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."

Nice try!

Like I'm going to click on a link with the word 'goat' in it.

Re:Nice try!

It's never a good sign when the words "reveal" and "Monty" are in the same sentence.

Re:Nice try!

on the goats-and-car paradox...

In journals where this phenomenon is referred to frequently, the paradox is frequently abbreviated to goats-x

Hmmm....

1) " Psychologists Don't Know Math" is a rather inflammatory, inaccurate, braindead headline, even by local standards.

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....

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.

One would think, but as it turns out, there are too many complexities. You see, you have to consider the socio-economic background of the monkeys, their upbringing, and their inherent biases to figure out if they like green, blue or red M&M's best. You see, the monkeys have an inherent bias toward green, but only if they have been captured from the wild (where presumably green would be comforting, the color of trees and whatnot). And of course there is the political bias associated with red and blue, so it depends on whether the monkey's political biases. These are especially hard to sort out as monkeys tend to just throw feces at the other side, at every opportunity, so you can easily separate the two groups, but rarely can you tell which is which. Its difficult to determine if they like to eat blue M&M's because they themselves are blue (or feel blue, as depressed monkeys have a significant bias toward the blue M&M's) or because they are red as it were, and feel like eating the blue ones to get back at the other side.

Seems to make sense

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".

There are several problems with all of this. The original experiment does not appear to have any control group, it is unclear if the population sampled was genuinely random, the size of group tested seems to have been extremely small for a meaningful statistical study, and (perhaps most important of all), it assumes that mammalian vision is uniform greyscale AND that the candy was monochromatic.

(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.

To be fair, mathemeticians didn't know math either

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.

Re:To be fair, mathemeticians didn't know math eit

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.

Re:To be fair, mathemeticians didn't know math eit

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.

We're being played

According to this site [scienceblog.com], 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.

Re:We're being played

Indeed. There's considerable evidence in favor of reductions in cognitive dissonance as a motivating psychological force from other types of studies and other disciplines. For instance, in my field of political science, the evidence is pretty overwhelming that citizens systematically misperceive candidates' positions to make them more similar to the citizens' own preferences. Voters often engage in "projection," believing that candidates' they prefer hold positions like the voters' own, even when those aren't the positions the candidates actually hold. The opposite process also occurs, where voters believe that candidates they dislike hold positions those voters dislike regardless of the candidates' true preferences. My own dissertation research on voters for the British Liberal Party in the 1960's and 1970's also confirmed these hypotheses.

Indeed

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.)

Pot, Kettle, Black

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.

You know who can't do math?

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.

Cognitive Dissonance

Amusingly, cognitive dissonance theory predicts that psychologists will rationalize their error and insist that it doesn't invalidate their conclusions.

Sadly, not as wrong as shown

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.

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 > .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.

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.

Re:Ummm, I don't get it.

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.

Put it into more physical/visual terms

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.

Re:Ummm, I don't get it.

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 :-)

Re:The problem is a fallacy

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...

This is analogous to observing that a lottery ticket can either be a winning ticket or a losing ticket, and then concluding that the odds of winning the lottery are 1 in 2.

Re:The problem is a fallacy

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.

Re:The problem is a fallacy

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.

Re:The problem is a fallacy

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.

Re:Real World & Monty Hall Problem

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).