favour.

Reality subjective or objective? I'm not even gonna dance to that tune...

[DD]

I think this is a load of old toot: What is the actual data? How many of the punters who switched actually won compared to the ones that didn't switch and won? If this is true we would expect to see a wide margin between the two groups (2:1): The snag is that for it to be statistically significant we'd need an "n" in the region of 2000. I don't think the show has been going that long, but I may be wrong as we don't get it over here anyway... Following the logic if the punter switches doors then switches back shouldn't they win 3/3 times? Nads. Quantum physics is just a construct designed to keep the monkeys dancing. I used to work in a Physics Department, and my opinion of physicists is that they are all a bunch of objectionable bastards anyway. Probably because they cannot get girlfriends - yes they were mostly men (9:1), and only one of them I met was openly gay (he left to be come a merchant banker, so clearly did not have his heart in being a physicist anyway).

[Old Ovarian]

One assumes that the Goat remains silent throughout, not giving away his location.
Way things are going though a Goat might be more useful than a Car.

Old Ovarian (OO)

[michael rowe]

This is something i am quite interested in -

DOES it only work if Monty knows which door the car is really behind? (as Scott and most posters have said)

or does it even work if he doesnt know? (As a few people have said)

I think the former but can't prove it to my own satisfaction.

What do you guys think of this thought experiment?

Say that there were 2 contestants, instead of one. Say John picks box A, and Jim picks box C. Then the wind, (or Monty, without knowledge) knocks over box B, to reveal a goat. Is it plausible to say that now BOTH of them should switch with each other and thus increase their probability? Surely not!

We can also argue in a similar way that in the case of the single contestant picking box A, (and box B is knocked over to reveal goat WITHOUT Monty knowledge) it is incorrect to split everything in two "buckets" (as Barry Kelly, and maybe others, said) with "Bucket A" containing the probablility that your original choice of Box A was correct (1/3), and "Bucket B" containing the probability that it was incorrect (2/3) - and therefore you should switch from Box A to Box C

This surely is a false (although seductive) argument, because it could equally well be applied as follows - "Bucket C" contains the probability that Box C is the one with the car (1/3), and "Bucket D" contains the probability that it doesnt (2/3).

Now that it certainly isnt in Box B, which has been knocked over by Monty, revealing a goat, this doesn't change the fact that "Bucket D" has a 2/3 probability as per Barry's argument above, and hence we should stick with "Bucket D", which means sticking with the only Box left in "Bucket D" - Box A.

So the same argument which tells us to switch to Box C, also can be applied in the opposite way to tell us to stay with Box A!

(This doesnt apply in the case of Monty's knowledge, see if you can work out why - the whole argument works differently in the first place, and in such a way that it isn't "reversible" like this)

Although all i have done is demonstrated the apparent falsity, without actually putting my finger on exactly where the fallacy is, i have a feeling that the fallacy lies in the discounting of the 1/3 chance of the game being ended early. although that didnt happen in this case (Monty knocks over Box B WITHOUT knowledge to reveal goat) it means we are in a scenario of 2/3 already, (cos 1/3 of times knocking over Box B ends game by revealing car) and it is incorrect to "use" the original 2/3 - 1/3 statistics to "bolster" "Bucket B"'s chances over "Bucket A's"

anyone want to comment or question to me directly, i'm on mcr4444 at yahoo dot com

(PS: there is a MUCH more interesting one of the $5 / $10 envelope and you can switch with the host... if any one of you knows it you'll know what i mean...)

Michael Rowe

[Warfreak2]

This doesn't work, because Monty will end up showing you the door with the car 1/3 of the time, and you are only allowed to choose from the two closed doors (which will then both have goats).

If you want knowledge changing probabilities, no need to use such a complicated example as Monty Hall: roll a dice. Before you roll it, the probability that it is a 6 is 1/6. After you roll it and know the outcome, the probability is 1 or 0 (depending on whether it was a six or not). This isn't philosophically profound; probabilities are an abstract invention of humans, they aren't part of actual reality.

Avoid getting the words "probability" and "statistics" confused (statistics is an application of probability to the real world, a statistic is a measurement but a probability is not).

[Colin Janke]

Probably adding to hordes of posts so far:

Statistics is based on knowledge in the form of selection criteria - if there is a 1 in 100 chance than a random woman you pick of the street earns over x thousand a year, that dos not mean that there is a 1 in 100 chance of a random person who earns over x thousand being a woman. The Montey Hall 'problem' is in many ways an extension of this. The chance of the ca being behind one of two random doors is 50%. The chance of you not having got it right on the first try is 2/3.

The Double slit experiment is based on the idea of Really Small Things (tm). To measure something, you have to interact with it. If you bounce a photon off a billiard ball (why is it always billiards? Why not pool or snooker?) it's not going to move, and you can say where it is. I you bounce a photon off an an electron, the electron's in London by the time you do the maths to figure out where it was, so you've disrupted anything the electron was going to do after you measured it.

I don't think the two things here do come from the same source, though they are two nice examples of how the world doesn't conform to what we learn to expect from day to day experiences.

[Kirado]

was going to say.. that sounds like quantum physics.. oh wait it is.

[Bugs]

For those who disagree with the (correct) answer to the Monty Hall problem in Scott's post, I don't like the explaination he gave either. Try looking at it this way:

Lets say you choose one door from ten. Your chances of picking the car are 1/10.

Now, the car is either behind your door (1/10) or behind one of Monty's nine doors (9/10). If given the choice between opening just your door or all of his, the best choice would be to open all of his, right?

By opening all but one of his doors and allowing you the chance to open the other, Monty is effectively saying that if the car was originally behind any of his his doors (9/10) and you switch, you're guarunteed to get it. So at the end it isn't a choice between two doors; it's a choice whether the car was more likely to be behind your one door (1/10) or Monty's set of 9 doors (9/10).

Note that Monty opening the "empty" doors makes absolutely no difference to the game, it's just a distraction from the choice between sets. Equally, if Monty doesn't know where the car is but manages to open his empty door(s) by sheer fluke, the maths still works. Monty's (lack of) knowledge makes absolutely no difference to the probabilities.

Subjective nature of reality? It's a stretch. It does demonstrate that humans without training are astonishingly bad at dealing with probabilities. There's a whole branch of science dedicated to studying our irrationalities; much of it is put to use by stock market traders to predict and exploit other investors' instinctive yet irrational decisions. IIRC Scott has an economics degree, so he was probably taught some of it as an undergrad.

This has absolutely no link with the double-slit experiment (which also doesn't require anyone to be watching or know what's going on) other than that they're both counter-intuitive.

[Nicholas]

The car is behind the door someone put it behind. It doesn't move. In effect, the game allows you to pick one door (if you don't change) or two doors (if you do).

[Alex]

I am not a statistician, nor indeed a mathematician of any sort, but there appears to me to be something to examine on the Monte Hall scenario. The diagram in Wiki shows 3 possible outcomes, based upon the choices made, and demonstrates a goat in one case and the car in two. However, there are two ways to achieve the goat outcome - Monte could choose Goat A or he could choose Goat B. So in other words, when combining the Player's choice and Monty's available choice, you get 4 possible outcomes, 2 of which lead to goats and two of which lead to cars. The thing is that we never get to see the outcome of the choice which Monte doesn't make in the situation where the player has picked the car.

Now I don't doubt that this doesn't affect the 2/3 probability of the player choosing the car if he switches. But it probably goes some way to explaining why there remains an instinct in us that there's a 50/50 probability hidden in there somewhere.

The Monte Hall scenario isn't evidence that our reality is subjective - it's evidence that few of us have a really deep understanding of probability (me included !)

[RichardT]

Scott, this might appear a bit pendantic but are you not just confusing Probability Theory with Statistics?

The game show set up is an exercise in probability theory. Why should it surprize you that the formula is affected by the introduction of additional variables? Probability Theory is a series of mathematical models designed to mirror reality. If you make the model more complex then it (should) mirror reality more appropriately. Hence the introduction of the vairiables does monte know? & do I know monte knows, should by definition change the outcome. If not then the model is not suitable for the job in hand.

Statisitcs is the analysis of empirical data in order to find patterns. You might well choose to achieve this analysis using a tool, say a model taken from Probability theory, but the two are not the same.

Reality remains objective it is just that the tool you choose to employ is becoming more sophisticated. Hence the outcome is improved.

At the risk of insulting your arguement with an absurd analogy "This piano plays a great tune if you use the white keys, but it sounds even better if you use the black keys as well!".

[ShaunL]

Why the hell has Monty Hall got so many goats? Freak.

[Marklar]

This problem is not a statistical one, it's all in the wording. Imagine instead that you choose a door and the host says "You can either stick with that door or open BOTH of the others". That is your choice. The fact that he opens one of them before asking you to change does not change the choice.

[Sacha]

I'm sure someone already noted this but the results has nothing to do with the "knowledge" but with the "action" of telling you where for sure there is one goat...

From the beginning you've had 2/3 of possibilities of choosing a goat... Hall is obliged to show you the other and therefore the logical thing is to switch (if not you'd be playing with the 1/3 of having initially chosen the car)

thanks for this mind exercises ... they are refreshing :-)

[Steve Bosman]

As a mathematician (in a former life) it used to freak me out that so many people didn't get that switching was the optimal strategy.

Having read this post I'm now freaked that because you spelt it Monte (i.e. with an 'e') lots of people are copying you rather than checking their facts - it is Monty (with a 'y').

[William]

I'm surprised that no-one has pointed out the most obvious and persistent criticism of statistics as a science, namely that it is only really completely predictive in two situations:
1) Where the set approaches infinity
2) Restrospectively

In other words if I flip a coin twice only an idiot would say that I will get one heads and one tails, since its too small a sample for statistical averages to start to apply.

The more times I flip the coin (and the more coins that are flipped, to negate bias from coin weighting, etc), the more average the results become until (near an infinite number of flips), the result become an even balance of heads and tails (ignoring the odd 'on its side' result). It is however, impossible to say is flip number 3 000 001 or flip number 3 000 002 will be heads or tails, since they might both be heads, and the balance only restored 10 or 10 000 flips later by a run of tails.

Once the coin is flipped and the result is observed then you can say that you are 100% that the coin that was just flipped would have produced that result. Before somene points out that this is both obvious and facetious please remember that statistics are often used to retrospectively justify courses of action.

Anyway, the point I'm rambling towards is that the Monte Hall example posits a single situation, not a near infinite number of situations. If you had to make the choice an infinite number of times, then the advantage presented by the Monte Hall example might actually be of some use... and you'd also have a LOT of cars, so who would really care about a few more.

Either way it is simple irrelevant if applied to a single situation, which is in all likelihood all you would receive.

[Chris]

Monte's knowledge of the goat positions is only required so that he doesn't pick the car; this does not prove that knowledge affects reality.

The photons in the double-slit experiment are altered not by our "knowledge" of their paths but rather by the means that we acquired that knowledge. In order to observe their path, we must scan them with some form of detector that by its very nature must interfere with their motion. This causes a breakdown in the order of their probability wave function.

[knyghtfall]

"You only improve your odds by switching doors if Monte Hall knows what is behind each door. If he simply got lucky and opened a door with a goat behind it, your odds are unchanged. In other words, your odds are changed by Monte’s knowledge, and your knowledge that Monte has that knowledge."

It is not accurate to say that the odds are changed, actually - That implies that there is only a single scenario, where the odds are changed due to some additional influence that was factored in. ACtually, you have described two scenarios, 1 in which Monty has knowledge of what are behind the 3 doors, and 1 where he doesn't. The odds are merely different for each scenario, but not changed per se.

Odds are merely a representation of the way we evaluate the situation we are in, and it is logical that knowledge influences odds. For example, if you have perfect knowledge of the winning numbers for the next draw, your odds of winning the lottery (should you choose to buy these numbers) become 1 out of 1. The reality that you are implying seem to refer to physical things - the 3 doors, the 2 goats and Monty Hall himself. Hence, it does not make sense for you to group "odds" into the same class of physical objects, and term them all as "reality".

[Chris]

Thanks a bunch Scott...my wife had to go about proving me wrong about the whole Monty Hall problem and she still believes that she is right! She can't quite figure out why it isn't a 50/50 chance even though it is AS PLAIN AS DAY!!!! Ugh, your entry today was very misleading and weasel like. Your powers grow stronger every day

[Dave Oblad]

One more Comment here from yours truly:

I read many comments that indicate some problem in understanding this problem.

It's simple:

You have 3 doors.. A,B,C
You pick one.. let's say it's door "A"
Now you easily understand your chances of being right are 1/3. (Right?)

Now suppose Monty offered you to keep that single door "A"

or...

You could have BOTH other doors. Would you not jump at the increase in odds
by selecting BOTH the other doors (B+C) instead of your single door (A)?
Your odds would now be 2/3 at winning.

The rest is just smoke...lol

Of course one of those doors (B+C) in the double choice would be No Good and Monty
shows you the Bad specific door. Big deal, the 2/3 odds have not changed for you.

Make the switch! Always choose two doors over one!

Best wishes from Dave :^)

[jennifer]

what does dmd mean? scott, you should post a glossary of your coined acronyms.

If the host got lucky and opened a door with a goat, its still better to switch. The math works exactly the same way, as the door you chose has a 1/3 chance of being right, and the only other possibility is that the other unopened door is the one with the car, and thus it has 2/3 chance of being right.

The problem with him now knowing which door has the car, is that if he opens it, he fucks up the game. If he gets lucky and opens a goat