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
Don't tell the Scientologists... You'll only arm them!
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"
What the hell are you talking about?
If you wanna get rich, you know that payback is a bitch
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
Shoes for Industry. Shoes for the Dead.
From an older article by the same author article:
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?
-- Political fascism requires a Fuhrer.
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.
I got my facts from the infinitely more trustworthy Wikipedia.
http://en.wikipedia.org/wiki/Monty_hall_problem
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.)
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
Anyway, here is the simple explanation that I've found helps people realize their error in thinking: The problem is a lot easier if you think about it in an "outcome" based fashion.
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.
(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)
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 3
If he really wants to assume all the probablities of each table entry is equi-probably.
My stat terms may be off but that's the flaw I see.
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.
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 No, the four possibilities here are not equally likely. If the initial pick is random, then the probability that case 1 occurs is 1/3, the probability of case 2 is 1/3, and the probability that EITHER case 3 or case 4 occurs is 1/3.
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
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 choice you did in the previous scenario. Monty's opening of his door has no effect on the outcome or the probabilities. So your probability of winning the car has to still be 1/3rd.
Final step, you make your choice and Monty opens his door, but now he offers you the chance to change your selection. Before you decide, your situation is exactly the same as in the previous scenario. That means your probability of winning has to be the same, 1/3rd. But since Monty showed you one door with a goat behind it, so there's only one door left. Since the total probability has to be 1, the probability that that door is the one with the car behind it is 1 minus 1/3, or 2/3rds.
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.
Statistics are tricky and generally counter-intuitive. As my stats professor said, often the best mathematicians are among the worst statisticians.
Les Miserables Volume 1 now up with my reading of
I remember a magazine called parade here in the UK when I was in my teens. To be honest I don't recall seeing any maths articles though, but to be honest I only looked at the pictures :D
Another way to look at the problem is to consider what you'd do if there were 99 doors, you picked one, then Monty opened 97 of the rest of the doors (leaving your door, and one other). Obviously in this case you'd switch.
The reason people don't switch may be related to regret theory. If you switch and lose, you'll feel really bad because it will feel like you just lost a car (even though you didn't technically have it to begin with). So people stick with their current choice.
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
See this link for the solution:
/. trying to describe it and a simple picture was the thing that helped me get it.
http://en.wikipedia.org/wiki/Monty_Hall_problem#Solution
Look at the picture and be amazed.
Honestly, 100s of comments on
Knowledge is power. Knowledge shared is power lost.
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.
Do you get to keep the goat?
sic transit gloria mundi
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 probabilities for each outcome, and you get the actual result.
What's purple and commutes? An Abelian grape.
My experience at a the University of Edinburgh ("a good uni") was that Psychologists really don't know math. I spent ~6 months being subjected to lectures on statistical theory about chi-squared and normal distribution that frankly didn't make any sense: "Why do we add +1 here?" "Because it works"
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.
Python coder | PyQt Applications | Writer
Evidently, psychologists prefer base 9.1. It's not that they are bad at math, but that the world at large doesn't understand base 9.1. As for why psychologists prefer base 9.1, I haven't the faintest clue. I'm not a psychologist. I'm one of those bad at math organic chemists.
99 (base 9.1) + 10 (base 9.1) = [(9 * 9.1^1) + (9 * 9.1^0)] + [1 * 9.1^1] (base 10) = 100 (base 10)
http://en.wikipedia.org/wiki/Monty_Hall_problem#History_of_the_problem
It's not a bug, it's a lepidopter!
- 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.Reduce, reuse, cycle
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 someone with a degree in psychology from an accredited university never studied any stats.
That said, there are many sub-disciplines in psychology. I studied cognitive psychology, and there was a fair bit of maths involved. Someone who wanted to be a clinical psychologist would not need be devoted to statistics, just as I was not devoted to learning how to help clients via talk therapy and the medical model.
I went to a conference where the cognitive psychologists and clinical psychologists reviewed the same case study and made suggestions on how to help a client who was an alcoholic and suffered from bouts of severe depression. The clinicians believed that they needed to identify and resolve the root cause of the depression in order to end the alcohol dependency, whereas the cognitives believed that the client needed to stop drinking first, because alcohol is a depressant.
At the time, I could not help but recall the story of the Petit Prince, and the episode in which he met the drunkard.
*** Where are we going? And what's with this handbasket?