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February 20, 2015 6:05 PM Subscribe

The Time Everyone “Corrected” the World’s Smartest Woman
posted by brundlefly (290 comments total) 62 users marked this as a favorite

Totally knew this was going to be about her Monty Hall column before I clicked. :)
posted by weston at 6:17 PM on February 20, 2015 [30 favorites]

It's so easy to dismiss something when it seems intuitively obvious to you. When someone who clearly isn't a misinformed goober asserts something that may not immediately conform to your comfortable intuitive ideas, you should at least look into what they're saying before acting like a smug dick.
posted by GoblinHoney at 6:18 PM on February 20, 2015 [10 favorites]

You know, her mailbag selections were actually less sexist than I expected. Some of them were, of course, but there was nary a rape or death threat in the lot. Wonder how come that is.
posted by Countess Elena at 6:19 PM on February 20, 2015 [27 favorites]

Seeing people get the Monty Hall problem wrong makes me want to bash my head into a wall just as a spectator. vos Savant deserves a freaking Nobel Peace Prize for handling that stupidstorm as well as she did.
posted by Holy Zarquon's Singing Fish at 6:19 PM on February 20, 2015 [4 favorites]

Wow, those letters. How far up one's own ass do you have to be to sit down, get out your typewriter and your paper, write a letter like that, address it, stamp it, and mail it without thinking twice about it?
posted by bleep at 6:20 PM on February 20, 2015 [21 favorites]

Ah, there's nothing like the Monty Hall problem to bring smart people to fisticuffs!
posted by Blue Jello Elf at 6:21 PM on February 20, 2015 [6 favorites]

I remember the original shitstorm. I love this problem because the proof is so easy to tabulate, yet most people would rather jump straight to the snark. Ah, human nature.
posted by ZenMasterThis at 6:22 PM on February 20, 2015 [5 favorites]

I usually avoid comments and suggest others do the same, but there's a certain sick joy in watching the exact same kind of pedantry run rampant in the comments section as is described in the article.
posted by averysmallcat at 6:22 PM on February 20, 2015 [8 favorites]

The comments are worth reading for entertainment value.
posted by desjardins at 6:23 PM on February 20, 2015 [2 favorites]

What's the proof, zen master? I mean...religiously speaking
posted by clockzero at 6:24 PM on February 20, 2015

If you pick a goat and switch, you'll get a car every time.
If you pick a car and switch, you'll get a goat every time.
There are two goats and one car.

Thus, 2:1 odds.
posted by Holy Zarquon's Singing Fish at 6:26 PM on February 20, 2015 [49 favorites]

You know, her mailbag selections were actually less sexist than I expected. Some of them were, of course, but there was nary a rape or death threat in the lot. Wonder how come that is.

The letters were actual letters back then. Most of the more aggregious sexist trolls are too cowardly to send traceable paper letters.
posted by EmpressCallipygos at 6:26 PM on February 20, 2015 [36 favorites]

Having watched Let's Make A Deal, I can say that your odds of winning drop to zero unless you come to the studio dressed as a six-foot ham sandwich or something of similar dignity.
posted by delfin at 6:26 PM on February 20, 2015 [42 favorites]

This problem fails to consider the crushing environmental problems associated with owning a car and the sheer bliss of riding a goat.
posted by Lemurrhea at 6:29 PM on February 20, 2015 [111 favorites]

Even Cecil Adams was wrong initially about this one but at least he was big enough to admit it.
posted by inthe80s at 6:30 PM on February 20, 2015 [4 favorites]

DAMN YOU MONTY HALL
posted by Sticherbeast at 6:32 PM on February 20, 2015

The beautiful thing about this is that it shows what self-regarding fools a lot of people who make a big fuss about being intelligent or logical are.
posted by thelonius at 6:34 PM on February 20, 2015 [12 favorites]

Does the host know what's behind the door he opens? Does that change the odds? The possible outcomes in the article suggest a chance that the host would reveal a car, but that doesn't make sense to me.

In the original Monty Hall Problem as posed by vos Savant, the host knows where everything is and will only open goat doors. There are variants of the problem where the pre-trade door opening is random, and I think that's supposed to change the odds somehow but I don't see how that could be, because if the host reveals a car then you know your odds of winning are zero and just take your goat.
posted by Holy Zarquon's Singing Fish at 6:35 PM on February 20, 2015 [4 favorites]

The best verbal explanation I've come up with is to factor the choice about switching / staying put out of the problem and put it up front as a strategy. So you have two kinds of players, the "switchers" and the "stayputniks". A stayputnik wins if-and-only-if he picks the prize on his initial guess. A switcher wins if-and-only-if he does *not* pick the prize on his initial guess. So the stayputnik wins 1-out-of-3 and the switcher wins 2-out-of-3.

But the really sad thing about the whole Marilyn kerfuffle, or about anyone arguing and yelling about this problem, is that it is so utterly trivial to test your answer to the puzzle. Any of those people who wrote letters could have tested their conclusions about as quickly as they wrote their letters and saved themselves a bunch of embarrassment. A couple of minutes at most with either dice or a buddy or a ten line computer program. It's a great example for talking about theory versus experiment. You should always test one versus the other, and if your theory and your experiment disagree then something is borked and you need to think harder.

If these guys were rocket scientists, I think they would have written down their equations and built their rockets without testing anything, and then happily watched the rockets crash while pointing at their nifty pieces of paper.
posted by madmethods at 6:36 PM on February 20, 2015 [39 favorites]

You know, her mailbag selections were actually less sexist than I expected. Some of them were, of course, but there was nary a rape or death threat in the lot. Wonder how come that is.

Oh, I think they were probably sent. But newspapers were pretty used to throwing away crank letters.

This problem does give me severe cognitive dissonance as someone who is not great at math, the chart in the article really helped.
posted by Drinky Die at 6:37 PM on February 20, 2015 [2 favorites]

The beautiful thing about this is that it shows what self-regarding fools a lot of people who make a big fuss about being intelligent or logical are.

I think this could be called the I Fucking Love Science problem.
posted by ambrosen at 6:37 PM on February 20, 2015 [11 favorites]

The only probability math I can get my head around is the 1:1 probability that I will not get it right.

"Never tell me the odds." - Han Solo
posted by wabbittwax at 6:41 PM on February 20, 2015 [4 favorites]

In the original Monty Hall Problem as posed by vos Savant, the host knows where everything is and will only open goat doors. There are variants of the problem where the pre-trade door opening is random, and I think that's supposed to change the odds somehow but I don't see how that could be, because if the host reveals a car then you know your odds of winning are zero and just take your goat.

Yes, the host knowing is a necessary part of the problem. They never would open the first door to reveal the big prize. There is therefore zero probability that the first reveal would show the car or trip or whatever, so it is a non-random event and doesn't change the probability that you picked the wrong door. If the host didn't know what door the big prize was behind, then opening the first door would even out your odds, as I understand it, because there would be a possibility that the prize was behind it.
posted by middleclasstool at 6:41 PM on February 20, 2015 [7 favorites]

I remember this one and the kerfuffle. When I first read it, I jumped right to the "intuitive" 50-50 chance solution. When she posted that she was indeed correct, I had to back up and re-approach the problem...and finally the light bulb went on (with her help, admittedly).

Like a lot of math, you need to check your work/solution.

I wonder where the various Ph.D. nastygram writers are these days? Needs folo.
posted by CrowGoat at 6:41 PM on February 20, 2015 [1 favorite]

Seeing people get the Monty Hall problem wrong makes me want to bash my head into a wall just as a spectator.

Aw, man. As someone who couldn't wrap my head around the Monty Hall problem for like the first 20 years of my life, that seems pretty over the top. I hope you're not a stats & prob teacher.

The reasoning that finally made me understand is similar to Holy Zarquon's, above - let's say you picked the car first (expected 1/3 of the time). If you switch you lose. Now, if you chose a goat first (2/3 of the time), you cannot choose the other goat if you switch because he just showed it to you! So you are guaranteed to win if you switch in that situation, i.e., 2/3 of the time.
posted by Joey Buttafoucault at 6:44 PM on February 20, 2015 [49 favorites]

The key to understanding this problem is:

1) It's true that given two choices and no other information either choice gives you a 50% of being right.
2) We have "other information" in the Monty Haul problem so the above does not apply.
posted by Justinian at 6:44 PM on February 20, 2015 [7 favorites]

I have sympathy for some of the writers because I was a Monty Hall doubter myself for a long time. It's all in how the question is framed. From the perspective of the player, is the second offer (I removed a wrong door, now will you switch) a modification of the first offer or is it a new, completely independent choice to make? If you view it as the latter, discounting whatever machinations Monty has up his sleeve, pick A or pick B certainly APPEARS to be a toss-up. Your first pick doesn't matter because Monty changed the situation and now you only have two choices, and how can that NOT be a 50-50 bet?

The logical extension is to imagine there are 1,000 doors instead. You pick #147, Monty says "okay, I will open every door except #147 and #842. The doors I'll open are all goats. Will you switch?" The same end result -- pick A or pick B -- but for you to keep your original choice and win, you'd have to have beaten 999-to-1 odds on your initial pick. So you are choosing either one door or, in effect, all 999 others. And so with three doors, the fact that one door is a known loser does not change that Monty is effectively offering you either your one door or both of the others, and since you had a 1/3 chance of getting it right from the start, the other 2/3 have only one place to go.

Now, those who have watched the show know that not all of the "zonks" were, in fact, goats. Sometimes there were intermediate-level prizes that were worth having in and of themselves. Sometimes he'd get you to the edge of making your decision and say "how about this, you can switch, you can keep your door OR I'll give you $2,000 right now and you can go home" just to mess with you. But figuring out the psychology of a carnival barker like Monty Hall requires an IQ of a lot more than three digits.
posted by delfin at 6:44 PM on February 20, 2015 [18 favorites]

If the host didn't know what door the big prize was behind, then opening the first door would even out your odds, as I understand it, because there would be a possibility that the prize was behind it.

Except that if there are only two options - big prize or booby prize - and the host reveals the big prize, then by definition every other door is a booby prize and the game ends there and then with no further decisionmaking on the player's part. So even with a blind host, in every scenario where you have a choice to make, the odds should still be 2:1 in your favor if you switch doors.
posted by Holy Zarquon's Singing Fish at 6:45 PM on February 20, 2015

weston: "Totally knew this was going to be about her Monty Hall column before I clicked. :)"

Well, 66% chance I would get it right. But did I?
posted by Samizdata at 6:46 PM on February 20, 2015 [6 favorites]

I had to look up E Ray Bobo, now retired from Georgetown. Dr. Bobo. She could have been snarky and been all like,"Hey Dr Dodo" hehe
posted by discopolo at 6:46 PM on February 20, 2015 [4 favorites]

I'll admit that I got this wrong the first time I encountered it (it is not vos Savant's invention). The first explanation that penetrated the dim crevasses of my brain was based around the idea that, after a door is opened, the contestant has information that he did not have before - there's a goat behind the door that was opened. This seems trivial to the incorrect solver at first - you know that the car is behind the door you picked or the one that stayed closed, so you have, they reason, a 50% chance of being right, and switching is pointless. But Hall was constrained: he couldn't open the door with the car. So, in 2 out of 3 cases, the door he did not open has the car, and switching is better.
posted by thelonius at 6:48 PM on February 20, 2015 [5 favorites]

Sticherbeast: "DAMN YOU MONTY HALL"

STICHERBEAST IS NOT PLEASED
posted by Samizdata at 6:49 PM on February 20, 2015 [2 favorites]

This reminds me of all the times I was 100% convinced I was 100% right and then it was pointed out to me that I was wrong.
posted by night_train at 6:50 PM on February 20, 2015 [18 favorites]

The logical extension is to imagine there are 1,000 doors instead. You pick #147, Monty says "okay, I will open every door except #147 and #842. The doors I'll open are all goats. Will you switch?"

Hrmm... Well an interesting thought occurred to me. Let's say you think of the problem this way: yes, you pick 147. But then can't we say Monty has picked 842? And if 147 has a 1/1000 a priori chance of having the money behind it, wouldn't 842 have exactly the same chance? And if so, why should the probabilities contained behind the doors being opened "redistribute" to 842 instead of to 147?

I know they do -- there's an asymmetry -- but what is it?
posted by shivohum at 6:51 PM on February 20, 2015

I have this weird affection for the Monty Hall problem. Partly because it's so fun to watch someone finally understand it.
posted by rmd1023 at 6:52 PM on February 20, 2015 [4 favorites]

The really interesting question with this problem is exactly WHY people's intuitions are so wrong. I think there's a little unstated assumption in people's minds as to what's going on and why. I can't quite figure out what that is, however.

It seems like the most common wrong intuition is this:

1. I have a 1/3 chance of getting the car from the outset.
2. One door is opened, and thus I can remove that element from the equation (so to speak)
3. Since there are only two unopened doors now, and I know one has the thing I want and the other doesn't, I have a 50/50 chance no matter what I do.

But this is incorrect, precisely because it doesn't treat the situation as a probabilistic one. And that points out the weakness we have congenitally for thinking probabilistically: we are geared to pattern-match, as humans, and reasoning backward from effect to cause is one way in which we do that. Sometimes outcomes affect odds, but sometimes they don't, and we frequently get tripped up trying to figure out which kind of situation we're dealing with. I think that's why this particular problem is such a stumper: both the choice to switch and the reveal of new information happen at the same time, so you're then trying to figure out what kind of problem it is while simultaneously being asked to bet on which type it is.
posted by clockzero at 6:52 PM on February 20, 2015 [11 favorites]

> the I Fucking Love Science problem

Personally, I don't see this as a problem. I'd rather live in a world where people argue passionately about science, even when they are wrong, than one where "loving" science is considered a personality flaw.

This isn't to excuse asshole-ish behaviour, obviously, but we hardly need science to supply the fodder for that.
posted by Poldo at 6:52 PM on February 20, 2015 [5 favorites]

Monty always has to give you a choice between a goat and a car. So if you picked a goat and swap, you will always get a car. If you picked a car and swap, you will always get a goat. So the chance that swapping will give you a car is the same as the chance that your initial pick was a goat.
posted by Holy Zarquon's Singing Fish at 6:54 PM on February 20, 2015 [7 favorites]

I also got Monty Hall wrong the first time I encountered it (which was through vos Savant's column and the ensuing controversy)... and I also was smug enough to think I was right and she was wrong. My only defense is that I was about 9 at the time, unlike all those PhDs who wrote in to "correct" her.

What troubles me is that, up until I saw this post, I remembered this event as "the time 'the world's smartest woman' got that probability puzzle wrong" -- although I now know (and have known for a long time) the correct solution to the puzzle. My memory had vos Savant arguing that the probability was 1/2! Ugh, I'm a bit disgusted with myself right now.
posted by aws17576 at 6:56 PM on February 20, 2015 [4 favorites]

Hrmm... Well an interesting thought occurred to me. Let's say you think of the problem this way: yes, you pick 147. But then can't we say Monty has picked 842? And if 147 has a 1/1000 a priori chance of having the money behind it, wouldn't 842 have exactly the same chance? And if so, why should the probabilities contained behind the doors being opened "redistribute" to 842 instead of to 147?

The simplest reason of all -- as the host, Monty gets to cheat and look up the answer BEFORE he makes his pick. Monty's chances of getting it right are 100%.
posted by delfin at 6:57 PM on February 20, 2015 [1 favorite]

The best verbal explanation I've come up with is to factor the choice about switching / staying put out of the problem and put it up front as a strategy. So you have two kinds of players, …

This really helped, thanks.
posted by migurski at 6:59 PM on February 20, 2015 [1 favorite]

This problem fails to consider the crushing environmental problems associated with owning a car and the sheer bliss of riding a goat.

On a related note...
posted by Shmuel510 at 6:59 PM on February 20, 2015 [4 favorites]

Conditional Probability.
posted by ZenMasterThis at 7:00 PM on February 20, 2015 [2 favorites]

The really interesting question with this problem is exactly WHY people's intuitions are so wrong. I think there's a little unstated assumption in people's minds as to what's going on and why. I can't quite figure out what that is, however.

To me, it comes from the somewhat subtle fact that Monty will always open a door with a goat behind it as part of his reveal before asking you whether you want to switch. When he does so, he is giving you extra information that you didn't have back when you made your initial selection. That's crucial, because getting extra information means reconsidering your probabilities in light of that new information. That's what Bayesian Statistics is all about and people have applied it to the Monty Hall Problem if you like theory. In many tellings of the Problem, that detail is fairly subtle and easy to miss. It also doesn't appear particularly important. That's the simplest explanation: you think you and Monty are playing the same game, but really it's rigged and he's the puppetmaster controlling the strings.

Now there's an interesting thing about Bayesian problems: we're really really freaking bad at using intuition with them. Eliezer Yudkowsky explains this all better than I could, so we'll just use one of the classic examples:

1% of women at age forty who participate in routine screening have breast cancer. 80% of women with breast cancer will get positive mammographies. 9.6% of women without breast cancer will also get positive mammographies. A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?...

Next, suppose I told you that most doctors get the same wrong answer on this problem - usually, only around 15% of doctors get it right. ("Really? 15%? Is that a real number, or an urban legend based on an Internet poll?" It's a real number. See Casscells, Schoenberger, and Grayboys 1978; Eddy 1982; Gigerenzer and Hoffrage 1995; and many other studies. It's a surprising result which is easy to replicate, so it's been extensively replicated.)

Most doctors said 70-80% (I haven't researched this, but perhaps better education in the last 20+ years has improved the situation?). The correct answer is 7.8%. We're really bad at intuiting anything about conditional probability. Even if you realize that Monty is giving us new information when he opens that door, you probably don't intuitively know what to do with that new information in light of your existing views of the probabilities.
posted by zachlipton at 7:00 PM on February 20, 2015 [33 favorites]

YOU BLEWWWW IIIIIIIIIIIITTT
posted by Evilspork at 7:01 PM on February 20, 2015

By the way, I wholeheartedly endorse Persi Diaconis's comment that probability is one area where even experts can easily be fooled. This was demonstrated to me in grad school when my advisor, addressing a roomful of mathematicians, posed this problem:

Person A flips a coin repeatedly, stopping the first time two heads in a row appear.
Person B flips a coin repeatedly, stopping the first time a head and then a tail appear in a row.
Who will flip the coin more times on average -- A, B, or is there no difference?

He let everyone think for a moment, then took a show of hands. Almost everyone got it wrong.
posted by aws17576 at 7:02 PM on February 20, 2015 [6 favorites]

This problem does give me severe cognitive dissonance as someone who is not great at math, the chart in the article really helped.

Yeah, I can't get it without the chart. And possibly I get confused by people not wanting to get a free goat. I mean, come on, people!
posted by Alvy Ampersand at 7:03 PM on February 20, 2015 [4 favorites]

Actually, a sa kid, riding a goat didn't seem like a bad deal at all: see Johan et Pirlouit (et Biquette).
posted by Vindaloo at 7:04 PM on February 20, 2015 [1 favorite]

Thelonious: "But Hall was constrained: he couldn't open the door with the car." ... The key here is that he's ALSO constrained by not being able to open the door you've chosen.
posted by Xyanthilous P. Harrierstick at 7:08 PM on February 20, 2015 [8 favorites]

This is a tough one for intuition! I think the table of possibilities is the real way to solve it, but I can rationalize it this way:

Three doors. Each as a 1/3 chance of being a car.

Consider door #1 vs (#2 and #3). There's a 2/3 chance the car is behind (#2 or #3).

Door #3 is opened to reveal a goat. That doesn't change the fact that the chance of (#2 or #3) is still 2/3.

Therefore the chance of #1 is 1/3 and the chance of #2 is 2/3.
posted by Ansible at 7:09 PM on February 20, 2015 [4 favorites]

Here's an interesting story, though I'm not sure of the broader moral statement we should take from it.

Take this question: "There are two contestants (instead of one); Monty Hall will always reveal one of the contestants to have chosen the wrong door (that contestant is then eliminated), and then give the other the chance to switch. You are the contestant who was not eliminated. Should you switch?"

The answer is the opposite as in the traditional problem, for exactly the same reason as in the original problem. The people who get the answer to the original problem right however, will often get this problem wrong. I'm not sure why.
posted by The Notorious B.F.G. at 7:13 PM on February 20, 2015 [3 favorites]

Person A flips a coin repeatedly, stopping the first time two heads in a row appear.
Person B flips a coin repeatedly, stopping the first time a head and then a tail appear in a row.
Who will flip the coin more times on average -- A, B, or is there no difference?

I like this one but I've never quite been sure how to form a general principle from it. (I guess thinking about how many steps back failure takes you?)
posted by PMdixon at 7:18 PM on February 20, 2015

When I read this in the Straight Dope oh so many years ago, I jumped on the bandwagon. I don't think I said anything horribly sexist, but I didn't call anybody else out on saying it, for which I feel bad.

I have to say, a big part of the issue for me was somebody calling themselves "Vos Savant" and claiming to have the highest IQ ever recorded.

None of which makes the claim any less true, which was a good lesson for me in my 20s.
posted by lumpenprole at 7:20 PM on February 20, 2015 [1 favorite]

Gotta say, whatever math cred MvS gained on Monty Hall, she lost almost immediately with her book about Fermat's Last Theorem.
posted by escabeche at 7:21 PM on February 20, 2015 [9 favorites]

Oh God, the shitstorm over this took *forever* to clear on sci.math (back in the days when rn and trn ruled the earth). But as discussed above, a lot of it has to do with the way the question is framed. If you don't make it clear that the host always reveals a goat --that is: Monty's choice is *not* made randomly-- the entire analysis is flipped on its head. I'm sure that this is what was trapping a lot of the respondents in the initial wars.

Not all, however: seems there's often a reflexive desire to prove someone else's assumptions wrong in probability problems. In particular, to embrace that "gotcha" moment when you catch someone forgetting that probability is memoryless.
posted by pjm at 7:22 PM on February 20, 2015

yeah, you people are so smart, but it's DREW CAREY that opens the door now!!

i declare you all invalid and dated
posted by pyramid termite at 7:24 PM on February 20, 2015 [5 favorites]

oh my god i finally get it
posted by KathrynT at 7:26 PM on February 20, 2015 [32 favorites]

There are variants of the problem where the pre-trade door opening is random, and I think that's supposed to change the odds somehow but I don't see how that could be, because if the host reveals a car then you know your odds of winning are zero and just take your goat.

Thinking about this for a minute, I think it doesn't change the odds (except that there's now a chance that the whole thing gets cancelled because the car was revealed). I said earlier that the idea that got through to me was that Hall is constrained, but I think that's a bit misleading. The essential thing is that the player started picking from 3 doors, but now has a situation of picking from 2 doors.

Suppose the player has to leave the room and return. There's an open goat door when he gets back, and he can choose to switch or stay. He doesn't know if the door was opened at random, or if someone who knows where the car is opened it. The logic of switching still holds, I think.
posted by thelonius at 7:30 PM on February 20, 2015 [1 favorite]

Monty always has to give you a choice between a goat and a car.

Was this always in the original wording? Does anyone have a copy of the wording?

The simplest reason of all -- as the host, Monty gets to cheat and look up the answer BEFORE he makes his pick. Monty's chances of getting it right are 100%.

Given the above two comments, I think I'm starting to see the real problem. I think people -- or maybe just some people, maybe just me -- have an incorrect intuition about the selection procedure. It's probably somehow embedded in a common-sense understanding of language.

Imagine that people have the intuition that the selection procedure worked this way:

Contestant chooses Door #1.
Hall then actually RANDOMLY chooses a door -- any of them. And let's say he has no requirement to show you a door with a car in it if you've chosen a door with a goat. And let's say he only offers the switch depending on the door he's chosen.

So in perhaps a hazily imagined, common-sense selection procedure, there are actually three possible worlds:

Universe 1: you've picked the car (33%)
In world 1 (33%) chance, Hall picks your door - and doesn't offer you a switch.
In world 2 (67%) chance, Hall picks one of the other doors with a goat in it and offers you a switch.

Universe 2: you've picked a goat (67%)
In world 3 (33%) chance, Hall picks your door - and doesn't offer you a switch.
In world 4 (33%) chance, Hall picks one of the other doors with a car in it and offers you a switch.
In world 5 (33%) chance, Hall picks one of the other doors with a goat in it and offers you a switch.

So perhaps people reason this way: Given the fact that Hall has opened another door with a goat in it, what are the odds that you've chosen the door with the car in it?

Probability (Universe 1 | Hall picks a door you didn't and which has a goat) = P(Hall picks a door you didn't and which has a goat | Universe 1) * P(Universe 1) / P(Hall picks a door you didn't and which has a goat)

= (.67)*(.33) / (.33*67) + (.67*33) = .5

So it's 50/50 based on that.

And I think that's where the confusion comes from.
posted by shivohum at 7:32 PM on February 20, 2015

lumpenprole: I have to say, a big part of the issue for me was somebody calling themselves "Vos Savant" and claiming to have the highest IQ ever recorded.

Vos Savant was her mother's surname. Apparently she's an advocate of boys taking their father's surname and girls their mother's.
posted by brundlefly at 7:32 PM on February 20, 2015 [7 favorites]

Not all, however: seems there's often a reflexive desire to prove someone else's assumptions wrong in probability problems. In particular, to embrace that "gotcha" moment when you catch someone forgetting that probability is memoryless.

For me, the "memorylessness" of probability was what tripped me up the first time I encountered the problem. It's easy to understand that your probability of picking the car on the first round is 1/3, and it's easy to understand that of the two doors you're left with in the second round, one is the goat and one is the car. The unintuitive part is that you retain those 1/3 odds if you keep your first pick, but instead have the 1/2 odds if you switch.
posted by kagredon at 7:32 PM on February 20, 2015 [1 favorite]

Except that if there are only two options - big prize or booby prize - and the host reveals the big prize, then by definition every other door is a booby prize and the game ends there and then with no further decisionmaking on the player's part. So even with a blind host, in every scenario where you have a choice to make, the odds should still be 2:1 in your favor if you switch doors.

No, it still matters very much whether Monty is choosing a door with a goat, or just choosing randomly. If Monty isn't acting with insider information, he can't convey any information to the contestant by his actions.

Or think of it another way: In your initial choice, there's a one-third chance you choose right. If you choose wrong and Monty can only open a door with a goat, then Monty gives you information about the remaining door. If Monty chooses at random, then he has a 50-50 chance of revealing the car, in which case you lose immediately.

So in the random scenario there are three equally likely outcomes:

1/3 of the time you will choose right initially
1/3 of the time you will choose wrong initially, and Monty will reveal a goat. (2/3 probability of wrong * 1/2 probability of revealing goat = 1/3)
1/3 of the time you will choose wrong initially, and Monty will reveal a car. (2/3 probability of wrong * 1/2 probability of revealing car = 1/3

So if Monty has revealed a goat, you can throw out the third possible outcome, leaving you with two equally likely outcomes, only one of which is a win for the switching strategy, for a 50-50 chance of winning by switching.
posted by firechicago at 7:33 PM on February 20, 2015 [1 favorite]

Of course, the right answer is the one Diaconis gives. When I teach this problem in class, I use it to make exactly this point.

“[But] the strict argument would be that the question cannot be answered without knowing the motivation of the host."

vos Savant was not right, nor were her critics. The answer to the question isn't determined unless you specify what rules Monty is playing by. If Monty doesn't know where the car is, and picks door 2 or 3 at random, the situation is as follows. There are six equally probable options:

car 1, Monty 2
car 1, Monty 3
car 2, Monty 2
car 2, Monty 3
car 3, Monty 2
car 3, Monty 3

There's a 1/3 chance that Monty reveals the car when he opens the door, and a 2/3 chance that he does not. The observation that the door Monty opens has a goat behind it eliminates the third and sixth items on the list, leaving

car 1, Monty 2
car 1, Monty 3
car 2, Monty 3
car 3, Monty 2

In half of these scenarios, the car is behind the door you've chosen; in the other half, it's behind the other closed door.

Of course, this is not the problem vos Savant had in mind; she is using a different decision rule for Monty, where he knows the location of the car and is certain not to reveal it. And this makes a difference!

This is always a difficult point for students to grasp, maybe the most difficult in the whole course: that the correct determination of the probability that the car is behind door 1 depends not only on what you observe, but on what you might have observed but did not.

Diaconis gets this. It's not totally clear to me that either vos Savant or her angry critics did.

(xp: OK, three other people just said exactly this, but I'm leaving this comment up anyway, having taken the time to write it....)
posted by escabeche at 7:35 PM on February 20, 2015 [4 favorites]

Vos Savant was her mother's surname. Apparently she's an advocate of boys taking their father's surname and girls their mother's.

I wonder if that would still be the case if her mom's surname was Dipstick or Numbnuts.
posted by Alvy Ampersand at 7:36 PM on February 20, 2015 [9 favorites]

Hall then actually RANDOMLY chooses a door -- any of them. And let's say he has no requirement to show you a door with a car in it if you've chosen a door with a goat. And let's say he only offers the switch depending on the door he's chosen.

Hall never picks the door that you picked, though, and that is clearly stated as part of the problem: he reveals one of the other doors. Even the hypothetical where Hall has no prior knowledge of the doors only changes things in that you have a chance (1/6) of Hall revealing the car and an automatic goat prize.
posted by kagredon at 7:38 PM on February 20, 2015

I wonder if that would still be the case if her mom's surname was Dipstick or Numbnuts.

At Ellis Island all the Dipsticks became Drapers and the Numbnuts became Nortons. It's not an issue here in the good ol' U.S. of A.
posted by Pater Aletheias at 7:38 PM on February 20, 2015 [4 favorites]

The only thing more improbable than a long Numbnuts family line is a long Numbnuts family line on the distaff side.
posted by aws17576 at 7:39 PM on February 20, 2015 [2 favorites]

The Connecticut Numbnuts' or the New Hampshire?
posted by axiom at 7:41 PM on February 20, 2015 [1 favorite]

Is this something I'd have to own a goat to understand?
posted by not_on_display at 7:49 PM on February 20, 2015 [1 favorite]

Who will flip the coin more times on average -- A, B, or is there no difference?

I like this one but I've never quite been sure how to form a general principle from it. (I guess thinking about how many steps back failure takes you?)

I like to think of it in terms of clustering: in a series of coin flips, HH and HT events occur equally often in the long run but HH events tend to occur in bursts (since they can chain together), so those bursts are farther between. To take a less abstract example of the same thing, it might rain 36 days a year (which is 1 out of 10 days) in L.A., but if you pick a random date to start waiting for rain, your average wait will be much longer than 10 days.

(Hopefully this isn't too much of a derail, but if you like this problem, a harder version is to work out how long it will take on average for a monkey typing letters at random to type out ABRACADABRA. There's a pretty amazing way to solve this -- the "standard solution" mentioned here.)
posted by aws17576 at 7:50 PM on February 20, 2015 [1 favorite]

An exercise proposed by vos Savant to better understand the problem was soon integrated in thousands of classrooms across the nation. Computer models were built that corroborated her logic, and support for her intellect was gradually restored. Whereas only 8% of readers had previously believed her logic to be true, this number had risen to 56% by the end of 1992, writes vos Savant; among academics, 35% initial support rose to 71%.

To me, this is the really interesting part of the article and I wish they'd included even more information. 71% seems like a high figure, and then you start thinking about the remaining 29%.

Total anecdata, and I'm aware of all the basic mathematical and cognitive reasons we tend to estimate similar splits that actually have nothing to do with the real data (same with Pareto), but damned if it doesn't feel like "the reliably crazy 30%" is totally a real phenomenon. At least with the stats for "laypeople", you can quibble over the idea that maybe even with the results in front of them, they aren't capable of parsing it or making the connections, but with academics, that falls apart.
posted by The Master and Margarita Mix at 7:50 PM on February 20, 2015 [1 favorite]

vos Savant was not right, nor were her critics. The answer to the question isn't determined unless you specify what rules Monty is playing by.

Except, of course, that the question vos Savant replied to did specify those rules.

From TFA:

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.

In order to make your claims that she "was not right" and "it's not totally clear" that she grasped the need for this rule—that the host's selection is not random, but will always reveal a goat—you need to ignore the rather clear wording given, or postulate an entirely different universe, in which the rules weren't specified.

This is precisely the sort of hogwash that is the subject of the article.
posted by Shmuel510 at 7:59 PM on February 20, 2015 [21 favorites]

I tried to explain to my wife that the odds are not 50/50. Not only does she still think it's 50/50, she's mad at me and I'm not allowed to bring it up again.
posted by tommyD at 8:02 PM on February 20, 2015 [5 favorites]

Alvy Ampersand: I wonder if that would still be the case if her mom's surname was Dipstick or Numbnuts.

Well, I would be in favor of it myself.
posted by brundlefly at 8:02 PM on February 20, 2015

Imagine if vos Savant had said something really provocative, like "0.9999... = 1" !
posted by ssr_of_V at 8:03 PM on February 20, 2015 [5 favorites]

This is like every knucklehead watching F-News feels smarter than POTUS, like any Phd is smarter than a woman, but the first descriptor that lept to mind when reading this was, what a bunch of goobs. Then the second or third post down specifically mentioned goobers. Intuitively I knew there would be goobs. I don't get the math, even with the chart, but I knew there would be goobers.
posted by Oyéah at 8:07 PM on February 20, 2015

The door can be opened randomly, and, if it reveals a car, everyone forgets the whole thing and goes home. This failure possibility plays the same role as does Hall knowing where the car is, and not opening that door. If, in the random door variant, you get as far as choosing to switch or stay, you are in exactly the same position that you are in if the host knowingly avoided the door with the car: there is a 1/3 chance that you were right originally, and a 2/3 chance that the car is behind the unopened door.
posted by thelonius at 8:07 PM on February 20, 2015

I really only just now got this. And I paid off my house by advantage gambling, ha.

When you pick your door, it might be a car or a goat, but Monty always picks a goat.

If you are already on the car, which happens 1/3 of the time overall, Monty has picked one of two goats leaving you with a 50/50 shot.

But if you are on a goat, which happens 2/3 of the time overall, Monty has locked up the other one, and jumping gives you the car every time.

Most of the explanations of this thing really suck.
posted by localroger at 8:12 PM on February 20, 2015 [6 favorites]

Explaining it shouldn't be hard.

2/3 time you = goat
1/3 time you = car

Monty always eliminates 1 goat.

So you by switching:
You = goat, you = car!
You = goat, you = car!!
You = car, you = goat, why u so stoopid
posted by OnTheLastCastle at 8:14 PM on February 20, 2015 [4 favorites]

If, in the random door variant, you get as far as choosing to switch or stay, you are in exactly the same position that you are in if the host knowingly avoided the door with the car: there is a 1/3 chance that you were right originally, and a 2/3 chance that the car is behind the unopened door.

Why doesn't escabeche's summary prove this wrong?
posted by shivohum at 8:16 PM on February 20, 2015

I've always had a shit ton to say about people's abilities AFA probability goes but...

I'd fucking love to sit down with an array of CEOs on the planet and set out the Monty Hall problem for them. I already know they objectively lose a sense of time by being in power, but their grasp on probability I'm hesitant to say is any better than the average person.

I think coaching on vertical, creative and lateral thinking would be helpful to them.
posted by OnTheLastCastle at 8:16 PM on February 20, 2015 [1 favorite]

Oh, I read up. Monty in the problem ALWAYS reveals a goat. That is explicit, you all.

In the real game show, he did not have to. He could offer money to stay.
posted by OnTheLastCastle at 8:18 PM on February 20, 2015 [1 favorite]

shivohum: And if so, why should the probabilities contained behind the doors being opened "redistribute" to 842 instead of to 147?

In my amateur way, I dislike the "redistribution" way of thinking about this. It seems to me that many people who are confused by the problem have a deep-rooted intuition, an invisible-to-them assumption, that there is a redistribution of odds when some number of doors are taken out of the running. There isn't any redistribution, though. That's exactly why switching works better. That's exactly why some people are confused. It seems to me that it's the crux of the problem.

I've always thought about it this way: The player's selection divides n doors into two groups: the Selected (1 door, 1/n odds of winning) and the Rejected (n-1 doors, (n-1)/n odds of winning).

The winning odds of those groups, selected some moments before the host starts opening booby prizes, cannot change because of some later event (like the opening of doors). The present doesn't alter the past. "Redistribution" of the odds is impossible without re-shuffling the prizes.

However, once the host of the game eliminates n-2 booby-prize doors, it is suddenly possible for the player to choose the much-higher winning odds of the entire Rejected group by choosing a single door.

That's a consequence of the way the rules (pick only one, optionally change after elimination of booby prizes) interact. If the player were instead allowed to switch his choice to the complete set of n-1 Rejected doors, no one would be confused about why it's better to switch. The host wouldn't even have to bother opening any booby prizes. But that is the situation the player really faces.

I not sure the host's opening of the doors adds any information; the player doesn't actually have to know which doors were eliminated to say "yes" when asked if he'd like to switch.
posted by Western Infidels at 8:20 PM on February 20, 2015 [1 favorite]

thelonius: The door can be opened randomly, and, if it reveals a car, everyone forgets the whole thing and goes home. ... you are in exactly the same position that you are in if the host knowingly avoided the door with the car.

Just to be clear: this is true if by "opened randomly" you mean that Monty randomly chooses a door from the three (including the one that you picked initially). It's not true if Monty's choice is restricted to the two doors that you did not choose.
posted by firechicago at 8:22 PM on February 20, 2015 [1 favorite]

If you pick a goat and switch, you'll get a car every time.
If you pick a car and switch, you'll get a goat every time.
There are two goats and one car.

Thus, 2:1 odds.

I'm generally shit at math and this explanation got the idea across very clearly. Thank you!
posted by Pope Guilty at 8:23 PM on February 20, 2015 [5 favorites]

but what if the goat is on a treadmill, but with a strong head wind
posted by DoctorFedora at 8:24 PM on February 20, 2015 [25 favorites]

Everyone should read escabeche's link about her criticism of the proof of Fermat's Last Theorem; vos Savant's grasp of math was good enough to get her to the bottom of the Monty Hall problem alright, but imploded like a Christmas ornament in the Marianas trench when she tried to come to grips with Wile's proof of FLT.
posted by jamjam at 8:24 PM on February 20, 2015

there is a redistribution of odds when some number of doors are taken out of the running. There isn't any redistribution, though.

There is, though. The odds from door 2 being a car get shifted to door 3 -- doors 1, 2, 3 each had a 1/3 chance, and now door 2 is out, so that 1/3 had to go SOMEWHERE.

The thing I always found weird was people who were only convinced by a simulation where they tested 10,000 trials. That doesn't feel like it explains why switching is good.
posted by jeather at 8:28 PM on February 20, 2015

Here's the part that I always love to wrap my head around. Say you're playing the standard game, but as you're about to pick your door for the first time, Monty lets it slip that there's a goat behind door #2. Thankful for the hint, you pick door #1 or #3 and you have a 50/50 chance of getting the car.

At first glance, that seems like exactly the same situation as the standard game. In either case, the host is giving you the additional information that there's a goat behind one of the three doors. The fact that you've already picked your door when you get that hint shouldn't matter. The catch is that, when you pick before the hint, Monty isn't randomly dropping that hint, he's giving you a hint constrained to the two doors you didn't pick. It seems like it shouldn't matter, but it actually makes his hint more useful.
posted by zachlipton at 8:32 PM on February 20, 2015 [7 favorites]

I've come to the conclusion that the only way to win the Monty Hall Problem is to not play it in the first place.

Either you're in the majority (and wrong), or you're in the minority (and pissing off the majority to no end).
posted by GhostintheMachine at 8:39 PM on February 20, 2015 [1 favorite]

Okay but here is the real question: if there is a chance that the goat behind the door Monty picks is dead, and he doesn't know whether or not the goat is dead until he opens the door, does the goat exist in a dead/alive superposition of states until the door is opened?
posted by BlueJae at 8:46 PM on February 20, 2015 [4 favorites]

Okay but here is the real question: if there is a chance that the goat behind the door Monty picks is dead, and he doesn't know whether or not the goat is dead until he opens the door, does the goat exist in a dead/alive superposition of states until the door is opened?

Maybe.
posted by Pope Guilty at 8:46 PM on February 20, 2015 [2 favorites]

Also, if logic problem A involves two goats and logic problem B involves one cat, which logic problem is cuter?
posted by BlueJae at 8:47 PM on February 20, 2015 [2 favorites]

To me, I understand this based on, it's *not* rolling dice. The thing behind the door does not change from choice one to choice two. So for me it works intuitively. (Though I'm sure if I think about it too long / hard I'll get confused.)
posted by ClaudiaCenter at 8:53 PM on February 20, 2015

BlueJae, are they baby goats?
posted by Pope Guilty at 8:59 PM on February 20, 2015 [2 favorites]

automatic goat prize

automatic goat prize

automatic goat prize
posted by clockzero at 9:01 PM on February 20, 2015 [9 favorites]

Monty Hall is one of the great problems for finding out just how good of a teacher someone is.

Even if you both agree on the right answer, the ability to effectively explain why it works the way it does is fairly rare. It requires not only clear thought on the teacher's part but the ability to understand how the other person perceives the problem.

That said, I've seen many people try to convince others over the years and two constants have emerged:

1) If the disbeliever believes they have grasped the problem mathematically and are talking in terms of numbers and probabilities, you have already started down the rathole and should let it go.

2) The single biggest game changer I have seen (already mentioned a few times in this thread) is pointing out that in choosing which door to show you the host injected more information into the system.

My brother uses the problem as an example of the unreliability of common sense for his freshman psych students. Most years he also ends up holding a special skeptics session afterwards where they do a walkthrough of a round of the game with the students playing the role of the host. Usually it's at the moment that when they realize that the hosts' choice of doors is constrained that most of them say "Ohhhh...."
posted by Tell Me No Lies at 9:14 PM on February 20, 2015 [6 favorites]

Pope Guilty: "If you pick a goat and switch, you'll get a car every time.
If you pick a car and switch, you'll get a goat every time.
There are two goats and one car.

Thus, 2:1 odds.

I'm generally shit at math and this explanation got the idea across very clearly. Thank you!"

Seconded!
posted by Samizdata at 9:27 PM on February 20, 2015

I once tried to teach someone studying to become a math teacher the Monty Haul problem over sushi on a first date. By the end of dinner it didn't matter which door I picked.
posted by meinvt at 9:28 PM on February 20, 2015 [17 favorites]

OnTheLastCastle> Oh, I read up. Monty in the problem ALWAYS reveals a goat. That is explicit, you all.

Really? Where is it explicit that Monty Hall always opens another door with a goat behind it? Here is the phrasing of the problem from the link:

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?

I don't see a condition anywhere in the phrasing of the problem that she printed that Monty Hall always opens a goat door and offers the contestant a choice to switch. And in her initial column, she absolutely did not make it clear that this was a necessary condition for her answer. And Monty Hall made it explicit in the followup to all of this that he doesn't always offer a choice to switch.

I have no doubt that quite a few of her letter writers got the wrong answer while still assuming that Monty Hall always reveals a door with a goat behind it. I also have no doubt that there were some letter writers among the 10,000 who correctly pointed out why Marilyn vos Savant was wrong.

The Monty Hall problem is interesting, even as stated in vos Savant's initial column: it's interesting because ultimately (most) people's intuitions are more correct than the pat answer that you double your chances of getting the car if you switch. Your intuition would tell you that it can't possibly be so easy to double your chances of winning a good prize instead of a goat, otherwise the obsessives who populated the Let's Make a Deal studio audience would quickly figure out that you should always, always, always choose the other door and you'd get something resembling a real-life version of Gold Case.
posted by UrineSoakedRube at 9:49 PM on February 20, 2015 [3 favorites]

Except, of course, that the question vos Savant replied to did specify those rules.

Oh, OK, didn't remember that. Sorry, I've been in about 1000 discussions of this problem and it's usually underspecified. And the fact that the answer depends on the specification is the interesting part!
posted by escabeche at 9:51 PM on February 20, 2015

automatic goat prize

What if you can only drive a manual goat?
posted by yoink at 10:04 PM on February 20, 2015 [14 favorites]

The Monty Hall problem is interesting, even as stated in vos Savant's initial column: it's interesting because ultimately (most) people's intuitions are more correct than the pat answer that you double your chances of getting the car if you switch. Your intuition would tell you that it can't possibly be so easy to double your chances of winning a good prize instead of a goat, otherwise the obsessives who populated the Let's Make a Deal studio audience would quickly figure out that you should always, always, always choose the other door and you'd get something resembling a real-life version of Gold Case.

No, the reason why this wasn't a winning strategy has nothing to do with people's incorrect intuitions about the Monty Hall Problem. As multiple people have pointed out, on the show the deal-making didn't usually take the form illustrated in the problem; instead, it would be something like offering a sum of money in lieu of opening one of the remaining doors.

I once remarked to my SO, as we walked through a casino on the Vegas strip, that I would be curious to see an aggregated data set of offers and plays on a show like Deal or No Deal, where (from my non-quantitative observation during the times when my stepmother was watching it and I was around) the settling offer presented at several points during the game was nearly always a small percentage greater than the expectation value of continuing to play, and yet most contestants continued to play. I'm not sure there's any other game show where the underlying odds were so unobscured, and it made for (IMO) a fairly boring show to watch, though given it's popularity it seems like most other people's mileage varied.
posted by kagredon at 10:05 PM on February 20, 2015

kagredon> No, the reason why this wasn't a winning strategy has nothing to do with people's incorrect intuitions about the Monty Hall Problem. As multiple people have pointed out, on the show the deal-making didn't usually take the form illustrated in the problem; instead, it would be something like offering a sum of money in lieu of opening one of the remaining doors.

Even then, if the phrasing of the problem in Marilyn vos Savant's initial column were to take into account that aspect of the real show, there'd still be the fact that her solution would rely on an unstated assumption that the contestant was always offered a deal; a deal presumably better than a goat, or whatever the goat-equivalent was on the show.

And most people's intuitions would tell them that you can't reliably increase your chances by any significant amount, let alone a factor of two, of getting the better deal by either taking the cash or by sticking by your initial choice. Otherwise the show would be less interesting than it was.
posted by UrineSoakedRube at 10:16 PM on February 20, 2015

(What if Monty picks one of the two doors at random and opens it):

Thinking about this for a minute, I think it doesn't change the odds (except that there's now a chance that the whole thing gets cancelled because the car was revealed). I said earlier that the idea that got through to me was that Hall is constrained, but I think that's a bit misleading. The essential thing is that the player started picking from 3 doors, but now has a situation of picking from 2 doors.

I'm fairly sure that this does change the game into one where switching makes no difference.

Sometimes, as you say, the game gets cancelled. The problem is that the game will never be cancelled if you initially picked the correct door and will only be cancelled if you picked one of the incorrect doors. Remember that switching will only let you win if you picked an incorrect door. For the original Monty Hall problem that's great - you are twice as likely to have picked the incorrect door as the correct one, but for this modified one you have a problem. Although you are twice as likely to have picked the incorrect door (and would thus win if you switched), there is also the chance that, if you did that, the game will be cancelled. This eliminates your advantage and puts you back at a 50/50 chance for switching.

tl;dr - It's vitally important that Monty knows where the prize is and opens one of the non-prize doors.
posted by It's Never Lurgi at 10:20 PM on February 20, 2015

Has anyone on Deal or No Deal simply chosen the boxes at random without employing the interminable woo that infects that fucking program.
Actually, has anyone done that and survived the ensuing mob?

Related, Game Theory is fun!
posted by fullerine at 10:20 PM on February 20, 2015 [1 favorite]

I don't see a condition anywhere in the phrasing of the problem that she printed that Monty Hall always opens a goat door and offers the contestant a choice to switch

I am overtired, which may be affecting my thinking, but I'm getting confused by people bringing up this point. We know Monty Hall knows what's behind the doors. If he picks the door with the car, there would be no game, so he effectively does have to pick a door with a goat, because why wouldn't he? You pick a door with no idea of what's behind it, hoping to win a car. To add another dimension to the game, the guy who knows what's behind the doors will open another door and give you a chance to swap. He then opens...the door with the car. Game over, you lose. It's not even a game if he doesn't pick a door with a goat.
posted by triggerfinger at 10:22 PM on February 20, 2015 [11 favorites]

Just say you're on a game show but the host, Monty Hall, always lies, while his beautiful assistant always tells the truth. You are given the choice of opening three doors: behind one is an Island populated by blue-eyed inhabitants, behind the others are Pascal and a hotel manager called David Hilbert holding a bill for $25. You pick #2, but then 100 pirates invade the studio, each demanding a different share of your gold depending on their rank. Monty Hall quickly runs up "n" stairs (taking only one or two steps at a time) and disappears behind #3, and only then do you notice that everyone left in the room has the same birthday. Given all this, what are your chances of voting #1 quidnunc kid? I'm asking for 0.999... of a friend. Who always lies.
posted by the quidnunc kid at 10:25 PM on February 20, 2015 [55 favorites]

Well, that assumes that you didn't "protect" the car door by picking it first. If you didn't then presumably Monty would save the car for the show. That guy was a dick!
posted by axiom at 10:28 PM on February 20, 2015

triggerfinger>We know Monty Hall knows what's behind the doors. If he picks the door with the car, there would be no game, so he effectively does have to pick a door with a goat, because why wouldn't he?

Because Monty Hall can -- and has explicitly stated that he did -- choose not to open any other door whatsoever from time to time. And the point of the show wasn't always to have a "game" -- Let's Make a Deal was only a game show in the loosest sense.
posted by UrineSoakedRube at 10:32 PM on February 20, 2015

What if you can only drive a manual goat?

The automatic goat is easier to drive than the manual goat. And before you ask, yes, you can drive automatic goat if you can drive automatic sheep. It's actually the same chassis anyway, most people don't know that. They just make the sheep in Mexico and the goats in Michigan. Shit, man.

In respect of the problem, a radical solution. Now maybe this is just me having been up for the past 18 hours writing a comprehensive exam, but I'm pretty sure that Monty Hall himself is one of the horcruxes.
posted by clockzero at 10:42 PM on February 20, 2015 [10 favorites]

And because this just occurred to me:

"If you pick a goat and switch, you'll get a car every time.
If you pick a car and switch, you'll get a goat every time.
There are two goats and one car.

You only know that the first premise is true because Hall reveals one goat at the beginning; if he didn't, this wouldn't work. But because he does, this has to be true.
posted by clockzero at 10:45 PM on February 20, 2015 [2 favorites]

That Hall reveals a goat is foundational to the problem, isn't it? It's part of what makes it what it is.
posted by Pope Guilty at 10:49 PM on February 20, 2015 [4 favorites]

Yes, if he doesn't goat a goat, goat will goat and then all is goat.
posted by Brocktoon at 11:07 PM on February 20, 2015 [5 favorites]

That Hall reveals a goat is foundational to the problem, isn't it? It's part of what makes it what it is.

Olympic-level condescension, or charitable acknowledgment of someone stating the obvious? Probably the latter, but the former is fun to imagine.
posted by clockzero at 11:17 PM on February 20, 2015

Pope Guilty> That Hall reveals a goat is foundational to the problem, isn't it? It's part of what makes it what it is.

I think it's foundational to one version of the problem. And it's a problem worth posing, because it's useful for people to think through the basic probabilistic aspect of the problem where Monty Hall always reveals a goat behind a door the contestant doesn't choose. But there's another version of the problem -- the one Marilyn vos Savant answered -- where there's something else that's foundational to the problem.

There're the probability problems you learn early on in school, and as I said, those are useful. But it is important to make the leap to figure out how well those problems model reality.

Nicholas Nassim Taleb is apparently a favorite whipping boy on Metafilter, but I give him points for getting to this point as concisely as I've ever seen. In The Black Swan, he presents 2 fictional archetypes: the straightlaced Dr. John and the street-smart Fat Tony (yeah, I know). He then asks both of them a question:

NNT (that is, me): Assume that a coin is fair, i.e., has an equal probability of coming up heads or tails when flipped. I flip it ninety-nine times and get heads each time. What are the odds of my getting tails on my next throw?
Dr. John: Trivial question. One half, of course, since you are assuming 50 percent odds for each and independence between draws.
NNT: What do you say, Tony?
Fat Tony: I'd say no more than 1 percent, of course.
NNT: Why so? I gave you the initial assumption of a fair coin, meaning that it was 50 percent either way.
Fat Tony: You are either full of crap or a pure sucker to buy that "50 percent" business. The coin gotta be loaded. It can't be a fair game. (Translation: It is far more likely that your assumptions about the fairness are wrong than the coin delivering ninety-nine heads in ninety-nine throws.)

Fat Tony would be wrong about the odds if the coin tosser can cheat and pick the outcome at will -- all bets are off, as it were -- but the point remains that you have to be really careful if someone claims that there is a surefire way to go from a 1/3 chance of hitting a jackpot to a 2/3 chance. You have to carefully look at the assumptions underlying the game. And more generally, you have to be sure that you correctly understand the real probabilities behind events even if there isn't a "house" or a Monty Hall on the other side of things.

And Marilyn didn't do it in her first column, claimed it was a minor assumption in her follow up, and that none of the 10,000 letter writing "professional critics" who said she was wrong managed to mention that the unstated assumption was behind her conclusion. Herb Wiskit, author of the old "Marilyn vos Savant is Wrong" website, discusses this in quite a bit of detail. Unfortunately, he was prohibited from reproducing significant portions of her column by Parade's lawyers; you can read what he says about that further down the main page.
posted by UrineSoakedRube at 11:23 PM on February 20, 2015

but what if the goat is on a treadmill, but with a strong head wind

No, no, no, you ignoranimus, the goat is in the car, and the car is on a conveyor belt. With wings.
posted by FelliniBlank at 11:25 PM on February 20, 2015

NO, WAIT: what if Monty Hall is just the goat's imaginary alter ego?
posted by FelliniBlank at 11:28 PM on February 20, 2015

Because Monty Hall can -- and has explicitly stated that he did -- choose not to open any other door whatsoever from time to time. And the point of the show wasn't always to have a "game" -- Let's Make a Deal was only a game show in the loosest sense.

Okay, but I was just taking the problem at face value - the rules are as stated in the problem, regardless of how things were or weren't actually done on the show (I've never seen Let's Make a Deal).

Here's what I have trouble wrapping my head around - why can't we eliminate door 3 once we know it has a goat?

I understand the answer to the problem. A person picking one door out of three at random has a 33.3% chance of getting the car and a 66.6% chance of getting a goat. The odds don't change regardless of new information we get. So, the contestant is in the 33.3% group with door #1. There are two doors in the 66.6% chance group. The host shows that one of the doors in this group is a goat, which means the odds on that particular door (door #3) having a car drop to 0%. But as a group, doors 2 & 3 still retain the 66.6% chance that the car will be behind one of them, it's just that now that we know that the chances of a car behind door #3 is 0%, the 66.6% odds on the group have moved to door #2 (instead of being split equally between the two).

What I have a hard time wrapping my head around is WHY do the odds have to stay the same when we know one of the options can be disregarded. Once the third door is opened, my mind wants to eliminate that door as an option - just get rid of the door entirely so now we have just two doors and a 50/50 chance. Why can't we do that?
posted by triggerfinger at 11:29 PM on February 20, 2015 [2 favorites]

Because it's a puzzle of semantics and sequence as much as mathematics.
posted by GallonOfAlan at 11:35 PM on February 20, 2015 [2 favorites]

Fat Tony would be wrong about the odds if the coin tosser can cheat and pick the outcome at will -- all bets are off, as it were -- but the point remains that you have to be really careful if someone claims that there is a surefire way to go from a 1/3 chance of hitting a jackpot to a 2/3 chance. You have to carefully look at the assumptions underlying the game. And more generally, you have to be sure that you correctly understand the real probabilities behind events even if there isn't a "house" or a Monty Hall on the other side of things.

You realize that Marilyn Vos Savant isn't hustling tourists on a boardwalk somewhere with this, right

What I have a hard time wrapping my head around is WHY do the odds have to stay the same when we know one of the options can be disregarded. Once the third door is opened, my mind wants to eliminate that door as an option - just get rid of the door entirely so now we have just two doors and a 50/50 chance. Why can't we do that?

That's exactly where your intuition is meant to blunder, like some poor blindfolded goat.

Think of it this way: from a probabilistic perspective, formally, this situation is like trying to toss a ball into one of two buckets of different diameters. In this example, if you switch doors, you're tossing the ball into a bucket whose diameter is bigger, so you'll get it in more easily; and if you conducted more than one trial, you would also get it in more often. There's only one trial here, but there's a "bigger space of possibility" if you switch doors. If you closed your eyes when aiming and just asked someone to point you at one bucket or the other, you wouldn't have any idea about the comparative likelihood of making it into one or the other, but your lack of perspective doesn't mean you have an equal chance.
posted by clockzero at 11:47 PM on February 20, 2015 [1 favorite]

And most people's intuitions would tell them that you can't reliably increase your chances by any significant amount, let alone a factor of two, of getting the better deal by either taking the cash or by sticking by your initial choice.

I should rephrase that, because I made a subtle error: I think that Monty and his producers set things up such that you can't reliably increase your average expected return by any significant amount by the use of any mathematical calculation.

If you're very good at reading poker tells, on the other hand, a psychological strategy might work for you.
posted by UrineSoakedRube at 11:48 PM on February 20, 2015

BUT WHAT DID THE DOG SEE?
posted by wam at 11:49 PM on February 20, 2015 [1 favorite]

clockzero> You realize that Marilyn Vos Savant isn't hustling tourists on a boardwalk somewhere with this, right

Of course not -- her hustle was entirely different.
posted by UrineSoakedRube at 11:49 PM on February 20, 2015

Metafilter : Olympic-level condescension, or charitable acknowledgment of someone stating the obvious?
posted by fullerine at 11:54 PM on February 20, 2015 [4 favorites]

Think of it this way: from a probabilistic perspective, formally, this situation is like trying to toss a ball into one of two buckets of different diameters. In this example, if you switch doors, you're tossing the ball into a bucket whose diameter is bigger, so you'll get it in more easily; and if you conducted more than one trial, you would also get it in more often. There's only one trial here, but there's a "bigger space of possibility" if you switch doors.

I understand this. This is exactly how I'm visualizing it - the smaller options of doors 2 and 3 joining together to become one group (or bucket) with an overall higher (joined) probability. What I don't understand is if we determine that any portion of this group (in this case door #3) is invalid, why it can't be eliminated. I think it has something to do with conditional probability but I can't put my finger on the exact reason.
posted by triggerfinger at 12:03 AM on February 21, 2015

I was arguing about this with my wife a number of years ago. Finally, I decided the easiest thing was just to write a program to try it experimentally. I wrote it for the Windows Phone (it's in the Windows Phone Store). It's what you call a veridical paradox, which is one that produces a result that appears absurd but is demonstrated to be true nevertheless. I finally won the argument, btw.
posted by Xoc at 12:06 AM on February 21, 2015 [2 favorites]

triggerfinger, you've actually analyzed the problem correctly. Everyone else is missing that important element of probability theory ... including the smartest woman alive.
posted by kanewai at 12:08 AM on February 21, 2015

I understand this. This is exactly how I'm visualizing it - the smaller options of doors 2 and 3 joining together to become one group (or bucket) with an overall higher (joined) probability

Maybe this way will help: assume that you've chosen a goat in door 1 (you have a 2/3 chance of this assumption being correct.) There's a 50/50 chance that door 2 is the car and door 3 is the other goat, and a 50/50 chance of the converse. What you are doing is forcing Monty to tell you which of these 50/50s is valid.
posted by kagredon at 12:12 AM on February 21, 2015

just get rid of the door entirely so now we have just two doors and a 50/50 chance. Why can't we do that?

You can only do that if Monty has no information about what is behind the doors, and chose them randomly. But that also means that there is a 33% chance you don't get a second choice - since his revealing a car would end the game.

But, that is not the scenario posited. Monty MUST NOT reveal the car. So, when he opens a door, you know either that both unchosen doors are goats and so he chose at random (33%), or one of the doors is the car, and he HAD to show the door that does not have a car (66%).

In other words, you are no longer operating on pure chance, but have possibly discerned something about the state of the system. If your first guess was incorrect - which it probably is (~66% chance to be wrong, after all), he tells you which of the remaining doors is the correct choice.
posted by Pogo_Fuzzybutt at 12:13 AM on February 21, 2015 [2 favorites]

A billion doors. You pick one to be opened (but leave it shut). Monty picks one to remain closed. All other doors are opened. The chance you picked the door with the prize: slim. The chance he picked the one with the car: if ain't behind yours (1 in a billion), it is guaranteed to be his. Now choose which of these two doors you really want to open.
posted by five fresh fish at 12:16 AM on February 21, 2015 [3 favorites]

Try this.

Pick one of three doors. Let's call them a, b, and c

Round one, there are three potential outcomes:

a
b
c

Round two there are nine potential outcomes:

a>a (i.e., you choose a and don't switch)
a>b (you chose a first, then switched to b)
a>c
b>a
b>b
b>c
c>a
c>b
c>c

Now, let's say you chose 'a' on the first round. And Mr. Hall reveals that 'c' is eliminated.

There are only two potential outcomes remaining:

a>a
a>b

The flaw in vos Savant's (et al.) math is that in she maintains that there are three remaining potential outcomes ... she doesn't remove enough from the equation. She bases her probability calculations on the original set, not on the final set.
posted by kanewai at 12:18 AM on February 21, 2015

that's...not even wrong.
posted by kagredon at 12:22 AM on February 21, 2015 [5 favorites]

Wow, I see from a quick check that my twenty-year-old and what was the first page on the web about the MHP has been relegated to quite a ways down the Google search results when for many years it was the top result and has been very widely linked. That's what leaving a twenty year-old webpage alone to remain very much a 1990s-era page will do to your search ranking. Not that I really care enough to do anything about it. (I'm still there in the citations for the MathWorld entry on it, in the company of Martin Gardner, so I'm happy.)

As it's fallen in the search ranking and as there's been an explosion of other and better resources on the web for it (at this point, the Wikipedia entry is extremely thorough, probably excessively), I've gotten less and less email over the years. But, even so, I've now had twenty years of correspondence with people who feel compelled to write to me the same way that people wrote to Marilyn vos Savant. So I'm somewhat of an authority on how those folk think and, more generally, how people think and approach the MHP. I was drawn to the problem and motivated to make the web page in 1995 (it later made the front page of Yahoo in mid-1996) because of my abiding interest in both how people's intuition leads them astray about the natural world and the pedagogical methods for working against this tendency.

One of the most memorable exchanges I had was with a Houston Chronicle reporter who was thinking about pitching a story on the MHP to his editor and wrote to me about it because although he found the controversy interesting, he was certain that the 50-50 answer was correct and I was wrong. At some point I practically begged him to call up a math professor at Houston or Rice or somewhere or, better yet, just sit down with some dice and a friend and empirically test it. To that he responded, with some exasperation, that really it wasn't that important and he was pretty sure he was right anyway. This was a major city newspaper reporter who presumably reported on these kinds of topics but couldn't be bothered to discover the correct answer. This validated every complaint I've ever had about mass media science reporting.

As I write on my page, I have vivid memories of sitting on my couch and reading that Sunday paper's copy of Parade and vos Savant's column and, like those PhD's (I'm not a PhD, but I am educated), thought about the gambler's fallacy and (cursorily) asked myself if the contestant had gotten new information, which I answered to myself with "no", and concluded that both unopened doors are equally likely to win. Even so, I was surprised that vos Savant said otherwise and so because it was trivial I walked over to my PC and wrote a short program to test it. And the result was that it was better to switch. Then I sat down and worked out why that was the case.

What's really quite amazing to me is how resistant so many people are to test their intuitions in some fashion especially when there's all these authoritative voices that are telling them that it's better to switch. For twenty years I've been frankly flabbergasted at the arrogance of people who don't bother looking up citations leading to authoritative sources that are claimed to prove them wrong because they just know that they're right, or will look up and consult numerous such sources and continue to claim that they know they are certainly right and still refuse to just do a ten minute experiment to fucking check.

I've also found that comprehension of the MHP is ... ambiguous. A large number of people will believe that they "understand" it because they've in some tentative sense accepted the correct answer, but that understanding is so shallow as to be mostly an illusion. I don't mean to sound contemptuous because the whole point is that I'm not -- to some degree and at various times we all think we "understand" something merely because we accept as truth some factual representation of it. A very large portion of our culture's fundamental scientific knowledge has this character for most people -- people confuse being aware of a fact with comprehension of what that fact represents; and this is a problem because this is what Plato called "true opinion", as opposed to "true knowledge". I don't accept his greater epistemological framework, but I find that distinction quite useful. We can be right, but for the wrong reasons, and that actually does matter because true opinion is light as a feather and blows away in a mild breeze.

But also the point is that comprehension is really hard work. I've found that my interlocuters over the years have pushed me into both a large number of different ways to think about the MHP, and a large number of different approaches I've used to explain it to them. And the result is that I've discovered that even decades of familiarity don't mean that I can reliably assume that I have the intuitive grasp of the problem that I'd like to think I have, and that a novel variation on the problem will often force me to work through the very same errors of intuition I had to work through 24 years ago.

The MHP for me became kind of a personal long-term study of the intersection of the issues of intuition, pedagogy, epistemology, and what it means to be someone with intellectual self-confidence while protecting and cultivating doubt and an awareness of fallibility. The problem became an arena where I could watch how these things play out in other people, and reflect on how I navigate this terrain.

"But there's another version of the problem -- the one Marilyn vos Savant answered..."

No, she answered the problem where Monty always opens a losing door. It implicitly follows both from the particular words she chose to describe it, and that she's presenting the problem at all. This parsing relies a lot on context, but so does reading almost any English sentence and, as it happens, pretty much every mathematical problem ever stated in English. If you're not a native speaker of English and also lack experience with the context of people discussing math puzzles, then I'm totally okay with your misreading of the problem statement as one that doesn't require that Monty open a losing door. But I suspect that's not true in your case and I'm certain it's not true in the vast majority of the cases where people criticize her on this basis.

"that's...not even wrong."

But typical. Tell Me No Lies's #1 case applies. He's quite right about this, and I say so with some considerable authority.
posted by Ivan Fyodorovich at 12:28 AM on February 21, 2015 [45 favorites]

How about imagining two "rounds" and thinking in terms of losing rather than winning:

The first round, you have a 1/3 chance of selecting the right door but more importantly a 2/3 chance in selecting the wrong door. The host doesn't tell you if you picked the right door, but does eliminate one of the wrong doors. Two remain.

Now, it's round two. As the naysayers say, you have a 50/50 chance of getting it right if you decide to play "another round," but by saying the odds are the same by remaining in place rather than choosing the other door, they are saying that you have an equal chance of winning by "not playing" this 50/50 game (keeping your original choice, which which you know had a 2/3 chance of being wrong in the previous round) vs. "playing" the 50/50 odds.

You can only "play" this 50/50 game by selecting the other door, because the prize remains unmoved from round 1. It would be like trying to win a dice roll by letting it sit idle, or winning a second coin toss without actually throwing the coin.

If you were to "shuffle" the doors, remaining in place would give you 50/50 odds and it would be a new "game" rather than a new "round." Since you are not "shuffling" them, remaining in place is forgoing a chance at playing a 50/50 game.

The only move you can make to play a 50/50 game in this scenario is to pick the other door. But statistically, your chance of being right by choosing the other door is 2/3 which can be intuitively hard to grasp without running some simulations or grokking the conditional probability that comes with the fact that the prize can't move between "rounds."

Your choice to remain still in the second round is a choice to forego a chance at playing with 50/50 odds and to only play the 1/3 game, where you had a 2/3 chance of being wrong. Choosing the other remaining door is like re-playing round 1 with a 2/3 chance of being right.

You have to pay to play...like if you picked a lottery number sequence of 5-16-30-46-60 for a chance at winning $100,000,000 and Monty said "well, guess what. Either you got it exactly right when you pulled that number out of the air, or you'll win the $100,000,000 by going with this other number scribbled in the palm of my hands. One of them is right, all of the other combinations are wrong. Which one would you like to go with?"
posted by aydeejones at 12:40 AM on February 21, 2015

Ivan, I followed the links and found a perfect example of the mathematical flaw in the Berkley professor's letter to the editor.

In his probability chart he lists this as one outcome:

keys in box a, contestant chooses a, monte opens b or c, contestant switches a for b or c: contestant loses

and this as one outcome:

keys in box a, contestant chooses b, monte opens c, contestant switches b for a: contestant wins

and he continues in this vein, resulting in six potential wins and three potential loses if the contestant switches.

The problem is, that first option contains two outcomes, not one, and the actual number of outcomes results in an even chance of winning and losing.

I'm also a bit confused by the claims that using probability theory to solve a probability problem is flawed.
posted by kanewai at 12:48 AM on February 21, 2015

But as a group, doors 2 & 3 still retain the 66.6% chance that the car will be behind one of them,

No, it's 100%.
posted by five fresh fish at 12:50 AM on February 21, 2015

Ivan Fyodorovich> No, she answered the problem where Monty always opens a losing door. It implicitly follows both from the particular words she chose to describe it, and that she's presenting the problem at all.

Which is not the problem that was asked. You simply can't sweep a significant assumption -- that Monty Hall always opens a door -- under the rug by claiming it's contextual and should be treated like "pretty much every mathematical problem ever stated".

For one thing, if we're going to insist on context, there's one game show that would spring to mind in 1990 if you asked anyone in their 20's or older a question in which you open a door and find a goat. That game show was Let's Make a Deal, and the host said straight out that he didn't always open another door and offer contestants the choice to switch. So that's one piece of context.

And the other thing is: who says it has to be considered a mathematical problem as opposed to a logic problem or puzzle? There are any number of questions like that one that turn on the fact that the person trying to solve the problem assumed something that wasn't in the problem.

You can try to claim that somehow every native speaker of English must read the question the same way you're insisting on doing. I'll note that it's particularly ironic given your study of, among other things, "the awareness of fallibility", and call it a night.
posted by UrineSoakedRube at 12:54 AM on February 21, 2015

Here is a simulation written in python that you can run right in your browser: http://repl.it/b79/4
posted by Pyry at 12:57 AM on February 21, 2015 [5 favorites]

Another way of putting it is that given three choices, they aren't equally likely to be correct if one of the parties pre-selected the right choice all along. It's not a coin toss, dice roll, or whatever, because someone has deliberately picked a number ahead of time and and we have a limited range to select from. If they eliminate all possibilities but two in the second round, we would be foolish to keep our original choice. But since there are only three doors instead of 196 million, it doesn't intuitively make sense that your first choice is a dumb thing to hold on to.
posted by aydeejones at 12:59 AM on February 21, 2015

UrineSoakedRube: "Which is not the problem that was asked."

How do you know this? What is the exact wording of the problem? I can't see it anywhere. Ivan apparently read it when it came out, so I tend to go with his side of it, since you're apparently just taking the word of a guy who runs a "Marilyn is Wrong" web site. (I can't find the original wording anywhere there, either.)

So - have I missed something? Is the original wording somewhere here or in the links? Or are we just arguing vaguely via old memories and website testimonials?
posted by koeselitz at 1:00 AM on February 21, 2015

Why doesn't escabeche's summary prove [my analysis of random door-opening] wrong?

Well, shit, it probably does. I hadn't seen his remarks last night.
posted by thelonius at 1:06 AM on February 21, 2015

Also:

UrineSoakedRube: "You simply can't sweep a significant assumption -- that Monty Hall always opens a door -- under the rug by claiming it's contextual and should be treated like 'pretty much every mathematical problem ever stated'."

Imagine I have two balls in my hands. One is green, the other is blue. If you choose one ball, what are the chances it will be blue? You might say: 50/50, an even chance. But you'd be wrong! Because the balls in my hand aren't always blue or green. Sometimes I'm holding a red or a purple ball. I have a dozen different colored balls. Why did you assume the balls were always either blue or green?

Why? Because that's how the problem was stated, and in these problems the stated parameters are regarded as static. Otherwise, no question with parameters has any meaning at all, because you can always say that any one of the parameters might change, and therefore it's unwarranted to accept them.

If the question as originally stated says that the game show host opens the a door with no prize, that can be regarded as a stated condition that happens every time.
posted by koeselitz at 1:15 AM on February 21, 2015 [11 favorites]

How do you know this? What is the exact wording of the problem? I can't see it anywhere.

It's quoted in the FPP.

“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?”

What Monty may or may not have done in other instances of the show aren't terribly material. The question doesn't require developing a general strategy for winning in all possible instances of Lets Make a Deal. The question asks "in this particular instance, with these particular circumstances, what is the best strategy ?"
posted by Pogo_Fuzzybutt at 1:15 AM on February 21, 2015 [5 favorites]

Never mind who gets the goat, who gets the PhDs?.
posted by Segundus at 1:16 AM on February 21, 2015

I can take away one wrong answer: it's not the woman who is the most intelligent person in the world.
posted by Segundus at 1:18 AM on February 21, 2015

Reading the letters to vos Savant is way more fun if you hear them in the voice of the Sicilian from Princess Bride.
posted by AAALASTAIR at 1:36 AM on February 21, 2015 [5 favorites]

And in her initial column, she absolutely did not make it clear that this was a necessary condition for her answer.

Eh, I'd say the bit where she writes "Then the host, who knows what’s behind the doors and will always avoid the one with the prize [...]" absolutely makes it clear that it's a necessary condition for her answer.
posted by effbot at 1:54 AM on February 21, 2015 [3 favorites]

I'm going to remember "Monty Haul".
posted by telstar at 1:58 AM on February 21, 2015

What Pogo_Fuzzybutt said. In the stated context, your odds of winning are 2/3 if you switch. However, that doesn't mean you could go on LMAD and expect to win.

In the context of LMAD, the answer cannot be solved using just mathematics or logic, because Monty Hall is an agent, not a pawn. He knows where the car is and knows where the goats are. If he opens a door it will have a goat behind it, but he is not actually constrained to open any door. He may choose to attach other conditions (e.g. money for sticking, money for switching). He may reveal that there is a second, medium-value prize behind one door. He is certainly talented at "reading" people, and also at influencing them - it goes with the job. Monty also has people talking in his ear who know the odds. Also, he is presenting a TV show, so what he does will be designed to generate interest and excitement in the contestants and watchers.

You pick a door. Monty looks at you carefully, listens to his earplug, and - opens another door to reveal the goat. He's looking at you.

Good luck!
posted by Autumn Leaf at 2:12 AM on February 21, 2015

Okay, the cards example in this previous post helped me to visualize it better (it's the same as the doors example but the fact that it was cards made it less abstract in my head).

The winning card is the ace of hearts. Monty Hall gives a contestant one card (placed facedown in front of the contestant) and keeps the remaining 51. The odds are very high (51/52) that Monty holds the ace of hearts.

Monty Hall then takes his 51 cards and deliberately removes 50 cards which are NOT the ace of hearts (showing the contestant all of them as he removes them) and then asks if the contestant would like to switch their card with the single remaining card that Monty holds. Obviously YES, because the odds of the ace of hearts being in Monty's hand from the beginning were 51/52. Those odds don't change when Monty removes 50 non-ace-of-hearts cards from his hand, but it does become much more likely that the one he has left IS the ace of hearts.
posted by triggerfinger at 2:14 AM on February 21, 2015 [1 favorite]

Given all this, what are your chances of voting #1 quidnunc kid? I'm asking for 0.999... of a friend. Who always lies.

I tried to vote #1 quidnunc kid, but a bee kept flying between me and the voting station! When I got there, almost everyone looked at each other, and committed mass suicide. Along with a few lucky blind veterans, I was able to escape down a manhole with a lifted corner, but with only one flashlight to cross a narrow bridge it was slow going and I kept running out of bananas. I hope I have toggled a lightbulb on 100% for someone! Please learn from my story and DO NOT vote #1 quidnunc kid.
posted by nom de poop at 2:36 AM on February 21, 2015 [4 favorites]

The simplest framing that explains the importance of the switch that I've found is: when you're first asked to pick a door, you are choosing one out of three (say, door 1); then, when you are asked to potentially re-pick, by switching you are effectively choosing two out of three (doors 2 and 3, the other closed door PLUS the goat door* Monty's opened for you). So, applying madmethod's excellent terminology, the stayputnik sticks with 1/3, the switcher opts for 2/3.

*This is the scenario as described in the MvS's stating of the problem; to hypothesize Monty offering the switch after opening a car door is just kinda peeing in the punchbowl.
posted by progosk at 2:41 AM on February 21, 2015 [2 favorites]

Meantime, who's worked out the probability of world's smartest person being named vos Savant?
posted by progosk at 3:05 AM on February 21, 2015 [3 favorites]

In theory, there is no difference between theory and practice. In practice, there is.

(opens door, gets goat)

Fuck.
posted by ostranenie at 3:06 AM on February 21, 2015 [2 favorites]

What I don't understand is why it matters whether Monty knows where the car is.

Leave Monty out entirely. Pretend the rules are that first you choose one door and then have the option of trading it for both of the other doors. You'd always switch, and win 2 out of 3 times, right?

Now say the rules are that first you choose one door and then have the option of trading it for one of the other doors. It's still true that 2 out of 3 times you've chosen the wrong door, but you don't switch because even if you've chosen wrong, you don't know which of the other doors is right.

But! Imagine that after you make your initial choice, a random lightning strike destroys one of the other two doors, revealing a goat. This random event doesn't change the 2/3 chance that you'd initially chosen the wrong door, but it does tell you which of the other two doors must have the car if you'd chosen the wrong door. It's the same as the first scenario where you get to trade your first pick for both of the other two doors. So you ought to switch.
posted by straight at 3:10 AM on February 21, 2015

Things change if the opened door reveals a car. There'd be no advantage in switching at that point.
posted by progosk at 3:23 AM on February 21, 2015 [3 favorites]

straight - that's exactly the intuition i had and that, I am being told, is refuted by eschabe above....
posted by thelonius at 3:33 AM on February 21, 2015

Metafilter: you know your odds of winning are zero ... just take your goat
posted by chavenet at 3:36 AM on February 21, 2015 [3 favorites]

I was maybe twelve or thirteen when I read the Savant column in the local paper. It was confusing, but seemed simple enough to try, so I got a penny and some cups and worked through it, and I was astounded that it worked. So I told the smartest person I knew: my grandfather, who was a chemical engineer at Goodyear. "Nonsense," he said. I couldn't believe he didn't just grasp it immediately, since he knew so much other math.

It took me (and eventually my step mother helping) about an hour and a half of demonstrations and discussion, and actual physical evidence to prove to him that the article was right. And that was the first time it occurred to me that you could be very smart, and very wrong.
posted by heyitsgogi at 4:03 AM on February 21, 2015 [9 favorites]

I can't drive and I like goat's cheese. I want to play the version with a million doors and kidnap the goats and become a goat baron.
posted by howfar at 4:06 AM on February 21, 2015 [4 favorites]

I usually go directly to the "duh, this is obvious, just look at the 100-door case" phase, but then a tiny voice pops up and asks "but hey, he just opened one more door in the 3-door case, so why would he open more 98 more doors in the other case instead of just one?"

At that point I usually tell the tiny voice I don't like it when he drops Lego pieces all over the floor, and that he should go and brush his teeth since it's way past bedtime.
posted by effbot at 4:19 AM on February 21, 2015

But what if Monty has two heads, one of which always tells the truth and one of which always lies, and one of the goats is a were-tiger and there is another prisoner of this insane gameshow who gets freed if you get eaten?
posted by robocop is bleeding at 5:18 AM on February 21, 2015 [3 favorites]

yeah, you people are so smart, but it's DREW CAREY that opens the door now!!...
posted by pyramid termite at 10:24 PM on February 20 [+] [!]

oh my god i finally get it
posted by KathrynT at 10:26 PM on February 20 [+] [!]

I doubt that KathrynT's comment was intended as a response to the comment by pyramid termite which immediately preceded it, but I like to imagine that it was.
posted by DevilsAdvocate at 5:38 AM on February 21, 2015 [7 favorites]

straight, the difference is in whether or not this fictional lightning bolt always opens a door with a goat, or whether it opens a random door that in the instant case happens to have a goat.

Of course, to the (first) individual player it's impossible to tell the difference. The first formulation is equivalent to the Monty Hall problem - it doesn't matter that MH knows, it matters that he always opens to a goat. The random formulation would be 50-50 odds.

Also, I realllllllllly want to create a fake web-accessible version of this in python or something. Point it at people and be like "run it for yourself! 50-50." Code examples only work if the user can verify that the code does what it should...which is honestly tricky.
posted by Lemurrhea at 5:41 AM on February 21, 2015 [1 favorite]

But! Imagine that after you make your initial choice, a random lightning strike destroys one of the other two doors, revealing a goat. This random event doesn't change the 2/3 chance that you'd initially chosen the wrong door, but it does tell you which of the other two doors must have the car if you'd chosen the wrong door.

Imagine there are 100 doors. Yours is protected from the raging thunderstorm by a lightning rod, but the other 99 are vulnerable. 98 of those are reduced to smoking splinters by random thunderbolts. All 98 have goats behind them.

If the car weren't behind the door you chose, it would be pretty unlikely for the lightning to just happen to miss the right 98 doors. On the other hand, if the car is behind your door, it's a certainty that all the lightning-destroyed doors have goats between them. So what you've observed is actually pretty strong evidence in favor of you having chosen the right door in the first place.

When Monty is intentionally picking the goaty doors, of course, it's a different story; in that scenario, the opened doors are certain to reveal goats in either case, so you're not getting any evidence towards the question of whether your door has a car.
posted by escabeche at 5:48 AM on February 21, 2015 [1 favorite]

Eh, I'd say the bit where she writes "Then the host, who knows what’s behind the doors and will always avoid the one with the prize [...]" absolutely makes it clear that it's a necessary condition for her answer.

I see that on her web page, but I guess I had the impression that she added that for clarity to the original column? Otherwise this doesn't make sense:

"The problem is not well-formed," Mr. Gardner said, "unless it makes clear that the host must always open an empty door and offer the switch. Otherwise, if the host is malevolent, he may open another door only when it's to his advantage to let the player switch, and the probability of being right by switching could be as low as zero." Mr. Gardner said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch.

Ms. vos Savant acknowledged that the ambiguity did exist in her original statement. She said it was a minor assumption that should have been made obvious by her subsequent analyses, and that did not excuse her professorial critics. "I wouldn't have minded if they had raised that objection," she said Friday, "because it would mean they really understood the problem. But they never got beyond their first mistaken impression. That's what dismayed me."

And vos Savant is right here -- the people writing to yell at her mostly seemed not to have grasped that the central issue was Monty's knowledge; they just thought "two doors, has to be 50-50."
posted by escabeche at 5:56 AM on February 21, 2015 [4 favorites]

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.

Because again -- what's important is not that Monty picks a goat, what's important is that Monty could not have not picked a goat.

I agree that this is the most natural interpretation; it's the one my mind naturally goes to, at least. But I'm not gonna say somebody's bad at math for picking the other one.

I apologize if I'm repeating myself, but when you use this example to teach probability, this is the critical point. You wish you lived in a world where you could determine probabilities by making observations. But you can't, not unless you know what the universe of possible observations is that the actual observation was drawn from. My goal in life is to teach people this when they're 18 so they can be better at interpreting p-values for the next sixty years.
posted by escabeche at 6:00 AM on February 21, 2015 [7 favorites]

You know, her mailbag selections were actually less sexist than I expected. Some of them were, of course, but there was nary a rape or death threat in the lot. Wonder how come that is.

Because people had to sign their name to it.

The sociological lessons of people's reactions to being told they are wrong should not be overlooked here.
posted by dry white toast at 6:02 AM on February 21, 2015 [1 favorite]

Paul Erdos got it wrong too, so the deniers are in good company.
posted by Pararrayos at 6:05 AM on February 21, 2015 [1 favorite]

Today on reddit:
Monty's Wager Caprinal. The explanation as a Gilbert and Sullivan song.

"Explain, in song, the Monty Hall Problem like you are Wayne Brady on "Whose Line is it Anyway?" and I, Colin Mochrie, just gave you an impossibly weird musical style and song title to use."
posted by jjj606 at 6:11 AM on February 21, 2015 [2 favorites]

The lesson here is that strange men lying in studios distributing goats is no basis for a system of government.
posted by delfin at 6:14 AM on February 21, 2015 [17 favorites]

keys in box a, contestant chooses a, monte opens b or c, contestant switches a for b or c: contestant loses

and this as one outcome:

keys in box a, contestant chooses b, monte opens c, contestant switches b for a: contestant wins

and he continues in this vein, resulting in six potential wins and three potential loses if the contestant switches.

The problem is, that first option contains two outcomes, not one, and the actual number of outcomes results in an even chance of winning and losing.

Sorry for the big quote. kanewai, what you're leaving out is that in the way you've expressed it, not every outcome has equal probability.

I think this is easier to think about conditioning on what the contestant chose, so for the case where the contestant chose Door 1 and where Monty just flips a coin if he has two goats available to reveal*:

Case 1: There's a one-third chance the car is behind Door 2 (pr=1/3) and Monty reveals the goat in Door 3 (pr=1). You should switch.

Case 2: There's a one-third chance the car is behind Door 3 (pr=1/3) and Monty reveals the goat in Door 2 (pr=1). You should switch.

Case 3: There's a one-sixth chance that the car is behind Door 1 (pr=1/3) and that Monty reveals the goat in Door 2 (pr=1/2). Don't switch.

Case 4: There's a one-sixth chance the that car is behind Door 1 (pr=1/3) and that Monty reveals the goat in Door 3 (pr=1/2). Don't switch.

You're still better off switching 2/3 of the time. The cases where the contestant initially chose Door 2 or Door 3 are the same with the names swapped around.

*How Monty decides which of two available goats to show you in the case where you initially picked the car doesn't actually matter -- Case 3 and Case 4 have to add up to the 1/3 probability that you picked the car right off the bat.
posted by ROU_Xenophobe at 6:28 AM on February 21, 2015

Because people had to sign their name to it.

I used to believe this until a ton of websites started using Facebook for article comments and people kept on keepin' on.
posted by Pope Guilty at 6:29 AM on February 21, 2015 [11 favorites]

I see that on her web page, but I guess I had the impression that she added that for clarity to the original column?

I spent a couple of minutes looking for scans, but only found this one about her readers disagreeing with her about another paradox. But that post led me to another article with the same quote, though, from a site titled "Marilyn is wrong" (because you know, what if the host is unreliable? hah, caught her!) which makes me think that he wouldn't doctor the quotes to make her look less wrong if he could avoid it (but what if that author is unreliable? hah!)
posted by effbot at 6:37 AM on February 21, 2015

Another way of looking at it is: You pick your first door at random, but by opening one of the other two doors Monty picks the door you will switch to if you switch.
posted by localroger at 6:44 AM on February 21, 2015 [5 favorites]

Also, that bit is clearly what she's referring to when she says "it was a minor assumption that should have been made obvious by her subsequent analyses". Read the entire column, and it's clear what assumptions she made. Stop somewhere before the end of the column and insert your own assumptions instead, and you can prove her wrong in multiple ways. Classic Internet logic, from a time before the Internet was widely used :)

(Why was every PhD in the US reading a magazine for "celebrity news, videos, entertainment, food, recipes, health tips, fitness, and games" in the early nineties, btw?)
posted by effbot at 6:51 AM on February 21, 2015 [1 favorite]

So I told the smartest person I knew: my grandfather, who was a chemical engineer at Goodyear. "Nonsense," he said. I couldn't believe he didn't just grasp it immediately, since he knew so much other math.

This whole thing highlights something that it is good to remember: people typically badly overestimate the truth of their intuitions about probability.
posted by thelonius at 7:30 AM on February 21, 2015 [3 favorites]

What if it's Marilyn vos Savant playing, and Monty Hall has read this MeTa thread? What then, mathletes?
posted by ostranenie at 7:45 AM on February 21, 2015

Imagine there are 100 doors. Yours is protected from the raging thunderstorm by a lightning rod, but the other 99 are vulnerable. 98 of those are reduced to smoking splinters by random thunderbolts. All 98 have goats behind them.

If the car weren't behind the door you chose, it would be pretty unlikely for the lightning to just happen to miss the right 98 doors. On the other hand, if the car is behind your door, it's a certainty that all the lightning-destroyed doors have goats between them. So what you've observed is actually pretty strong evidence in favor of you having chosen the right door in the first place.

I fell asleep last night trying to think through something functionally identical to this random-strike example, hoping that it would work out that sticking with the original door would then have greater odds than switching, but -- and I'm not 100% confident about this -- I think this modified 100-door-with-random-strikes example does result in 50/50 odds.

Indeed, unless you've picked the good door to begin with, it's extraordinarily unlikely for the random lightning to miss the good door. And, indeed, if you have picked the good door to begin with, the random lightning has a 100% chance of missing the good door, revealing only bad ones. But it's also extraordinarily unlikely for you to be in that scenario, since it's unlikely you did pick the good door to begin with, and -- oddly, to my sense of intuition -- it seems those unlikelihoods balance out completely.

Here are the numbers I'm working with (please correct any errors!):

You pick a door. The lightning randomly strikes 98 of the 99 remaining doors.

-- Your chance of picking the good door to begin with: 1/100 = .01 = 1%
-- Lightning's chance of striking only bad doors among the 99 remaining: 100%
-- Total chance of this scenario: 1% x 100% = 1%

-- Conversely, your chance of missing the good door to begin with: 99/100
-- Lightning's chance of sparing the good door among the 99 remaining, revealing only bad ones: 1/99.
-- Total chance of this scenario: (99/100) x (1/99) = 1/100 = 1%

[-- For completion's sake: total chance of missing the good door to begin with, and then the lightning revealing the good door, ending the game: (99/100) x (98/99) = 98/100 = 98%]

[I think the reason this modified version wasn't intuitive at first is that the math is complicated if you think through all 98 random strikes in the second example. The full series (98/99) x (97/98) x (96/97) x [...] x (2/3) x (1/2) must be equal to 1/99, right? (I don't remember how to calculate that easily.)]
posted by nobody at 7:45 AM on February 21, 2015 [2 favorites]

What must be amazingly sad is that at this point would anyone going on that show NOT switch. I mean, how could anyone planing to be a contestant not google information about the experience and have this bit of famous probability knowledge in their face?
posted by sammyo at 7:48 AM on February 21, 2015

CONTESTANT: Now, a clever man would switch doors, because he would know that only a great fool would reach for the goat he was given. I am not a great fool, so I can clearly not choose the door in front of you. But you must have known I was not a great fool, you would have counted on it, so I can clearly not choose the door in front of me.

MONTY: You're just stalling now.

CONTESTANT : You'd like to think that, wouldn't you? I've beaten your would-you-rather-have-fifty-dollars-or-what's-in-this-box game, which means I'm exceptionally lucky, so you could've put the car behind your own door, trusting on your knowledge of this probability problem to save you, so I can clearly not choose the door in front of me. But, I've also reached the Deal of the Day, which means I must have studied, and in studying I must have learned that switching gives me better odds of winning, so you would have put the car as far from the most likely outcome as possible, so I can clearly not choose the door in front of you.

MONTY: Look, 2 and 3 are both goats and the car behind 1 gets shitty gas mileage anyway. Get the fuck off my stage.
posted by delfin at 8:14 AM on February 21, 2015 [20 favorites]

The plane totally takes off.
posted by Camofrog at 8:45 AM on February 21, 2015 [4 favorites]

The full series (98/99) x (97/98) x (96/97) x [...] x (2/3) x (1/2) must be equal to 1/99, right? (I don't remember how to calculate that easily.)]

The mathematical term you're looking for is "orgy of cancellation."
posted by escabeche at 8:52 AM on February 21, 2015 [5 favorites]

Why was every PhD in the US reading a magazine for "celebrity news, videos, entertainment, food, recipes, health tips, fitness, and games" in the early nineties, btw?

It came free in the Sunday newspaper, so it was pretty much ubiquitous. Either that, or the PhDs of America were drawn to the advertisements for creepy dolls that Parade has been running since at least the early nineties.
posted by corey flood at 8:53 AM on February 21, 2015 [1 favorite]

a. You stick with your original pick, revealing the car (1/3 of time)
b. You switch doors, revealing the car (2/3 of the time)
c. It doesn't matter what you do because the second goat got bored and drove away
posted by gottabefunky at 9:33 AM on February 21, 2015 [1 favorite]

Why was every PhD in the US reading a magazine for "celebrity news, videos, entertainment, food, recipes, health tips, fitness, and games" in the early nineties, btw?

I was amazed and somewhat relieved to find that Parade in the US was a general interest magazine, because growing up in Britain in the 80s and 90s, it was a synonym for porn magazine.
posted by ambrosen at 10:02 AM on February 21, 2015 [1 favorite]

literal viking
posted by cortex at 10:11 AM on February 21, 2015 [2 favorites]

damned if it doesn't feel like "the reliably crazy 30%" is totally a real phenomenon

It's funny that in this case, no actual "belief" is required — you can exhaustively list all the cases and see the 2/3 for yourself. Yet the level of disbelief is somewhere between (for the US) heliocentrism (18%) and evolution (30–40%). So-called "probability theory" is all lies, damned lies from the pit of hell!
posted by mubba at 10:39 AM on February 21, 2015 [1 favorite]

I am dumb at this stuff and was so confused, and this comment from delfin helped:

The logical extension is to imagine there are 1,000 doors instead. You pick #147, Monty says "okay, I will open every door except #147 and #842. The doors I'll open are all goats. Will you switch?" The same end result -- pick A or pick B -- but for you to keep your original choice and win, you'd have to have beaten 999-to-1 odds on your initial pick. So you are choosing either one door or, in effect, all 999 others. And so with three doors, the fact that one door is a known loser does not change that Monty is effectively offering you either your one door or both of the others, and since you had a 1/3 chance of getting it right from the start, the other 2/3 have only one place to go.

Your option to switch is not an option to switch from one of two doors to another. It is Monty Hall saying: "you can switch from your one door to all the other doors, and I'll throw in a freebie by eliminating one of the wrong choices from the universe of all other doors."
posted by univac at 10:40 AM on February 21, 2015 [2 favorites]

Kanewai wrote:
The problem is, that first option contains two outcomes, not one, and the actual number of outcomes results in an even chance of winning and losing.

That's a pretty common misconception. Here's an example: say you have two coins to flip. What are the possible outcomes?

Many people would assume three:
1. Both heads
2. Both tails
3. One head and one tail
So, the odds of flipping one particular combination must be 1 in 3, right?

Nope, there are actually FOUR combinations:
1. Both heads
2. Both tails
3. Coin A is heads, Coin B is tails
4. Coin A is tails, Coin B is heads

So yeah, the odds of getting both heads are only 1 in 4, not 1 in 3. And the odds of "one of each" are actually 1 in 2.
posted by ShutterBun at 10:42 AM on February 21, 2015

>How Monty decides which of two available goats to show you in the case where you initially picked the car doesn't actually matter

It does matter, because you can extract information from it. In the extreme case, Monty might always select the leftmost available door with a goat. If he skips a door, you know you should switch. If he doesn't, you're back to 50-50.
posted by eruonna at 10:58 AM on February 21, 2015

Aha. Thanks escabeche and nobody. I was missing the fact that lightning randomly revealing one of the other doors is a goat would be more likely if I've chosen the door with a car.

It's interesting how much more intuitive everything in the scenario is when you convert it to a version with 100 doors. I wonder how generally you can apply that hueristic. It doesn't seem to help with the HH vs HT coin-flipping puzzle...
posted by straight at 11:32 AM on February 21, 2015 [1 favorite]

I apologize if I'm repeating myself, but when you use this example to teach probability, this is the critical point. You wish you lived in a world where you could determine probabilities by making observations. But you can't, not unless you know what the universe of possible observations is that the actual observation was drawn from. My goal in life is to teach people this when they're 18 so they can be better at interpreting p-values for the next sixty years.

This is a wonderfully lucid explanation. Thanks, escabeche.
posted by clockzero at 11:33 AM on February 21, 2015

the odds of getting a particular card on the river or turn being fifty percent--it's either going to happen or it's not

"I'm not sure that's how probability works, Walter."
posted by effbot at 11:38 AM on February 21, 2015 [1 favorite]

Thanks for this great thread. My husband would like you all to know that the answer is definitely still your chances are 50/50, despite me reading half of the responses here to him. Yikes.
posted by roomthreeseventeen at 11:41 AM on February 21, 2015 [1 favorite]

Dress him up as a five-foot ham sandwich and offer him a box. See if that helps.
posted by delfin at 11:44 AM on February 21, 2015 [2 favorites]

We even did it with cards! He said I got lucky.
posted by roomthreeseventeen at 11:47 AM on February 21, 2015 [1 favorite]

Thanks for this great thread. My husband would like you all to know that the answer is definitely still your chances are 50/50, despite me reading half of the responses here to him. Yikes.

Something I've thought about for years is a restating of the problem such that a) it causes the same cognitive hiccup and b) it is really easy to take someone's money by playing off it.
posted by Tell Me No Lies at 12:43 PM on February 21, 2015

the odds of getting a particular card on the river or turn being fifty percent--it's either going to happen or it's not

That's how some doctors I know think about probability sometimes when thinking about medical studies. After all, studies are all about the statistical meaning of group differences. How do you know which particular group any particular individual falls into? You don't, and there's no probability figure you can attach. So in some sense, the fate of any particular individual may often be -- 50/50.
posted by shivohum at 12:49 PM on February 21, 2015

room317, let me see if I can take a different tack with it. The illusion that gets most people (and got me for years) is that you're looking at what happens after you switch. Instead, let's consider what happens if you decide you're going to stay or switch before the game even begins.

Stay: Your chances are 1 in 3, or 0.33, no matter what Monty does.

Switch: There are two possibilities; you either start out on the car, or you don't.

There is a 0.33 chance you start on the car, in which case you won't get the car, because you're switching away from it.

But there is a 0.66 chance you start on a goat. Since Monty always shows you a goat, it's always the other goat, and switching will always get the car.

So by planning to switch, your chance of getting the car is 66% instead of 33% before the game even begins.
posted by localroger at 12:53 PM on February 21, 2015 [5 favorites]

Just a question, is anyone annoyed about Carl Sagan's last name? I heard him say that it came from his ancestor's hometown in Germany. How 'bout that, a town ith a history of sages?
posted by Oyéah at 1:26 PM on February 21, 2015

Your option to switch is not an option to switch from one of two doors to another. It is Monty Hall saying: "you can switch from your one door to all the other doors, and I'll throw in a freebie by eliminating one of the wrong choices from the universe of all other doors."

That is completely incorrect. The only reason this is an interesting "game" (or study of human mis-intuition) is that all but one other door remains closed. Monty is not giving you a freebie by opening one door: he is challenging your intuition by leaving you with two doors. The fewer doors you start with, the more challenging it is to refute your intuition: that's why there are three doors, not a hundred.
posted by five fresh fish at 1:30 PM on February 21, 2015

Since Monty always shows you a goat, it's always the other goat, and switching will always get the car.

No, you will not. You do not have a 100% chance of getting the car by switching. You have a (100% - your initial odds) chance. With three doors, switching gives you 100-33=66% chance. With 100 doors, a 100-1% chance=99%.
posted by five fresh fish at 1:33 PM on February 21, 2015 [4 favorites]

That's a very intuitive way of putting it, localroger.

My partner is having trouble with insomnia right now, and it helps her sometimes if I can offer her a mid night puzzle to occupy her mind long enough for her to relax and go to sleep. Last night after reading this thread I used the Monty Hall problem, and it worked. She got it, but I could tell by the way her brow crinkled before sleep smoothed it that she wasn't completely happy with the explanation. Tonight I'll try yours, and I think it'll work much better than mine did.

I really wonder whether it took this experience of being right when so many highly educated and credentialed people, apparently mainly men, were telling her how wrong she was, to embolden vos Savant sufficiently to publish that embarrassing and ridiculous book on the proof of Fermat's Last Theorem.

There's a cautionary tale in there somewhere.
posted by jamjam at 1:35 PM on February 21, 2015

Since Monty always shows you a goat, it's always the other goat, and switching will always get the car.

No, you will not. You do not have a 100% chance of getting the car by switching. You have a (100% - your initial odds) chance. With three doors, switching gives you 100-33=66% chance. With 100 doors, a 100-1% chance=99%.

Switching will always get you the better odds of getting the car, since clearly there's no strategy that guarantees success.
posted by clockzero at 1:48 PM on February 21, 2015

Huh, kinda neat that what finally convinced me was Ivan's actual simulation (which is what I thought "Monty Hall problem" formally referred to: An unintuitive probabilistic answer revealed by brute force calculation) which I saw when I was in high school.

And it's somewhat embarrassing to admit that even though I know the answer, when the problem comes up I usually have to reason through it again to be confident enough to talk about it (and certainly any variations mean that I go to scrap paper).

It's one of those weird foundational things that made me interested in learning about computer programming at all, to be able to test weird math assumptions that I'm not bright enough to reason through purely through logic without comparing them to actual data.
posted by klangklangston at 1:58 PM on February 21, 2015 [1 favorite]

Just a question, is anyone annoyed about Carl Sagan's last name? I heard him say that it came from his ancestor's hometown in Germany. How 'bout that, a town ith a history of sages?

"Sage," when it refers to wisdom, comes from the same Latin word that "sapient" does. I'm not sure what toponym Carl may have been referring to, but it probably wasn't in the same etymological clade as "sage," though it might have been related to the German verb "Sagen" which means "to say" or "to tell."

However, it occurs to me that the Latin "sapĕre" could also have an even further-back connection with an Indo-European root that also produced that German word for speech, since "capable of sense/wisdom" and "to say" could be pretty semantically close. Since I don't have access to whatever German's equivalent of the OED is, I can't explore that any further.
posted by clockzero at 2:02 PM on February 21, 2015

The Monte Carlo method is what you call it when you run repeated simulations to approximate an answer.
posted by achrise at 2:05 PM on February 21, 2015 [3 favorites]

The only reason this is an interesting "game" (or study of human mis-intuition) is that all but one other door remains closed.

Ack. I mean, all but one other door is opened. D-oh! Sorry.
posted by five fresh fish at 2:06 PM on February 21, 2015

No, clockzero, German sagen leads via OHG to the indo-European sek, whence also the English say, which appears to be cognate with sehen/seeing (so sagt Duden).
posted by progosk at 2:21 PM on February 21, 2015 [1 favorite]

Whereas Latin sapere goes to iproto-Indo-European sap, to try/research.
posted by progosk at 2:25 PM on February 21, 2015 [1 favorite]

I decided to try out another 100-door-with-random-lightning permutation, because my intuition told me that the odds of success between staying and switching can't remain at 50% if we change the rules to allow the 98 lightning strikes to hit any of the 100 doors, including the one you picked initially. (The assumption is that the game would end prematurely if the lightning hits either the one good door or if it hits the door you initially picked.)

(Again, let me know if you see any mistakes here.)

So the rules now are:

You pick one out of 100 doors.
98 random lightning strikes hit 98 of them, including possibly yours.

1/100 chance you're right with first pick
2/100 chance for 98 random lightning strikes to then not end the game by ever striking your (correct) door [ (99/100) x (98/99) x ... x (3/4) x (2/3) = (2/100) ]
total odds of reaching this scenario: (1/100) x (2/100) = (2/10000) = .02%

99/100 chance you're wrong with first pick
-- to not end the game, we need the 98 bolts of random lightning to neither strike your picked door nor the correct door. So the probability series of not ending the game early, as we go through each lightning strike, is (98/100) x (97/99) x (96/98) x (95/97) ... (4/6) x (3/5) x (2/4) x (1/3)
-- and so the "orgy of cancellation" [thanks escabeche!] I think leaves us with (1/100) x (1/99) x (2) [someone should probably check my work here.]
-- and so, assuming I haven't made a mistake, the total odds of reaching this scenario is: (2/100)x(1/99)x(99/100) = 2/10000 = .02%

What I like about thinking this through is how it shows that while I feel I have a firm grasp on the original Monty Hall problem from a bunch of different directions, I obviously don't have a good intuitive understanding of this sort of probability problem in general. It seems so weird that the difference between this permutation and the previous one I worked through -- removing the lightning rod that protects your initial chosen door -- doesn't end up having any effect on the stay/switch question. (It does, of course, mean it will probably take thousands and thousands of more trials before you ever reach one that even presents you with the stay/switch choice.)
posted by nobody at 2:43 PM on February 21, 2015

Five Fresh Fish, in the sub-case where you have both decided to switch before the game begins and you have landed in the 66% chance of starting on a goat, in that sub-case once Monty shows you the other goat you have a 100% chance of getting the car if you switch. That sub-case represents 66% of the total use case, so 100% of it is 66% overall, and thus the switching strategy gets you the car 66% of the time.
posted by localroger at 3:20 PM on February 21, 2015 [1 favorite]

I've known the correct answer for over a decade, and I can think of a handful of different proofs for it, but it still seems wrong to me.

Also, I wonder if there was an element of jealousy in various academics over the "smartest" label she got, and people were just waiting to pounce.
posted by mantecol at 3:22 PM on February 21, 2015 [3 favorites]

"Huh, kinda neat that what finally convinced me was Ivan's actual simulation (which is what I thought 'Monty Hall problem' formally referred to: An unintuitive probabilistic answer revealed by brute force calculation) which I saw when I was in high school."

Yeah, per achrise, maybe you conflated the MHP with the Monte Carlo Method? The latter is specifically a technique you use when solving the problem directly is simply not possible and solvable simplification that is good enough is also not possible (for example, three body problems in Newtonian gravity where a full calculation is not feasible but a simplified analysis is good enough even for rocket science), so you're left with running a large number of probabilistic simulations and measure the distribution. As I understand it, this was first developed rigorously as a technique during the Manhattan Project in Los Alamos by the very impressive Stanislaw Ulam. The Wikipedia article that archrise linked to is pretty good.

In a way, the MHP in this context is sort of an inversion of what Ulam was faced with. If you have the math, completely and correctly solving the MHP is pretty easy (but a formal solution requires conditional probability and not just classical probability). But most people don't have the math but a bunch of people (including myself in 1990) have the minimal programming skills and access to computers to simulate the problem and look that resulting distribution. That won't tell us why the results are what they are; but if we wrote the program correctly (Lemurrhea's earlier point is a good one) then the valid results will confirm or disconfirm our intuition and we can work from there.

Ulam's situation was the reverse. He was one of the best mathematicians in the world, but the results he needed weren't directly calculable. And computers were rudimentary -- he and others conceived of fully automating the necessary simulations, but this wasn't possible at the time. They simply did most of this painstakingly by hand, as I and others suggest people do with the MHP and dice or whatever (but you really need quite a few iterations of the MHP to get significant results, although fewer will be enough to give you a clue -- the Money Carlo simulations that Ulam did were just a punishing amount of tedious work. But it was necessary.)
posted by Ivan Fyodorovich at 3:31 PM on February 21, 2015 [2 favorites]

Localroger: my apologies, you are correct and, on re-read, were clearly talking the sub case. Sorry.
posted by five fresh fish at 4:48 PM on February 21, 2015

"Yeah, per achrise, maybe you conflated the MHP with the Monte Carlo Method?"

I think I did, yeah. Probably because the MCM you ran was what convinced me on the MHP.

I LEARNED IT FROM WATCHING YOU DAD
posted by klangklangston at 5:05 PM on February 21, 2015 [1 favorite]

That's OK FFF, I see on both sides of this (having now been on both sides of it myself) how FRUSTRATED you get TEARING YOUR HAIR OUT because those IMBECILES just wont' see what is SO FREAKING OBVIOUS!!!111!! that you just have to start skimming what you're replying to because, hey, high blood pressure is a dangerous thing. :-)
posted by localroger at 7:10 PM on February 21, 2015

By the way, in case anyone is curious about the hh/ht problem, here is a sketch of why the expected number of flips you'll get before HH is 6:

First, the number of sequences of length n that end in HH and where that ending HH is the only HH in the sequence is F{n-1} (the n-1 th fibonacci number). You can get this through induction but I won't prove it here. So, for example, for sequences of length n=4, there should be F{3} = 2 sequences that end in HH and have no other HH (these two sequences are TTHH and HTHH if you're curious).

Then, the probability of stopping after exactly n flips is given by
P_stopping(n) = F{n-1} / (2^n)
because there are F{n-1} sequences ending in HH and without HH elsewhere out of the total 2^n sequences of length n.

Then it's simply a matter of computing an expectation over all the possible stopping points:
E(HH) = Sum_{n=2 to inf} n * P_stopping(n)
E(HH) = Sum_{n=2 to inf} (n * F{n-1}) / (2^n)

Which I won't go into the details of actually computing this infinite sum but it's six.

Here's a python implementation that'll both simulate the HH and HT scenarios, as well as numerically estimate the above infinite sum in case you don't trust me that it sums to six.
posted by Pyry at 9:23 PM on February 21, 2015 [1 favorite]

I feel like I am literally in an insane asylum when I hear people say that it is not IMPLICIT in the problem that Monty chooses a goat. It says it right there! He chooses a goat!

This is like hearing Scott Walker talk about basically anything.

Also, Ivan is my hero. The failure of intuitive thinking is why diagrams are important. If you draw this out for someone, it is so mind numbingly-obvious that they'll probably just quietly mutter "huh."

I'm all for intuitive thinking, I've based my livelihood on it, until it comes to probability because humans are so bad at it.
posted by OnTheLastCastle at 9:33 PM on February 21, 2015

I feel like I am literally in an insane asylum when I hear people say that it is not IMPLICIT in the problem that Monty chooses a goat. It says it right there! He chooses a goat!

This is understandable, though, since some phrasings of the problem make it ambiguous as to whether Monty just happened to choose a goat or whether he must necessarily choose a goat. If Monty only happened but didn't have to choose a goat, then the 50%'ers are right (you can confirm this with simulations including the one I linked earlier).
posted by Pyry at 9:40 PM on February 21, 2015

I find that the hardest thing to get past is the intuition that Monty opening a door is just irrelevant or uninformative.

My favorite way of undermining that intuition is the one given by Clark Glymour, which is based on the collider principle in causal reasoning. Suppose A, B, and C are random variables. And suppose that C is a common effect of A and B, which are otherwise unrelated.

The structure looks like this: A → C ← B.

Conditioning on the value of the common effect induces an association between A and B. A simple illustration (I think due to Pearl) has you imagine a car that (1) either has or doesn't have gas, (2) either has or doesn't have a charged battery, and (3) starts iff it has both gas and a charged battery.

The structure looks like this: Battery → Starts ← Gas

Suppose the state of the battery and the state of the gas tank are independent: Knowing the battery is charged doesn't tell you whether or not there is gas in the tank. But now, suppose you go to start the car and it *fails* to start. And suppose that after the car fails to start, you check the battery and find that it is charged. You now know for sure that the gas tank is empty. Learning the value of a common effect -- whether the car started or not -- induced an association between previously independent variables.

In the Monty Hall problem, which door Monty opens is caused by two otherwise independent variables: the door that you initially choose and the location of the prize.

The structure looks like this: Initial Choice → Monty's Act ← Prize Location

So, conditioning on Monty's act induces an association between your choice and the prize location, which makes clear how Monty's act of opening a losing door can be informative.
posted by Jonathan Livengood at 10:16 PM on February 21, 2015 [2 favorites]

I won't lie, I kind of want to create a quick little python implementation of the MHP, obfuscate the code a bit, and have it spit out a 50/50 answer. And then link it all over the place. Just to mess with people.
posted by Lemurrhea at 5:14 AM on February 22, 2015

OnTheLastCastle: "So you by switching:
You = goat, you = car!
You = goat, you = car!!
You = car, you = goat, why u so stoopid"

I've understood it before (by writing out every possible permutation), but you're right, the answers are all seriously longwinded. You could much more easily explain with just the following ten lines (based on what you said):

"Don't switch" approach:
You pick the door with Goat 1. Monty shows you Goat 2. You don't change. YOU LOSE!
You pick the door with Goat 2. Monty shows you Goat 1. You don't change. YOU LOSE!
You pick the door with the car. Monty shows you a goat. You don't change. YOU WIN!
End result: You win 33% of the time!
"Do switch approach":
You pick the door with Goat 1. Monty shows you Goat 2. You change, thereby picking the car. YOU WIN!
You pick the door with Goat 2. Monty shows you Goat 1. You change, thereby picking the car. YOU WIN!
You pick the door with the car. Monty shows you a goat. You change, thereby picking a goat. YOU LOSE!
End result: You win 66% of the time.

All the "You've increased the amount of knowledge in a closed system through the input from Monty, so whereas you initially had a 33% chance of incorrectness, the correctness case has been reduced by 50% while the incorrect ratio remains constant, meaning that..." stuff is great to work through and try to parse once you've already accepted the correct answer and are just trying to figure out why it's correct, but it's so often used to try to explain which answer is correct, and it rarely seems to weork.
posted by Bugbread at 7:03 AM on February 22, 2015 [3 favorites]

Here's how it makes sense to me after reading this (excellent) thread. Those who know better, is this a flawed way of describing it, from any standpoint?

(If you originally picked a goat) When Monty shows you a goat, you have learned which door he had to choose, in order not to reveal the car.

(If you originally picked the car) When Monty shows you a goat, you have been mislead into thinking Monty revealed the above.

You have a 2/3 chance of having picked a goat.

Hence, in 2/3 of cases Monty has shown you where the car is.
posted by howfar at 9:42 AM on February 22, 2015

I think the best way to convince someone is to play it for money. I'll be Monty. You try it 10 times and don't switch, and I'll pay you a buck if you get the car; you pay me a buck if you get the goat. Then you be Monty and I'll switch every time. I think losing money convinces people.

My bone to pick with Marilyn was the time she flat-out stated that always, the faster you drive a car, the more gas it burns. (This had something to do with the 55mph speed limits). While I'm sure that I use more gas going 75 than going 55, I'm pretty sure I get less mpg driving at 5mph than 50mph.
posted by MtDewd at 1:40 PM on February 22, 2015 [1 favorite]

"(If you originally picked the car) When Monty shows you a goat, you have been mislead into thinking Monty revealed the above."

I wouldn't say that.

You're on the right track, though. One way I explain it when someone's on the track you're on (they accept and understanding a few crucial things already) is that when you actually play the game and you've already made your initial choice of a door, you're in one of two universes -- one where Monty absolutely had no choice whatsoever in which door to open and so by opening the door he does open he's basically pointing to the other door and saying "this is the door with the prize"; and one where Monty can open either of the other two doors, whichever he likes, and he's not telling you anything in doing so. In the latter universe, you've learned nothing and your initial choice is the best you're going to do. In the former universe, you know exactly which door has the prize.

Unfortunately, you have no way to distinguish those two universes apart.

But fortunately for you, it's twice as likely that you'll end up in the universe where Monty has to tell you where the prize actually is than the one where he doesn't tell you anything. So if you just assume that he's told you where the prize is and make your choice accordingly (by switching to the door he didn't open), you won't get the prize every time, but you'll get the prize two-thirds of the time as compared to only one-third of the time by staying with your original choice.
posted by Ivan Fyodorovich at 2:06 PM on February 22, 2015 [1 favorite]

The correct answer is 7.8%

zachlipton, no it isn't! I read your example and thought to myself that can't possibly be right, it has to be over 8% so I checked with a calculator and it is 8.4%.

Not sure whether it is a deliberate trap in the article you cited, but the example is stated with the false positive rate as 9.6% and then she runs the numbers using 9.5%
posted by lastobelus at 2:45 PM on February 22, 2015

Unfortunately, you have no way to distinguish those two universes apart.

Thank you. Yes, what you say makes sense. I think I probably do largely conceptually grasp it, at this point, although it's very much a sliver of soap, and prone to slipping away.

I considered whether to use "misled" (although not how to spell it, I now note), over a description closer to yours, and settled for it because it seemed, to me, to relate more closely my sense of Monty as a communicating agent, but, as you point out, I know that I don't know which universe I'm in, and I also know the degree of uncertainty I have about it, so "misled" is probably, um, misleading.
posted by howfar at 3:39 PM on February 22, 2015

I feel like I am literally in an insane asylum when I hear people say that it is not IMPLICIT in the problem that Monty chooses a goat. It says it right there! He chooses a goat!

I don't think anyone is saying that it's not implicit that Monty chooses a goat. In fact, as you say, it's explicit. But what's not necessarily implicit in the problem statement is that Monty would have chosen a goat if you'd chosen a different door. And that's an important assumption, because otherwise we could be dealing with Gardner's Evil Monty. If you've chosen the door with a car behind it, Evil Monty will open one of the other doors (with a goat behind it), in an attempt to get you to switch to the remaining door (which also has a goat). But if you've chosen a door with a goat behind it, Evil Monty won't open any other doors, or even offer you the option of switching. So if you're dealing with Evil Monty and he opens one of the other doors, showing you a goat, then you should not switch.
posted by klausness at 3:44 PM on February 22, 2015 [1 favorite]

lastbobelus I ran the numbers in Excel and come to 7.8%.

http://i.imgur.com/5yQCuKb.png

Bayesian probability was one of the more useful stats modules I did in university. Actually, nearly all the stats modules I did were good, I pretty much did a Math degree with a focus on stochastic probability.

The Monty Hall problem was covered in a programming module where the naive students were asked to create a simulation of the Monty Hall problem as an assignment - without being told what the theoretical solution was supposed to be! - and the first few times I ran the simulations (10,000 iterations) I was so sure I had made a mistake in programming it because the answer didn't match my intuition... sneaky. I can only imagine the professor cackling in glee as he set this assignment...
posted by xdvesper at 3:49 PM on February 22, 2015

Newcomb's Problem is also good for starting nerd war. There should be some ways to give Evil Monte's show some Newcomb-like quality, where he reveals goats or cars based on his perfect knowledge of your strategy.
posted by thelonius at 4:33 PM on February 22, 2015

zachlipton, no it isn't! I read your example and thought to myself that can't possibly be right, it has to be over 8% so I checked with a calculator and it is 8.4%.

It took me a while, but I think I figured out what happened. Try this handy calculator with the values from the problem: P(H)=0.01, P(D|H)=0.8, P(D|H')=0.096. You'll get 0.0776 = 7.8%.

So what's the deal with 8.4%? Well the calculator also turns up that value as W1, the posterior odds of cancer given a positive test result. So the question is whether you want the answer as a percentage or in terms of odds.
posted by zachlipton at 7:20 PM on February 22, 2015

2) The single biggest game changer I have seen (already mentioned a few times in this thread) is pointing out that in choosing which door to show you the host injected more information into the system.

Yes, this was what finally made it click for me when this first blew up in the wake of VS's column. The way I made it make sense for me was to imagine a scenario in which there are some arbitrarily large number of doors: let's say 1000. You pick one, and Monty Hall opens 998 of the other doors, all with their sad-faced goats behind them, leaving just one unopened door. Would you switch? Of course you would. It would be like playing minesweeper when you click on one little square and a vast field of mines disappears leaving just one isolated little square in the middle. Of course there's a mine there and, similarly, of course the car is behind the one unopened door of the 999 doors you didn't initially pick (o.k., one in a thousand times you'll be wrong--but who cares).

Change the numbers around and the fact that a 1/2 chance is different from a 1/3 chance suddenly becomes intuitively graspable.
posted by yoink at 1:56 PM on February 24, 2015

I think the reason the solution doesn't make sense to me is I just can't see it as anything other than two separate random choices dressed up like they are related.

Monty always shows you a goat and asks if you want to switch regardless of what door you chose first. The whole point of that is to make the game more interesting - it feels like a game of skill because you're given this additional information to consider, and asked if you want to change. But that information has absolutely no relationship to your first choice.

Pick the car first, Monty shows you one of the goats. Pick a goat first, Monty shows you the other goat. The second choice will always be between one car and one goat, with a 100% chance the car will be behind either the door you picked first or the one you didn't. The third door is irrelevant because it is not included in the second choice.
posted by InfidelZombie at 2:37 PM on February 24, 2015

I try not to look at the second choice as a random decision at all. You pick a door, and then Monty offers you the chance to swap whether that choice was a win or a loss.
posted by Holy Zarquon's Singing Fish at 2:47 PM on February 24, 2015

"But that information has absolutely no relationship to your first choice."

But it absolutely does. In the generalization, the door that Monty chooses to open is constrained by your choice, there is a dependency there. They're not independent events.

Specifically, it's true that when you pick the car, Monty is free to choose between either of the other two closed doors. But when you pick the goat, Monty has no choice at all. He is forced to open the door he knows doesn't contain the car. Your choice determined his actions (in combination with the fact that he does know where the car is and the rules specify that he can't open a winning door).

You don't know if this particular iteration of the problem is one where Monty's choice isn't constrained by your choice, or if it is one where it is. What you do know is that the latter is twice as likely as the former.
posted by Ivan Fyodorovich at 5:23 PM on February 24, 2015 [1 favorite]

But Monty's choice is pretty meaningless when it's his to make, right? One goat is the same as any other, and he always reveals a goat.

As a player in round 1 you don't learn anything about what's behind the door you picked, or the door that is left, from the goat Monty shows you. You make your guess in round 2 with the same information as before just one fewer door.
posted by InfidelZombie at 6:58 PM on February 24, 2015

Maybe it's easier to think about this as cards:

The dealer has three cards, two of which are "Goat" cards and the third is the "Prize" card. The dealer gives one card to Player A and two cards to Player B. Neither player can look at their cards. Then, the dealer looks at Player B's cards and throws away one "Goat" card. Then both players reveal their cards and the one holding the "Prize" wins.

Would you rather be Player A or player B?
posted by Pyry at 7:57 PM on February 24, 2015

InfidelZombie, as I pointed out above it's a lot easier to think about this if you consider that you've made your choice of strategy before the game even begins.

If your strategy is STAY, you have a 33% at the car, since nothing Monty does matters.

But if your strategy is SWITCH, there are two further possibilities:

1. 33% you land on the car in your iniital choice. In this case you've LOST the car, because your strategy switches you away from it.

2. 66% you land on a goat in your initial choice. In this case you will get the car, because Monty is going to show you a goat, and the only goat he can show you is the OTHER one. So when you switch, you will be switching to the car.

So STAY gives you 33% chance of car, and SWITCH gives you 66% chance of car.
posted by localroger at 8:09 PM on February 24, 2015 [2 favorites]

Ok, I think that finally makes sense for me- thanks localroger
posted by InfidelZombie at 9:54 PM on February 24, 2015

Monty's Choice
posted by clockzero at 9:03 AM on February 25, 2015

Hobson's Goat
posted by cortex at 9:05 AM on February 25, 2015 [1 favorite]

Pick the car first, Monty shows you one of the goats. Pick a goat first, Monty shows you the other goat. The second choice will always be between one car and one goat, with a 100% chance the car will be behind either the door you picked first or the one you didn't. The third door is irrelevant because it is not included in the second choice.

See, this was what struck me as intuitively clarifying about multiplying the number of doors. Because what you're describing is the basis of the cognitive error that makes people opt for the wrong answer, but it seems to me that that cognitive error dissipates if we imagine some arbitrarily large number of doors in the sample, and yet the reasoning you'd be relying on would be identical.

Let's say you are faced with a million closed doors and Monty tells you there's a car behind one of the doors and asks you to pick a door. You pick one, but you're pretty damn certain there isn't a car behind it. After all, what are the odds? One in a million, right?

Then Monty opens 999,998 of the doors to reveal goats. So you have the door you initially picked--the one you're 999,999/1,000,000 doesn't have a car behind it, and one other door. That door, you're 999,999/1,000,000 sure does have a car behind it!

Monty offers you a chance to switch doors. Do you switch? Of course you do. Why wouldn't you?

And yet, the apparent reasoning against switching remains identical with the original Monty Hall problem. The case has reduced to an apparent 50-50 split. You know the car is behind one of two doors and you've chosen one of the two--how could changing to the other of those two doors improve your odds?

Pick the car first, Monty shows you 999,998 of the goats. Pick a goat first, Monty shows you the other 999,998 goats. The second choice will always be between one car and one goat, with a 100% chance the car will be behind either the door you picked first or the one you didn't. The 999,998 other doors are irrelevant because they are not included in the second choice.

The reasoning is identical but now it's self-evidently (and intuitively) false.
posted by yoink at 9:51 AM on February 25, 2015

The reasoning is identical but now it's self-evidently (and intuitively) false.

Right. You have to ask yourself: why did Monty show me these 999,998 doors and then stop with this one? It's easy to imagine him going down the long line of doors, opening all of them to reveal the world's largest goat farm, and finally stopping at the very last one, but that's not what happens in the problem as stated. What really happens is that Monty works his way down the doors, opening all of them except for one in the middle that he just happens to skip. We also know that if he reaches a door with a car behind it, he won't open that door. That's the door with the 999,999/1,000,000 chance of a car.

That's the thing that always bugs me a bit about the problem. If you really clearly state what's going on w.r.t. Monty only opening goat doors, it's an interesting little puzzle, but nothing to get violent over. If you hide that fact in the problem description, the whole thing turns into an obnoxious riddle that relies on wordplay, and that's no fun for anybody.
posted by zachlipton at 10:07 AM on February 25, 2015

If you hide that fact in the problem description

This thread is the first place I have ever seen anyone raise the idea that Monty might randomly open a door with the car behind it. In no discussion of the problem I've ever read (including the original one) was it remotely unclear that Monty knows which door the car is behind and is not in the business of giving away cars by accident.

Really, the problem remains a puzzle for most people even when they fully understand that Monty will always and only open the door behind which he knows there is a goat. That has always been absolutely implicit to the structure of the problem.
posted by yoink at 10:21 AM on February 25, 2015 [5 favorites]

I'm too lazy/dumb to math it out further, but in short, person A is always in danger of flipping tails and losing their streak, whereas person B only needs to get heads once, and then after that they will either get heads again, leaving them no worse than before, or else get tails and stop.

Yes, I guess that's a nice intuitive way to solve the problem: for A, getting a T is always "total failure" for A, that is they have to start from the beginning, whereas B only has to start over if they get T on their very first flip: if they keep getting H, then they're already halfway through winning.

I think it's also nice to solve the problem quantitatively because it's not at all obvious but it's easy to generalize once you see it. Besides, our intuition is often wrong even if we know about conditional probability.

Let's call E(HH) the average number of flips needed to get HH. Now consider what happens after the first flip. If we got T then we've "wasted" one flip, so in this case we start over: now we need E(HH)+1 flips. This happens with probability 1/2.

If on the other hand we get H, then consider what happens after the second flip. If get get H again we are done, so the number of flips was 2. If we get tails we have wasted two flips, so now we need E(HH)+2. Both of these possibilities happen with probability 1/4 (since we've done two flips now).

To get E(HH) we add up these three events, weighted by their probabilities. So we get the equation E(HH) = (1/2)*(E(HH) + 1) + (1/4)*2 + (1/4)*(E(HH)+2), which gives E(HH) = 6.

The case of E(HT) is a little more complicated. If we get T on the first flip then it's the same as before: we need E(HT)+1 flips, and if we get H on the first flip and T on the second we're done. But what if we get HH? In this case we can't say that we've wasted two flips, since we would be done if we flipped T next. But it's also not correct to say that we've wasted one flip; that's essentially the same fallacy that people often apply to the Monty Hall problem.

The right approach is to use our knowledge of past flips, which we didn't need in the case of HH because we were always starting over when we got T. Let E(HT|H) be the average number of flips needed to get HT assuming that the first flip was H. Applying the same reasoning as before we get E(HT|H) = (1/2)*2 + (1/2)*(E(HT|H) + 1), which gives E(HT|H) = 3. Going back to the original problem, we have E(HT) = (1/2)*(E(HT) + 1) + (1/2)*E(HT|H) = 2 + E(HT)/2 which gives E(HT) = 4.

(You could of course use this approach in the HH case, but, since we were always starting over in case of failure, conditional probabilities don't really play a role: E(HH|T) = E(HH).)
posted by maskd at 11:17 AM on February 25, 2015 [1 favorite]

This thread is the first place I have ever seen anyone raise the idea that Monty might randomly open a door with the car behind it.

I promise, if you were a mathematician, and you had been in dozens and dozens of conversations about this, you would be very familiar with this idea. It's one of the main things you talk about when you talk about this problem.

But you're absolutely right that lots of people, including the people who wrote angry letters to MvS, interpret the question to mean "Monty is not allowed to reveal a car" and then get that question wrong.
posted by escabeche at 12:31 PM on February 25, 2015 [1 favorite]

What if they open door #3 and both goats are there because one has chewed through the wall? What does that make the odds?

Also I still don't get it. Having three doors and saying there is a goat behind two of them and then opening one of the doors and showing that there's a goat there is basically just saying there are two doors, and one has a goat behind it, and the other doesn't. No amount of legal mumbo-jumbo changes that. And what kind of car is it? Maybe I would prefer the goat? Nobody seems to have asked this important question.
posted by turbid dahlia at 2:40 PM on February 25, 2015

What if they open door #3 and both goats are there because one has chewed through the wall? What does that make the odds?

Confusing.

Also I still don't get it. Having three doors and saying there is a goat behind two of them and then opening one of the doors and showing that there's a goat there is basically just saying there are two doors, and one has a goat behind it, and the other doesn't.

First you pick a door at random. 1/3 chance it's a car, 2/3 chance it's a goat.

Next, Monty opens a goat door that you didn't pick. But -- and this is the key to the problem -- his doing so doesn't change the odds that you picked a car one iota. There are two goats and you only get to pick one door, so (barring wall-eating shenanigans) there will always be at least one door you didn't pick that has a farm animal behind it. Showing you a goat door adds no meaningful information to the game that you didn't already know from the basic rules. The odds are still 1/3 that your door is the car, and 2/3 that it's a goat.

At this point, Monty offers you the chance to stay with your original door or switch. If you stay, you get whatever you picked in the first place. If you switch, you get the opposite of what you picked in the beginning -- assuming the opposite of "goat" is "car."

There's still a 2/3 chance you picked a goat to start out, and if you picked a goat and switch, you always end up with the car. So switching gives you 2/3 odds to win the game.
posted by Holy Zarquon's Singing Fish at 3:14 PM on February 25, 2015

That reminds me of how my high school stats teacher tried to emphasize that it's almost always better to think through problems based on not winning rather than winning — from the "does switching leave me with the goat" perspective, I think the problem's a little more clear.
posted by klangklangston at 3:44 PM on February 25, 2015

Yeah, going with Holy Zarquon's Singing Fish's example, and I'm sure my numbers are probably wrong, but they do give a quick-and-dirty sense of why switching is good:

When you first pick, you have a 66% chance of picking a goat and therefore losing.
Monty opening a door doesn't change your chance of having picked a goat at the start.
If you stay with your first choice, you still have a 66% chance of losing.
But, kinda like your instinct tells you, if you pick the other door, you've got a 50% chance, right?
So if you stay with your first choice, you have a 66% chance of losing. If you switch, you have a 50% chance of losing. So switching is better.
posted by Bugbread at 5:11 PM on February 25, 2015

That reminds me of how my high school stats teacher tried to emphasize that it's almost always better to think through problems based on not winning rather than winning

This is true of most gaming / chance / probability problems. In fact, it's usually best to think the problem through every possible way, and regard it as a huge alarm if the results don't converge.

A really excellent example is what you should do if you're playing blackjack and you get dealt 16 against a dealer 10. Years of computer analylsis agree that you should take the hit, even though you will probably bust. The problem is that you have, statistically speaking, most likely already lost, but you will lose less if you take the hit than you will if you sit there and let the dealer try to make any hand at all. I know people who made a couple of million dollars by knowing stuff like that.
posted by localroger at 5:21 PM on February 25, 2015

I promise, if you were a mathematician, and you had been in dozens and dozens of conversations about this, you would be very familiar with this idea

I think that's a classic "assume a spherical cow" problem, though. I mean, no one in any real world, non-academic case of having the setup explained to them fails to connect the fact that Monty Hall knows which doors have goats behind them with the fact that he's not going to open a door with a car behind it. This just sounds like Déformation professionnelle, but not an interesting thing about the problem (or the confusion it causes) as it exists out in the wild.

Whatever strange inability mathematicians have in grasping implicitly clear points about the way game shows work, what remains interesting about the problem is that it reliably generates confusion in people who fully understand the rules Monty Hall is supposed to be playing by. In other words, the idea that the confusion is caused only because some crucial part of the set up is deliberately obfuscated is just not the case (except--I will take your word for it--among some over-literal-minded mathematicians).
posted by yoink at 10:01 PM on February 25, 2015 [3 favorites]

Yeah, in the larger scheme of things MHP, it's a minor issue. My twenty years of experience corresponding with people about the MHP (which counts for something) is that the vast majority understand that Monty will always open a losing door and that of the minority who get caught up on this supposed ambiguity, at least half of those are just retreating to a fall-back position because they can't admit they were wrong.

Of that small portion who really and truly did interpret the problem this way, they're personality types and/or academic types who are predisposed to a strict literalness -- two things which are often true about mathematicians.

I don't doubt for a second that some non-trivial number of academic mathematicians will interpret it this way in complete sincerity, while at least as many of them will willfully misinterpret it this way because, hey, the wording of the problem isn't clear.

I tend to think of the latter group as like those barometer question smart-ass types who, really, ought to be considered to have simply gotten the question wrong because -- I'm going to repeat this -- any English statement of a problem, even a well-defined mathematical problem when it's expressed mathematically, is going to rely upon a huge amount of context in order to evaluate. Being literal minded and genuinely not understanding the intent of the problem is fine; arguing that the intent isn't clear because it's not explicitly stated isn't fine because the presumption of such an argument obfuscates the truth that all such word problems require contextual interpretation and are not, in fact, on their own formal problems.
posted by Ivan Fyodorovich at 11:14 PM on February 25, 2015 [2 favorites]

This thread is the first place I have ever seen anyone raise the idea that Monty might randomly open a door with the car behind it. In no discussion of the problem I've ever read (including the original one) was it remotely unclear that Monty knows which door the car is behind and is not in the business of giving away cars by accident.

On the other hand, this thread is where I first heard the idea that Monty's real-world behavior might have been to only open a door revealing a goat as an enticement to switch in cases where the contestant had picked the door with the car.

So yeah, it's ridiculous* to suppose a version where there's a possibility he could have randomly revealed the car, but it's actually a very good point to say that the traditional formulation of the problem doesn't tell you whether Monty always opens a door or not. (And I think if that's true, the problem can't be decided by probability, unless you have access to records of Monty's past behavior.)

* but a worthwhile mathematical exercise, as my intuition about it turned out to be wrong.
posted by straight at 8:02 AM on February 26, 2015

it's actually a very good point to say that the traditional formulation of the problem doesn't tell you whether Monty always opens a door or not.

That's neither here nor there. You are not being asked to judge the problem on the basis of real-world knowledge about the historical game-play of the specific TV program. You are being asked to solve a strictly delimited situational case that is simply modeled on aspects of the game play of the TV show. The problem specifies in so many words that

A) there were three closed doors.
B) you picked one of those doors, hoping a car was behind it.
C) Monty, knowing which doors had goats behind them and which had the car then opened one of the two doors you did not pick.
D) that you were then faced with the choice of switching to the remaining closed door or sticking with your original pick.

It is quite clear, then, that we are only being asked to consider situations in which you have a 1-in-3 chance of guessing correctly on the first round and a 1-in-2 chance on the second.

Bringing up the historical Monty's past behavior is equivalent to bringing up questions like "what if the dice aren't perfectly cube-shaped" in a statistical puzzle involving dice--it's bringing in real-world specificity that is clearly not part of the puzzle being posed.
posted by yoink at 9:12 AM on February 26, 2015 [1 favorite]

No, I only bring up Monty's real-life behavior to show it's completely reasonable to ask whether or not Monty always opens a door when offering the chance to switch. It's not some strained, "Are the dice loaded or not?" technicality. The puzzle doesn't say whether opening the door or not was an option for Monty, and it's a crucial bit of information, without which the puzzle should actually be considered unsolvable.
posted by straight at 1:14 PM on February 26, 2015

I found the solution of the Monty Hall problem pretty difficult to understand, but I have to say, I'm also finding people's inability to understand the scenario also pretty difficult to understand. It's just:

Scenario: “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?”

What's next, "The question doesn't state that you are aware that there are two goats and a car. Maybe you think there's only one goat"? "The question doesn't state the sizes of the doors. What if the other door is really tiny, so you know there's no way there could be a car behind it"? "The question doesn't say that you want the car. Maybe you want a goat"?

straight: "it's actually a very good point to say that the traditional formulation of the problem doesn't tell you whether Monty always opens a door or not."

You'll have to point out the comment in question, because the only ways I can think to parse this are:
1) He only opens another door if you picked the car. This is no longer a logic puzzle, it's a trivia question. "The answer is, actually I watch the show every week and Monty always opens the other door when your initial choice was correct, so you should never change the door"
2) He only opens another door if you picked the goat. See #1 above.
3) He has a greater statistical likelihood of opening a second door depending on whether you picked the goat or the car. This is no longer a logic puzzle, this is a statistics question with no data provided. You may as well ask vos Savant "If I roll a die a number of times, how many times will I roll a six?", without stating how many times you rolled the die.
4) The probability of him deciding to open another door is totally random. This does not affect the answer, so vos Savant is still right.

The first two scenarios are ridiculous trivia questions, and the third doesn't have enough information to answer, let alone to declare that vos Savant's answer is wrong. The disagreement on the Monty Hall question breaks down pretty cleanly into "vos Savant was wrong" and "vos Savant was right" camps. The size of the "she could totally be right, or totally wrong, but without a sufficiently large data set to analyze, it's impossible to determine" camp is so vanishingly small I think it's fair to say that it's completely ridiculous, not completely reasonable.
posted by Bugbread at 3:09 PM on February 26, 2015 [1 favorite]

4) The probability of him deciding to open another door is totally random. This does not affect the answer, so vos Savant is still right.

This was proven wrong up above in the discussion of lightning and 100 doors.
posted by straight at 3:17 PM on February 26, 2015

"This does not affect the answer, so vos Savant is still right."

It does change it, but in a way that's arguably slightly more difficult to analyze and which would be very unlike a puzzle in a newspaper Sunday supplement.
posted by Ivan Fyodorovich at 3:31 PM on February 26, 2015

straight: "This was proven wrong up above in the discussion of lightning and 100 doors."

The lightning and 100 doors discussion was a metaphor for Monty Hall always picking a door, but his choice of which door to open being random. My #4 was about Monty Hall sometimes choosing to open another door (but always picking the goat), and sometimes choosing not to open any doors at all. Totally different issue.
posted by Bugbread at 3:54 PM on February 26, 2015

I mean, it states in the question that he knows what's behind the door. The whole digression about lightning bolts was premised on "let's ignore an explicitly stated part of this question". If you're going to do that, why not just ignore the part that says there's a car, or the part that says that you can switch doors?
posted by Bugbread at 4:08 PM on February 26, 2015 [1 favorite]

Sorry, I wasn't specific enough. Nobody's analysis of what if Monty opens any door at random, possibly including the contestant's choice, is functionally equivalent to your option #4 (in that you could swap the event of Monty not opening any door for the event of Monty opening the player's door). And you can see that in that case the answer to the problem would change.

As to why people would ask that, I was just asking, in order to get a better understanding of the statistics behind the problem, whether the answer depends on Monty knowing where the car is. Wouldn't the player still have the same reasons to switch if a random event opened a door and it happened to be a goat? As you can see above, my intuition was wrong and the answer is no.
posted by straight at 10:02 PM on February 26, 2015

Man, I really, really, really hate maths.
posted by Too-Ticky at 1:18 AM on March 6, 2015

Oh, I think they were probably sent. But newspapers were pretty used to throwing away crank letters.

No, I grew up around newspapers and worked with them years ago, and during that time one of the things I sometimes had to do was deal with the mail. Before email and comments, we got crank letters, but they were super rare and by very obviously crazy people. There was no real equivalent to online comment hate before there were anonymous/pseudonymous online comments. Most newspapers would not publish a letter without a name attribution (and we also called to verify identity so some asshole wouldn't prank you by writing a letter to the editor in your name), so people did not really bother writing that many stupid, nasty anonymous letters to the press because they would never get an audience outside of the sniggering copy desk.
posted by Miko at 8:08 PM on March 6, 2015

Happy Pi Day Everyone!
posted by jeffburdges at 6:08 AM on March 14, 2015

"3.14.15"? Come on, nobody's using "MDY" order with dots in the date (and very few countries use that order at all, because it's kind of silly). IFLS, misinformation for the masses since 2012.
posted by effbot at 8:07 AM on March 14, 2015

and very few countries use that order at all, because it's kind of silly

I agree MDY is a silly order, but it's the standard order in the US, so it's an entirely natural way to think of this for a very, very large number of English speakers.
posted by yoink at 9:09 AM on March 14, 2015 [1 favorite]

Plus it would be kind of hard to have pi day at all if it needed to be in month 14.
posted by localroger at 9:29 AM on March 14, 2015

The non-American Pi Approximation Day is 22/7, which works out to 3.142857. It's as close to the mark as 3.14 is in most years.
posted by Shmuel510 at 9:45 AM on March 14, 2015

22/7 isn't a great choice. I think we'll stick to March 14 and April 31.
posted by effbot at 11:24 AM on March 14, 2015

"I agree MDY is a silly order, but it's the standard order in the US, so it's an entirely natural way to think of this for a very, very large number of English speakers."

Do other anglophones use this ordering when they give dates with words and not numbers? I'm going to report my birthdate as "November the eighth, nineteen sixty-four". Isn't the MMDDYY ordering a reflection of this? I do know that some people would say "Eighth of November" and maybe that's the standard non-American formulation.

I agree, though, that there are advantages to ordering the numerical format differently, either biggest units to the smallest, or the reverse. I idiosyncratically do a lot of quick-and-dirty text data manipulation using text editor search and replace and sorting, and the MMDDYY is annoying.
posted by Ivan Fyodorovich at 2:51 PM on March 14, 2015

I am British and always write little-endian, e.g. "15 March 2015". I also think that that is probably the more common way to express dates in speech. "It's the fifteenth of March today" is, I am fairly confident, more common that "It is March fifteenth today". I think "It is March the fifteenth today" would be rarest of all.

It depends, though. I think that, in the past tense, that intermediate "the" is more common, e.g. "It happened on March the fifteenth". But even there I think " It happened on the fifteenth of March" would be more typical.

Does anywhere apart from the US use middle-endian dates? I know that the French (among others) seem to tend to favour big-endianism, which is sometimes more useful for sorting text.

Whenever I am forced to use middle-endian dates (sadly often due to the state of IT procurement where I work) I find it completely baffling. It seems objectively harder to read, requiring the eye and brain to make more shifts than any other system. But the US still measures carpets in feet and inches, and the UK still measures roads in yards and miles; logic is often much less important, even from a practical perspective, than custom.
posted by howfar at 5:23 PM on March 14, 2015

Does anywhere apart from the US use middle-endian dates?

When I get invoices in Quebec about 50% are big-endian, 45% little-endian (sometimes written as 15-MAR-15), and the remaining 5% are middle-endian. The problem is that I can never remember which invoices are which when I am doing the bookkeeping.
posted by jeather at 5:33 PM on March 14, 2015 [1 favorite]

Little-endian dates, like metric measurements, make a lot more sense. But mmddyy is, like avoirdupois measure, the standard in the US and good luck getting us to change it.
posted by localroger at 6:13 PM on March 14, 2015

"It's the fifteenth of March today" is, I am fairly confident, more common that "It is March fifteenth today". I think "It is March the fifteenth today" would be rarest of all.

No, not in the US. "It's March 15 today" or "let's meet on March 15th," or "my birthday is March 15th" is the standard, common way to say it ("March the fifteenth"would sound a little weird).
posted by Miko at 8:51 PM on March 14, 2015

No, not in the US.

I was responding to Ivan's question about non-US anglophones.

"It's March 15 today"

Seeing it written down like that, for some reason, makes me think about direction of causality. I don't conclusively know whether the numerical convention follows the dominant speech convention or the other way around.

It's an interesting phenomenon, in any case. It does seem to be a potential example of a sort of conservatism of dominance. I think one could possibly put together an interesting argument that one factor in the demise of empires and hegemonies is an unresponsiveness to global change as a direct result of global dominance. Metrication is the most obvious issue in the case of the US. It is hard to estimate the precise economic cost of not metricating, but it is clearly significant for those industries with a significant reliance on import and export. The potential economic costs of middle-endianism are much less apparent, mainly due to a lack of a universal non-US standard, but it seems reasonable to posit that there are some costs.

It could be a useful starting point for a broader consideration of the structural factors that limit and control the longevity of geopolitical dominance.
posted by howfar at 2:01 AM on March 15, 2015

Month-Day makes more sense in speech when you take the year for granted. You're working from the general to the specific. You know just saying the 15th could be confusing because for a lot of purposes the 15th of several months are all equally plausible, so you specify the month first, then the date. March 15.

But it's much less likely that there will be ambiguity about the year, so it would be very redundantly redundant to go around saying 2015 March 15 every time you talk about a date. It's really only in writing where you need to do that, where it's kind of an afterthought tagged onto what we would do in speech. We met on March 15, oh yeah, in 2013.

The pure convention 15 March doesn't work as well when abbreviated with no year in spoken speech. And of course, people love to abbreviate our dates, which is why we're back to using 2 digit years a lot so soon after the Y2K mess.
posted by localroger at 5:53 AM on March 15, 2015 [3 favorites]

15 maart works just fine for me. I don't see why it would be more logical to go from the general to the specific; also, since every month has a 15th, one could also see '15' as more general. Which 15? Oh, right, the one in maart.

I think you are just used to what you're used to, and so that's what you are used to. I specify the day first, then the month. That's what I'm used to.
And it works great. No one thinks I mean 'maart 2015', for example. With 'march 15', there might be confusion. And 'march 15 15' sounds just odd to my non-US ears.

Mostly, it all comes down to what one is used to.
posted by Too-Ticky at 6:10 AM on March 15, 2015

I think the proof is in how the pattern of speech evolved before business communications started making the year a thing. Europe has always had a bit more success forcing the issue when common usages weren't optimal which is why America is still using inches and pounds as well as mm-dd-yyyy.

Incidentally, I think it's a bit funny that one of the other fliers in the global date game is China, which uses yyyy-mm-dd, doubling down on the general to specific idea.
posted by localroger at 6:22 AM on March 15, 2015 [2 favorites]

Month-Day makes more sense in speech when you take the year for granted. You're working from the general to the specific.

I can see the logic you're using here, but like all "rational" explanations for matters of linguistic usage it's pretty much post hoc. The human mind just doesn't really absorb spoken sentences in that "let me analyze each independent bead of meaning in turn" kind of way: that is, I don't hear you say "March" and think "good, good, I've narrowed it down to the month of March...now, I'm ready to hear which day in March we're talking about: the fifteenth, eh? O.K. now we're ready to get on to what happens on that day" or what have you. We take our sentences (let alone the individual phrases) in in pretty large chunks, and there's no real processing speed advantage to "fifteenth March" over "March fifteenth" or vice versa.
posted by yoink at 7:06 AM on March 15, 2015 [2 favorites]

YYYY-MM-DD is the only one that works if one wishes to have a list of dates which naturally sorts into the correct order. So I tend to prefer it, at least as a database programmer.
posted by koeselitz at 7:18 AM on March 15, 2015 [3 favorites]

"I can see the logic you're using here, but like all 'rational' explanations for matters of linguistic usage it's pretty much post hoc."

Yeah, I think that's right. People are weird about arguing how their usage is, of course, the one that is rational and useful. But language communities use what they use, and it works for them.

There's a lot of different conventions for speaking and writing dates. I think it's wrongheaded to argue that some are "better" than others.

The exception for this, I think, is the specific usage we're discussing -- the XX/XX/XX format -- within any global context like the internet, because there's a huge amount of ambiguity especially given that anglophones themselves differ about it. Unless there's a good reason to use that particular convention -- such as a need to use numbers for sorting or something -- it's probably better to use the month names, at least.
posted by Ivan Fyodorovich at 7:35 AM on March 15, 2015 [1 favorite]

I can see the logic you're using here, but like all 'rational' explanations for matters of linguistic usage it's pretty much post hoc.

Especially when the only argument is "I do this, so therefore it's better," completely ignoring that lots of other people do it differently. It's like hearing americans argue why it's obvious you cannot get rid of the penny coin or actually start using the dollar coin, or that all bills must have exactly the same size, or else civilization will collapse after a day or two :-)

(btw, my actual issue with the IFLS link was that she's using MDY order with dots as separators. At least stick to the slashes if you're going to use US date order...)
posted by effbot at 8:08 AM on March 15, 2015 [1 favorite]

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