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July 20, 2004 9:12 AM   Subscribe


 
Heh, I love the Monty Hall problem. It's amazing how some people will just completely refuse to believe the solution, no matter how you explain it or prove it.
posted by reklaw at 10:10 AM on July 20, 2004


My favorite way of demonstrating the solution is this :

Imagine you have 100 doors. You pick one. Then I open all but 2 one you've picked and the one that has the car. Do you switch? Of COURSE you do. The likelyhood of you picking the car on the first try is one in 100. The likelyhood on the second try is one in 2. EASY DECISION!
posted by psychotic_venom at 10:16 AM on July 20, 2004


My favorite way of demonstrating the solution is this :

Imagine you have 100 doors. You pick one. Then I open all but 2 one you've picked and the one that has the car. Do you switch? Of COURSE you do. The likelyhood of you picking the car on the first try is one in 100. The likelyhood on the second try is one in 2. EASY DECISION!

Oh, and by the way, on the "More Info" page, some of the links are out of date and NSFW
posted by psychotic_venom at 10:17 AM on July 20, 2004


Actually, p_v, the odds for the second pick is 99 to 1, not 1 in 2. That's what the paradox is about.
posted by michaelonfs at 10:34 AM on July 20, 2004


Exercise 3.9.[2, p.61] The three doors, earthquake scenario.
Imagine that the game happens again and just as the gameshow host is about to open one of the doors a violent earthquake rattles the building and one of the three doors flies open. It happens to be door 3, and it happens not to have the prize behind it. The contestant had initially chosen door 1.
Repositioning his toupee, the host suggests, `OK, since you chose door 1 initially, door 3 is a valid door for me to open, according to the rules of the game; I'll let door 3 stay open. Let's carry on as if nothing happened.'
Should the contestant stick with door 1, or switch to door 2, or does it make no difference? Assume that the prize was placed randomly, that the gameshow host does not know where it is, and that the door flew open because its latch was broken by the earthquake.

[A similar alternative scenario is a gameshow whose confused host forgets the rules, and where the prize is, and opens one of the unchosen doors at random. He opens door 3, and the prize is not revealed. Should the contestant choose what's behind door 1 or door 2? Does the optimal decision for the contestant depend on the contestant's beliefs about whether the gameshow host is confused or not?]

Hey, try and figure it out before looking at the solutions!
posted by sonofsamiam at 10:39 AM on July 20, 2004


I didn't understand the logic behind it until I looked at this matrix, now it's crystal clear. Neat!
posted by loquax at 10:56 AM on July 20, 2004


You posted this just to spite me, didn't you? Dammit.
posted by Ethereal Bligh at 10:56 AM on July 20, 2004


This is good. Thanks!
posted by jammer at 11:04 AM on July 20, 2004


Psychotic_venom's misunderstanding is illuminating. Huh. I've kept my MHP email all these years, it might be interesting to try to categorize/summarize the different ways that people misunderstand the problem. (In p_v's case, maybe misunderstanding the solution?)

I've always thought the key to successfully explaining it was in making it clear that Monty knows where the prize is, and his action (in opening a losing door) is constrained by this knowledge in a way that forces him to involuntarily tell you something you didn't know when you made your pick. You know something about the door he doesn't open that you don't know about the door you picked. Those two closed doors aren't the same.

Although I'm not sure how many people actually are confused on this point, it's also important to be clear that Monty never opens the door you've initially picked to reveal that it's a losing door.
posted by Ethereal Bligh at 11:15 AM on July 20, 2004


sonofsamiam: ooh - that's my PhD supervisor's book!
posted by Singular at 11:15 AM on July 20, 2004


This problem feels like it's little more than elementary statistics and probability, but it's fascinating to a dork like me. I mean, I absolutely see the logic and the probabilities behind switching, but still it feels like a magic trick everytime I change my mind.

Marvelous.
posted by chicobangs at 11:31 AM on July 20, 2004


Forgive me if this is redundant from any of the linked posts above, I am trying to finally work this thing through...

I have a 1 in 3 chance of picking correctly up front. (33%)

Since Monty knows where the prize is and eliminates one loser door.

And he will never pick my door first.

There is 66% shot that I picked wrong initially, and we also know that one of the two loser doors has just been eliminated. So, by switching my odds of winning increase because I was probably wrong up front, and the only option left is likely to be the winner.....
posted by szg8 at 11:31 AM on July 20, 2004


Since I'm often accused of being a smartass, I'll reveal the limits of my intelligence and admit that I'm still thinking about sonofsamiam's mention of MacKay's version of the problem. My ability to suss out the correct answer has everything to do with my innate intelligence and nothing at all to do with my familiarity with the MHP—since I'm probably as familiar with the MHP as pretty much anyone else, anywhere. I've thinking and writing about it for a decade.

I have two trains of thought on it. And this shouldn't give anything away to anyone. The first is from "constrained action"/information perspective I describe above. This version of the problem is clearly designed to apparently eliminate the notion of Monty revealing information. It seems pretty persuasive, but that makes me suspicious. (The first and a half thought is the single-event objection.) The second—and I'm not through figuring out what I think about this—is to take a step back and think about all the possibilities for doors you've chosen and doors jostled open by earthquakes. This particular chain of events (that the earthquake would jostle open a door you did not pick and a door that is a losing door) is relatively unlikely. Does that mean anything?

Well, in writing this out, I have to say that I don't believe that it does. And this is because I think, for example, the strong (and even the weak, to a degree) anthropic principle is utter nonsense. I don't think you can draw any conclusions on the basis of a single event that is unlikely. So, I don't think it matters which door you pick.

On Preview: szg8, yes, that's a concise explanation of the problem. I think you understand it.
posted by Ethereal Bligh at 11:34 AM on July 20, 2004


As I just posted on Crooked Timber, who also linked to this site today:
Here is my MHP page, transposed to my new domain. Please keep in mind that I wrote this ten years ago, and have hardly changed anything since then. (This feeling that it's now inadequate is why I've wanted to just redo the whole thing from scratch. You know how it is.) And, um, this is from an archive from 2001 I scrounged up—I think there are some uncorrected typos in there.
...this is the io.com/~kmellis/monty page that is referenced (or linked to) on many of the MHP sites, including ddonovan. I think his and mine were the first two web pages on the MHP, both ~1994.
posted by Ethereal Bligh at 11:49 AM on July 20, 2004


Nope, I still don't believe the 2/3rds, 1/3rd solution. And that coloured dial diagram only reinforces my stubbornness.

If I pick door #1 and the prize is behind door #2, that wheel shows Monty can pick either door #3 or... door #3, and counts that as two distinct possibilities. An incorrect initial guess constrains Monty's choice.

It still comes down to me picking one of X doors, someone else knowingly eliminating X-2 incorrect doors, then asking me to pick between the two of them. The initial choice is irrelevant. The game decision comes once the incorrect options are gone. The initial odds may be 1/3, but the basis of that calculation is taken away in the second situation.

And as an irrelevant note, I picked door #1 and stuck with it, even after door #2 was revealed. And where was the prize? Door #1. So obviously I'm right.
posted by GhostintheMachine at 12:11 PM on July 20, 2004


I think my goat is cuter.
posted by perplexed at 12:12 PM on July 20, 2004


I've never been 100% comfortable with this, but like loquax, seeing it in various matrices and tables is greatly clarifying. Thank you!
posted by Songdog at 12:21 PM on July 20, 2004


Here's another way to think about it that I just now made up, so take that for what it's worth:

Let's say you have 100 doors, and you pick door #1. So you have a one in 100 chance that it's that door, and 99 in 100 that it's in another door. Now when Monty opens one of the other doors, he doesn't change the probabilities. What he's effectively saying is, "There's a 99/100 chance that it's in another door—but not this door!" And when he opens another door, he's what's implied is, "There's a 99/100 chance that it's in another door—but not in either of these two doors!"

So after he's opened 98 doors, all with goats behind them, what's implied is, "There's a 99/100 chance that the prize is not in door #1. And since I've opened 98 out of the 99 others, the prize is definitely not behind those doors. So there's a 99/100 chance that the prize is behind the door you didn't pick. Switch, you idiot!"
posted by Khalad at 12:24 PM on July 20, 2004


Ghost—the initial choice isn't irrelevant because the initial choice has only a 1 in 3 chance of being correct. Period. It can't possibly have a 1 in 2 chance of being correct. Monty opening a losing door has no affect on the odds of that first door you chose. Think about it.

If you're sure that it doesn't matter, then you'd need to be arguing that the door Monty doesn't open also has a 1 in 3 chance of hiding the prize.

Does it? No.

It doesn't because:

A) That door cannot be the door you picked; and
B) Monty can't open a winning door.

Two times in three, you'll pick a losing door. That means that of the two remaining doors, one of them is a winner. Monty cannot open that winning door. He knows which door is it, and he cannot open it. He is forced to open the other one—he has no choice in what door to open. Two times in three, Monty is forced to avoid the winning door. One time in three—the one time in three your first choice is correct—Monty can open either one of the losing doors and he tells you nothing. So you can think of the problem this way: a third of the time, your first choice is right but Monty doesn't tell you this and you don't know it. Two thirds of the time, you're wrong, and Monty tells you so by avoiding the door that is correct. You, of course, can't tell if this is the 1 in 3 case or the 2 in 3 case. But a strategy of staying will win, because of this, 1 in 3 trials while a strategy of switching will win, because of this, in 2 of 3 trials.
posted by Ethereal Bligh at 12:29 PM on July 20, 2004 [1 favorite]


Khalad: if Monty did it that way, people'd figure it out much easier. In fact, the winning door should be ostentatiously avoided. Like, you pick a door, and then he starts to open all the other doors in sequence—but someone runs up and stops him from opening one of them. He leaves that door alone and finishes opening the rest of them, excepting yours and the one mysterious door.

It would be clear to anyone, I think, that unless the prize is behind the door the person initially picked (which is very unlikely, but possible), it must be behind the door Monty so obviously avoided.

The real MHP is the same problem, except without 97 of those 99 other doors.
posted by Ethereal Bligh at 12:36 PM on July 20, 2004


Well written, Ethereal Bligh. Point B cannot be stressed enough:

B) Monty can't open a winning door.

No matter what, he will remove a wrong door. The error lies in thinking that once a door is removed ones odds have improved somehow from 1/3 to 1/2, which is clearly impossible.
posted by cohappy at 12:41 PM on July 20, 2004


The explanation that finally made sense of this (for me) starts the same (one in three chance) but then instead of opening a door, Monty offers to let you trade your one door for the remaining two doors. Which is in essence what Monty is doing in the classic problem, with the added bit of distracting information that one door is a loser (since both doors cannot be winners).

Thinking about it with a deck of cards makes it even more obvious. Monty spreads out a deck of cards face down and bets you can't find the ace of spades. After you draw a card, he looks through the rest of the deck and throws away fifty cards. He then says to you, "Either the card you're holding is the ace of spades, or the card I'm holding is the ace of spades. Do you want to trade?"
posted by turaho at 12:50 PM on July 20, 2004


EB -- why is your page no longer up? Do you need someone to host it?
posted by weston at 12:51 PM on July 20, 2004


EB, the initial choice only has an initial 1 in 3 chance of being correct. From that perspective you're correct. But the situation changes the odds.

Put another way, what's the chances of rolling a 4, 5, or 6 on a die after you've just rolled a 4, 5 or 6? You're arguing that it's still 25% (the odds of rolling said sequence before the sequence begins). That's the difference that makes the initial choice irrelevant. You've already satisfied the initial condition and you've moved on to a secondary condition. The odds in the secondary condition are different than the initial situation.

The problem with the problem is the assumption that you're faced with the initial choice all over again. But you're not. You're presented with an entirely different situation. You're suggesting the odds of my first choice being right are only 1 in 3, and that's correct. But the odds of my second choice (should I stay or should I switch) are 1 in 2.

Imagine you're kept in a booth while your friend plays the game. After Monty reveals the guaranteed loser, you and your friend switch places. You have a door. There is one other door to pick from. One of them holds a prize. Do you stay, or do you switch? From this perspective, your odds are the same. Numeric sleight of hand aside, that is the real question of the game. The game doesn't start at the beginning... it starts when you're asked to switch. That's why it doesn't matter what door is picked initially. I can guarantee that one of the other two doors is bogus, and I can guarantee that a bogus door is revealed.

It's an interesting problem, but it's one that has been overtaken by statistical proofs which ignore the reality of the situation.
posted by GhostintheMachine at 1:00 PM on July 20, 2004


I could not wrap my brain around this concept until I read Ethereal Bligh's post. Thanks, EB.
posted by GeekAnimator at 1:06 PM on July 20, 2004


GhostintheMachine, it's one thing to not get the proof, or to be skeptical of the reasoning behind it. But claiming that we're ignoring the reality of the situation? All you have to do is try it. That's exactly what the FPP link does; it lets you try the thing out empirically, and as you can see from the results table, switching is better than staying. You may not agree with the reasoning for why it's better to switch, but you can't argue with the fact that it is indeed better to switch.

Look at it this way: let's say you pick a door. Then Monty blindfolds you, does some stuff you can't see, and then asks: do you want to switch, or stay with the door you picked? Of course, you'd switch.

Are you saying your odds change depending on what Monty does while you're blindfolded? Are you saying that he can do something (besides cheating) to change the fact that you have a 1/3 chance of picking the door correctly on your first try?

As I see it, he can open doors, close them, build new doors, stand on his head, whatever... none of that makes it your initial guess more or less likely.
posted by Khalad at 1:15 PM on July 20, 2004


the fact that you have a 1/3 chance of picking the door correctly on your first try?

Understand, I'm not disputing this point. From the outset, you do only have a 1 in 3 chance. But the outset is not the game itself. The game begins once Monty's revealed a door that does not contain the prize.

We're arguing two different points here. Yes, nothing changes the odds of your initial guess being right. The question really is, what are the odds of choosing the right door of three, once one of the doors has been revealed to be incorrect? What are the odds starting at that point? I'm saying that the problem is in the problem itself.
posted by GhostintheMachine at 1:25 PM on July 20, 2004


Ghost, in your "in a booth" scenario, you're right that it would be 50-50. That's because you (the one who was in the booth) can't differentiate between the door that was chosen first and the door that was left alone. That's why the problem doesn't start with the "stay or switch" choice, it starts with the initial choice from among three doors.

The two situations are not independent of each other (the first choice, and the choice when there's only two doors left). There is a link between which door you choose and which door Monty leaves alone. His choice of what door to leave alone is dependent upon the door you initially choose.

If there were three doors, and you chose one of them. Then Monty opens all three of them, revealing the door that hides the prize (say it's not the one you initially chose), then, um, what are the odds of your winning for staying with your original choice or switching to the winning door? You still have a stay or switch choice. It just happens that Monty's told you where the prize is. It's not fifty-fifty, obviously. It's zero if you stay, certain if you switch.

In the real MHP, Monty doesn't actually tell you this much information. Or, rather, most of the time he does and the rest of the time, he doesn't. You don't know which is which. You do know that the "most of the time he tells you which door is the winning door" is exactly the same times in which you've chosen a losing door. You can be confident that you're more likely in that case than the other, and so if you switch, you're more likely to win than if you don't switch.

On Preview: you wrote: "...once one of the doors has been revealed to be incorrect". But see, it's not just that any one of the doors has been revealed to be incorrect. Specifically, it's never the case that the door you initially picked is revealed to be incorrect. And that makes all the difference.
posted by Ethereal Bligh at 1:29 PM on July 20, 2004


GhostintheMachine: I'm sympathetic to your position because I argued it on a few occasions myself before being won over by the deck of cards analogy. Suppose you start out in the booth and I'm given one chance to draw the ace of spades from a deck of cards. I pick my card and it stays face down. Now, Monty will look at the rest of the deck and flip over 50 cards that are not the ace and leave one face down that may or may not be the ace.
Next, you come out of the booth and are asked to pick the ace of spades by either keeping the card that I picked, or switching to Monty's card.
You may not hear me from the booth but, having watched Monty flip over those 50 cards, I'll be screaming for you to switch.
posted by Zetetics at 1:31 PM on July 20, 2004


The key to understanding for me is that Monty gives you a little bit of information. If he chose the door randomly, and could potentially open the prize door, then you would always have a 1/3 chance of winning. But he doesn't choose randomly, he purposefully avoids the winning door. This is also crucial to understanding the earthquake variant.
posted by sonofsamiam at 1:33 PM on July 20, 2004


The game begins once Monty's revealed a door that does not contain the prize.

That's simply not true. Your point of view assumes that there's a randomization following the revelation of the losing door. There is no such randomization, however: the location of the prize remains fixed, as does the probability that you've selected it.

Let's look at this simplified situation: After you choose a door, Monty offers you the option of either keeping your initial choice, or taking everything that's contained behind both of the other two doors. You'd choose the two doors, right?

This is the same thing, really. Of the two doors you do not initially choose, Monty will always eliminate one concealing a goat. So if there's any prize behind either of those two doors, you'll get that prize by switching.
posted by mr_roboto at 1:39 PM on July 20, 2004


Weston: well, you can see that I threw up an archived version just today after the Crooked Timber post and this MeFi post.

I just got my own domain at the end of last year, and then eventually shut down my old ISP web/shell account at IO. It's a bit of a performance anxiety thing. That page has existed for ten years, was in '96 a Yahoo pick of the week, and has essentially been the main MHP page on the web. It was only recently that I saw that Mathworld cited me.

Anyway, so it's real old, I wrote it a long time ago when I was, um, younger. I've exchanged many, many emails with people about the problem. I've had lots and lots of ideas about how I should recreate the page. And should I give it it's own domain, maybe? Should I write a server-side app to demonstrate it like this person has? A java app? And, really, I should just rewrite the whole thing from scratch. Lots of things make me wince when I read it now.

So, when my old account was turned off, I hesitated in moving the page to the new site. I had planned to Google all links to it and inform the linkers about the new URL.

But I haven't done it. You know how it is. And letting the whole thing disappear into the ether—well, I don't know what that's about. A sort of depressed nihilism, maybe? I'm proud of that page, but also a bit embarassed that it's nothing like what I think it should be. So, performance anxiety induced procrastination?

I was pretty nervous just now about putting the old page up. I guess I fear criticism from the Crooked Timber academics, whom I respect quite a bit, and fear criticism from here, as well. Because, frankly, I'm obviously proud of it. When I look at my name in that list of citations on Mathworld, and it's among Martin Gardner, for example, it freaks me out a little.

That's more information than you, or anyone, could possibly have wanted. Um, sorry.

GeekAnimator: you're very welcome. It makes me very happy to successfully explain the problem to anyone.

On preview: sonofsamiam, aha. That's very interesting. A clue. I was about to lay down for a bit and think about the earthquake scenario.
posted by Ethereal Bligh at 1:41 PM on July 20, 2004


Ah. You can see what the answer is to the earthquake variation if you recast it in the analagous 100 doors version. But I'm still having grappling with developing a true comprehension of what's happening in this variation.
posted by Ethereal Bligh at 2:11 PM on July 20, 2004


While our choice looks the same in both variants, there were different histories which brought us to that choice.

In the original variant, Monty does some computation and informs us of it. In the second variant, we find ourselves in a situation where no such computation or information transfer has occurred. There was a random series of events that brought us to this point, not a "purposeful" series of events, as in the original.

At least, it makes sense to me that way!
posted by sonofsamiam at 2:24 PM on July 20, 2004


Ethereal Bligh, having an earthquake open a door is exactly the same as if Monty had randomly chosen a door and only happened to open one with a goat behind it. It's entirely possible that the earthquake could have opened a door with a prize behind it, and so if we assume that it could have done so with equal probability, we learn nothing from the fact that there's a goat behind door #3 except that there's a goat behind door #3.

Put it another way. The earthquake's "choice" of a door to open isn't dependent on your choice, so it doesn't give you any information about whether your choice is good or not. Monty Hall, on the other hand, bases his choice on your choice, so if he doesn't choose a door randomly then he has given you information about the quality of your choice.,
posted by Khalad at 2:39 PM on July 20, 2004


perplexed, that is, admittedly, a very cute goat.
posted by chicobangs at 2:42 PM on July 20, 2004


Yeah, that was my first thought, as explained previously. Then the 100 door version made me think differently. Now I realise that I was right to begin with. Heh. You have to ask: what would it mean if an earthquake knocked open 98 losing doors out of a hundred...and one of the two unopened doors was the door you initially picked? What does that mean? I'm thinking that it doesn't mean anything. But I don't completely grok it yet.

On preview: I want to say that the earthquake can't tell you anything because it doesn't "know" anything—which seems very correct. But then, um, I think about if the earthquake opened the winning door. Well, gee, in that case it's told you a lot. :)

I'm still wandering around, partially understanding the variant from different perspectives. Different arguments each seem compelling. I'm not totally convinced of any one of them because I don't truly understand the problem yet. It's a nice recapitulation of what it's like to try to understand the MHP the first time. And also why, in my opinion, so many people think they understand it when they really don't. They settle on the first explanation that makes good sense, not recognizing that they really don't grasp the problem, yet.
posted by Ethereal Bligh at 2:46 PM on July 20, 2004


I want to say that the earthquake can't tell you anything because it doesn't "know" anything—which seems very correct. But then, um, I think about if the earthquake opened the winning door. Well, gee, in that case it's told you a lot.

It's true that a real earthquake wouldn't contain any information about the state of the system. In the earthquake variation of the MHP, the "extra information" (analogous to Monty's knowledge of the location of the prize) is embedded in the statement of the problem itself. In sonofsamiam's problem statement, "one of the three doors flies open. It happens to be door 3, and it happens not to have the prize behind it." The apparent randomness of the earthquake is belied by this statement of happenstance. In fact, it is not a random earthquake, or an earthquake with random results: it is an earthquake that opens a door without a prize behind it.

By constraining the earthquake in this manner, the problem statement embeds within the earthquake information about the location of the prize.

I think. IANAIT (I am not an information theorist).
posted by mr_roboto at 3:01 PM on July 20, 2004 [1 favorite]


Nice statement of that point of view, mr_roboto. I'm sorta of flipping back and forth between that and the other, in a metaphorical figure/ground perceptual sort of way.
posted by Ethereal Bligh at 3:10 PM on July 20, 2004


Okay, I've got another explanation. Let's say every time a person is on Monty Hall, they pick a door. Then an earthquake occurs that opens 98 of the 99 remaining doors, completely randomly.

Now, 98 out of 99 times, the earthquake is going to reveal where the prize is, giving you a guaranteed win by switching. Or, equivalently, it made your initial choice a loser. Now your initial choice had only a 1 in 100 chance of being correct, so the earthquake actually hasn't told you very much since your chances went from 1/100 to 0/100, not a very big drop.

However, in the very, very unlikely chance the the earthquake doesn't reveal the prize, it's really given you a lot of information. It's given you enough information, in fact, to bump up your chances of being right from 1/100 to 1/2.

Here's how you can convince yourself that the information-giving is a correct analysis of the situation: 98 out of 99 times, your probability of being correct goes from 1/100 to 0/100. 98 * (1/100 - 0/100) is 49/50. However, 1 out of 99 times, your probability of being correct goes from 1/100 to 1/2. 1 * (1/100 - 1/2) is -49/50.

Notice how the numbers cancel out. This is crucial: most of the time the earthquake will reveal you were wrong and should switch. This is not big news, since you already knew you were highly likely to be wrong. In the fluke case, it gives you the opposite information in one big lump sum: "You very well could be right," says the earthquake. "I randomly opened 98 doors, and purely by chance, none of them had the prize. In this case, your odds of having chosen right are much better than 1%—they're 50%."

In the regular Monty Hall formulation, in which he deliberately chooses a bad door, those 98 out of 99 cases never occur. It's impossible for Monty to open a door with a prize behind it. That means there's nothing to balance out the one remaining case, where he opens a door with a goat behind it. Since there's nothing to balance out that case, his revealing a goat gives you no new information. Since you have no new information, you still have a 1/100 chance if you stay and a 99/100 chance if you switch.

With this explanation, it's clear to me that I was wrong to say before that Monty gives you information. In fact, the key to the problem is that he doesn't give you any new information. When you receive information, probabilities can change. If you receive no new information, the probabilities cannot change.

The earthquake, unlike Monty, does give you new information. Sometimes it reveals the prize, sometimes it does not. Monty, on the other hand, never reveals the prize. You knew he was going to show you a goat. He always reveals a goat. You learn nothing new from his revealing a goat. And if you learn nothing, your odds don't change.
posted by Khalad at 4:05 PM on July 20, 2004


I'm pretty sure mr_roboto has the right idea here. Pardon my use of equations, but, well, IAAIT.

Let X = random variable representing location of prize
Let Y = random variable representing player's initial guess
Let Z = random variable representing door revealed by earthquake
Let X* = random variable representing player's final choice

The key is to figure out what problem we're trying to solve. A naive answer is that we want to choose a strategy to maximize Pr(X* = x | X = x), the probability that we choose the correct door.

But that's not what is asked for. What we want is to maximize the conditional probability Pr(X* = x | X = x, Z != x, Z != Y), the probability that we choose the correct door given that the earthquake chooses an incorrect door that differs from our first guess.

After writing down the proper problem statement, you can decompose by the total probability theorem:
Pr(X* = x | X = x, Z != x, Z != Y) = Pr(X* = x | X = x, Y = x, Z != x) Pr(Y = x) + Pr(X* = x | X = x, Y != x, Z != x, Z != Y) Pr(Y != x), where Pr(Y = x) = 1/3 and Pr(Y != x) = 2/3. As in the original MHP, the "switch doors" strategy maximizes the sum, yielding a 2/3 probability of success.
posted by Galvatron at 6:34 PM on July 20, 2004


This is a splendid problem, and I am glad to see the spirit of philosophical inquiry alive at Metafilter. (/adjusts toga)

I have 2 boxes. One contains nothing. The other contains either $1000 or $1,000,000. You, the very next poster to complete this sentence,....
posted by crunchburger at 7:15 PM on July 20, 2004


will choose a box.
posted by crunchburger at 7:17 PM on July 20, 2004


I will give you a a choice of box - but , I always already know which box you will pick.
posted by crunchburger at 7:19 PM on July 20, 2004


let's make a deal
posted by crunchburger at 7:22 PM on July 20, 2004


For the earthquake situation, with 100 doors:

If the earthquake happens to open the door with the car in it, the earthquake did inded reveal lots of information. However, it is impossible to win in this situation, because you don't have the option to switch to one of the open doors. You can only switch to the one other closed door, which you know does not hold the car.

when you pick your door, there is a 99% chance the car is in one of the other doors. There are 99 possible combinations of doors the earthquake can knock down, and in 98 of those situations the earthquake will reveal the car and you lose. so thats (99 / 100) * (98 / 99) = 98% chance you will lose because the earthquake opened the door with the car.

If you picked the wrong door initially (99 times out of 100), and the earthquake failed to reveal the car, 1 time out of 99 (different ways the remaining doors can fall) the final remaining door will indeed contain the car. But that's (99/100) * (1 / 99) = 1% chance.

The last situation is that you picked the correct door initially, which is 1 / 100 = 1%.

So if you find yourself in a situation where the earthquake failed to open the door with the car (100 - 98 = 2% chance), there is an equal chance that the door you chose and the remaining door contain the car (1%).
posted by mfbridges at 7:30 PM on July 20, 2004


To put the earthquake situation into the card analogy, lets say I shuffle a deck of cards, and you pick one. Then I pick one *randomly* from the remaining cards. Do you switch? Both selections were random, and have the same probability of being the ace of spades.
posted by mfbridges at 7:37 PM on July 20, 2004


crunchburger: Newcomb's paradox is interesting, but my response to it has always been that its notion of foreknowledge is rationally incompatible with its notion of choice. There is a response to this, however. The CT folks had a nice discussion about this problem.

I'm still uncertain on the earthquake problem, but my thinking before I nodded off to sleep was mfbridges's, not Galvatron and Khalad's.
posted by Ethereal Bligh at 7:42 PM on July 20, 2004


re: the original mh scenario, does it make any difference if you randomly determine (by flipping a coin say) whether or not to switch compared to if you just switch every time?
posted by juv3nal at 7:49 PM on July 20, 2004


Criminy, you're making this more difficult than it has to be.

If the goody is behind A, then:
If you pick A, Monty shows you B or C, and you should stay.
If you pick B, Monty shows you C, and you should switch.
If you pick C, Monty shows you B, and you should switch.

If the goody is behind B, then:
If you pick A, Monty shows you C, and you should switch.
If you pick B, Monty shows you A or C, and you should stay.
If you pick C, Monty shows you A, and you should switch.

If the goody is behind C, then:
If you pick A, Monty shows you B, and you should switch.
If you pick B, Monty shows you A, and you should switch.
If you pick C, Monty shows you A or B, and you should stay.

Thems is all the possibilities there are. The probabilities of it being behind A, B, or C are equal, so you just need to add up instances.

2 times out of 3, you're better off switching.

(see? solving things through brute force can be fun)
posted by ROU_Xenophobe at 7:55 PM on July 20, 2004


juv3nal, suppose you choose to switch with probability p (between 0 and 1) and choose not to switch with probability 1-p. Then your probability of success is p*2/3 + (1-p)*1/3, which is a value between 1/3 and 2/3. So randomly switching is a better strategy than never switching, but it's not as good a strategy as always switching. This can be shown using the total probability theorem as in my previous post.
posted by Galvatron at 7:59 PM on July 20, 2004


Here's yet another way of explaining it. There are two groups of doors, the one you picked and the group you didn't pick. There is always a 1/3 chance that the prize is behind the door you initially picked. And there is always a 2/3 chance that the prize door is in the group you didn't pick. After Monty shows you a losing door there is only one door in the group you didn't pick. The remaining door now carries the full 2/3 probability of the group so you should switch.

Following this reasoning the answer is the question is the same whether Monty Haul or an earthquake opens the losing door.
posted by euphorb at 8:23 PM on July 20, 2004


mfbridges, you have chosen an earthquake that behaves rather strangely, because it will never knock down the door you selected. Consider the following disjoint events that represent all possible outcomes:

A: you choose correctly, and earthquake does not reveal prize
B: you choose incorrectly, and earthquake reveals prize
C: you choose correctly, and earthquake does not reveal prize

As you say, for this rather odd earthquake P(A) = 0.01, P(B) = 0.98 and P(C) = 0.01; since A and C are equally probable, it doesn't matter whether you switch or not.

Now consider a more rational earthquake that randomly opens 98 of 100 doors, possibly even the door you selected, leading to a different set of possible outcomes:

A': you choose correctly, and earthquake does not reveal prize
B': you choose correctly, and earthquake reveals prize
C': you choose incorrectly, and earthquake does not reveal prize
D': you choose incorrectly, and earthquake reveals prize

P(A') = (1/100) * (2/100) = 2/10000
P(B') = (1/100) * (1 - 2/100) = 98/10000
P(C') = (99/100) * (2/100) = 198/10000
P(D') = (99/100) * (1 - 2/100) = 9702/10000

If your choice is between events A' and C', clearly you're much better off (99/100) going with C' and switching doors. The earthquake you described imparts an unnaturally large probability to event A.

From a conditional probability point of view,
Pr(final choice is correct | earthquake did not reveal prize) = Pr(final choice is correct | earthquake did not reveal prize and initial choice was incorrect) * Pr(initial choice was incorrect) + Pr(final choice is correct | earthquake did not reveal prize and initial choice was correct) * Pr(initial choice was correct) = Pr(you decide to switch) * (99/100) + (1-Pr(you decide to switch)) * (1/100); therefore, your best strategy is to always switch, giving you 99/100 odds of winning the prize.
posted by Galvatron at 9:32 PM on July 20, 2004


Sorry, event C should read:
C: you choose incorrectly, and earthquake does not reveal prize
posted by Galvatron at 9:43 PM on July 20, 2004


Speaking of paradoxes (well, in the Alanis sense of misusing a word), I read this post earlier today then came home this evening to finish _The Curious Incident of the Dog in the Nighttime_. I was surprised to find this exact problem cited at length by the autistic, math-loving narrator of the book (on page 62 of the Doubleday Canada edition.)

It even seemed to have very similar wording to examples on one of the pages linked on this thread though I can't for the life of me find it now. Possible plagarism or just strange coincidence? Which, come to think of it, is a better word to describe my experience today...
posted by Jaybo at 1:47 AM on July 21, 2004


Some people who do not accept the proof seem to be commiting this (or it's inverse)
posted by ed\26h at 3:00 AM on July 21, 2004


Continue discussing the statistical proofs all you wish, you're missing my point.

There is something called "the Monty Hall Paradox", which has very specific wording, that given a specific starting point (prior to the initial choice), the statistically better decision at the secondary point (after Monty's reveal) is to switch. That can be mathematically proven.

But when you look at the problem in reverse, the situation changes. You're faced with a choice between two doors, one which has a prize and one which does not. One of them is "yours", but you have the option to switch. What has gone on before does not matter. Call it the logical solution as opposed to the statistical one.

Let's see if I can explain it another way. You're suggesting the base chance of success is 1/3 (you're only picking one door of three). But that's not true. You know, up front, the whole situation. You know one (or all but one, in expanded versions) of the other answers will be eliminated. So you start with two of the three doors on your side. Since there's only one prize, you know one of the two you have is incorrect, and it's shown to be incorrect. Should you now trade the two doors you have for the third, unknown? There are two doors in your favour right from the start - the one you pick, and the one wrong door Monty reveals. Does knowing which one of the two in your favour is definitely incorrect change your odds? No, because at least one has to be incorrect.

The third door is a ruse. It's a distraction. It's never a factor in the equation, and although there are ways of mathematically including it, that's based upon the false assumption that there's a reason to consider it. I understand the math involved (or at least, I understand it enough to accept it as mathematically correct), but this isn't a question of mathematics or statistics. In the same way the third door distracts the player, the numbers distract the mathematician.

Forget the third door. It doesn't exist, and it has never existed. You have a door, and Monty has another. One of them holds a prize. That is all you need to consider.
posted by GhostintheMachine at 5:46 AM on July 21, 2004


Ah youth! I remember when I thought my decisions didn't matter either.
posted by wobh at 6:31 AM on July 21, 2004


GhostintheMachine is actually pretty representative of a certain personality type's response to the MHP. I find this type to be fascinating, though often maddening. It's closely related to, for example, young Earth creationism. No amount of argument, no amount of evidence to the contrary, this type of person is absolutely sure if his conclusion because, well, they're sure of it. They feel it.

In the case of the MHP, I've come to believe that this level of certainty reflects a deep misunderstanding of the problem whereby they are correctly understanding a different problem. This is why they are adamant, and, in a way, justifiably so. This part I can understand.

The part I can't understand is why they don't test their judgment empirically. That's what's baffled me for a long time. I suppose it's because I can't imagine being so sure of oneself, so without doubt, that one wouldn't verify one's judgment in the face of presentation of journal citations, simulations, etc. As I write on my MHP page, the very first thing I did, after assuring myself that the answer must be 50-50, was write a computer program to simulate the problem and verify my reasoning. The program showed that it was better to switch. That was a surprise. Then I sat back and tried to understand where I had gone wrong.

I've been fascinated with the MHP for ten years because I think it exemplifies how we understand and, more often, misunderstand the world around us. It exemplifies that "understanding" is also a social construction, something negotiated. It's also an emotional thing, burdened by pride and shame. It shows how we can be right about a conclusion even though our reasoning is faulty and wrong about a conclusion even though our reasoning is correct. It also shows how many misunderstandings are misapplications of learned rules where they don't apply.

My search for the most effective pedagogical approach to successfully explaining the MHP is, I suppose, a quixotic attempt to defeat irrationality itself. Which is, of course, deeply ironic. Excuse me, I have a windmill to slay.
posted by Ethereal Bligh at 8:51 AM on July 21, 2004 [1 favorite]


As I write on my MHP page, the very first thing I did, after assuring myself that the answer must be 50-50, was write a computer program to simulate the problem and verify my reasoning.

...and right there you're STILL missing what I'm saying.

In order to write that computer program, you'd have to accept the inherent assumptions of the question itself.

Switching is the best answer to the MHP. That's a given, and not something I'm disputing. But the MHP is a blind. It requires assumptions be made, and one of them is that the first decision is important. I'm disputing that.

The second choice is always presented as "what are your chances your initial guess was right?", and in that context the better option is to switch. But that's the sleight of hand. You don't "keep" the door. It's not your door. There is always a 50-50 chance you'll be right on your second guess, because it's not connected to your first guess. Monty revealing one option has changed the initial conditions. He's modified the conditions of the experiment. The MHP requires the assumption that the conditions are unchanged, when they're fundamentally altered.

You have to throw out the results of the first experiment (the one in three proposition) and start afresh with the second experiment (two unknown doors, one known). That makes the second choice "knowing what you know now, which door do you think holds the prize?". There are only two doors. Are you seriously suggesting that, UNDER THAT NEW SET OF CONDITIONS, one door has a greater than even chance? No, you admitted as much (far, far) above.

Philosophy students love to bring arguments back to our inability to prove anything external exists. It's a wonderful theoretical exercise. But it's also completely unimportant.

And windmills can drag you into the mud, or lift you to the stars, remember. That's the part most people forget, and the true lesson to be learned.
posted by GhostintheMachine at 10:44 AM on July 21, 2004


Continue discussing the statistical proofs all you wish, you're missing my point.

GhostintheMachine, I'm having a very hard time seeing your point; it's like you're speaking a different language. I'm an engineer. From the engineering perspective, performance is what matters--and if you and I have a 100-round MHP contest, my strategy will sure as hell outperform your random selection. Yes, yes, with some confidence interval that I'm too lazy to compute.

On preview:
You have to throw out the results of the first experiment (the one in three proposition) and start afresh with the second experiment (two unknown doors, one known).

Why? That's an entirely different, and much less interesting, problem. The MHP has a history that affects the likelihood of the outcomes.

Maybe you would feel better with this restatement:
There are three doors, one of which contains a prize. A robot comes and randomly stands next to one of the doors. Monty opens another door, revealing that it contains no prize. Do you want to open the door with the robot, or the door without the robot?

Now your first guess has been removed from the problem... but I'm sure as hell picking the door without the robot.
posted by Galvatron at 11:06 AM on July 21, 2004


GhostintheMachine, if someone walks into the room just as Monty opens a losing door, then they can't tell the difference between the door initially picked and the one left closed. Since, for them, the two doors are undifferentiated, it wouldn't matter which they choose. But you, who knows which door was initially picked and which door was left closed by Monty, can tell the difference. And you know that the door Monty leaves alone is twice as likely to contain the prize as the door you initially picked.

I think I already wrote this, but here's another variation: say that Monty asks you to pick from among three doors. You pick a door. He then open all three doors, revealing where the prize is, and then closes all three doors. Say that you saw the prize was behind a different door than you picked. Should you switch or should you stay?

You should switch. You can't say that this is a new problem in the sense that the initial choice is irrelevant, because the whole notion of "switching or staying" is dependent upon that prior event (your initial choice). So, in this variation, as far as you could tell, before Monty opens all three doors, your choice had a one in three chance of being correct. When he opens all three doors, you discover that your initial choice now has zero chance of being correct, and switching to the door with the prize is a certain win. You can't disentangle the initial choice of door from the "second" stage of the problem, because the whole problem is about differentiating between that door and the other door(s).

Just so the Monty Hall Problem.

The two unopened doors are not equally likely to contain the prize.

Just because there's two closed doors doesn't mean that they're equally likely to hide the prize. You know something that differentiates the two doors in terms of how likely each is to hide the prize.

When you talked about rolling a die a second time, someone corrected you and pointed out that, for example, the second die roll (in this analogy) is not like a brand new die roll. For one thing, the number you rolled on the first roll is now removed from the die. So, just right there, the two rolls are different and the second roll is different in a particular way dependent upon the outcome of the first roll. Also for the same reason, of the remaining numbers possible to roll, the die becomes, metaphorically weighted to favor one of the numbers. The two events are connected. They are connected. The two events are connected. The configuration at the beginning of the second event is dependent upon the particular outcome of the first event.

Monty knows where the prize is. You select a door. The combination of Monty's knowing where the prize actually is and his inability to also select the door you selected and the requirement than he open only a losing door, mean that his actions are very constrained by yours. Your action which was your initial choice.

Thinking about the one in three of the initial choice is confusing you. That value is correct—but it's confusing you so stop thinking about it. Just know that because Monty is constrained by your choice and the rules of the game, that of the two doors left unopened (your choice and one other), the other door has twice the likliehood of hiding the prize than your choice.

Where you are also possible going astray is misapplying what you understand about the "gambler's fallacy". In this case, these two events are not independent. The conditions of one are depedent upon the outcome of the first. They are a chain of connected events.

What trips people up is the insistence on looking at two doors and saying, well, the prize could be behind either one of them so it's 50-50. Which is silly, really. Because if you know where the prize is, then of two unopened doors it's 0-100. Probability is all about information. Someone walking into the studio after the initial choice is made, who can't tell the difference between the doors—to them, it's 50-50. To you, who knows the difference but doesn't know exactly where the prize is, it's 33-66. And to Monty, who does know exactly where the prize is, it's 100-0 (or 0-100). Yet it's the same two doors, with the same single prize, at the same moment in time. Probability is about what you know and what you don't know.
posted by Ethereal Bligh at 12:18 PM on July 21, 2004 [2 favorites]


That's an entirely different, and much less interesting, problem.

Thank you. This, oddly enough, explains my point far more than all my rambling.

There's the MHP. It's an interesting problem, with an odd and counter-intuitive solution. But it relies on a strict reading of the problem. Take that strict reading away, and it's a much more mundane problem, with a straight 50-50 chance.

Kind of like this thread. It's far less interesting than it appears, based upon the amount of attention it has received.
posted by GhostintheMachine at 1:35 PM on July 21, 2004


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