Psychologists Don't Know Math 566
stupefaction writes "The New York Times reports that an economist has exposed a mathematical fallacy at the heart of the experimental backing for the psychological theory of cognitive dissonance. The mistake is the same one that mathematicians both amateur and professional have made over the Monty Hall problem. From the article: "Like Monty Hall's choice of which door to open to reveal a goat, the monkey's choice of red over blue discloses information that changes the odds." The reporter John Tierney invites readers to comment on the goats-and-car paradox as well as on three other probabilistic brain-teasers."
Nice try! (Score:5, Funny)
Re:Nice try! (Score:5, Funny)
Re:Nice try! (Score:4, Funny)
Re:Nice try! (Score:5, Funny)
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Re:Nice try! (Score:5, Funny)
In the puzzle, I clicked on the car instead to avoid goat links. However, the car had a huge ugly rusted gaping hole in the back bumper, dripping oily sludge. It was horrible! I'll never look at cars the same way again. The humanity!
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It's like a trained reflex, when you have a link that starts with "goat" and also contains the letters "c" and "x"..
I think this was a pretty interesting psychology experiment to get people to click on the link. Even just reading the phrase out loud brings people to a halt!
They don't know math? (Score:3, Insightful)
Re:They don't know math? (Score:4, Insightful)
In my experience (and I have a fair bit of exposure to and experience with the medical psychology) psychology is only good when the practitioners ignore their trade and just act like friends to their patients. That has nothing to do with the fact that they are psychologists, and more to do with the fact that they are good people. The world needs more good people, not psychologists.
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Wait... You don't have any psychological problems (other than replying to my post ^.^) You've never had sessions - you only run into this psychologist because she rents the space next to where you work. And you definitely haven't given her any money.
Do you mean to say, sir, that you are guilty of stealing psychology?!
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Re:They don't know math? (Score:5, Insightful)
After all, I have plenty of friends, and I'm in complete contact with my family, but they have no idea how to help me get through a bout of depression in anything approaching a concrete manner. Just being there isn't enough.
And I noticed further down where you market your experience with psychology. I'd just like to remind you, your personal evidence isn't any sort of justification for such sweeping statements.
I'd also like to remind you that your concept of "good people" seems a little skewed to me. I think you need to dwell a bit on how to remove so much of your personal bias from your opinions on general topics. You have no basis for positing that the world is shy of good people, because you only know a vanishingly small fraction of them.
Re:They don't know math? (Score:5, Informative)
Fortunately, there's some other kinds of psychologists that actually do stuff that works. I'll discuss a trifle about them below. Before that, though:
Any psychologists have a couple of things going for them, even without the "working method of psychotherapy" part. Going to a psychologist will make a patient regularly think about their problems, and will make them feel that they are in a process with the problems, and this seems to lead to change. It also makes the person deal with the problems in contact with a stranger, which makes for a more neutral setting than with a friend or family member. With a friend or family member, the relation in other contexts will very often intrude.
So, any psychotherapy will usually have *some* effect, though it may be very restricted, and for some kinds of problems it does not work at all. There are some forms that have more effect, chief among them behavioral therapy (with most research having gone into the cognitive behavioral version of this, but with very little evidence the cognitive part add effectiveness.) This is mostly "common sense" put into a system. Some examples: If a person is depressed and sitting at home, make them go out and do stuff, starting with small enough stuff that they're able to do it ("Behavioral Activation"). If the person is afraid, have them go through the fear in small enough parts that they can handle it, exposing them to situations they are afraid of and let them learn that they can be safe there, waiting until the fear dies down. If they have OCD, expose them to the situation that makes their obsessive response come forth, and prevent/delay the response. ("Exposure and Response Prevention.)
The good thing is that the psychologist knows that this common sense works, and can put the weight of both experience and theory behind the words to make the person feel that it can work.
Most psychotherapy works better without drugs; drugs interfere with the learning process.
Eivind.
Scientologists Paradox (Score:4, Funny)
- Scientologists, cockroaches, all your money
- Scientologists, all your money, cockroaches
- All your money, Scientologists, cockroaches
In two out of these three cases, you prefer Scientologists to all your money, so your best course is to join the church immediately.Re: (Score:3, Insightful)
I went to a counseller for 4 years and it did me good. I agreed goals with my counsellor, not exactly something you'd do with a friend. Our relationship was very productive.
It was a wierd feeling, recounting my life to her, and in the process relearn
Hmmm.... (Score:5, Insightful)
2) The issue seems easy enough to settle empirically, given a few monkeys and a bag of M&Ms, besides the fact that it seems to have been empirically settled decades ago anyway.
3) This is, though, a good opportunity to ridicule "21" for completely botching the Monty Hall problem, along with pretty much everything else relating to math, gambling and Boston-area geography.
Re:Hmmm.... (Score:5, Funny)
Inaccurate? (Score:5, Funny)
To the psychology researcher, it's more about getting the "story" right than actually quantifying anything.
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There are many sub-disciplines of psychology (Score:3, Interesting)
When I took my degree (double major: CS and Psych) all psychology undergrads were required to take courses in statistics and scientific methodology. I find it hard to believe that s
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Re:Inaccurate? (Score:5, Interesting)
It's kind of depressing, so I try to make light of it when I can.
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I haven't found this in my experience but then it might depend upon what kind of circles you move in. If it's the "lower end" of the scale in terms of ability, then yes I would agree but the same probably goes for any subject. Otherwise, I would disagree. I majored in psych and did a PhD in the topic. My external examiner for my viva was a Cambridge statistician and ex-math olympian (Alan Dix if you're curious) who focused his interests in human-computer interaction I think because of the challenges he got
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Re:Hmmm.... (Score:5, Interesting)
Suppose Monty Hall gives you a choice of two envelopes. Each envelope contains a check, and one of them is written for TWICE the amount of the other. So you pick an envelope.
Now, Monty gives you the chance to switch envelopes. (Assume Monty always gives you the chance to switch.) Logically, since your envelope contains X, the other envelope can contain either 0.5X or 2X, with 50% probability... So the expected value of switching envelopes is 50% (0.5X + 2X), or 1.25X. So, you should switch.
But here's the tricky part: Monty now gives you the chance to switch back! Since your new envelope contains Y, then by the same logic as above, the expected value of switching back is 1.25Y... So you should switch back. Right?
Clearly, something is wrong with this chain of thinking. Can you figure out what it is?
Re:Hmmm.... (Score:5, Informative)
G = 50% * (Gained if we were holding X) + 50% * (Gained if we were holding 2X)
= 0.5 * (2X - X) + 0.5 * (X - 2X)
= 0
So switching envelopes doesn't change the expected value.
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Re:Hmmm.... (Score:5, Interesting)
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True. But not all schools of economics try to make themselves a science. It's a difference in methodology. The Austrian School is a notable example, because they specifically reject scientific positivism. The Neoclassicists are obsessed with deriving mathematical formulas, and the Monetarists are obsessed with scientific predictability.
I sympathize with the Austrians, but realize that the Neoclassi
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And true... (Score:4, Interesting)
Seriously.
At the end of the course we were given a summary lecture that (shock horror, ladies fainting at the back) gave us a FORMULA that explained the whole point of what we'd been taught. I wasn't the only person who, at this point, suddenly realised wtf they had been blabbering on for the past 2 months... and more to the point, how much crap they'd been talking. Psychologists were taking formulae based on reason and using them to support conjecture. That's not inflammatory, it's fact.
Re:Hmmm.... (Score:5, Funny)
I wonder what does a true scotsman do?...
Seems to make sense (Score:5, Interesting)
The psychologists were claiming that if you choose X over Y then you are more likely to choose Z over Y because your *choice* causes bias against Y. (This fits the observed data).
The new suggestion is that if you choose X over Y then you are more likely to choose Z over Y because the choice indicates prior bias against Y. The important part being that this holds even if the bias against Y is so small that it is hard to detect. The only thing required is that there is a fixed "preferred order" of the three.
At least, that's what I understand from the article. Given the field, I also understand that I am most probably wrong :)
I dislike things that "seem". (Score:5, Interesting)
(That last pair of points are important. Monkeys do not see all colours with equal clarity. Neither do humans, which is why monitors actually have more real-estate set aside for blue than for anything else. Complicating things, colours are usually the product of mixing. They are not "pure". We don't know what the monkeys saw, therefore cannot tell if their decision was influenced by their ability to even see the treats.)
Personally, I have developed a skepticism of such observational science. Too many possible explanations, yes, but more importatly too little experimentation to eliminate alternatives. If an explanation is put forward and then acted upon, especially in an area like psychology where those being acted upon are likely vulnerable groups, it's important to make sure the explanation is likely to be correct. Likely to be possible isn't good enough.
What would I suggest? Well, in the 1950s through to the last few years, options have been limited. These days, though, you can take fMRIs, MRIs and CAT scanners into the field. During the Chernobyl accident, it was fairly standard procedure for MRIs on trucks to be used to scan farm animals for contamination. See the brain in action as it makes the choices. See when the choice is made and which neural pathways were involved. Much better than speculating about what's going on. If you want more data, scientists decoded the optic fibre transmissions of cats ten years ago, or thereabouts. We can literally see if that plays a part in the decision.
You still end up doing statistics, sure, but with far more numbers that have far more meaning behind them and far less room for interpretation.
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You could equally well run the experiment with three types of treat - say peanuts, brazil nuts and pecan nuts - as long as individual monkeys have p
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There is another problem.
Not to brag, but I have very acute taste buds. So much so, that when I was in high school, I would put M&Ms in my mouth with my eyes closed and be able to tell which color it was with nearly 100% accuracy.
The reason I could do this is that the dyes actually taste different.
Now that they have added pure red and pure blue (n
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I agree it would be interesting to have some links, so I hope GP isn't just talking out of his ass.
Re:I dislike things that "seem". (Score:4, Funny)
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"...Tversky and Kahneman offered each subject a bet. They would roll a fair die with four green (G) and two red (R) ones, and the subject made a choice between betting that the seque
To be fair, mathemeticians didn't know math either (Score:5, Interesting)
Re:To be fair, mathemeticians didn't know math eit (Score:2)
Re:Martin Vs. Marilyn (Score:4, Informative)
Re:To be fair, mathemeticians didn't know math eit (Score:5, Insightful)
She does seem to be brilliant, but everyone makes mistakes, and calling them on them will educate them if they were wrong, and educate you otherwise.
Re:To be fair, mathemeticians didn't know math eit (Score:5, Funny)
But there's a more-than-50% chance that 9 is prime!
I test primeness by dividing the test-number by all integers, from 2 through the test-number's square root, looking for a zero remainder. So, first, I divided 9 by 2. I worked on this for a while, and ended up with a nonzero remainder. So far, 9 looks prime, and I've already tested half of the potential divisors! In fact, there's just one more potential divisor to try: the number 3. I'm almost done, and everything rides on this final calculation. There's a lot of uncertainty here.
What are the chances that 9 is just going to happen to be divisible by the very last potential divisor that I try? I'll grant you that the chances are non-zero; there really are some composite numbers out there. But the chances aren't one, either. For example, when I was testing 17 for primeness, the last potential divisor I tried was 4, and it didn't work. This last calculation could go either way.
So here we are, having tested half of the possible divisors, and so far 9 is looking prime and there's just one more divisor to test against. So, I ask you: do you want to bet 9's primeness/compositeness on this last calculation? I'll make it easier for you: I tell you right now, that 9 is just like 17, in that it is not divisible by 4. And then, I'll even give you an option: we can finish the calculation by dividing 9 by 3, or you can change your candidate divisor to 5, now that you know 4 doesn't work. Well.. what'll it be?
Re:To be fair, mathemeticians didn't know math eit (Score:2)
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Re:To be fair, mathemeticians didn't know math eit (Score:5, Informative)
The problem can be easily misunderstood. If it is a known rule of the game that after we choose a door, a door with a goat is opened, then it always pays to change our choice: as TFA indicates, it raises our odds to 2/3.
If, however, opening a goat door is the host's choice, then we are entering a poker-like situation. For example, if the host only chooses to reveal a goat when we choose correctly, then changing our choice will cause us to loose every time! And in general, for each strategy that a host might employ, there is an optimal counter-strategy.
In the latter scenario, it may be our goal simply to preserve our initial odds. If so, it pays to toss a coin on the second choice. This way, quite regardless of the host's strategy, we will have our odds at 1/3 or above.
Re:To be fair, mathemeticians didn't know math eit (Score:5, Insightful)
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Don't worry. (Score:4, Funny)
http://en.wikipedia.org/wiki/Monty_hall_problem [wikipedia.org]
We're being played (Score:5, Informative)
In truth, the 1956 experiment may have had flaws (though Chen's paper doesn't prove this), but many subsequent ones have upheld the original findings, and are not subject to the alleged problems.
Re:We're being played (Score:5, Interesting)
Re:We're being played (Score:4, Interesting)
The relationship is obviously bi-directional. Determining the direction of causality is thus a difficult matter, and one that preoccupied folks in my discipline for quite some time. One method is to use an "instrumental variables" approach (see any advanced econometrics text for details), but perhaps a more accessible answer comes from my own research.
The Liberal Party was often seen as "between" the Conservative and Labour monoliths. I focused my attention on the preferences of voters who switched from one of the major parties to the Liberals between elections. (We have "panel" surveys where the same people are interviewed over time which helps to eliminate problems of misperceived past voting behavior.) Now it turns out that voters who switched to Liberals usually saw them as taking positions in opposition to the party from which the switchers came. Sometimes those views were, in fact, contrary to espoused Liberal positions. For instance, on the question of entry into the European Economic Community, the forerunner of today's EC, former Conservative voters who supported entry were more likely to switch to the Liberals, while former Labour voters who opposed entry made the same switch. This pattern recurred across a number of issues. The most parsimonious explanation is that voters who disagreed with their normal party for whatever reasons were more likely to defect to the Liberals, using them as a instrument to express displeasure regardless of the Liberals' true position. (In the case of the EC the Liberals were consistently pro-Market; the other parties tended to waver.) Voting Liberal was "easier" than moving all the way over the opposition major party. That meant that voters would tend to "project" their own views on the Liberals rather than being persuaded to support the Liberals because of agreement with that party's positions.
Most of the traditional literature on American voting behavior focuses on the role of "party identification" as a primary determinant of issue opinions rather than the other way round. Voters often seem not to tote up the various stances of parties and candidates as a method of determining which party to support. Many people have Democratic or Republican partisanships because of family and social factors. People "inherit" partisanships from their parents or adapt to conform to the social roles they adopt in adulthood. These prior partisan dispositions then color their interpretations of events and campaign issues.
Let me tell you a story about my grandmother. She emigrated from Ireland in the late 19th century and lived outside Boston for the rest of her life. Despite the fact that most Irish Catholics living around Boston voted Democrat in her lifetime, she was a stolid Republican for the entire time I knew her. Her Republicanism wasn't based on support for that party's positions; it originated in the 1928 Presidential election when the Catholic (and "Wet") Al Smith ran as the Democratic candidate. Smith lost that year because anti-Catholic "Drys" in the Southern states defected to the Republicans. My grandmother felt that the Democrats failed to work hard enough for Smith because of his Catholicism, and so she started voting Republican. She was unfazed by the rather substantial evidence that showed that the Democrats in this period supported policy positions much closer to her own views. By the way, after Kennedy was shot in 1963 she claimed she had voted for JFK in the 1960 election, but we all knew she'd voted for Nixon.
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Chen didn't try to prove that the experiment was definitely flawed - he showed that their own reasoning for why it was correct was not valid. That is there w
Dude (Score:4, Funny)
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Yeah, no kidding.
:(
The "Monty Hall problem" link in the summary informed me that I need Flash to understand the problem.
However, on that page they then offer "Need to know more? 50% off home delivery of The Times."
This confuses me terribly - if I now pick the home delivery choice, does the probability of learning about the Monty Hall problem go down 50%?
Damn - I should have picked the Flash answer from the start.
What? (Score:2)
Ummm, I don't get it. (Score:3)
door 1 - door 2 - door 3
I pick door 1, monty shows me what's behind door 3 - a goat. Door 1 might have a goat or a car, door 2 might have a goat or a car. Sounds like 50/50 to me - I don't see the benefit of changing my choice. I don't have any evidence of a goat or car behind 1 or 2. I picked 1, and without evidence, I don't see how changing my choice will make it better.
I don't think this has anything to do with cognitive dissonance at all. It's a question of probability. There were 3 - my odds of success were 1 out 3. Monty shows me that one of them is bad, so now my odds are 1 out of 2. In any particular Monty event, the odds will always be 50/50. If you ALWAYS pick door 1, and if Monty ALWAYS shows you door (not 1) is a goat, then your odds will always be 50/50, assuming the assignment of the car or goat to door 1 or 2 is always truly random and fair.
What am I missing?
RS
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Re:Ummm, I don't get it. (Score:5, Insightful)
When you choose one door out of three, and one of those three was pre-chosen randomly to be "the winner", your chance of having picked the right door is 1/3. At least one of the other two doors is not the winner, so the fact that Monty can show you that one is not the winner doesn't change your chance of having chosen the winner.
HOWEVER, now your chance is the same (1/3), but the chance of either the door you chose or the remaining door closed door being the winner is 100%. Therefore the chance that the remaining door is the winner is 2/3. Switch doors to double your chances.
I have a BS in math (not statistically oriented, but I had the normal discrete math sequence) and I still had to think about this a lot before I switched answers from the wrong one to the right one
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Re:Ummm, I don't get it. (Score:4, Informative)
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One door has a car, 990 have nothing.
You pick a door. Monty opens 998 of the other doors showing nothing. Which door would you pick? He essentially gave away the location of the car - you only had a 0.1% chance of winning, but he eliminated 998 incorrect choices. The chance of the car being behind the last remaining door is 99.9%. This way it actually is somewhat intuitive.
The 3-door game is just the same game but
Re:Ummm, I don't get it. (Score:5, Informative)
Suppose the car is behind door number one.
If you pick door number one, then Monty has a choice of picking door number two, or three. If you switch, you lose.
If you pick door number two, then Monty must open door number three. If you switch, you win.
If you pick door number three, then Monty must open door number two. If you switch, you win.
Monty's choice of which door to open is constrained in two out of three choices. Pick the door he didn't open, and you'll win two out of three times.
But the problem assumes that Monty has to offer you that choice. On the game show, he didn't.
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Your original odds were 1/3. Monty has a 2/3 chance of having the right one. Monty's odds of having the right one is greater than your odds of having the right one so statistically you should switch.
Look at it by way of cards (in the article).
You need to pick the ace of hearts. Monty will then go through the deck and pick the ace of hearts or a random card. He will then show you the other 50 "goats" and ask if you want to trade. You have a 1/52 chance of picking it. Monty then h
Re:Ummm, I don't get it. (Score:4, Informative)
If 1 has the car, he can pick either door. If you switch, you lose. Prob 1/3
If 2 has the car, Monty *has* to open 3. If you switch, you get the car, Prob 1/3
If 3 has the car, Monty *has* to open 2. If you switch, you get the car, Prob 1/3
Thus, there's a 2/3 chance of getting the car when you switch.
The other way to think about this is that Monty is revealing no information about *your* door when he opens one of the other two. Thus, the probability that your door has the car must be 1/3 both before and after Monty opens one of the other doors. Since there's only one closed door left, the car is behind it with prob = 2/3.
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You have a 1/3rd probability of choosing the car initially and a 2/3rds probability of choosing the goat. If you do not switch, you have a 1/3rds probability of having the car. After one of the other doors has been revealed to be a goat, however, the following is true: If you originally picked a car, you will get a goat. If you originally picked a goat you will get a car. Since you had a
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Because your initial probability of picking the car isn't 50/50, it's 2:1 against the car. You choose from 3 doors, remember, not 2. So initially the probability is 1/3rd that you've chosen the car, 2/3rds that the car is behind one of the doors you haven't chosen. Then Monty opens one of the doors you haven't chosen. He's constrained to open a door with a goat behind it, but the fact that he's opened a door doesn't change the initial probabilities. So the probabilities remain 1/3rd that you've chosen the c
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It has nothing to do with cognitive dissonance. The cognitive dissonance experiment has been show to contain a similar type of error, that is all. I don't think you really read the article.
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If you pick car first (1/3) and don't switch, you win. If you pick goat first (2/3) and don't switch, you lose.
Better yet, imagine that there was 2,001 doors, one car and 2,000 goats, and then when you picked a door 1,999 other goats were revealed. Now you know almost for certain that you picked a goat, so y
Put it into more physical/visual terms (Score:5, Insightful)
I realized that the door analogy wasn't working as it didn't help them visualize 'possession of the odds'
Instead I explained it as follows:
We're going to play the game with 10 boxes - 9 boxes are empty and 1 box contains a prize.
My wife is asked to pick a box and she is handed the box that she chose.
Then my step-son is handed the other 9 boxes.
I then ask both my wife and step-son what each ones odds are of having the prize is. The agree on :
Wife : 1 in 10 (or 10%) chance of having the prize
Step-Son : 9 in 10 (or 90%) chance of having the prize
At this point I explain the physical-ness of my son 'holding the odds' - It is clear to both that he is in possession of 90% of the odds.
I ask my wife, at this moment, with her holding 1 box and he holding 9 boxes, if she would like to switch possession and trade her 1 box for his 9
She of course says 'heck yeah!'
They both have an 'ahah!' moment and I don't really have to go any further, but I did for completeness.
I make a statement that my step-sons 90% is evenly distributed across the boxes he posses - currently 9 of them.
Now I start opening my step-sons boxes, one at a time - Boxes guaranteed NOT to contain the prize
After opening one of the 9 boxes, leaving my step-son with 8 boxes, I point out that he is still in possession of 90% of the odds, but now those odds are distributed between the 8 remaining boxes.
Then you remove one more box, along with explanation, and they see the pattern - The odds stay the same, and are still in my step-son's possession, but are continuously distributed among fewer boxes.
Finally both my wife and step-son are each holding one box.
I bring back the fact that my step-son is still in possession of 90% of the odds, but that entire 90% is wrapped up in that one single box.
With a final closing - that they were patient enough to listen to, since they asked me to explain after all - I point out to my wife that, since she was willing to trade 1 box for 9 boxes earlier, she must certainly be willing (if not eager) to trade her 1 box for my step-son's 1 box.
They really connected the dots pretty fast once I placed the prize in a box and had them each holding the boxes - Putting a physical location to the odds.
Article title misleading? (Score:4, Insightful)
Since she gave her [correct] answer [to the Monty Hall Problem], Ms. vos Savant estimates she has received 10,000 letters, the great majority disagreeing with her. The most vehement criticism has come from mathematicians and scientists, who have alternated between gloating at her ("You are the goat!") and lamenting the nation's innumeracy.
Since some math PhDs got it wrong too, isn't it a bit disingenuous to claim its the psychologists are the issue as the article title states?
Indeed (Score:5, Interesting)
Some researchers involved in pchycology (social behaviour etc.) came to high schools and drew up the friendship graph of the class. (Maybe school works differently where you live, we had a class of size 30-40 students attending exactly the same lectures.)
They assumed friendship to be mutual (if not, than it was not considered friendship). One clever cookie made the observation that almost always there is a group of 6 students who all friends to each other (a clique), or alternatively a group of 4 students, who do not like each other.
There were excited discussions among the researchers what social forces are the reason that one of the above situations always seemed to occur.
They were somewhat disillusioned when our math teacher explained them Ramsey's theorem. Since R(6, 4) is between 35 and 41, indeed one can expect either a frienship or hateship clique to appear with quite high probability... (This does not mean that properties of the frienship graph worth not examining, but one needs to know the math to do it properly.)
Pot, Kettle, Black (Score:5, Funny)
One should remember what happens when you put 50 economists in a room - you get 100 opinions - one for each hand.
I recognize that the author of the article may be correct, I just couldn't help commenting on the first sentence.
Re:Pot, Kettle, Black (Score:5, Funny)
But if they reveal their opinions, should you switch hands?
You know who can't do math? (Score:5, Insightful)
If you are sick on a Friday or Monday, they assume you are 'taking a long weekend' even though there is a 2/5 chance someone will be sick on those work days. 40% of the time it would be Monday or Friday. More so for a 4 day work week.
Cognitive Dissonance (Score:5, Funny)
Sadly, not as wrong as shown (Score:5, Interesting)
I've been told by "superiors" to perform certain analyses because "everyone does", and they gave me references which supposedly showed these were proper. When I looked these up, the authors not only made no claims supporting their necessity, but both stated that the researcher should know enough about what they're doing to know what analyses to perform. I took my instructions to the statistics consultant for our department, and without showing him the references he made the same claims as both authors, contradicting the rationale given by those who gave me the instructions. I've seen many cases of psychologists performing statistical analyses based on their knowledge of how to use SPSS et al., rather than any fundamental grasp of the maths required by the design. Perhaps the most egregious error is their faith in fMRI analyses via statistical probability mapping, when the correction factor required by the 10^4 to 10^5 simultaneous T-tests makes any one result within the traditional collective p >
On the other hand, some of us can apply such analyses as tensor calculus and Gabor transforms to dendritic electrical fields, showing where each of those are correct and where each fail, and can correctly apply nonlinear, N-dimensional statistical testing of time/frequency maps produced by continuous wavelet transform. But of those of us who can do these things, I know of none who learned of them, much less how, within the confines of a psychology department. (Well, except for the Gabor stuff, as used and taught by Karl Pribram, that being the only case I know of).
"Everything I Needed To Know I Learned At The Santa Fe Institute". No, not everything, but that'd make a hell of a book.
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However, your example from fMRI speaks to complete ignorance of the field, and I'd like to force you to defend it. Thousands of fMRI experiments have been carried out, and this standard for significance is often met. When you say "very unlikely to actually exist," I can't imagine what you're thinking, since this statement is so easily falsified (in f
A Simple Explanation of the Monty Hall Problem (Score:4, Informative)
Anyway, here is the simple explanation that I've found helps people realize their error in thinking:
In other words, what are the three possible outcomes given that the person always switches their door?
[car] [goat] [goat]
Choose door 1. Host reveals door 3. Switch to door 2. NO CAR.
Choose door 2. Host reveals door 3. Switch to door 1. CAR.
Choose door 3. Host reveals door 2. Switch to door 1. CAR.
What are the three results? NO CAR, CAR, and CAR. In other words, always switching your answer results in a 2/3 chance of getting a car.
If we repeat this process but we never switch our door, you get:
Choose door 1. Host reveals door 3. No switch. CAR.
Choose door 2. Host reveals door 3. No switch. NO CAR.
Choose door 3. Host reveals door 2. No switch. NO CAR.
Now we only have a 1 in 3 chance of getting the car.
Re: (Score:3, Informative)
The possibilities are not equally likely. (Score:3, Informative)
Count: 4 wins, 4 loses
True. But each of those four wins and losses are not equally likely.
You have two (equally probable) possibilities under case "I choose 1" -- "Host chooses 2" and "Host chooses 3". The probability of "I choose 1" is a third, so each of the two possibilities have probability a sixth.
But each of the other cases -- I choose 2 or 3 -- only have one possibility each. So since the probability of choosing each of the cases is a third, each of the possibilities have probability a third.
Count up the prob
Let me be an asshat (Score:4, Informative)
http://en.wikipedia.org/wiki/Monty_Hall_problem#Solution [wikipedia.org]
Look at the picture and be amazed.
Honestly, 100s of comments on
Explanation glosses over the most important point! (Score:4, Funny)
Re: (Score:2)
Re: (Score:3, Insightful)
Re: (Score:3, Insightful)
This one has been debated over and over, and is a classic example of lies, bloody lies and statistics. The fallacy lies in stating that before Monty opens the door and shows the goat, your chance of picking the car is 1/3. It is NOT, because as Monty will always pick a door with a goat behind it, your choices are always going to be two ... one with a goat, and one with a car, because the one Monty opens is taken out of the equation - in fact it was never IN the equation in the first place. You only ever had two options ... one with a goat and one with a car.
Thus your chance of picking the door with the car are 1/2, they were 1/2 at the start, and they are STILL 1/2 after Monty opened his door. The odds do not change "in your favour", because they simply do not change AT ALL. Ergo, there is no advantage or disadvantage in changing doors.
A careful mathematical analysis of the problem proves that you're wrong. There are many computer simulations of the problem that show that you're wrong. The only thing you have going for you is your intuition, and your intuition is wrong.
Re:The problem is a fallacy (Score:5, Insightful)
Re:The problem is a fallacy (Score:5, Insightful)
"It is NOT, because as Monty will always pick a door with a goat behind it, your choices are always going to be two"
Your argument *only* works if Monty opens a door *before* you pick. *And*, you get to pick *twice*. First time from three doors, second time from two doors.
You pick, from a choice of three, giving Monty a choice of two.
Your argument is based on the reverse, Monty being able to pick from three doors, and you only get two.
Do you see it now? You 'lock' a door, precluding Monty from choosing it.
Remember, since you have first pick, your chances of getting a goat are 2/3. Meaning you most likely picked a goat. Meaning when Monty reveals a goat, the remaining door is most likely a car.
Re: (Score:3, Informative)
Sorry, but your truth table is a fallacy. Though I doubt anything anyone says is going to convince you of that.
For everyone else: where he's going wrong is assuming that each of the 24 table entries is equally probable.
They're not. The table is assymetric.
Such a table can't have repeated entries in (for example) the column labeled "you" and still provide equi-probably outcomes for each.
In other words, where he has (going down the 'You' column):
He actually needs:
1 1 2 2 3
Re:The problem is a fallacy (Score:5, Informative)
Let's define some events:
TDC = Contestant chooses door with car
TD1 = Contestant chooses door with goat #1
TD2 = Contestant chooses door with goat #2
MG1 = Monty reveals goat #1
MG2 = Monty reveals goat #2
Here are the possible game outcomes, under the switch strategy:
Outcome A (TD1 MG2): Contestant chooses door with goat #1, Monty reveals goat #2, WIN
Outcome B (TD2 MG1): Contestant chooses door with goat #2, Monty reveals goat #1, WIN
Outcome C (TDC MG1): Contestant chooses door with car, Monty reveals goat #1, LOSE
Outcome D (TDC MG2): Contestant chooses door with car, Monty reveals goat #2, LOSE
Now, we will establish some conditional probabilities:
P(X|Y) means "the probability of X given that Y has already occurred"
P(MG2|TD1) = 1 (Monty MUST reveal goat #2 if contestant chooses goat #1; he cannot reveal the car or the door the contestant selected)
P(MG1|TD2) = 1 (Monty MUST reveal goat #1 if contestant chooses goat #2; he cannot reveal the car or the door the contestant selected)
P(MG1|TDC) = 1/2 (Monty reveals goat #1 or goat #2 with equal probability if the contestant selects the car)
P(MG2|TDC) = 1/2 (Monty reveals goat #1 or goat #2 with equal probability if the contestant selects the car)
Now, some simple probabilities for the initial choice:
P(TD1) = 1/3 (Contestant chooses any door with equal probability)
P(TD2) = 1/3 (Contestant chooses any door with equal probability)
P(TDC) = 1/3 (Contestant chooses any door with equal probability)
Now, using the law of conditional probability:
P(MG2|TD1)=P(TD1 MG2)/P(TD1) -> 1 = P(TD1 MG2)/(1/3) -> P(TD1 MG2) = 1/3
P(MG1|TD2)=P(TD2 MG1)/P(TD2) -> 1 = P(TD2 MG1)/(1/3) -> P(TD2 MG1) = 1/3
P(MG2|TDC)=P(TDC MG1)/P(TDC) -> 1/2 = P(TD2 MG1)/(1/3) -> P(TDC MG1) = 1/6
P(MG2|TDC)=P(TDC MG2)/P(TDC) -> 1/2 = P(TD2 MG1)/(1/3) -> P(TDC MG2) = 1/6
So, let's review the outcomes now that we know their probabilities:
Outcome A (TD1 MG2): Contestant chooses door with goat #1, Monty reveals goat #2, WIN (Probability 1/3)
Outcome B (TD2 MG1): Contestant chooses door with goat #2, Monty reveals goat #1, WIN (Probability 1/3)
Outcome C (TDC MG1): Contestant chooses door with car, Monty reveals goat #1, LOSE (Probability 1/6)
Outcome D (TDC MG2): Contestant chooses door with car, Monty reveals goat #2, LOSE (Probability 1/6)
Let's find the probabilities of winning and losing:
X Y means EITHER X or Y occurs.
P(X Y) = P(X)+P(Y) if X and Y are mutually exclusive (this is a probability theory axiom)
All four of our outcomes are mutually exclusive (they CANNOT occur at the same time)
P(WIN) = P(A B) = P(A)+P(B) = 1/3 + 1/3 = 2/3
P(LOSE) = P(C D) = P(C)+P(D) = 1/6 + 1/6 = 1/3
Under the "switch" strategy, you have a 2/3 chance of winning and a 1/3 chance of losing.
Everything I just wrote is basic probability and set theory, usually taught within the first 2-3 weeks of a college-level introductory probability course.
Re:The problem is a fallacy (Score:5, Insightful)
Re:The problem is a fallacy (Score:5, Informative)
Redraw the entire truth table with branches instead of separate rows for each possible outcome. Drawn this way, there are three starting points (CGG, GCG, GGC) and 24 outcomes.
For each starting point, write "1/3" above it. That is the probability of it occurring, since each is equally likely. Step down each node, and for each one, multiply the previous denominator by the total number of branches that could have been taken at the last node. So, for example, you'd have written 1/3 above CGG, and for each of the three branches coming from it (door 1, door 2, door 3), you'd have a 1/9 above it. You'll soon see that in the "Monty" column, when he has no choice about what door he could have picked, you'll have a 1/9 above the node, but when he could chose from two doors, there will be a 1/18 over each (this assumes that his choice of the two doors is random. If it isn't, it doesn't matter, because the probabilities above each choice will sum to 1/9, even if they aren't equal).
Proceed down each branch this way to the end, but don't branch on choosing "switch" or "don't switch." Since we want to see the results if we had picked either "switch" or "don't switch" in every possible situation, just write down the results as if you had picked "switch." We'll logically NOT the results later to simulate picking "don't switch".
When you finish the last column, you'll see that not every outcome has the same probability of occurring. Some of the outcomes will have probabilities of 1/9, and there will be outcomes that have probabilities of 1/18, because there was an extra decision branch involved in Monty picking the door.
Finally, sum up the probabilities of each outcome. "Win" will be 2/3, and "lose" will be 1/3. Obviously, if we logically NOT all the results to represent picking "don't switch" each time, the results invert, so "lose" has 2/3 probability and "win" has 1/3 probability.
Branching on each decision fixes the "problem" of a truth table like yours making it look like each outcome is equally probable.
Re:TFA Is Wrong (Score:4, Insightful)
Pick #1, Monty opens #2 (switch = win)
Pick #2, Monty opens #1 (switch = win)
Pick #3, Monty opens #1 (switch = lose)
Pick #3, Monty opens #2 (switch = lose)
50/50
Re: (Score:3, Informative)
No, you're wrong.
Start with the initial case: you choose from 3 doors, 1 of which has a car and 2 of which have goats behind them. Now, suppose Monty just opens all the doors on the spot, revealing whether you won or not. What's the probability that you chose the car? 1/3rd. It has to be, only 1 door out of three had the car.
Next step, you make the same choice. Monty opens a door but doesn't give you the option of changing your selection. Now, what's the probability of your winning? You made the same choi
Re:Real World & Monty Hall Problem (Score:5, Informative)
WHEN YOU ARE ON A GAME SHOW, YOU ONLY GET ONE ATTEMPT!
Hehe, oh, and there's a bigger fact that is left out of the classical form of this problem, one that was revealed when someone asked Monty Hall himself what he thought of the eponymous probability problem.
The problem assumes that Hall always offers you the choice to switch, but this was not the case! He did not necessarily have to give you the choice (which kinda makes that part of the game show boring), and in fact said that he mostly only offered the choice to switch when the player had chosen the correct door, in order to lure them away from it!
So in the math problem version, switching is the best choice (by a factor of 2 even), while in the real-world version, staying was the better choice (by some unknown factor, but maybe a lot more than 2 depending on how evil Mr. Hall was).