# Behind Door Number One...April 8, 2008 6:10 PM   Subscribe

The Monty Hall Problem has struck again, and this time it’s not merely embarrassing mathematicians. If the calculations of a Yale economist are correct, there’s a sneaky logical fallacy in some of the most famous experiments in psychology." The NY Times' John Tierney reports on new research into cognitive dissonance as examined through the famous Monty Hall Problem. [A previous MetaFilter thread about the Monty Hall Problem: Let's Make A Deal!]
posted by amyms (119 comments total) 37 users marked this as a favorite

Interesting, a very impulsive read
Suppose instead we expand the number of doors to 52 and play it using a deck of cards. Your goal is to pick the ace of hearts from the deck in Monty’s hand. You start by picking one card from the deck, but you don’t get to look at it. You leave it face down on your side of the table. Then Monty looks at the 51 cards remaining in the deck and deliberately picks out 50 of them that are not the ace of hearts. He lays down these 50 cards so you can see them. Then he gives you a choice: Do you want to stick with the original card you chose, or switch to the one card in his hand that he hasn’t yet shown you?
So If I pick at random and use frequency conception of probability, I have 1/52 chance of picking that ace.

I take one, so now the set is 51.

50 cards are shown to me (?) that are not the ace I am looking for

I am left with 2 unknown cards, the one I have and the one not know to me. Isn't it obvious I have 1/2 prob of picking the right one, because the unknow set has been reduced from 51 to 2 ? Or am I missing something ?

posted by elpapacito at 6:22 PM on April 8, 2008

Very interesting—thanks for the post!
posted by languagehat at 6:26 PM on April 8, 2008

Yes but your chances of picking the right card initially are very small. Which means, therefore, that the chances of Monty holding the right card in his hand right now are very high -- much higher, in fact, than the chance that you picked the right one on your first try.
posted by Avenger at 6:27 PM on April 8, 2008 [7 favorites]

elpapacito, you might want to read that earlier thread—it's as good an education as you're going to get in the Monty Hall Problem.

*misses EB*
posted by languagehat at 6:27 PM on April 8, 2008

Isn't it obvious I have 1/2 prob of picking the right one

Umm, unless I'm missing something I think you would ALWAYS pick the card they leave face down because basically you're left with your card (with odds of 1/52) and their card (odds of 51/52). No? I am horrible at math by the way.
posted by rooftop secrets at 6:31 PM on April 8, 2008

"... Play enough rounds and the best strategy will become clear: You should switch doors.

This answer goes against our intuition that, with two unopened doors left, the odds are 50-50 that the car is behind one of them. But when you stick with Door 1, you’ll win only if your original choice was correct, which happens only 1 in 3 times on average. If you switch, you’ll win whenever your original choice was wrong, which happens 2 out of 3 times.
"
posted by flatluigi at 6:34 PM on April 8, 2008 [1 favorite]

Even Marilyn vos Savant (aka "The World's Smartest Woman") had trouble explaining the problem to people. Well, she didn't have trouble explaining it, it's just that people had trouble understanding it.
posted by amyms at 6:35 PM on April 8, 2008

*ReadsTFA* Oh. You can redact that comment for being redundant
posted by rooftop secrets at 6:37 PM on April 8, 2008

I think the key part is remembering the INITIAL likelihood that you've picked the right door/card. It is 50-50 if you have 2 cards/doors. The issue is that it is far more likely that your original choice was wrong.

Imagine a 1000000 doors instead of 52/3. The chance of picking the right one are 1/1000000 . The chances that the remaining one is better are 999999/1000000, precisely because it was left.
posted by olya at 6:40 PM on April 8, 2008

What with the swiftly coming collapse, you'll wish you picked the goat. You can get milk or meat from a goat.
posted by TheOnlyCoolTim at 6:40 PM on April 8, 2008 [9 favorites]

As languagehat said above, reading the previous post that I linked to is a great education about the problem, and on other people's perceptions of it.
posted by amyms at 6:40 PM on April 8, 2008

Even though I feel like I understand why switching always makes sense, this problem always gave me a little bit of the heebie-jeebies. I feel like someday, there's going to be some chaotic quantum wave collapse that destroys the universe with an ever expanding wave of zombie cats.
posted by BrotherCaine at 6:42 PM on April 8, 2008 [2 favorites]

The NYT comments section (scroll down body of post to see it) is full of excellent questions and answers.
Clever find, clever article!
posted by Dizzy at 6:50 PM on April 8, 2008

I love, love, love breaking people's minds with this one. Very, very smart people consistently get it wrong, and when they get it the twitching is worth every second spent explaining it. Humans are terrible at probability measurements.

If you don't believe it, write out all the possible outcomes:
Win -> switch -> Lose
Lose -> switch -> Win
Lose -> switch -> Win
vs.
Win -> stay -> Win
Lose -> stay -> Lose
Lose -> stay -> Lose

posted by Skorgu at 6:55 PM on April 8, 2008 [6 favorites]

Oops, I meant to link to Marilyn vos Savant's official page devoted to the problem (and her readers' arguments) when I mentioned her earlier.
posted by amyms at 6:58 PM on April 8, 2008

Remember, though, the real Monty Hall would cheat you if you tried this shit with him.
posted by Mr. President Dr. Steve Elvis America at 7:01 PM on April 8, 2008

It sounds like this might be relevant to any kind of research in which people are asked the same question more than once, and inconsistent answers are thrown out- a standard technique in opinion polling, I believe.
posted by gsteff at 7:04 PM on April 8, 2008 [1 favorite]

See, the trick is that the goats and car rest on a complicated system of silent, high-speed conveyor belts...

BTW, if you ever needed greater proof that Yahoo! Answers is mostly populated by people that only pretend to know what they're talking about: Link
posted by hjo3 at 7:14 PM on April 8, 2008 [5 favorites]

metafilter: monkeys choosing m&m's
posted by quonsar at 7:15 PM on April 8, 2008 [1 favorite]

This feels to me like one of those things that's not obvious at first, but once explained(Monty's choosing gives you information to guide your choice), then it's stunningly obvious.

If he's right about how people were excluded from the choice studies, then it's similarly obvious that the studies contain an error, and no at length analysis is required.

What am I missing?
posted by Mr. Gunn at 7:15 PM on April 8, 2008

Ooh I think I figured it ....

I am told in the MH problem that there is a car behind one door while GOATS lurk behind the other two.

If I see a goat, I think there is now a goat and car , because I saw one goat and I conclude we must have one car and one goat left, so the odds are of picking car are 1/2.
But the distribution didn't change at all. Given that I am told the car is behind ONE door, and I have THREE doors, therefore it follows there's a probability of 1/3 of picking the car door ; if I immediately pick the right door, I win and game is over.

If I don't, the game doesn't change (I only know there's a goat behind one door) , my _perception_ of it does. Indeed my initial doorchoice is accurate 1/3 of times , and wrong 2/3 of times (because I can only pick one ok door out of three) ; but if I am wrong 2/3 of the times, then it only follows I should change door, because I am right only 1/3 of the times.
posted by elpapacito at 7:21 PM on April 8, 2008

Of course the plane will take off.
posted by maxwelton at 7:22 PM on April 8, 2008 [9 favorites]

amyms> Even Marilyn vos Savant (aka "The World's Smartest Woman") had trouble explaining the problem to people. Well, she didn't have trouble explaining it, it's just that people had trouble understanding it.

Actually, she did have trouble explaining it. What she failed to mention the first time is that her result depends on Monty Hall always opening a door with a goat behind it and offering the contestant the choice to switch to the other unopened door. Then again, maybe I'm just "suffering from a Smartest Babe burr in his smarty pants" like Herb Wiskit apparently is.
posted by UrineSoakedRube at 7:24 PM on April 8, 2008

So Monty Hall did this in every single episode of Let's Make a Deal?

Lucky the internet wasn't around then, or else the odds would always have been stacked in the contestants' favour.

That is, assuming they didn't get their strategy from Yahoo Answers.

Assuming only two possibilities (right answer & wrong answer) if you ask a question of Yahoo Answers, then you present yourself with a choice of either sticking with the Yahoo answer, or switching to the other possible answer, you must be 66% better off by switching, right? Right?
posted by UbuRoivas at 7:27 PM on April 8, 2008 [2 favorites]

BTW, if you ever needed greater proof that Yahoo! Answers is mostly populated by people that only pretend to know what they're talking about: Link
Geeze. All my suspicions are confirmed. Thanks for reminding me why anonymous voting is NOT a good indicator of correctness.
posted by verb at 7:27 PM on April 8, 2008

That's an interesting link, UrineSoakedRube, thanks.
posted by amyms at 7:32 PM on April 8, 2008

This answer goes against our intuition that, with two unopened doors left, the odds are 50-50 that the car is behind one of them. But when you stick with Door 1, you’ll win only if your original choice was correct, which happens only 1 in 3 times on average. If you switch, you’ll win whenever your original choice was wrong, which happens 2 out of 3 times."

That's the clearest explanation I've read yet.
posted by Bookhouse at 7:32 PM on April 8, 2008 [7 favorites]

I have an intellectual crush on Laurie Santos.

Apparently this type of poor Monty-Hall-style reasoning infects "some of the most famous experiments in psychology." Which ones?
posted by painquale at 7:35 PM on April 8, 2008

Actually, she did have trouble explaining it.

This is how I remember it when the question was raised in her column lo those many years ago (I was in high school. No, you may not call me grampa). It's wasn't until I worked it out on paper and came to the realization that Monty Hall will chose a different door depending on the one you picked and where the prize is that the problem made sense. She never raised that obvious point, nor did she use the "deck of cards" analogy, which is the best demonstration.
posted by deanc at 7:39 PM on April 8, 2008 [2 favorites]

I highly recommend that folks, once they stop thinking about the Monty Hall problem, read (at least the introduction to) the working paper by Chen. The introduction is much easier to understand than the convoluted reasoning in the NYT article (which seems more aimed to tie the research to the Monty Hall problem than to explain Chen's argument).
posted by ssg at 7:45 PM on April 8, 2008 [1 favorite]

I'm a moron. I actually just signed up for Yahoo Answers, so strong was my desire to correct people who are wrong. But I guess you can't answer a question that's already been "solved." Can we have that as a feature in AskMe?
posted by roll truck roll at 7:45 PM on April 8, 2008

The Yahoo! Answer answer that hjo3 posted is instructive because it highlights precisely where everyone goes wrong, namely in thinking that the problem resets to 50-50 odds after Monty reveals a goat. This answer is wrong for a variety of reasons, but the intuitive reason it is wrong is because Monty revealing the goat you didn't provides valuable information that is being ignored when you reset the problem.

There are n doors. 1 of them has the prize. You get to pick one, leaving (n-1) doors unknown. There are only two sets of doors now: first set: the lonely single door you picked initially, and the second set of all the doors you didn't pick initially, i.e. the rest of the doors. In the first set, there is only one door. In the second set, there are n-1 doors. Monty then asks the question "Do you want to switch" but he's effectively asking you "Is the prize behind a door in the first set of doors, or a door in the second set?" Obviously, the chances that it's in the second set is much greater, because there's only one door in the first set, and all the other doors in the second.

In other words, if the prize is behind any of the doors in the second set, then you win by switching. The odds are obviously much higher that the prize is behind n-1 doors than just 1 door.

Monty opening the doors does nothing but confuse you. Revealing the no-prize doors in the second set really just distills the second set down to a single high probability choice, instead of n-1 different choices. So when Monty asks, "Do you want to switch?" he's giving you a binary choice. Yes or no. First set or second set (with the crap now removed from the second set). There are only two choices, but their probabilities aren't the same. The first set has probability of success of 1/n. The second set has probability of n-1/n, which is far higher.
posted by Pastabagel at 7:47 PM on April 8, 2008 [1 favorite]

amyms> That's an interesting link, UrineSoakedRube, thanks.

You're welcome. I should note that a lot of people made the same initial assumption she did and still got the answer wrong. But she pulls a fast one when she says
So let's look at it [the Monty Hall problem] again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There's no way he can always open a losing door by chance!) Anything else is a different question. [emphasis mine]
in her initial posing of the problem:
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

Craig F. Whitaker
Columbia, Maryland
(Hat tip to Herb Wiskit, who pointed out her error) Well, that is a different question, and it is a different question than the one she answered, and answered incorrectly.

I get the distinct impression that she got some letters pointing out that she had gotten the problem wrong because she made the unstated assumption that the host always opens a door with a goat and then offers to let the contestant switch. She should have just admitted that she goofed instead of printing dozens of angry letters from people who thought she messed up the math.
posted by UrineSoakedRube at 7:54 PM on April 8, 2008 [1 favorite]

Monty opening the doors does nothing but confuse you. Revealing the no-prize doors in the second set really just distills the second set down to a single high probability choice, instead of n-1 different choices.
That's interesting, because it also highlights the fact that the more 'doors' there are, the more compelling the case to switch.
posted by verb at 8:05 PM on April 8, 2008

So Monty Hall did this in every single episode of Let's Make a Deal? Lucky the internet wasn't around then, or else the odds would always have been stacked in the contestants' favour.

I don't think Monty Hall actually did do this every time, which renders it a completely different problem as pointed out. Without knowing the mechanics of how and if Monty Hall decided when to offer the contestants the choice of switching we can't draw a conclusion on whether switching would have been advantageous to contestants. My hunch is that, in order to make the show as exciting as possible, there was no particular pattern. They probably offered the switch only sometimes, and of those times the contestant probably had picked correctly half the time and incorrectly half the time to avoid just this sort of advantage.
posted by Justinian at 8:23 PM on April 8, 2008

On the other hand, it's not like other game shows (like that NO WHAMMIES one) haven't screwed up and let players figure out a pattern resulting in breaking the bank.
posted by Justinian at 8:24 PM on April 8, 2008

Here's my explanation of the Monty Hall problem.

Assuming a goat, a car, and another goat, hidden behind three curtains at a game show, as I would randomly pick the car 1/3 of the time overall, if Monty shows me a goat behind another door, then by switching, I will always be wrong (but only by 1/3 of the time overall). However, when I pick a goat 2/3rds of the time overall, and Monty then only shows me the other goat each time, then if I switch each time, I will get the car 2/3rds of the time overall. Therefore, I should switch each time to be safest.
posted by Brian B. at 8:26 PM on April 8, 2008 [1 favorite]

I applaud the NYTimes' Flash designer. That interactive app with the doors, goats and cars -- and the explanation -- is probably the best intuitive explanation of this problem I've seen yet.
posted by dylan20 at 8:57 PM on April 8, 2008

Fwiw, here's my articulation of it . . .

What are the chances that you picked correctly the first time? 1/3. So when Monty Hall shows you the goat, what are the chances that you picked the right one, and that he's showing you one of the two remaining goats? Still 1/3. Nothing has changed, it's just another way of framing the same question. So, it's a still a 1/3 chance that you picked the right one -- that is, it's still a 1/3 chance that you picked the right one.

But the revelation of the goat makes you feel -- at least at first glance, and especially under pressure to make a quick decision -- as if the chance that you picked the right one is 1/2, not 1/3. So you have a 50/50 chance of staying with your first choice or choosing the remaining door.

If half the people choose the remaining door, and half the people stay with their current selection, then the house still wins -- because those that choose to stay are only right 1/3 of the time, not 1/2.
posted by treepour at 9:14 PM on April 8, 2008

Er, clearly I had some trouble with the last sentence of the first full paragraph of my comment. On the other hand, maybe my stumble underscores the peculiar obviousness of this problem . . . ?
posted by treepour at 9:18 PM on April 8, 2008

Here's (perhaps) the most simple way of thinking about this problem.

You choose a door. Then, Monty asks you if you want to keep your door, or trade it for BOTH of the other two.

Would you switch?

Again, with some numbers:

You choose a door (1-in-3 chance of winning). Then, Monty asks you if you want to keep your door, or trade it for BOTH of the other two (2-in-3 chance of winning).

If you think about it this way, it is quite simple, and rather undeniable, that you increase your odds from 1/3 to 2/3 by switching. That's what is really happening when he shows you a door... he is giving you both doors.
posted by Ynoxas at 9:18 PM on April 8, 2008 [3 favorites]

I think this, the Plane on a Conveyor Belt, and a few other similar things could form the basis of an interesting intelligence test of some sort. The test is not so much if you get the problem right or wrong when it's posed, it's how much the ignorance persists as the correct answer is illustrated in multitudinous ways.
posted by TheOnlyCoolTim at 9:57 PM on April 8, 2008 [4 favorites]

What trips people up about the MH problem is that the door that's opened is not random. If it were, changing wouldn't matter. The fact that knowing which door is opened tells you something about the initial state of the doors is what makes switching the correct choice.
posted by empath at 10:16 PM on April 8, 2008 [2 favorites]

Here's my articulation of the probabilities involved: do you want a goat? Probably not.
posted by davejay at 10:23 PM on April 8, 2008 [2 favorites]

Isn't it obvious I have 1/2 prob of picking the right one, because the unknown set has been reduced from 51 to 2 ? Or am I missing something ?

You're missing something: The set of 51 is unknown to you, but not to Monty. 51 times out of 52 the ace of hearts will be in that set, and 51 times out of 52 Monty will select the ace of hearts to give you as your choice if you switch. Monty "throwing away 50 cards that are not the ace of hearts" amounts to him simply "selecting" the ace of hearts to give you, except for that one time in 52 when you happen to get it.
posted by weapons-grade pandemonium at 10:56 PM on April 8, 2008 [2 favorites]

I don't think the plane on a conveyor belt is in the same category, as its answer depends on how you conceptualize the problem.

You can see the plane on a conveyor belt (POCB) in three broad ways:
1. The plane is pretext for a question about mathematics and geometry. Answer: no, the plane cannot take off because its relative velocity is defined in such a way that it has zero forward velocity by definition.

2. The plane is a pretext for a question about idealized physics. Answer: maybe, depending on what assumptions go into your model.

3. The question is asking whether a literal plane could take off from a literal conveyor belt, literally. Answer: yes, a real plane could take off from a conveyor belt because its forward thrust does not come from its wheels.

Perhaps the POCB could be useful in sorting mathematicians from physicists from engineers, but I don't see how it has much to do with intelligence.

However, the "does 0.99999... = 1" question is a good one for sorting out people who absolutely cannot admit being wrong, ever.
posted by Pyry at 10:57 PM on April 8, 2008 [1 favorite]

No it isn't.
posted by weapons-grade pandemonium at 11:21 PM on April 8, 2008

Plane on a conveyor belt is so yesterday. Helicopter on a Turntable is the new hotness.
posted by Rumple at 11:35 PM on April 8, 2008 [4 favorites]

"Registration required."

If you can't be bothered to find a registration-free version, then I can't be bothered with your post.
posted by mr. strange at 12:09 AM on April 9, 2008

I think the flaw in the psychology studies would have been a lot easier to understand if it weren't linked the Monty Hall Problem.
posted by Nattie at 12:26 AM on April 9, 2008

If you can't be bothered to find a registration-free version, then I can't be bothered with your post.

I think you're the only one getting that. Maybe it's time to clear some cookies?
posted by hjo3 at 12:27 AM on April 9, 2008

mr. strange: Go to Google News, search Monty Hall. It's the first hit.
posted by weapons-grade pandemonium at 12:47 AM on April 9, 2008 [1 favorite]

I'm sorta more interested in finding out what the implications of this are in psychology. The article makes it sound like these findings invalidate a good deal of prior research. Even if that's the case, is it really going to have much impact on the discipline of psychology as a whole?
posted by timelord at 12:50 AM on April 9, 2008

I seemto remember When I would watch Let's Make a Deal as a kid, I remember that sometimes instead of a goat, it was a donkey. Which I always thought of as a prize. As in they were TRYING to get the donkey. Because when you're seven, a donkey (or a goat) is way cooler than money or appliances.
posted by billyfleetwood at 1:58 AM on April 9, 2008 [3 favorites]

Twenty times through the test accessed via the linked article: with switching: 70% won; 20 times through without switching: 50% won.

I will now move on to remembering lustful thoughts of the lovely Carol Merrill.

And if you're wondering what's become of Carol: In 1985 Carol married Mark Burgess and four years later they moved to the Big Island of Hawaii. Carol spent several years working on environmental issues in Hawaii before joining her husband in his landscape/nursery business. She was recognized and honored for her environmental work when she received the Hawaii County Mayor's Award of Excellence and the First Lady's Outstanding Volunteer Award from the State of Hawaii.

Since selling their business in Hawaii two years ago, Carol and Mark have spent the majority of their time in a beach community in Australia. Carol's present day activities include growing fruits, veggies and herbs organically in containers, Ayurvedic cooking, staying in shape by doing a variety of exercises on the beach (while her husband surfs), and writing. She has written health-related articles for two on-line newsletters and is currently working on an autobiography. By sharing her life and what she's learned along the way, she hopes to inspire others to live healthier and happier lives.
posted by ambient2 at 2:11 AM on April 9, 2008

If you can't be bothered to find a registration-free version, then I can't be bothered with your post.

The probability that someone on Metafilter will bitch about site registration asymptotically approaches 1 as the post comment count approaches ∞.
posted by Blazecock Pileon at 2:15 AM on April 9, 2008 [1 favorite]

I... I finally understand the Monty Hall Problem.

My god, it's full of stars.
posted by Faint of Butt at 3:34 AM on April 9, 2008 [2 favorites]

geez, can you all just shut up about the Monty hall problem - its been mentioned before. If you don't get it you are stupid.

the intersting thing here is that a whole stream of psychology theory is based on an failed experiment. - an experiment that actually proves nothing.
posted by mary8nne at 4:23 AM on April 9, 2008

Huh. What with this thread, the previous thread and the NYT's excellent game/simulation, I think I finally get it enough to teach it to others (who are willing not to be asshats about it).
posted by kalessin at 5:45 AM on April 9, 2008

Nthing read the actual Yale paper; it's more clear than the article. Nice post!
posted by Kwine at 6:07 AM on April 9, 2008

I think I'll mention the St. Petersburg Casino game, as it's sort of in the same vein.

The game goes like this. A fair coin is flipped until "heads" shows up. If heads appears on the first flip, the house pays \$2. If tails appears on the first flip and heads on the second, the house pays \$4. If tails appear the first i-1 flips and heads on the i-th flip, then the house pays \$2^i.

Here's the catch. It costs \$1000 to play the game. Would you play it?

The natural thing to do would be to calculate the expected amount of money you should win. This is easy. The probability you win \$2 is 1/2. The probability you win \$4 is 1/4. The probability you win \$2^i is 1/2^i. If you remember your elementary statistics, this means that the expected amount of money you should win is:

1/2 * 2 + 1/4 * 4 + ... + 1/2^i * 2^i + ... = 1 + 1 + 1 + ...

That is, infinity.

Mathematics tells you should play the game. Would you?
posted by King Bee at 7:03 AM on April 9, 2008

Having already selected a door seems crucial, because if we frame the execise a different way, the relevant odds seem 50/50:

There is a car behind one of these three doors. Actually, behind door three there is a goat, so that's not it. Which one do you select? I would think the odds at this stage would be 50/50.

I guess it depends on how you treat the revealed door. What if we said: There is a car behind one of these two doors (odds 50/50). Also, here are 1000 doors which hide goats.

Maybe my examples have changed the problem enough so that they cannot be considered the same as the original, but I'm sure I'll be corrected (for being a n00b at teh Monty Hall Problem).
posted by preparat at 7:42 AM on April 9, 2008

Pyry typed "However, the 'does 0.99999... = 1' question is a good one for sorting out people who absolutely cannot admit being wrong, ever."

Yeah, that's true. The funny thing is that, in many cases, the people who simply refuse to give up are the ones who actually know a little bit about math, enough to think that their guts are always right.

This might actually be my favorite Metafilter discussion ever, with its odd juxtaposition of the .9 repeating discussion and the think vs. thing discussion, and the one guy who invented new number systems in order to refuse to admit defeat.
posted by roll truck roll at 8:09 AM on April 9, 2008

The deck-of-cards analogy is what really brought this home for me.

But what I want to know is, did anyone ever bring home the goat?
posted by adamrice at 8:20 AM on April 9, 2008

Here's a very similar (and just as counterintuitive) puzzle:

If I tell you that I have two children, and that one of them is a girl; what are the odds that the other one is a boy?
(Hint - It's not 50-50)
posted by rocket88 at 8:37 AM on April 9, 2008

Whoops, I figured out the pattern on the NYT game. 100% not switching and 40% switching (I switched every time until I figured out the pattern).

Suck on those statistics.
posted by subaruwrx at 8:42 AM on April 9, 2008

rocket88, 1/8

1/2+1/2+1/2=1/8
posted by subaruwrx at 8:44 AM on April 9, 2008

If I tell you that I have two children, and that one of them is a girl; what are the odds that the other one is a boy?
(Hint - It's not 50-50)

If "one of them is a girl" means one and only one, 100%.

If two girls are possible, 66%.
posted by TheOnlyCoolTim at 8:50 AM on April 9, 2008

That is, assuming children's sexes are iid Bernoulli, which I think is not quite true.
posted by TheOnlyCoolTim at 8:53 AM on April 9, 2008

The real scandal is that you didn't actually get to keep the goat.
posted by George_Spiggott at 8:54 AM on April 9, 2008

Yes, assume 50/50 boy/girl odds for any one child.
And you're right, Tim...it's 67%, or 2/3.
posted by rocket88 at 8:57 AM on April 9, 2008

Woke up thinking about this problem and why it seems so confusing and counterintuitive.

The reason, perhaps, has to do with two inter-related things:

a) metalogic (existential logic vs. ordinary logic)
b) the psychological, or specifically psychosemantic, ambiguity of what constitutes "choice"

I will address b first. In any game like this the only definition of "choice" that makes any sense within the given framework is this: a choice is ONLY a choice once it has consequence. If a choice is without consequence, strictly speaking it is not a choice.

The confusion in the Monty scenario partly stems from the ambiguity of what constitutes a choice. You only get one choice in the Monty scenario, but you are fooled (by yourself) into thinking you get two. Your "second" choice is really your first, as your first had no consequence--and was thus not, strictly speaking, a choice at all.

If someone says, pick a card, any card, from a card deck, and then says, are you sure that's the one you want?, the tendency is to think you are being given a SECOND choice. You are not being given a second choice, b/c stricty speaking there are no second choices in such a game: the only definition of choice here that makes any sense is one that has consequence. So, by definition in all such games, there is only a first choice (obviously this is different, if you are provided more than one "choice with consequence," i.e allowed to pick three actual cards). The first problem w/the Monty Problem (why both "right" and "wrong" answers are missing something crucial in our understanding of it) is that your "second choice" (your "switch") is really your first choice. The Monty Problem mistakes your first "choice" as an actual choice. It is not an actual choice, since it had no consequence.

Now here's where the metalogic comes in:

In metalogic, or "existential logic" as I conceive it, there are only EVER two choices to anything.

If you are shown the backs of 100 cards, 99 jokers and one ace, and asked to pick one, your metalogical odds are ALWAYS 50/50: 50% chance you will pick the ace, and 50% chance you will pick one of the 99 cards that is a joker.

Now it's true you can't gamble on metalogic, b/c it is so starkly existential. But the fact that you have a 1 in 100 chance in ordinary logic, does not negate the fact that in metalogic you only ever have a 50/50 chance on anything.

To tie these two things together: the Monty Problem problem makes us think we have two choices, when by definition in this game we only have one (one in three in ordinary odds), and the Monty problem blurs the line between our intuitive, inarticulate sense of metalogic and our more usual sense of ordinary logic. As a brain-buster, it rather devishly plays these two logics off one another.

**

(As a windy aside, if "metalogic" as conceived above asserts that all choices are both first and final choices, it seems to start over every time, and to inhabit a deviant and different modal universe--one that is re-made every instant--from our own. The tendency will be to reject metalogic as merely a tautology, but it is a perhaps essential tautology that allows us to clear something up: strictly speaking, there are only ever two choices, never more, and although ordinary logic appeals to our sense of infinite choice, it functions as something like an illusion.)
posted by ornate insect at 9:03 AM on April 9, 2008

Now, I have two children, and I tell you that the oldest is a boy. What are the odds the other is a boy?
posted by TheOnlyCoolTim at 9:04 AM on April 9, 2008

50-50
posted by rocket88 at 9:17 AM on April 9, 2008

If I tell you that I have two children, and that one of them is a girl; what are the odds that the other one is a boy?

Total possible, unordered offspring combinations: {Girl, Girl}, {Girl, Boy}, {Boy, Boy}

Possible combinations with one-girl, one-boy restriction: {Girl, Boy}

Odds = Possible combinations with restrictions / Total possible combinations = 1/3

What have I missed?
posted by Blazecock Pileon at 9:44 AM on April 9, 2008

For one thing, your unordered offspring combinations aren't equally likely.
There are four equally likely combinations of two children: listed as {older, younger}. They are:
{G, G}, {G, B}, {B, G} and {B, B}
The fact that at least one is a girl eliminates {B, B}, but the remaining three are still equally likely (1/3). Two of them have a boy as the 'other' offspring, so the odds of a boy are 2/3.
posted by rocket88 at 9:51 AM on April 9, 2008

Total possible, unordered offspring combinations: {Girl, Girl}, {Girl, Boy}, {Boy, Boy}

Actually it's {Girl, Girl}, {Girl, Boy}, {Boy, Boy}, {Boy, Girl}

There are two cases of a boy and a girl, not just one.

However, in Rocket88's later example, he specifies that the oldest is a boy, which eliminates one of the possibilities.
posted by George_Spiggott at 9:51 AM on April 9, 2008

On ill-timed preview, what rocket88 said.
posted by George_Spiggott at 9:52 AM on April 9, 2008

Actually it's {Girl, Girl}, {Girl, Boy}, {Boy, Boy}, {Boy, Girl}

I don't think so. The problem offers no information about the ordering of offspring, and it doesn't seem relevant, i.e. {G, B} is the same as {B, G} for the purposes of determining the complete set of unique combinations of offspring.
posted by Blazecock Pileon at 10:11 AM on April 9, 2008

That is, all we were told is that one of the children is a girl. That's all the information we have to cull the set.
posted by Blazecock Pileon at 10:14 AM on April 9, 2008

The first information you are given is that there are two children. That sets up the four-outcome scenario where each has 1/4 odds. (You're correct that order isn't specified and isn't important, as long as you understand that unordered {G, B} has 2/4 odds of occuring.)
The second information that at least one is a girl eliminates {B, B}.
posted by rocket88 at 10:24 AM on April 9, 2008

You're correct that order isn't specified and isn't important, as long as you understand that unordered {G, B} has 2/4 odds of occuring.

Okay, the prior assumption is that the parents will have 1:2:1 odds of having {GG}:{GB}:{BB} offspring. That's not given, but I can see how that's a reasonable assumption. Thanks for the explanation.
posted by Blazecock Pileon at 10:29 AM on April 9, 2008

Yes, but what does the gender of the revealed offspring have on the gender of the second? It's like saying I'm flipping a coin twice. One flip is heads. What is the other flip? It's 50-50 for heads or tails (or boy or girl).
posted by Doohickie at 10:45 AM on April 9, 2008

Let's look at it another way: Let's say for instance the odds are the following, listed in birth order:

(B,B), (B,G), (G,B), (G,G)

Now, you know that one of the two offspring is a girl, but you don't know whether she is the older one or the younger one. You have two cases:

Case 1. If she is the older one, you can be left with
(G,B) or (G,G), but NOT (B,G) or (B,B)
The chance of the other offspring being a boy is 50-50

Case 2. If she is the younger one, you can be left with
(B,G) or (G,G), but NOT (G,B) or (B,B)
The chance of the other offspring being a boy is 50-50

In EITHER case it's 50-50.

The knowledge that one is a girl does NOT have any effect on the gender of the other offspring. The revelation of the gender of one child eliminates two of the combinations. Without knowing who is older, though, you can't know which two combinations are eliminated, but two of the are in each case.
posted by Doohickie at 10:53 AM on April 9, 2008

This feels to me like one of those things that's not obvious at first, but once explained(Monty's choosing gives you information to guide your choice), then it's stunningly obvious.

Yes. Thank you, Metafilter, for finally helping me to "get" the Monty Hall problem. I had read about it before but just couldn't seem to wrap my head around it. Reading this thread and the former one switched on the lightbulb and now it's like, "of course."
posted by LeeJay at 10:54 AM on April 9, 2008

In the first line of my last post, it should say "possible combinations" instead of "odds".
posted by Doohickie at 10:55 AM on April 9, 2008

I missed that .9999=1 thread, and I have never, ever heard anyone say "You have another think coming". Are people being serious, or is this some sort of meta-joke in the brainteaser realm I'm not aware of?

I am familiar with plane on a conveyor, but I can't believe anyone would seriously believe the saying is "you've got another think coming". It's "thing". There's no such thing as "a think", so you can't have "another think" coming. Absurd.

Like the phrase "all intents and purposes". I just thought someone saying "all intensive purposes" was an attempt at a joke. Surely noone seriously thinks it is "intensive purposes". If they do, they have another thing coming.
posted by Ynoxas at 11:32 AM on April 9, 2008

Do people actually write noone when they mean no one?
posted by Doohickie at 11:34 AM on April 9, 2008

(In other words, cut us language slugs some slack, Ynoxas; after all, you're one of us! ;-)
posted by Doohickie at 11:35 AM on April 9, 2008

The .9999... = 1 thing became blindingly obvious when someone told me to add 1/3 + 1/3 + 1/3 in decimal (.333... + .333... + .333...)
posted by straight at 11:40 AM on April 9, 2008

What if of the remaining 51 cards, he only removes one (much like he only opens one door) and then asks you if you want to change? I'm not disagreeing with the "classical" Montyhall problem, but I have never seen it as "he opens all but one of the remaining doors". I have always seen it as "he opens one of the remaining doors, and then asks you if you want to switch." And in the case with 100,000 doors, the same thing; he opens one of the ones you didn't choose. How does this change the math?
posted by Eideteker at 12:12 PM on April 9, 2008

Let's look at it another way: Let's say for instance the odds are the following, listed in birth order:

(B,B), (B,G), (G,B), (G,G)

(Boy, Boy), (Boy, Goat), (Goat, Boy), (Goat, Goat)
posted by gordie at 12:17 PM on April 9, 2008 [1 favorite]

gordie--I think the goatboy is Pan.
posted by ornate insect at 12:26 PM on April 9, 2008

The problem people have with .9 repeating--and, I suppose, also the problem with Monty Hall--has to do with the very non-mathematical way in which we think about numbers. When we say things like ".9 repeating just gets closer and closer to one," we make the mistake of thinking of a number as a moving thing, and not as an abstract value.

As for you, rocket88, if you're going to continue with this silliness, please give us a Wikipedia link or something.
posted by roll truck roll at 12:28 PM on April 9, 2008

Do people actually write noone when they mean no one?
posted by Doohickie at 1:34 PM on April 9

Yes, but in my case, it is a stylistic choice. I get a little thrill everytime (you see that?) I defy those little red underlines from auto-spellcheck.

Don't fence me in!

Otherwise, I was being serious. I've simply never heard a person say "another think", and until today, had no idea there was any sort of controversy.
posted by Ynoxas at 1:05 PM on April 9, 2008

I think both "another think" and "another thing" sound ridiculous. "If you expect cake for dessert, you've got another expect coming!" "If you expect cake for dessert, you've got another thing coming!" Well, no, not if you're not getting any dessert.
posted by Eideteker at 1:10 PM on April 9, 2008

I posit that "nothing" is indeed "another thing" and is, in fact, coming.
posted by Skorgu at 1:46 PM on April 9, 2008

Skorgu thus demonstrates the Zen of Monty Hall.
posted by adamrice at 2:02 PM on April 9, 2008

Case 1. If she is the older one, you can be left with
(G,B) or (G,G), but NOT (B,G) or (B,B)
The chance of the other offspring being a boy is 50-50

Case 2. If she is the younger one, you can be left with
(B,G) or (G,G), but NOT (G,B) or (B,B)
The chance of the other offspring being a boy is 50-50

These are the cases when you know she is the older one or the younger one (the problem I gave.) Only knowing that at least one of the older child OR the younger child is G, you can only eliminate the outcome (B,B) which is eliminated in both cases. You do not have the information required to eliminate (B,G) or (G,B).
posted by TheOnlyCoolTim at 2:04 PM on April 9, 2008

Going back to King Bee's St. Petersburg Casino game, the wiki article gives a nice overview. The stuff about finite backers at the bottom is pretty neat.

Also, I think I've calculated that you win less than \$1000 with probability 511/512.
posted by matematichica at 9:17 PM on April 9, 2008

I think everyone should be aware that there is a school of economics out there that calls itself rational choice theory. They believe that every choice is either a maxmized utility or something that can be measured to demonstrate an objective consistency. Their opponents are game theorists who accept cognitive dissonance as part of the game. We might be seeing that debate going on here because they are framing cognitive dissonance as an extension of the monkey experiment.
posted by Brian B. at 7:11 AM on April 11, 2008 [2 favorites]

Sorry, TheOnlyCoolTim, but you are wrong. Just because you don't know whether you are in Case 1 or Case 2, you are undoubtedly in one or the other, but not both, so they can be viewed independently. So look at them one more time:

Case 1. If she is the older one, you can be left with
(G,B) or (G,G), but NOT (B,G) or (B,B)
The chance of the other offspring being a boy is 50-50

Case 2. If she is the younger one, you can be left with
(B,G) or (G,G), but NOT (G,B) or (B,B)
The chance of the other offspring being a boy is 50-50

Knowing the gender of one offspring has no effect on the odds of the gender of the other offspring.
posted by Doohickie at 7:22 PM on April 12, 2008

You must agree that, like getting two heads in a row in a coin toss, getting two girls happens 1/4 of the time, as does getting two boys. Getting one boy and one girl happens 2/4 of the time. Knowing one is a girl, you limit yourself to the cases "one boy and one girl" and "two girls". But you see that the former is twice as likely as the latter.
posted by TheOnlyCoolTim at 1:25 AM on April 13, 2008

Someone sent up the MHP Signal, so here I am.

First, I find it odd that no one is really discussing the link. It's pretty interesting.

Assume that I asked you to name three television shows in which you have a definite order of preference. Then, I choose at random two of those three shows and ask you which of the two you prefer. What do I know about the one you didn't choose and the third I didn't ask about? The answer is that I know that you're twice as likely to favor the third over the one you didn't choose.

This is counterintuitive. Our intuition tells us that knowing that one of the three is preferred over specifically one other of the three (A over B) doesn't tell us anything about the ordering of that dispreferred and the third (B and C). Couldn't it just as likely be either way (B over C or C over B)? No. It's twice as likely that we prefer the third over the dispreferred (C over B is twice as likely as B over C).

Why?

Well, here are all possible orderings:

A, B, C
A, C, B
B, A, C
B, C, A
C, A, B
C, B, A

Here are all possible orderings of the three where A is preferred over B:

A, B, C
A, C, B
C, A, B

We can see that of those possibilities, C appears before B in two cases while B before C in only one. Therefore, if we know that A is preferred over B, then it's twice as likely that C is preferred over B than is B over C.

The choice experiment takes three items and A) attempts to assure that there is no preference of any one over the others by the subject, B) randomly takes two of the three, C) forces the subject to choose one of those two over the other, and then D) forces the subject to choose between the one not chosen and the third.

The experiment assumes that if the choices were made truly at random that there would be no differentiation, over iterations of the experiment, in the distribution of the choices between the second pair. However, the results of actual conducts of this experiment show a strong preference for the third item and the interpretation of this result is that in having been forced to choose against something, people (and chimps) retain a bias against it that shows up in later choices.

Is the assumption that there would be no differentiation if the choices were truly random correct?

On the one hand, it's most assuredly the case that if there were a coin toss deciding the first choice, and another coin toss deciding the second choice, that the distribution on the second choice would be 50-50. On the other hand, if we assume that the consecutive choices are an implicit random ordering of preferences for all three items, then it's the case that the distribution of the second pair of choices would not be even.

So which is it?

In my opinion, strictly speaking, it's the former. I don't really see how the two coin tosses in sequence reveal an implicit ordering of all three. For precisely the same reason as that this problem is counterintuitive, the subject of the experiment is almost certainly not making his/her second choice with the awareness of all possible orderings of all three items, choosing one ordering randomly, and then making his/her choice between the second two items in accordance with this.

Stated more explicitly, it's the difference between thinking:

Hmm. I see two items, B and C. I will choose one of those two randomly.

...and

Hmm. Now that I've chosen A over B, I see three possible orderings of A, B, and C where A is preferred over C. I'll randomly choose one of those three orderings and then pick whichever of B or C is in accordance with that ordering.

If the subject did do the latter, he/she would randomly choose C over B twice as often as B over C.

But I don't think people, or chimps, do this. As far as I can tell, this is the "strong" criticism of this experiment. It seems to be claiming that the conduct of the experiment has within it an implied ordering of the three items at the outset, that the subjects are choosing randomly between orderings of three items. I just don't see how this is the case.

This leaves the weaker criticism of the experiment being that there are inevitably some undetected preferences that order all three items. That is to say, in any actual conduct of this experiment, the experimenter cannot rule out with absolute certainty that there is a preferential ordering of the three items. The best the experimenter can do is try to eliminate any preferential ordering to some degree of uncertainty. And, of course, the experimenter then evaluates the results of the experiment with that degree of uncertainty in mind. To properly evaluate the results, the experimenter has to properly evaluate how that uncertainty propagates through the experiment—if the experimenter doesn't understand the math above, then he/she will wrongly evaluate the results.

Any possibility that there are very small, undetected preferences that order all three items means that those preferences will be expressed in such a way that unbalances the second choice. You can't say how big the psychological choice effect is until you quantify that baseline. A slight preference in ordering in the first part of the experiment will express itself doubly in any preference of the third item over the unchosen item. So the experimenter will misstate the confidence interval for the result of the experiment if he/she thinks it's the same as the confidence interval for there being no initial preference.

How does this actually reflect upon this experiment as it's done in the real world? Well, it seems to me that if A) they've eliminated initial preference to any reasonable degree, and B) they're seeing a very strong preference for the third item over the unchosen item, then the experiments are still largely valid.
posted by Dances with Werewolves at 12:59 AM on April 14, 2008 [2 favorites]

With regard to the MHP itself...

I've been pondering recently (again) this whole "only if the problem explicitly states that Monty must open a losing door" defense of those who have offered the wrong answer. I've a couple of thoughts on this.

First, I remain a bit skeptical that most of these people truly began with an understanding that Monty may have revealed the winning door. Especially given that many of the same people offer the criticism that it only makes sense to think about the problem as a real Monty Hall would likely play it. The two are inconsistent as Monty would never play the game in such as way as revealing the winning door. That version of the game are as utterly unrealistic as it can reasonably get.

Having said that, I can imagine that someone might think about this problem as if the scenario were that you just happen to find yourself in the situation where Monty opens a losing door.

Put another way, is the lax problem statement the equivalent to the strong problem statement, or not? Is it the case that by virtue of specifying that Monty has opened a losing door the problem statement has required that Monty open a losing door? Is it the case that if Monty were allowed to open a winning door, but that you find yourself in this situation and Monty has opened a losing door, that this is the same thing as the classic (strong) version of the Monty Hall Problem?

The answer is "no". If you just happen to find yourself in the situation where Monty has opened the losing door, but was able to have randomly opened the winning door, then it doesn't make a difference if you switch or stay. Having "culled" (via the problem statement) all iterations where Monty opens the winning door, you've not created a version of the problem equivalent to one in which Monty must knowingly choose a losing door to open.

That said, I still have some difficulty with anyone thinking of the problem in terms of Monty being able to open the winning door. It seems to me to be a very unnatural way of thinking about it and I'm convinced that most (but not all) of those who object to the correct solution on this basis are doing so as a means of saving face, and not truly honestly. (Though, to be sure, when any of us go through the pains of trying to understand something as counterintuitive as the MHP we're not likely to be very aware of how and why we've come to the decisions we've made. That is to say, this means of saving face is basically trying to make sense of the initial confusion and by its very nature, making sense of that confusion is ambiguous. Whether it was a valid misunderstanding or not is difficult to prove, even when one has access to what's inside one's own head.)

"The test is not so much if you get the problem right or wrong when it's posed, it's how much the ignorance persists as the correct answer is illustrated in multitudinous ways."

Very much so and this aspect of the MHP truly fascinates and depresses me. This latest example of the MHP being in the news has led to a spike in traffic on my MHP website, and I've already gotten one email of the "I'm sorry, but you don't understand probability..." variety.

What blows my mind is that my page begins with the statement that the answer is very counterintuitive and that many mathematicians get the answer wrong; I go on to urge the reader to experimentally verify that I'm right (or prove me wrong, whichever); and my page includes several references and citations. I am dumbfounded that anyone could so self-confident to ignore all that and just go with their own belief that it doesn't make a difference whether one switches or stays.

As someone else mentioned with regard to the .999... = 1 thing, it takes a certain level of competency for someone to be so certain about this. People with little competency in math will either ignore the problem altogether, or simply accept whatever authoritative opinion is at hand, or will work through that opinion and come to a decision on it. Only someone with a certain minimal level of competency will see the correct solution and immediately dispute it. Those people are both interesting and dangerous because they've clearly illustrated the "Peter Principle": they've reached their own level of incompetency. And not only cannot see it, but they refuse to see it.

Now, we all illustrate the Peter Principle in some aspect, or another. None of us is perfect, certainly not me. I've stubbornly held onto wrong opinions out of both pride and an excessive degree of self-assurance. But those who are adamant about asserting the wrong answer to the MHP seem to me to be quite beyond what is normal. I keep repeating this here and elsewhere, but what just bothers me the most is that this is trivial to experimentally verify. Unlike the plane and the treadmill, one can just work it out with a friend or simulate with a few minutes of programming. When I first encountered the MHP in 1991 via the vos Savant fracas, I too believed that it couldn't make a difference whether one switched—but given the controversy, it seemed incumbent upon me to actually check and see if I were right. I wrote a horribly simple BASIC program in a few minutes which proved that I was wrong. It was trivial. And if I couldn't have programmed, I could have gotten my wife to work through, say, ten iterations of the problem. Given the claim of it being twice as effective to switch, this trend would show up easily.

Over the thirteen years that I've maintained the Monty Hall Problem webpage, I've had quite a few discussions with non-believers. The typical pattern is that they remain unbelievers. I'd say that maybe only a fifth finally comprehend the problem during our discussions; although, to be sure, there's no way for me to know how many people who just end the conversation have done so because they've realized they were wrong but can't bring themselves to admit it to me. Still, given that I've probably discussed the MHP with more people, and more often, than pretty much anyone else alive, and that I'm damn good at explaining it (I've collected dozens of approaches to it), you'd expect that more of these emailers would come to understand the problem. The explanation, I think, is that these people are self-selecting themselves to be incompetent at recognizing their incompetency given that they've jumped through a few hoops just in reading my page and still emailing me to insist that I'm wrong.

I'm convinced that these people represent a certain dangerous category of our population. A category that I think is unusually represented on the Internet in places like MetaFilter and other where people are self-selected to be somewhat above average in intelligence while also self-assured enough to be frequently opinionated. These are the global warming skeptics and the creationists who write about these topics frequently on the net. They are notably often engineers or (less so) physicists who are commenting on subjects outside their expertise. They are the sort of people to call other people "pseudo-intellectuals".

(I think engineers tend to this because they think that their competency in math and practical problem solving implies a generally high level of intellectual competency. Physicists are more prone to this than other scientists because, quite simply, they're inclined to think of all other sciences as being elaborations on physics and thus their own competency implies a general intellectual competency. This doesn't necessarily apply to you if you're an engineer or physicist—some of my best friends are engineers and physicists and, as a group, they're more inclined to be right about stuff than most other groups I can think of. But among, say, the lower ranks of these there's a noted tendency to greatly exaggerated sense of intellectual competency.)

One such email correspondent I've had was a guy who was, at the time (I don't know if he still is) one of the, or the, science correspondent for the Houston Chronicle. He came across the MHP and my webpage, and then emailed me to tell me that I was wrong and that it didn't matter whether one switched or stayed. After several emails and no success in explaining the correct answer to him, I asked him in frustration why he didn't consult the citations I listed on my webpage and, more importantly, call up a Houston math professor and/or simulate the experiment himself to determine the truth of the matter. He replied that "it wasn't that important to him". Note that it was important enough to him to have a) initially considered writing about the MHP or including something about it on the paper's website, and b) exchanged pages and pages of email argument with me about it.

This was someone people trust to explain difficult concepts in math and science to them.

Another thing that I find interesting about the Monty Hall Problem is that it reveals, I think, the ambiguity of what we think of as "comprehension". As is the case with many things, one can think one understands this problem right up until one attempts to explain it to someone else. I've watched quite a few people hear the explanation and say, "Oh, yeah, I see now" and then, in further discussion, reveal that they didn't really understand the problem at all. Or, alternatively, I've seen quite a few people have some sense that there's something going on with how Monty is constrained by the contestant's initial choice and not opening the winning door while still not quite getting it. And even more commonly, I've seen people who are pretty facile with the MHP and yet have trouble with variants or the same problem but in a different guise. In all these cases, what does it mean to say that one "understands" the problem and its solution?

In my own case, I still occasionally come across strange and subtle variations of the problem that I don't understand at first glance and yet I expect myself to, given how well I understand the MHP.

For example, it seems to me that anyone who truly and deeply understands the Monty Hall Problem, the boy/girl problem above should be trivial. But it wasn't to me when I first encountered (which was long after I was adept at the MHP). Neither was this criticism of the choice experiment.
posted by Dances with Werewolves at 2:12 AM on April 14, 2008 [2 favorites]

Dances with Werewolves: ... the subject of the experiment is almost certainly not making his/her second choice with the awareness of all possible orderings of all three items, choosing one ordering randomly, and then making his/her choice between the second two items in accordance with this.

I see no good reason to dismiss the idea. On the face of it, the belief that people have an ordering of items in terms of preference before the experiment starts seems pretty reasonable. The subject doesn't need to have an awareness of all possible orderings, they just need have their own preference ordering. To suggest that the subject will make their second choice in accordance with their pre-existing preference doesn't require that the subject be aware of all orderings and pick one randomly, which is what you suggest above.

Well, it seems to me that if A) they've eliminated initial preference to any reasonable degree, and B) they're seeing a very strong preference for the third item over the unchosen item, then the experiments are still largely valid.

According to the paper:

A) The experiments haven't eliminated initial preference.
B) The effect disappears entirely if you assume that subjects have set initial preferences (at least for the study the author discusses).

Finally, the paper deals with another similar set of experiments as well and concludes that they aren't valid either.
posted by ssg at 8:47 AM on April 14, 2008

"I see no good reason to dismiss the idea. On the face of it, the belief that people have an ordering of items in terms of preference before the experiment starts seems pretty reasonable."

You're absolutely right about that last part and I didn't intend to imply nor thought ("imply"? Well, I basically said it explicitly by giving the example) otherwise. If one has a preferential ordering, the effect will manifest even if the subject doesn't consider it consciously.

However, I'm not as convinced as you are that that an ordered preference of three items is more likely than not, especially when the experimenters have selected three items with a mind toward minimizing any reasons for having a preference. (Whether they are successful at this or not is their burden to demonstrate. I'm not convinced that choosing red, green, and blue M&Ms is minimizing the likelihood of preference.)

"A) The experiments haven't eliminated initial preference."

That's the crux of it, no? When you go on with this following statement:

"B) The effect disappears entirely if you assume that subjects have set initial preferences (at least for the study the author discusses)."

...you (and, presumably, Chen) are failing to account that, in reality, when trying to set up this experiment, researchers are only going to be able to eliminate preference to some degree of confidence. That is to say, neither will they entirely eliminate preference, nor will they likely end up with a complete failure to eliminate preference. Of course the effect disappears if you assume that the experimenters have entirely failed at eliminating preference; i.e., in every conduct of the experiment the subject has an ordered preference. In that unlikely event, the subjects will choose the third item over the unchosen item twice as often as not...a very large effect on the distribution. If there is such a thing as what these experiments have claimed to reveal, then it must be very large to not be obscured by this effect.

That second conclusion seems to me to be wildly uncharitable and unrealistic. Given that the experimenters have initially tried to eliminate preference (which was the initial requirement for their experiment, after all), to assume that they entirely failed to eliminate preference in every event is just silly. There is most likely preference acted upon in some of the events, but not in others.

How would you go about trying to ensure that there's no preference? One way would be to force subjects to repeatedly choose, in sequence, three items and given a number of experiments, quantify the randomness of their selections. You can't be sure that there's not some subtle preference for red over blue over green that only very occasionally reveals itself in how the subject chooses—you can only estimate that they are randomly choosing to some degree of confidence. If you have a subject choose three M&Ms in some order over one-hundred iterations, and they appear mathematically to not have a preference, you still only know that their lack of preference is necessarily below some amount, not that it doesn't exist.

But, yes, I ought to read the paper and I shall. But your A and B seem very strange to me. Why would Chen assume that either there's preference that manifests in every iteration, or that the standard for the experimenters in "eliminating" preference means certainty that never once is there preference? Saying that we have a "slight preference" is very meaningful because it implies that we won't exhibit it as regularly as we do a stronger preference. If I have a slight preference in a certain kind of food, that doesn't mean that I cannot choose against that preference when given a choice—after all, in my actual daily life such a choice truly does involve things I prefer over others. And yet, I occasionally choose the things I do not prefer. That's because the preference is "slight". And those are the preferences that experimenters are going to have a more difficult time detecting. A preference in which I am always consistent will show up immediately in any effort to detect it.
posted by Dances with Werewolves at 10:38 AM on April 14, 2008 [1 favorite]

I haven't read any of the studies that Chen is discussing, but Chen claims that the studies in question fail to account for initial ordering of preference at all. Just read the paper.
posted by ssg at 12:12 PM on April 14, 2008

Ah, well I was going by the linked NYT column which claimed that the design of the experiment eliminated initial preference. Tierney wrote (emphasis mine):

"The Yale psychologists first measured monkeys’ preferences by observing how quickly each monkey sought out different colors of M&Ms. After identifying three colors preferred about equally by a monkey — say, red, blue and green — the researchers gave the monkey a choice between two of them."

Either the studies attempted to eliminate preferences at all, or they did not. Chen's claim (as you state it) seems extreme.

(Pause a moment while I go ahead and browse the paper we're discussing...)

Now that I've looked over the paper (very briefly and tentatively), I think you're partially wrong in your characterization. Chen isn't claiming that the experimenters didn't attempt to eliminate preference; he is claiming that they attempted to but failed and (more importantly) there is a definite pre-existing preference implied if a subject chooses between two items:
"Then, three objects that are rated equally (say rated 4) are chosen for use in a second stage of the experiment. Note, importantly, that the discreteness of the scale leaves open the possibility that these items might not be perfectly equivalent; for example, a subject may 'truly' rate one of the items 4.1, one 4.26, and one 4.3."
Then again, if might be that the subject 'truly' rates both items as exactly 4. Chen isn't just claiming that there might be an undetected preference within the granularity of the initial stage of the experiment, he's going much further and asserting that there necessarily is an undetected preference, in all cases, in that initial stage of the experiment.

I think that Chen is quite strikingly wrong, here, and wrong in a way which has everything to do with the fact that he's an economist and the idea of revealed preferences is a central dogma. That is to say, Chen explicitly makes this claim:
"Note that the analysis above assumes that subjects are never completely indifferent between two options; that if pressed they can always decide which of two options they prefer. Economic theory suggests that this is by far the most likely case, but even if subjects can be indifferent (which a discrete rating can never show) the above analysis does not change; the computation simply becomes more difficult."
So, what's being overshadowed by his flashy critique of the math involved is that he denies the very basis of the experiment (that any subject can simply not have a preferred ordering of three items). Per the above, he denies this assumption with nothing more than "economic theory suggests that this is by far the most likely case", which, in my book, is at least as piss-poor science as that which he is criticizing.

Worse, that final sentence is weaseling, in my opinion, because what's really at stake here is that if subjects can be irregularly indifferent (a "slight" preference) then the magnitude of the mathematical effect will be anywhere from insignificant to quite distorting. Chen neatly elides the possibility that it can be insignificant and makes the claim that all these studies are unreliable because he, as he states above, believes that there are always ordered preferences if and when choices have been made.

Chen is claiming that if you are asked to choose between A and B, and then B and C, those choices necessarily reflect a pre-existing ordered preference of A, B, and C. That, I think, is an absurd assumption. It is also easily tested experimentally: if there is a definite, pre-existing preference that will always be reflected by a subject's choice, then that preference will be repeatable with no variance whatsoever. If a chimp chooses A, B, and then C one week, then he'll choose the same order the next.

(Okay, Chen might try to weasel out of this by claiming that the preference will have changed if the choices change. He might even claim that this happens in the same day, even each time the chimp is asked, one time right after the other. However, the problem with this is that if this is the case, and assuming there's no long-term pattern of preference, then the effect he's looking for won't appear; it will all cancel out.)

Of course, again, it is the psychologists' responsibility to demonstrate that they've begun with a lack of preference between the items to some given margin of error. Looking over Chen's paper, it appears to me that he's quite right (as I suspected) that the experimenters didn't factor in any confidence level for that supposed lack of preference—and of course the reason they didn't consider this was because they were not aware that the math of the situation makes this margin of error more important than they thought it was.

None of this is to say that I don't find Chen's critique relatively damning. Because the effect of an undetected preference results in twice the likelihood of a selection, it was absolutely crucial that the experimenters take this into consideration when they evaluate the results of their experiment. They did not. I don't have the statistical chops to try to evaluate the margins of error (with regard to their initial stage where they identify a lack of preference) in these experiments as they've been conducted, so I cannot evaluate how much it undermines their conclusion.

Note, however, that the first methology as Chen describes it, with this rating of numerous items and then taking three of the items that share the same ranking, is (in my opinion), a very flawed method. It quite leaves the door open to Chen's criticism because, as he says, three items otherwise rated at some discrete value in another context may, in fact, have a preferred ordering when taken as a group by themselves. Far better would have been something like I mention above, and that is to experimentally identify three items that people have demonstrated a lack of preferred ordering in X amount of iterations of being asked to order them.
posted by Dances with Werewolves at 1:37 PM on April 14, 2008 [1 favorite]

Chen isn't claiming that the experimenters didn't attempt to eliminate preference

I never said that he made any claim about what they attempted. I said: Chen claims that the studies in question fail to account for initial ordering of preference at all. I think that is quite accurately what he claims: that the subjects had initial orderings of the items that the experimenters selected and that the experimenters did not account for the initial orderings of the selected items. We don't really care what they attempted to do; this is science, so we care about what they actually did.

It is also easily tested experimentally: if there is a definite, pre-existing preference that will always be reflected by a subject's choice, then that preference will be repeatable with no variance whatsoever.

This might work for chimps, but I don't think it would work for people, who are likely to remember their previous choice. You'd have to throw a lot of obscuficatory testing in to have a chance that the subject will forget their previous choice. Chen suggests (and, according the NYT, is engaged in) a similar experiment that avoids this problem, though it is a tweak of a similar type of experiment, not of the three item experiment.

I think you are misstating Chen's position. You quote Chen writing: "Note, importantly, that the discreteness of the scale leaves open the possibility that these items might not be perfectly equivalent" and then write that "Chen isn't just claiming that there might be an undetected preference within the granularity of the initial stage of the experiment, he's going much further and asserting that there necessarily is an undetected preference, in all cases, in that initial stage of the experiment." Chen's "might not be perfectly equivalent" is entirely different from your "necessarily is an undetected preference, in all cases".

The problem isn't one of statistical calculation, as you suggest, but of experimental design. There isn't any way to determine what proportion of the subjects have an initial ordered preference from the data collected by the original experimenters.

Chen neatly elides the possibility that it can be insignificant and makes the claim that all these studies are unreliable because he, as he states above, believes that there are always ordered preferences if and when choices have been made.

Firstly, Chen doesn't state that there are always ordered preferences; he merely states that economic theory holds that there are always ordered preferences, while nothing that there need not always be ordered preferences for his critique to be valid. Secondly, his critique relies on the fact that we don't know what proportion of the subjects had ordered preferences. Unless one can somehow show that the effect due to ordered preferences is insignificant, then the confidence interval becomes too large to reach any conclusions.

Chen's argument is not that cognitive dissonance doesn't exist, but merely that experiments to date have failed to prove its existence. He argues strongly that the experiments failed to reach any conclusion, which is not at all the same as arguing that the conclusions reached by the experiment are false. In fact, he explicitly states in his conclusion: "This is not to say that cognitive dissonance does not exist".
posted by ssg at 4:20 PM on April 14, 2008

"I never said that he made any claim about what they attempted. I said: 'Chen claims that the studies in question fail to account for initial ordering of preference at all.'

Unfortunately, there's ambiguity in that wording with regard to "account" and "at all". But arguing about it is mostly a red herring.

More importantly, you're mistaken when you say that "Chen doesn't state that there are always ordered preferences". As I already quoted, he wrote:
"Note that the analysis above assumes that subjects are never completely indifferent between two options; that if pressed they can always decide which of two options they prefer."
...where "the analysis above" is his initial explanation of the difficulties of the experimental design. His heaviest criticism relies upon the assumption that there is always an initial preference if a subject has "proven" a preference by making a choice (which the experiment does, in fact, force).

Yes, he allows for the possibility—by arguing in the alternative, which has clear implications(1)—that there might not be preference and says that the "analysis does not change; the computation simply becomes more difficult". Unfortunately, that "the computation simply becomes more difficult" elides the fact that the degree of undetected preference determines the degree of the effect. I think this is a very dishonest sentence here, it makes me suspicious. Note that I have no dog in this fight—I have no preference for psychologists over economists. If anything, the reverse is true. But the fact of the matter is that if there is no initial preference then there is no effect he's describing(2).

"Unless one can somehow show that the effect due to ordered preferences is insignificant, then the confidence interval becomes too large to reach any conclusions."

The effect of ordered preferences, where they exist, is an entirely known quantity. It doubles the likelihood of the third item to be chosen over the disfavored of the initial two. The "effect due to ordered preferences" is not open to question.

What is not known is how often ordered preferences exist but are undetected by experimental design. How often preferences are undetected determines whether the mathematical effect ranges from nothing to doubling the third item's frequency of being chosen. I can't evaluate this. But I don't trust Chen, either, as he is assuming that there is always preference, with an argument in the alternative allowing for the possibility that sometimes there might not be preference. That's extremely biased towards the "undetected preferences" end of the spectrum. I see no justification for that other than his appeal to the economic theory of revealed preferences. So an equally valid assertion is that "unless one can somehow show that the effect due to ordered preferences is significant, then the confidence interval is not reduced enough to prohibit conclusions". (Where "effect due to ordered preferences" is understood to mean "the magnitude of undetected ordered preferences".)

I apologize for the scattered way in which I've responded to your comment. You write:

"I think you are misstating Chen's position. You quote Chen writing: 'Note, importantly, that the discreteness of the scale leaves open the possibility that these items might not be perfectly equivalent' and then write that 'Chen isn't just claiming that there might be an undetected preference within the granularity of the initial stage of the experiment, he's going much further and asserting that there necessarily is an undetected preference, in all cases, in that initial stage of the experiment.' Chen's 'might not be perfectly equivalent' is entirely different from your 'necessarily is an undetected preference, in all cases'."

Yes, but Chen goes on to make the claim that there are always preferences, as I've already quoted. Thus I argue in that latter sentence.

Chen's is a two-fold argument with regard to those two quotes. Firstly, he's arguing that the experimental design where items are rated equally in discrete quantities cannot, by itself, guarantee that there's no ordering of those equally rated items because a discrete rating of individual items will inevitably force items that are differentiated by an amount smaller than the unit to appear to be undifferentiated.

If he had stopped there, he wouldn't have any real basis upon which to estimate whether, as you say above, that the effect the researchers are looking for are swamped by the ordered preferences effect. He wouldn't be able to make that argument because he has no real way of knowing whether 4 truly does equals 4, or not.

But Chen doesn't want to stop there, and that's why he goes on to make the argument that there always ordered preference. If he assumes this, then he can identify the exact quantity the experiments were failing to account for and the compare that with the correct confidence intervals of the effect they were looking for. Which he does, and with which he concludes, as you say, that it swamps the effect they were looking for and they've proven nothing. Assuming there is always preference, and that the design is failing to account for that, is the maximal version of the criticism. But, again, I see little support for that maximal assumption.

Really, though, because the "equal rating" design is so flawed in having no ability whatsoever (as far as I can tell) to estimate the amount of undetected ordered preference, only if the effect they're looking for is very large would that design be useful. I agree that Chen's critique is devastating with regard to it.

But other designs, where we can estimate some level of undetected ordered preference, are only vulnerable to criticism when that level is large enough to swamp the effect being measured. For example, if there is undetected ordered preferences in only 1% of the iterations, then only if the suspected choice effect is very small will it matter.

"This might work for chimps, but I don't think it would work for people, who are likely to remember their previous choice."

I was initially puzzled by this comment—or, rather, I wasn't initially puzzled by it—because I assumed you were referring to my suggestion that the experimental design verify a lack of ordering directly (but checking for a random distribution of ordering of the three items over some number of iterations). But this follows directly a quote where I mention, instead, that Chen's assertion that there is always preference is an assertion easily tested empirically because subjects would, when asked to order three items, always order them the same.

Taken in that context, I'm very puzzled because it seems as if you are asserting an "initial preferential ordering" that isn't consistent. That seems to me like an initial preferential ordering that isn't. I think I know what you might be getting at, but I can only guess. Can you elaborate?

Finally, you write:

"Chen's argument is not that cognitive dissonance doesn't exist, but merely that experiments to date have failed to prove its existence."

I don't think I ever asserted that Chen's argument is that cognitive dissonance doesn't exist. If I have given that impression, or mistakenly wrote such a thing, it was entirely accidental.

1. The implication being that he doesn't believe this, but he's arguing it for the sake of completeness. "And even if my client were the killer, the witness couldn't possibly have recognized him leaving the house, as she claims." His previous sentence, and choice in arguing in the alternative make it pretty clear that he does believe, in fact, that there is always ordered preference. And even if it didn't, he relies upon there being such an ordered preference much more than he argues that they've failed at eliminating it when they otherwise could have done.

2. That he might be saying otherwise is one way to interpret "but even if subjects can be indifferent (which a discrete rating can never show) the above analysis does not change; the computation simply becomes more difficult."

If this is his claim—that there is de facto ordered preference and the relevant math applies even when there's no initial ordered preference of all three items—then he's making the sort of mistake of overapplying the Monty Hall Problem: two consecutive random choices between two items of three in the way described by the experiment does not mathematically equate to the probability distribution of choosing according to an ordered preference. If there is no ordered preference, then given the first two items the subject will choose randomly with an even distribution, and then given the second two items, the subject will choose randomly with an even distribution. The effect will not exist. You cannot claim that this implies an ordering, because it does not..

Now, I doubt that he's making that error, although you might be. Rather, I interpreted that quoted comment as meaning that sometimes there's preferences that are not accounted for and other times there are no preferences as assumed, given the experimental design. That is to say, sometimes 4 does equal 4 and sometimes 4 doesn't equal 4. That mix would, indeed, "complicate the math".

posted by Dances with Werewolves at 8:16 PM on April 14, 2008

3-color pairwise possibilities beginning with Red over Blue

R, B; B, G; (G, R)
R, B; B, G; (R, G)
R, B; G, B; (R, G)
R, B; G, B; (G, R)

The middle row shows even odds when not making assumptions about a positional ordinal ranking (which makes more sense when major differences are present). But we can't just assume that a monkey would stick with a logical ordering. Such a human bias is how Chen's counter-example ignores the fact that the M&M candies are the same taste and size, and that the monkeys want all of them, but must choose only one (which is consistent with cognitive dissonance). Chen must then assume ranked preferences for these choices, rather than prove them. See the fallacy of misplaced concreteness (defined here as "neglecting the degree of abstraction involved when an actual entity is considered merely so far as it exemplifies certain [preselected] categories of thought. (Note 11: Alfred North Whitehead, Process and Reality: An Essay in Cosmology. New York, 1929, pp. 11).
posted by Brian B. at 9:34 PM on April 14, 2008

You must agree that, like getting two heads in a row in a coin toss, getting two girls happens 1/4 of the time, as does getting two boys. Getting one boy and one girl happens 2/4 of the time. Knowing one is a girl, you limit yourself to the cases "one boy and one girl" and "two girls". But you see that the former is twice as likely as the latter.

TheOnlyCoolTim- Yes, when starting from the beginning you are absolutely correct. But.... once you've already revealed the results of one coin toss, what are the odds that the second toss is heads? It is exactly 50%. The previous flip has no influence whatsoever on subsequent flips. Assuming the gender of a child is determined by a 50-50 chance (an assumption we've been working with all along), there is no reason why knowing the gender of one child is going to affect the gender of another child.

Your profile states your a PhD engineer; think it through. Use some common sense.
posted by Doohickie at 9:22 AM on April 15, 2008

Go back to your original question:

Now, I have two children, and I tell you that the oldest is a boy. What are the odds the other is a boy?

Without knowing anything about your children, the possible combinations are as follows, with the older child listed first):

(B,B), (B,G), (G,B), (G,G), so there is a 25% chance that you have two boys, 50% chance of one of each, and 25% chance of two girls.

With the knowledge that the oldest is a boy, you lose BOTH the third AND fourth ordered pairs. You are left with two choices: (B,B) and (B,G). The odds that the other is a boy is 1 out of 2, or 50%.

If your oldest is a boy, you simply cannot keep the (G,B) combination in play like you keep trying to do.
posted by Doohickie at 9:30 AM on April 15, 2008

You now seem to be conflating the two scenarios?

1: I have two children, and at least one is a girl. What are the odds the other is a boy?

The answer here is 2/3. The statement "you've already revealed the results of one coin toss" is not true here. What's been revealed is something about the combined outcome of both coin tosses, but not the specific outcome of any one. This is why you can only eliminate the (B,B) case.

2: I have two children, and I tell you that the oldest is a boy. What are the odds the other is a boy?

The answer here is 1/2, because the result of one specific coin toss has been revealed, and you may eliminate two cases.

Wikipedia has more.
posted by TheOnlyCoolTim at 10:20 AM on April 15, 2008

Okay.... I didn't look back far enough in the thread and thought you were asking Question 1 from the Wikki (which I repeated in my last post). Further upthread, though, you also asked Question 2.

Having read the Wikki, I'm still trying to get my head around Question 2. Urrhhgh.
posted by Doohickie at 12:27 PM on April 15, 2008

"3-color pairwise possibilities beginning with Red over Blue..."

You know, I had to think about this awhile before I understood what you're getting at. At least, insofar as your chart and your first sentence. For other people who didn't get that comment, the chart presupposes that the preferences we're talking about here are not transitive. That is to say, they can exhibit a relationship such as the three items in the paper, scissors, rock.

Where there's no transitivity, then the possibilities are the same as where there are no preferences at all and there is no mathematical effect Chen is describing.

This does, I think, further weaken Chen's argument because it implies that it's theoretically possible at least that even when there are initial preferences, this experiment isn't vulnerable to this effect (not having been taken into consideration in the experimental design).

On the other hand, this is probably a moot point in reality because all the sets of three things I can think of where I have preferences that are intransitive, I couldn't really describe all three as being alike. It seems highly unlikely to me that any experimenters wouldn't choose items that are extraordinarily alike and thus quite likely to be able to be ranked, when preferences exist.

It seems to me, anyway, that the bottom line of all this discussion is that it's not so much that the experimenters need to account for the correct math involved where there's ranked preferences, it's that experimenters need to be very sure they've eliminated actual preferences from the experiment to a high degree of certainty. Then, to be rigorous, to the degree they are unsure they have eliminated preference, they need to assume transitivity of those undetected preferences—even if that's uncertain—to determine the maximum amount of uncertainty in their results.

Brian, isn't that fallacy the one that critics of economists would probably argue economists are most prone?

I dunno. I think that, overall, I'm moderately dismayed by the entire mess. Given all this, I'm not very confident that experimental psychologists will get the math right, and I'm not very confident that economists will get the psychology right.
posted by Dances with Werewolves at 7:09 PM on April 15, 2008

Brian, isn't that fallacy the one that critics of economists would probably argue economists are most prone?

I believe so. Here are some examples. I find it most in popular armchair analysis. The most famous one was a law in France where workers were given more time off in hopes of giving more people a job.
posted by Brian B. at 7:19 AM on April 16, 2008

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