Slashdot Mirror
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."
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"
Did a Scientologist write the title?
The protests against scientology is this Saturday in every major city around the world! Sunday for Philadelphia.
They can't move the goats or car during the game. Therefore, changing your door choice will change your chances of winning from 1/2 to 2/3.
What a fool believes, he sees, no wise man has the power to reason away.
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.
Look at it this way.
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 has a 51/52 chance of picking it. Obviously, statistically speaking you should switch.
Monty's choice of a door to open is not random -- he has to pick a door that doesn't have a car. Say you pick door 1. Here are the three equally-likely possibilities:
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.
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.
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.
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.
Uhhhh.... you DO know that Britain invented the MRI? That the MRI was invented in 1973 but Chernobyl went up in 1986? As for the rest of your claims, they sound more like sour grapes (Britain has more high-ranking Universities than any country other than the United States). If you're more interested in trolling than querying, you're doing a good job of it. As for "proof", since you're probably not going to consider the fact that I was a research assistant at the University of Manchester for the inorganic biochemistry group involved in this research... Well, many scientific papers are pay-to-view and I'm not buying the whole Slashdot readership access to all the journals articles were published in, just to satisfy one anonymous coward over one point. If you're that interested, and care that much, look it up yourself. Try looking for papers on radioactive sheep in Cumbria.
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)
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.
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.
All these "experiments" always remind me of one thing and one thing only: Cargo Cult Science (http://wwwcdf.pd.infn.it/~loreti/science.html).
Also, from reading the comments, the article ought to be tagged "/.ers also don't know math"
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.
"Lucent" is also a Latin verb meaning "They shine". It is the third-person plural present active indicative form of luceo, lucere, luxi.
Just thought yall would like to know that...
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.
Here is the easiest explanation to follow that I've heard. Extend the game to 1000 doors.
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 reduced in size.
http://en.wikipedia.org/wiki/Monty_Hall_problem#History_of_the_problem
As the parent noted, it goes wrong because every choice you make after you initial selection -- including the first switch -- is completely dependent on that initial selection. In other words, after you've made your initial selection, the entropy of the system drops to 0.
That is unless Monty is cutting quantum checks where the value isn't determined until you open the envelope. But even there switches have no impact, because the entropy of the system is constant until the envelope is opened.
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.
Rampant carbon sequestration destroyed the Dinosaurs' tropical paradise. I'm here to help repair the damage.
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.
I used to laugh at economists when they claimed to do science too. Then one of my friends at uni showed me the notes from their math course. As a physicist I like to think I can handle a few equations, but they do some serious math. After that, I kept quiet. Keep picking on the psychologists, it's safer.
It's not a bug, it's a lepidopter!
Not quite.
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 that's just the statistics. This ALSO fits the observed data. But it's a simpler hypothesis, because the effect falls naturally out of the equations.
The Monkey picking the M&M's could choose the 3 in these ways
X Y Z
X Z Y
Y X Z
Y Z X
Z X Y
Z Y X
If the monkey chooses X over Y, then he could have chosen
X Y Z
X Z Y
Z X Y
Of those 3, you can clearly see that TWO of the THREE options, show a natural preference for Z over Y
Thus, the monkey is twice as likely to pick Z instead of Y.
This experiment does not support the Cognizant Dissonance theory.