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Psychologists Don't Know Math
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."
Like I'm going to click on a link with the word 'goat' in it.
The Kruger Dunning explains most post on
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.
What I'm listening to now on Pandora...
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 :)
Marilyn vos Savant explained the problem in Parade magazine, and a whole bunch of math professors wrote in to tell her that she was wrong... turns out it's kind of a bad idea to play "gotcha" with someone who has an IQ of 228.
According to this site, Dr. Chen is being quite devious, seemingly in order to discredit a colleague.
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.
"Be light, stinging, insolent and melancholy"
I read one of Marilyn Vos Savant's books, and in it she listed 9 as a prime...
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.
It's quite simple.
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.
This reminds me the story my high school teacher told me:
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.)
No, changing your door choice changes your chances of winning from 1/3 to 2/3.
:-)
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
I started questioning this article before the end of the first sentence. An Economist, calling a Psychologist "wrong" about math?
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.
HR people.
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.
The Kruger Dunning explains most post on
You're missing something.
"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.
Amusingly, cognitive dissonance theory predicts that psychologists will rationalize their error and insist that it doesn't invalidate their conclusions.
TFA has been adequately refuted, so I'll forego more on that. And despite the inflammatory nature of the title and claims here, it is unfortunately too correct too often.
.05 significance level to have an individual p value in the 10^-6 to 10^-9 range. That's a hell of a requirement for a single test, and very unlikely to actually exist. "Figure the odds" applies, and they don't seem to grasp that they don't grasp it.
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.
"I may be synthetic, but I'm not stupid." -- Bishop 341-B
(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.
It's a small world and it smells funny; I'd buy another if it wasn't for the money; Take back what I paid (SoM)
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.
My wife and step-son asked me to clarify this probability after getting home from watching "21".
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.
Cube On! (http://stores.ebay.com/PuzzleProz)
That's equivalent to providing a table with all possible outcomes of a roll of two dice (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12) and saying that they are all equally likely just because each outcome has one entry in the table, except what you have done is the logical inverse. The example of the dice is combining multiple outcomes and pretending they are one - you are taking one possibility and branching it on a variable that has no effect on your outcome: the door that Monty picks if you picked the car to start with. If you pick the car to begin with, the number of the door that Monty picks has no effect on your outcome. To be more precise, the number of the door that Monty picks NEVER affects your outcome. If you want to keep the Monty column, you should replace the numbers with the word GOAT and then get rid of all of the duplicate entries, and the table will then represent the probabilities correctly.
However as a sometime game show contestant I know you have to take into account one fact that is left out of the classical form of this problem.
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).
The enemies of Democracy are
Give it up. Conditional probability supports the conclusion that you are better off by switching:
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.
As someone who majored in psychology, worked in two labs, and read countless psychology papers, I can tell you that 99% of psychologists avoid math when possible, and the other 10% try to use it but make obvious errors.
To the psychology researcher, it's more about getting the "story" right than actually quantifying anything.
Les Miserables Volume 1 now up with my reading of
One thing that I think needs to be pointed out, however, is that for the odds to increase from 1/3 to 2/3, the player must know for sure that the host will *always* uncover a goat after the player's first choice irrespective of initial choice of goat vs. car. If the host's decision to uncover or not to uncover a goat is related to the player's initial choice, one can't say anything about the new odds.
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?
"Believe me!" -- Donald Trump
You're almost there.
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.
You glossed over the part where most people don't know how to deal with problems that a psychologist is trained to handle. There's something to be said for the education.
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.
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.
Doubting the existence of evolution is like doubting the existence of China: It just shows that you're uninformed.