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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."
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"
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.
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.
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.
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.
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.
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.