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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
Don't tell the Scientologists... You'll only arm them!
I had to laugh when this came up in "21". I can't tell you how many times the outcome of this has been debated at my work, to the point where it's become an interview question.
And just for the record: Nick, you're still wrong. And bad at statistics.
*ducks*
In these troubled times, those of you with yards might want to get the goat.
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 :)
One out of four mathematicians already know that psychologists don't know math.
Sig this!
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
And who stole the entire problem, lock stock and barrel, from Martin Gardner without citation.
Hey janitors erm "editors" how about doing some editing and cleaning up that summary?
Only the State obtains its revenue by coercion. - Murray Rothbard
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.
It's also a bad idea to get your facts from Parade magazine.
Did a Scientologist write the title?
The protests against scientology is this Saturday in every major city around the world! Sunday for Philadelphia.
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.
Not that I ever bought that a small sample, however truly random, really does prove what the larger whole would do.
If brevity is the soul of wit, then how does one explain Twitter?
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
As long as psychologist is a good one, I see no problem with that.
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.)
- The Car
- Goat A
- Goat B
You have a 1/3 chance of picking the car from the beginning. Wikipedia explains it quite well: http://en.wikipedia.org/wiki/Monty_hall_problem#SolutionThere's no debate. Of course when you first choose among three equally likely options, your chances of picking the car are 1/3. If you can't figure that out, you're completely lost before Monty even shows the goat!
What a fool believes, he sees, no wise man has the power to reason away.
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.
I understand the math. I accept that over multiple trials the probabilities indicate that you are better off switching.
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!
I did the online game in the latter link. I switched doors.. and ended up with a goat.
Monty Hall said "Thank you for playing...Good Night!" and I left with my prize.
All the probability theory in the world doesn't make me feel better about not winning the car. (Which, of course I had actually correctly selected with my first guess).
The analogy of the "Pick the Ace of Hearts" in the deck of playing cards is a better illustration, but as the number of doors (or cards) gets smaller it becomes a lot less clear cut that you should switch, especially if your iteration set size is 1.
You're wrong.
I wrote a program to simulate this situation repeatedly. The contestant won in 2/3 of the cases where he switched, and 1/3 of the cases where he didn't.
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
I think your problem here is not that you don't understand the answer, but instead that you from the beginning don't correctly understand the problem.
Only after you pick your one of three doors does Monty reveal one of the remaining two doors which contains a goat.
to "Psychologists Don't Know Shit". Being a "proper" science student i once peeked into a friend's psychology book (non clinical i might add) only to find it was full of long convoluted words used to explain the most mundane boring common sense stuff. I quickly concluded that the couch analysis shrink students should be stood right along side the Sociologists and other such riff raff.
Physcologists and economists have a lot in common. They both use probability and statistics to legitimise what is, and clearly never can be, science. Each group of people tries to posit theories about the large scale behaviour of groups of people. Fine, because introspection and extrapolation don't work. But people aren't ideal gases in a lab. Godels theorem should be enough to see the system cannot examine itself, since the system (groups of people) examining the subject (groups of people) is introspective as a whole. Every experiment you design brings unavoidable biases, and pre-judgements, because we are people, studying people. And the very act of performing the experiment changes behaviours anyway. And even though increasing your sample size gives better results you still can't escape the self-reference. Psychology and economics would only make sense if conducted retrospectively, at the end of time, by a machine, with the hindsight of knowing everything about everybody who ever lived. Until that point they are mildly useful predictive yardsticks for humans, no more. Of course you could say that applies to all science, but you'd seriously have to believe the laws of physics might change at any moment. People, on the other hand, make bloody minded decisions, just to assert free will over determinism, or because they saw a bee on a flower yesterday, or because their boyfriend pissed them off two weeks ago. To their credit, economists recognise micro and macro scales better than psychologists who presume to extend their interpretations to a general scope. All said, both projects are probably strewn with methodological errors, so this article is no surprise, it only makes you wonder what grave mistakes the economists, whose theories direct the fate of entire nations, have to be ashamed of.
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.
If she had stolen it from someone less illustrious than the famed Gardner, would your opinion of her increase?
I watched the movie "21" last week. I have a BSEE but when that part of the movie (very short) went by, I didn't really understand the reasoning. Something just felt wrong, but it went by too fast (and since it was not on DVD there is no way to back it up, etc) I am glad to know that it is not just me.
Oh, and here's a link to a simulation written by others:
http://c2.com/cgi/wiki?MontyHallSimulation
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
This Just In: Apparently Economists Don't Know Psychology
The only in-class programming we ever did in high school was implementing the Monty Hall game in TI-BASIC. Helping the teacher explain IF-ELSE statements to my classmates was... fatiguing. Especially since as the sole owner of a TI-86 in a class full of TI-83 users, some of the stuff I knew was plain wrong. Still have that program (named GOAT) on my TI somewhere... I should reimplement it in Reverse Polish Lisp on the HP-48.
Klingon programs don't timeshare, they battle for supremacy.
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.
No, No and No ... your chances to choose the car are NOT 1/3, because monty will ALWAYS eliminate one goat ...
... the car (C) is behind one, and the goats (G) are behind the other two. Whichever one I choose, Monty is constrained to open one of the remaining doors that has a goat behind it. Then I may choose to change or not to change.
... now examine the last two columns.
...
... always was, always will be.
I get so tired of this one, so here's the truth table I always trot out.
We have 3 doors numbered 1,2,3
I apologise for the font, only way to preserve the formatting.
1 2 3 You Monty Change WIN/LOSE
C G G 1 2 NO WIN
C G G 1 2 YES LOSE
C G G 1 3 NO WIN
C G G 1 3 YES LOSE
C G G 2 3 NO LOSE
C G G 2 3 YES WIN
C G G 3 2 NO LOSE
C G G 3 2 YES WIN
G C G 1 3 NO LOSE
G C G 1 3 YES WIN
G C G 2 1 NO WIN
G C G 2 1 YES LOSE
G C G 2 3 NO WIN
G C G 2 3 YES LOSE
G C G 3 1 NO LOSE
G C G 3 1 YES WIN
G G C 1 2 NO LOSE
G G C 1 2 YES WIN
G G C 2 1 NO LOSE
G G C 2 1 YES WIN
G G C 3 1 NO WIN
G G C 3 1 YES LOSE
G G C 3 2 NO WIN
G G C 3 2 YES LOSE
So, a 24 state truth table
In 6 cases, you stay with your original choice, and you WIN
In 6 cases, you stay with your original choice, and you LOSE
In 6 cases, you switch your choice, and you WIN
In 6 cases, you switch your choice, and you LOSE
So
In the 12 cases where you stay with your original choice, 6 are WINS, 6 are LOSE
In the 12 cases where you switch your choice, 6 are WINS, 6 are LOSE
50:50
pls.
(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)
then why is she working for Parade magazine?
** No, this is not an ad hominim fallacy, but a genuine question. Why would a person with exceptionally high intelligence want to work for a magazine so utterly stupid?
I love this one; it's so devious. If Monty doesn't show you which door he opened, then you gain nothing.
As copyright owner of this comment, I authorize everyone to defeat any technological measure which limits access to it.
If you're going to plagiarize, at least do from someplace more obscure. Especially if you get paid to be clever for a living.
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.
Monte Hall Problem You pick a door, now given a choice of keeping what is behind that door or switching to get the better of the two prizes that are behind the other two doors, would you switch -- of course. That is what you are doing when you switch.
Watch for Tom Cruise quoting this article in his next interview.
"Nine times out of ten, starting a fire is not the best way to solve the problem." - my wife
The "Monty Haul" problem is where you place the choice, and Marilyn's problem was failing to state it plainly.
For anyone not familiar with it, the "Monty Haul" problem is simple: in a game you get to pick from one of three boxes, inside one of which is the prize. After you pick your box, but before it's opened, the game reveals which one of the boxes you didn't pick is empty. You are then offered the choice to switch to the other box. Mathematically, is it better to go with your first choice or to switch?
The answer: You get a 66% chance of success if you switch, because the only way you lose is if you hit the 33% chance of grabbing the prize box first.
Or, shorter: Switch, because you probably didn't pick the right box to begin with.
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.
You've made a basic mistake. You assume that each of the options in your table are equally likely which they are not.
In your notation CGG 1 2 and CGG 1 3 both have a chance of occurring of 1/18. A factor 1/3 for the car to be behind door 1, a factor 1/3 for you choosing door 1 and a factor 1/2 for Monty choosing either 2 or 3.
On the other hand CGG 2 3 and CGG 3 2 have a chance of occurring of 1/9. A factor 1/3 for the car being behind door 1 and a factor 1/3 for you choosing door 1. Monty doesn't have a choice any more.
Adding up the wins and losses with the proper weights will give you a 2/3 win ratio rather then a 1/2 one.
No but this is the whole point ... there are NO repeated entries ...
...
... monty can open door 2 OR door 3, and I can choose to either stick or change my door after that ... that leads to 8 possibilities for each car position, or 24 entries in the truth table ...
IF I have already chosen the door with the car behind it, Monty has TWO options for which door to choose the goat from
Example, I choose door 1, and it has the car
Please, examine it again, there are no repeated entries I assure you.
Your truth table is actually the fallacy, because all of your 24 possibilities are NOT equally likely.
Your table suggests your odds of picking the car in the first place are 1/2, when in fact your odds of picking the car are 1/3. Each one of the cases you take where the player chooses the car is half as likely as the cases where he picks the goat first. This is a common error in misapplying the "equally likely" theorem.
The Monty Hall scenario assumes I wouldn't rather ride...er, I mean pick the goat over the car.
Ceci n'est pas une sig.
and /. apparently doesn't like the "pre" tag anymore...I actually wrote:
No, the table doesn't have repeated entries. The table's entries aren't equally likely - as is often the case in probability that goes beyond the very basics. An intro college course in probability would assuage your doubts.
Your truth table reflects each possible outcome, but not the odds of each possible outcome. The fallacy can be seen if you remove all the columns but the first four. Take the first set of 8, where the car is door number 1 and doors 2 and 3 have goats behind them.
Why is picking the car represented four times and picking a goat represented four times? That's not right. Of that first scenario, where the car is behind door number one, if I randomly pick one of your 8 lines, I should have 2/3 probability that I pick a goat, and 1/3 that I pick the car, not 1/2 in both cases.
Here's why: each set of 8 should only be a set of 6. Picking the door with the car should be represented two times, not four. You cannot differentiate on the door that Monty picks because the *number* of the door that Monty picks doesn't matter, only what's behind it, and what's behind it is *always* a goat. Your truth table should remove the "Monty" column, because the only things that affect your outcome is what door you pick, and if you switch or not.
I always liked to think of it this way:
If I pick randomly, I have a 2/3 chance of picking a goat. That should be obvious.
If I pick a goat and switch, I win. Picking "switch" or "don't switch" doesn't have a random outcome unless I want it to by flipping a coin or something - I can make a conscious decision to always choose "switch."
Therefore, if I ALWAYS choose "switch," I have a 2/3 chance of winning.
So I think it best not to try anymore.
But look at the bright side...the "Intelligent Design" folks are hiring for a new spokesperson. You may have all the qualifications they're looking for.
Two ways to look at it. No matter what you do, your first choice is meaningless. Only your second choice matters. In that choice you have two doors, one win and one lose. 50/50.
The article obfuscates the common sense nature of it by incorrectly collapsing two separate possibilities in to one.
If the car is in #3, the _four_ possibilities are:
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
In reality, the first choice just obfuscates. The second choice, the one that determines if you won is a choice between one win and one lose.
I hate to bang on with this ...
It is a truth table with every one of the 24 possible states ... not 18.
Each set of choices is equally likely.
Just because CGG 1 is repeated twice, this is because Monty has two possible choices for which door to offer you ... It means that the chance of me picking the car is 8/24 or 1/3 in absolute terms, but in logical terms it is 12/24 ...
To evaluate all the possible outcomes, you have to consider all the possible multiplicative steps - switching from logic to odds when it suits you is what throws out your final evaluation.
There are three doors. There is a dollar behind one door, and nothing behind the other doors. I know what is where, you don't.
You pick a door. Before you open it, I may or may not open an empty door. In either case, you then have the chance to switch doors.
You open the door you picked originally or the one you switched to. You can keep the money if you find it.
This is not the game in the article, but it matches the description on the MvS website. So you always switch when shown an empty door.
What they didn't tell you was that I only show you an empty door if you picked the door with the money initially. So if I show you an empty door, and you switch, you will lose.
You are wrong. The problem with your "proof" is that you have assigned equal probabilities to each row in your table when they are not equally likely.
Take the four results where the car is behind door one and you always switch. One time in three you choose door 1. One time in two Monty chooses door 2 and one time in two Monty chooses door 3.
C G G 1 2 YES LOSE 1/3*1/2 = 1/6
C G G 1 3 YES LOSE 1/3*1/2 = 1/6
Now one time in three you pick door two. In this case Monty must pick door three every time:
C G G 2 3 YES WIN 1/3*1 = 1/3
Same situation if you pick door 3:
C G G 3 2 YES WIN 1/3*1 = 1/3
Now add up the probabilities. If you switch you lose 2 times in 6 and you win 2 times in 3. You can do the exact same analysis on the cases where you don't switch and you'll find you win one time in three and lose 2 times in 3. The cases where the car is behind doors 2 and 3 work exactly the same way.
If you still don't believe me then go the site and play the game. You should see a difference in success for switching vs not switching after only 10 or so trials.
All the excuses I make for my poor choices are based on cognitive dissonance! What the hell am I supposed to do now?
I prefer rogues to imbeciles because they sometimes take a rest.
I just went thru all the possibilities exhaustively.
:initial_chosen_door, :switched_to_door)
I look forward to seeing someone complain about my code, I probably did something mathematically wrong
% cat monty_exhaust.rb
class Round < Struct.new(:car_door,
def won?
car_door == switched_to_door
end
def switched?
initial_chosen_door != switched_to_door
end
end
rounds = []
doors = [1,2,3]
doors.each do |car_door|
doors.each do |initial_chosen_door|
(doors - [car_door, initial_chosen_door]).each do |montys_door|
round_noswitch = Round.new(car_door,
initial_chosen_door,
initial_chosen_door)
round_switch = Round.new(car_door,
initial_chosen_door,
(doors - [initial_chosen_door, montys_door]).first)
rounds << round_noswitch
rounds << round_switch
end
end
end
puts "Won when switching %: "
switching_rounds = rounds.select { |r| r.switched? }.uniq
puts switching_rounds.select { |r| r.won? }.size / switching_rounds.size.to_f
puts "Won when wasn't switching %: "
nonswitching_rounds = rounds.select { |r| !r.switched? }.uniq
puts nonswitching_rounds.select { |r| r.won? }.size / nonswitching_rounds.size.to_f
% ruby monty_exhaust.rb
Won when switching %:
0.5
Won when wasn't switching %:
0.333333333333333
Why not fork?
The problem with your truth table is here:
C G G 2 3 NO LOSE
C G G 2 3 YES WIN
C G G 3 2 NO LOSE
C G G 3 2 YES WIN
which actually should be:
C G G 2 3 NO LOSE
C G G 2 3 YES LOSE
C G G 3 2 NO LOSE
C G G 3 2 YES LOSE
because you chose "G"oat in the fist chance, and you lose right there, because don't get the second chance.
"This one has been debated over and over, and is a classic example of lies, bloody lies and statistics."
Wrong on all counts. Including the fact that it's not even statistics -- it's probability.
We know where leadership by an anti-intellectual "strongman" who scapegoats minorities and likes boisterous rallies goes
You should go play roulette at the local casino. After all, there are two possibilities, either you win or lose. Therefore, 1/2 of the time you win and 1/2 of the time you lose. They pay off 36 to 1 when you win -- you do the math.
Er, or the other hand, maybe you shouldn't do the math...
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.
I'm totally buying two tickets.
So who's the idiot?
Now that is funny. True, but funny.
The solutions that the psychologists came up with is correct under the assumption that the subject has either no preferences, or has no stable preferences. The new analysis is for the case where there is a pre-existing preference and it's stable across experiments, even if the preference is slight. That's a good and valid point, but it is a point about assumptions and consequences, not about getting the math wrong.
So, the psychologists didn't get their math wrong, but they made an assumption that has a significant chance of being wrong. However, until someone actually does the experiment, we simply don't know whether the original assumptions were right or whether the new assumptions are right. (I suspect it probably depends on the experiment.)
This sort of thing happens all the time in the sciences, including physics, chemistry, and biology. It's not a problem with people failing to understand mathematics, it's just that every experiment and every analysis needs to make a lot of assumptions that can't all be examined in detail, and sometimes even the seemingly most reasonable assumptions turn out to be wrong.
Can someone source this? This would be a nice counter-factoid for some of the copyright discussions.
My first Journal Entry ever, in 8 years! http://slashdot.org/journal/365947/aphelion-scifi-fantasy-horror-poetry-webzine
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.
Okay, I'll bow out gracefully on this one.
I realise now that IF you only have one chance at the game, yes, then it is better to switch.
BUT, if you play the game 24 times, using each of the possible combinations, then there is no advantage.
I guess this is what is so perplexing about this problem, in that when you start talking about odds, and possible outcomes, it then seems illogical to not consider EVERY possibility as equally possible. Play 24 games with my table, and see how much you win or lose. regardless of Monty's choice of door NEVER affecting the outcome, it still remains that they are 2 out of 24 possible scenarios that must be played out and considered in the whole.
Time for bed.
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
And, if it existed, it probably would also be a bad idea to get your parades from Fact magazine.
Actually, I found myself compelled to finally register with /. over this.
davime is quite correct in his conclusion although I would argue it a bit differently.
Monty's choice *cannot* change your odds of winning; his removal of one of the two goats simply informs you that your odds have just changed from 1/3 to 1/2. It simply cannot provide any additional information as to whether your choice was a strong or weak one.
I challenge some of the 2/3 thought processes as follows; if you you could pick two of the three doors and win if the car was in either of them, then and only then would your odds be 2/3. And this two selection Monty game is lower odds than a straight up two door clean option.
The whole Monty fallacy is that the first choice gives you more information. It never can and it never will, ergo the second choice always is a 50% *probability independent* choice.
T.
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
"Monty Hall". Did you NEVER watch "Let's Make A Deal"?
It's named off of a person, not a "haul" or anything.
My blog. Good stuff (when I remember to update it). Read it.
doesn't make it proof.
The Monty Haul problem is the kind of cognitive bug that can catch anybody and go unnoticed by people who should know better. Which is not to say that social scientists in general have as much math as they ought to.
If you want to talk about a mathematical fallacy that people ought to be able to detect, talk about the base rate fallacy. My daughter's pediatrician recommended a nasty test; after interrogating him I realized he had no idea of how to interpret the data.
Post may contain irony: discontinue use if experiencing mood swings, nausea or elevated blood pressure.
Sorry, I'm no economist... but if we're assuming Monty shows us a goat EVERY time we do this, doesn't that mean that we're only ever choosing from two possible choices when we pick the first door? So it's 50/50 whether we should switch or not?
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.
yeah, psychologists aren't pro mathmeticians, but how many of you math nerds know about BF Skinners theory of radical behaviorism. Barring a quick google search, the answer is 1 (Cowboy Neal, duh.)
Orbis terrarum est non altus satis
Some psychologists and one economist differ over the interpretation of experimental results, therefore psychologists don't know math?
If there is an inherent distaste for blue that is significant enough to cause the 2:1 preference of green over blue, and the monkey is initially exposed to all colors, why doesn't he also chose green over blue more often then?
Or can you recommend any reading on the subject?
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 can register on Slashdot and put as many words in a row as you like, but that doesn't magically make the choice to switch "probability independent".
Remember: once Monty opens that goat door, you know for a fact that the door you've already picked is twice as likely to have a goat behind it than it is to have a car. That's 2:3 odds, regardless of how many doors are left. In a "probability independent" choice, you would not have this information.
The problem is that you have too many options in your table. I'll simplify the case and assume car is behind door one.
Strategy 1, always switch:
Choose one, then switch: LOSE
Choose two, then switch: WIN
Choose three, then switch: WIN
Strategy 2, always stay:
Choose one, then stay: WIN
Choose two, then stay: LOSE
Choose three, then stay: LOSE
Your first choice is "which door?" (three options) and the second choice is "stay or change?" (two options). So, you have a total of six paths to winning or losing. Your mistake in the above table is assuming that "you choose one, monty chooses 2, you switch" and "you choose one, monty chooses 3, you switch" are distinct possibilities.
It's not an easy problem, read the wikipedia article a couple more times...
Your truth table assume the contestant has a 1/2 (4 of 8) odds of picking the winning door in every configuration, which should obviously show you that it cannot be right. The other posters already told you that you failed in assuming each entry in the table is equally likely, when they are not.
It also helps to group the tables according to the "Switch" option rather than the configuration, since afterall we're trying to get a decision about whether it is better to switch, regardless of the configuration.
So, try including the probabilities like these tables. The first table assumes we don't switch, and the second table is exactly the same except that it assumes that we do switch.
1 2 3 You/prob Monty/prob Comb Switch Win
C G G 1 (1/3) 2 (1/2) 1/6 No Yes
C G G 1 (1/3) 3 (1/2) 1/6 No Yes
C G G 2 (1/3) 3 (1) 1/3 No No
C G G 3 (1/3) 2 (1) 1/3 No No
1 2 3 You/prob Monty/prob Comb Switch Win
C G G 1 (1/3) 2 (1/2) 1/6 Yes No
C G G 1 (1/3) 3 (1/2) 1/6 Yes No
C G G 2 (1/3) 3 (1) 1/3 Yes Yes
C G G 3 (1/3) 2 (1) 1/3 Yes Yes
Similarly for the configurations G C G and G G C so those don't need to be shown.
The probability under "Comb" shows the combined probability of you and Monty picking the respective doors in each row. In each table this column adds to 1. The result under "Win" must be weighted according to the "Comb" value.
So in the first table where we don't switch there are 1/3 wins (1/6 times two). In the second table where we do switch there are 2/3 wins (1/3 times two).
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.
What you are missing is that Monty Injects Knowledge into the REMAINING set - and it is Not your set.
This is easier to see with a larger set, say 10 boxes. You pick one box, and it is moved to the side. From the remaining 9 boxes Monty reveals 8 goats. One box is left. Which would your instincts tell you to pick now?
That's right the remaining box from the 8/9 boxes - chance it is a winner 9/10. Odds are NOT 50/50 that you picked the right one at the beginning.
It's tricky to understand because were' dealing with a relatively small difference in chances. So let's stretch this thing wide open.
1: There are 100 doors, 99 goats, 1 car.
2: You choose a door, you know it only has a 1% chance of hiding a car.
3: The host opens 98 doors.
4: You now know that the other door has a 98% chance of hiding a car.
5: You still know that your door only has a 1% chance of hiding a car.
6: Do you change doors?
7: Yes!
I know it seems silly, but it's EXACTLY the same principle as the 3 door example.
http://www.dms.umontreal.ca/~andrew/PDF/VS.pdf
Won when wasn't switching %: 0.333333333333333 Well, I'm afraid you must have done something wrong, since whatever you think of the problem, those two probabilities should sum to one!
In fact, %Won when switching should be 2/3.
What's purple and commutes? An Abelian grape.
I knew I should have taken the green pill..
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.
Those of us who think they know everything annoy those of us who do.
Ah, so there is a 1% chance of neither succeeding nore losing. :)
So, is Deal or No Deal just a Monty Hall game then? Always best to switch if get to the end with two cases?
I'th thought of another argument to convince the ItDoesn'tMatterWhetherYouSwitch-ists.
When you first choose, the probability of there being a car behind the door you choose is 1/3. Now imagine that as soon as you've chosen, you stick your fingers in your ears, shut your eyes, ignore anything Monty Hall does, and sing very loudly, "I'm Sticking With This One, I'm Sticking With This One..." over and over again.
Clearly, if at any time during your singing you were to open the door, there is always a probability of 1/3 there being a car behind it. No matter what Monty Hall does.
Now consider that probabilities for mutually exclusive and exaustive options must sum to 1. Now, if you hadn't stuck your finger in your ear, you would have had an option to switch. Since the car is either behind your door or the door you could switch to, and the probability it's behind your door is 1/3 (probabilities don't change just because you've not stuck your fingers in your ears), the probability that it's behind the other door must be 2/3; so you should switch.
Clear?
What's purple and commutes? An Abelian grape.
Is it ethical to give potentially toxic stuff to monkeys, just to see what poison they like more?
.. paranoid crackpot leftover from the days of Amiga.
Do you get to keep the goat?
sic transit gloria mundi
C G G 1 2 NO WIN
C G G 1 2 YES LOSE
C G G 1 3 NO WIN
C G G 1 3 YES LOSE
C G G 2 3 NO LOSE
C G G 2 3 YES WIN
C G G 3 2 NO LOSE
C G G 3 2 YES WIN
You're missing four possibilities in each block. The ones where Monty picks the car and you lose instantly.
Can you be Even More Awesome?!
I think the Monty Hall problem is easier to understand probability-wise if you increase the number of doors from 3 to 100 and slightly change the way you lay out the problem. In the end it doesn't really change the probability, it just makes the case for changing doors much more apparent. If you start out with 100 doors, you only have a 1 in 100 chance of picking the right door. There's a 99 percent chance you picked the wrong door. If Monty Hall collectively removes 98 "wrong" doors and gives you a chance to pick between 2 doors, you don't have 50-50 odds, your odds are still 1 in 100 that you picked the correct door. By switching doors, 99% of the time you'll win.
With 3 doors, the odds that the correct door is in the other group is 2 in 3. Opening one wrong door in the other group doesn't change those probabilities. You should always change doors for better probabilities.
Good start, but your truth table is incomplete. You have removed the non-sensical rows in which Monty opens the door showing the car and you either switch or don't switch. It would seem that this doesn't matter, but it is skewing the results of your truth table because every line of a truth table is equally weighted. A truth table represents all possible inputs and then displays all logical (but not neccessarily "sensical") outputs based on some application of an operation.
Here is an example. Assume we have a random number generator, 0 though 7. What is the probability of any of the outcomes? 1/8. There are 8 rows to the truth table, which results all possible numbers generated in our system.
000
001
010
011
100
101
110
111
What is the probability of an even number? 4/8 = 1/2. What is the probability of an odd number? 4/8 = 1/2. So far, so good.
Now, let's apply a rule that says that any time the system generates an even number (include 0), we add 1 to make it odd (formally, the operation is outcome=input|0x001). The probability of the outcome of any given even number is now 0/8 and an odd number is now 2/8 = 1/4. The truth table still has 8 lines, but because of the rule applied to the initial uniform distribution of inputs, we no longer have a uniform distribution for the outcomes.
Note that I show the non-trivial application of the operation as ->
000 -> 001 = Odd
001 = Odd
010 -> 011 = Odd
011 = Odd
100 -> 101 = Odd
101 = Odd
110 -> 111 = Odd
111 = Odd
The initial uniform distribution has been altered (weighted) in favor of a new distribution by modifying certain random input to become an different outcome.
If one were to reduce the truth table by not showing the altered line, the truth table would be:
001 = Odd
011 = Odd
101 = Odd
111 = Odd
which clearly does not show all the initially generated values, but still gives an accurate probability due to the uniform application of an essentially linear altering / weighting operation.
Now, to match this up with the Monty Hall Problem more closely, change the modification rule to only add 1 to even numbers greater than or equal to 4 (if(input>=0x04) then outcome=input|0x01). This operation is applied uniformly, but the operation itself has a non-linearity in it (only greater than or equal to 4). Here is the new truth table showing the non-uniform distribution of outcomes based on a uniform distribution of inputs when a non-linear operator is applied:
000 = Even
001 = Odd
010 = Even
011 = Odd
100 -> 101 = Odd
101 = Odd
110 -> 111 = Odd
111 = Odd
Now, the probability of getting an even number is 2/8 = 1/4 and the probability of getting an odd number is 6/8 = 3/4, despite the initial uniform distribution. If the truth table is reduced simply to outcomes by collapsing input rows affected by the non-linear operation into input rows with the same output, then a deceptive truth table, similar to the one in the above post, is created:
000 = Even
001 = Odd
010 = Even
011 = Odd
101 = Odd
111 = Odd
In this case, it appears that the probability of there being an odd number is simply 4/6 = 2/3 and even is 2/6 = 1/3, which is incorrect, since our number generator was uniform, and the rule altered the 2 out of 8 of the initial numbers to become something else. The new truth table is no longer uniform in distribution because it is no longer representing all input rows.
In the truth table in the above post, only 8 outcomes are shown from an actual truth table of 12 outcomes (using only CGG, but similar for other permutations). 4 input rows have been eliminated, thus the error.
The actual truth table is (N=NO, Y=YES for the switch column):
CGG12N = Win
CGG12Y = Lose
CGG12N = Win
CGG12Y = Lose
CGG21N -> CGG23
It's si-LAY-us, you Silly Ass!
Bacon writes: "In the year of our Lord 1432, there arose a grievous quarrel among the brethren over the number of teeth in the mouth of a horse. For 13 days the disputation raged without ceasing. All the ancient books and chronicles were fetched out, and wonderful and ponderous erudition, such as was never before heard of in this region, was made manifest. At the beginning of the 14th day, a youthful friar of goodly bearing asked his learned superiors for permission to add a word, and straightaway, to the wonderment of the disputants, whose deep wisdom he sore vexed, he beseeched them to unbend in a manner coarse and unheard-of, and to look in the open mouth of a horse and find answer to their questionings." Of course, Bacon was attempting to adjudicate the debate between rationalists and empiricists. For all of our arm chair reasoning about the Monty Hall problem, perhaps the best way to resolve the matter is to perform an experiment...Click on the link to the Times article and run a few hundred trials. My results - 200 attempts, 133 cars = ~67%.
They would have a problem having an IQ over 200 since there aren't enough people in the world for that to be possible. Only about one in a million to one in a billion[0] people have an IQ of 200 but to get to 214 there would have to be more than 100 billion people on Earth. That is to say more than all the people who have ever lived.
Unless you are talking about kids in which a 250 IQ isn't hard to pull off. Finding a 9 year old that can test like a 24 year old isn't that hard. In most schools finding a 24 year old that can even get out of bed to take a test is hard enough.
[0] depends on which source you want to use a) University of London study by Vernon Sare in 1951 or b) Straight Talk About Mental Tests", The Free Press, A Division of the Macmillan Publishing Co., Inc., New York, 1981
Ascii artist &
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.
What happens if you buy three tickets? Does the universe implode?
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
The number of the door that Monty picks does not affect your outcome unless choose to let it. However, it is conditioned on your first choice. That's the extra information that you are missing.
I think you're thinking of the "Monty Hall" problem.
It's much more intuitive if you consider a lot of doors and condense the guessing. Say there are a million doors. You choose one. Now Monty offers to open 999,998 other doors, leaving only your door and one other. Now, clearly your odds of picking the right door on the first try were 1 in a million... but Monty has eliminated a bunch of choices for you and that other door is now almost certainly the correct choice... It's now a 1 in a million chance that it's *not* the other door.
Pat
Far be it for be to object to the vilification of psychologist, but the Monty Hall problem is a specific can where the release of additional information by an agent with perfect information, not done at random, result in additional information about the original decision.
Monty
A) Will never reveal a car
B) will never open the door you already picked.
If either of those factors don't apply, the counterintuitive probabilites of the Monty Haul problem disappear.
Neither of those seems to apply to the psychological cases used for cognitive dissonance. The error in this case doesn't seem to me to be the psychologists - the economist is applying an incorrect model based on a superficial similarity between a carefully reviewed experiment, and, ah, a game show.
Hard to believe, I know. But I would avoid getting too emotionally invested in this - Refusal to believe in the experimental evidence for cognitive dissonance because of bad logic like this would create an entirely new category of "Ironic Dissonance".
Do you really want to be responsible for an entirely new psychological discipline? Of course you don't.
Pug
Pug
An Invisible Entity of Vast Power whose existence must be taken on faith alone: Liberal Media
I think I'm going to have to agree with Cecil Adams on this one.
Very funny indeed.
Expanding a vast wasteland since 1996.
The point being, if you know the square root in advance and it is an integer then you would be able to conclude that your subject is non prime (or 1).
"If we knew what it was we were doing, it would not be called research, would it?" - Albert Einstein
If you read the article, the logic makes sense. The monkey is revealing a preference in the first experiment, and that changes the odds for the second experiment in a counterintuitive way.
I've been saying this for years. Psych students are bombarded by vast numbers of stats courses, yet they typically don't know what a derivative is, nor can they tell you what basic stats results like the Law of Large Numbers mean. They learn to do all sorts of tricks with time series, yet can't tell you what "linear" means.
My suspicion is that the academics desperately want psychology to be considered a science, and they think that quantifying everything is the answer, rather than addressing the issue of the Scientific Method. As a result they tend to use the most impressive tests and designs they can, regardless of applicability.
Psychologists tend to over-sample, over test and over analyse. They often perform multiple tests on the same sample, reducing the power of their results. They routinely perform massive multiple regressions on huge samples, ignoring the fact that with a sample large enough, and enough 'independent' variables, you can make any data set show a significant fit to any model.
They also seem to love the various measures of correlation, without asking what (if anything) that particular magic number might mean in the context under investigation (i.e. "r=0.76 which is significant at alpha=0.05 so food colour preference is related to hat size" is the kind of statement that seems to be highly regarded amongst them).
Much of the seminal work in inferential statistics was done by psychologists (Fisher, Pearson). These were people who could do maths. Today psychs don't seem to have the same level of mathematical literacy. If psychology does want to be considered a hard science, then it should train its people that way. Otherwise they should be content to be a social science, and realise that there is no real competition or ranking between social science and hard science.
Wait, am I being glib?
Have you been touched by his noodly appendage?
I have a math degree, too, and parent post is quite correct. However, the article has a point that it becomes a LOT easier to see when there are more doors:
Imagine that Monty has a 52-card deck and says that if you get the ace of hearts, you win. You draw a card, face down, and put it in front of you. Monty then lays 50 of the cards face up, proving that none of them are the ace. He then asks you if you want to trade him for his one remaining card.
If you still don't think it's best to switch, you must think that Monty is a con artist, or you're really bad at math.
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.
No, your odds with flipping the coin to choose between two doors will always be exactly 1/2, regardless of the host's strategy. (Think about it.) But if you know something about the host's strategy, then a non-random choice can give you better odds; 2/3 from switching in the classic version.
Weeks of coding saves hours of planning.
Once you have made your first choice, it's no longer a question of 1/3, as Monty's actions are determined by both your choice AND his knowledge of where the prize is.
The thirds are gone.
-- All your bass are below two Hz
dang.
-- All your bass are below two Hz
*aside* from issue of who chooses the original door...
...now the answer really become intuitive/obvious.
I once had a math teacher who explained the Monte Hall problem beautifully:
"imagine that there are one million doors, and that you have chosen one of them. Monte now opens 999,998 of the 999,999 other doors, leaving only your door and one other remaining."
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)
After just reading the article descriptions, I thought for sure that "the monkey's choice of red over blue" was going to have something to do with American Politics.
Anyways . . . fascinating stuff
She cannot have a measured IQ of 228 becuase she took the Stanford-Binet test which would only give her 167+.
Trying to install linux on my microwave, but keep getting a kernel panic...
http://en.wikipedia.org/wiki/Monty_Hall_problem#History_of_the_problem
Monty never actually played the Monty Hall game on Let's Make a Deal as the math problem has it.
Degaussing scares the bad magnetism out of the monitor and fills it with good karma.
import java.util.HashMap;
public class MontyHall {
public static void main() {
final HashMap<Boolean, Integer> map = new HashMap<Boolean, Integer>();
map.put(true, 0);
map.put(false, 0);
for (int i = 0; i < 1000000; ++i) {
final int winner = random3();
final int choice = random3();
map.put(winner != choice, map.get(winner != choice) + 1);
}
System.out.println(map.toString());
}
private static int random3() {
return (int)(Math.random() * 3d) % 3;
}
}
The best way to look at this problem is to take it to the extreme, and it also makes a lot more sense that way..
Say you have a million doors and have to pick one.. Chances of picking the car are 1 in a million.. Now all but your door and one other are opened reveling goats.. You know one of the remaining doors has the car and one has a goat.. You can pretty much guarantee you didn't pick the car, so swapping is the only option.
It isn't said that the host's choice to open the door is independent on whether the player chose the car. In the article version it is independent as it is constant.
This common bad problem description is also why some mathematicians got publicly tripped up by this problem
I dunno why I picked my random number the way I did. I'm tired, looking at it after I posted it I wonder what I was thinking.
Wurp is claiming it doesn't matter if Monty must open a door every time. This is wrong, and the fact that he must open the door every time is what makes the Monty Hall Problem a non-intuitive curiosity. Most people will intuitively assume that choosing between the remaining two doors is a 50-50 chance and therefore switching does not matter. But that is beside the point. The point is, that if Monty doesn't have to open a door every time, then he can just show a door when you choose correctly the first time. Then your chances of choosing correctly are not 2/3 but are 0. Now, you may counter by saying that, "then everyone will know they chose correctly and will stick with their choice." But that is missing my point still. If Monty doesn't have to open the door every time, he can still open the door with some small majority of the time being when you choose correctly the first time, thus deviating from your absolute probabilities of 1/3 and 2/3.
How many more years will slashdot have an off-by-one error on your Score in your profile?
You're right. I considered that the contestant always has the opportunity to switch, based on what I've seen of the show, but it may not actually be true, which changes all of the assumptions. But it also isn't necessary that they are always done after the first pick as the GP suggested (if they never had the opportunity, there wouldn't be a problem).
If you consider that Monty may actually be deciding whether to give the opportunity based on his knowledge (or his personal mood that instant), then the problem may not even have an optimal solution afterall.
Oh, I see that maybe Wurp wasn't suggesting that Monty's choice doesn't matter. He was replying to another poster who mentioned a 1/2 probability. I guess my comment threshold didn't allow me to see that post... (I'm new) But, my point still stands. This only works if Monty is required to reveal a choice every time, which apparently he is not. So, I suggest that the name of this dilemma should be changed.
How many more years will slashdot have an off-by-one error on your Score in your profile?
I did not say it doesn't matter if Monty must open a (losing) door every time. I didn't say anything about it either way (I should have).
Someone else already pointed this out (that it's important that Monty open a losing door every time) and I agreed with them.
http://www.questia.com/library/book/how-to-use-and-misuse-statistics-by-gregory-a-kimble.jsp was used in more than a few statistics intro classes in spite of having been written by an internationally recognized psychologist.
Where? Googling for this claim, I only came up with people asserting that she had made the blunder without saying where and other people responding to ask them where (book, page number, and context) she'd claimed 9 was prime. But the question never gets answered.
The only exception to this pattern I found was your post which had the claim but was missing the requisite challenge. Thus my post and query.
--MarkusQ
....Psychologists have known about this problem for a while.
Way to go Slashdot by making them all look like idiots with a sweeping general statement that doesn't exclude psychologists working in fields that have nothing whatsoever to do with Cognitive dissonance.
Psych will never get the respect it deserves as long as shit talk like this persists.
-5 Spreading ignorance, Zonk
There is more to science than physics!
www.iomalfunction.blogspot.com
If shes so smart how come she wasn't the one to solve Fermat's last theorem? If your the smartest person in the world, what would you do?:
A. Get rich.
B. Help humanity by solving hard problems.
C. Work for a second rate magazine company.
The only one that's not on my list is C.
Imagine it were not 3 doors, but 1000000
you pick one door, the moderator opens 999998 doors and says
the right door is EITHER the one you chose out of 1000000 OR the other one that I didn't open...
according to your argumentation that would still be a 50:50 chance...
I hope this more extreme example helps you understand why TFA is right afterall...
The MAFIAA is a bunch of mindless jerks who will be the first up against the wall when the revolution comes
Whoa, let's not get too advanced too fast. We're still trying to do an exposition to a general audience, after all.
So if you get down to 2 cases in Deal or No Deal (one has 5 cents and the other has a million) does that mean you should always switch cases because you only had a 1/26 (???) chance of picking the million initially? Does this mean you have a 25/26 chance of winning the million with the switch?!!!
Good method. Since the density of primes is higher than the density of squares, a number that is not divisible by anything less than its root will probably not have an integer root.
Number of primes less than 1,000,000,000: (approximated with x/ln(x))
48254942.433694647516792102101845
Number of squares less than 1,000,000,000:
31622.776601683793319988935444327
D'oh! I was wrong. Thanks, that was exactly what I needed!
No, your odds with flipping the coin to choose between two doors will always be exactly 1/2
When I said "host's strategy", I mean his strategy in regard to opening a goat door in the first place. If, for example, the host chooses to never open a door, then your odds will be at 1/3. Certainly, you cannot devise a strategy to improve above that, but deciding to toss a coin will let you remain at these odds.
I remember my professor returning in disgust after teaching 3rd year students. It turned out they didn't know what a standard deviation was. Average, ok, they knew that, but anything beyond that was a mystery to them, after two years of statistics classes. And these were students specializing in experimental psychology.
And don't get me started on a colleague who did a presentation on his research and was proud that he could tell that he finally had a obtained a significant result.
After 19 failures...
But he will offer you some money instead.
Monty puts a $30,000 car behind one of three doors. You pick a door. Monty reveal a goat behind a different door. You have a choice.
A. Stick with your original door.
B. Take a check for $12,000.
Go on, make a rational decision.
it seems she did take the door at random instead of always taking a dude door in her explanation
I would like to see that checked. Because if true, then INDEED she was wrong and all mathematician right.
C. Sagan : A demon haunted world:
http://www.amazon.com/gp/product/0345409469/
visit randi.org
I don't think that's how it works. It's more that human beings have a probability distribution of intelligence that centers around 100 IQ, not that the actual distribution of intelligence at the present moment in time is a smooth curve. If one in 100 billion people have an IQ of 214 that doesn't preclude such a person living today, it only suggests that on average, only one such person will live for every 100 billion other people in the world.
In Repressive Burma, it's not just your connection that dies. slashdot.org/comments.pl?sid=314547&cid=20819199
And, of course, I've made a usual mistake: "the odds" are an odd term which I almost never use. I was referring to P("winning the prize"), or the expected value, if you think of a car as one and goat as zero. For E = 1/3, the corresponding odds are 1:2.
Just simulate it. Serious. Get a friend and have them list out a series of door configurations at random. You pick a door, your friend shows you a goat, then you stick with your choice for 100 tries. Then do it with the switching strategy. I guarantee you you will win more often by switching.
Your truth table is flawed. Do the experiment, you will be enlightened.
Take the deck of cards version: you get to pick one card from a deck (without seeing the card). You are aiming to pick the ace of spades. Monty then shows you 50 cards which are not the ace of spades, so he has one unrevealed card left. Now, should you switch cards with him? Do you see that the card you picked still only has a 1 in 52 chance of being the ace of spades?
If you are still not convinced then go to the website and play the game.
It's not a bug, it's a lepidopter!
You now know the contents of BOTH envelopes. There's no statistical probability.
So if you pick the £100 envelope and change, if the change gives you £50, changing back is 100% likely to be a good choice. If the first change gives you £200, staying with this choice is 100% likely to be a good choice.
A mathematician should deduce the fact that Monty knows where the car is even if she is not told about it. An average person might not, but a mathematician should since it is part of her damned job!.
They didn't have blue M&M's in 1956. There's your problem right there!
Don't forget, she is a being from a higher dimension where 9 actually is a prime number. Sometimes she forgets she is manifesting in our pitiful 3D world where we still think pi is an irrational number and the square-root of -1 is imaginary...
----------------------------------- My Other Sig Is Hilarious -----------------------------------
I've always seen this problem as follows:
Since this is how the game plays out then you always have a 1/2 chance of getting it right, regardless of what door you always end up having to pick between two doors, one with a prize and one without a prize.
/Mikael
Greylisting is to SMTP as NAT is to IPv4
Seeing in how much detail this is all portrayed i beleive its pretty well thought tru. I consider myself a good conceptiual thinker and am not convinced by any explination i found!
Chances of winning change dynamicly in your favor as soon as Monty "disables" one losing option.
Conclusion: changing doors does not improves your odds. The odd have already changed in your favor.
In more detail:
At the beginning chances of winning equal 33%
Monty reveales a goat behind one door, and chances of winning become 50%.
You don't have to "re-choose" a door so the chances apply. In my view the chances already are 50%-50% at the start, since Monty always throws away one losing option.
Either there is a reason the reviews of this case in statistics doesn't include my reasoning, or i need a lesson why my view on all this is wrong.
Hivemind harvest in progress..
And that quote just proves that the problem was insufficiently specified.
I make the assumption that the game show host wants me to _lose_.
Therefore, the only reason the game show host opened a door with a goat behind it is because I chose correctly the first time. Therefore I should NOT swap.
Tim.
God said, "div D = rho, div B = 0, curl E = -@B/@t, curl H = J + @D/@t," and there was light.
It doesn't matter what Monty knows. All that matters is what is behind the door he opens. If it is a goat you need to switch to get a 2/3 odds for a car. It wasn't specified what happens if he opens the door with the car.
Thanks.. I'm so sick of this story being told wrong. TFA had it wrong again of course. People's intuition told them not to switch, and if one does not specify that Monty *always* opens a bogus door after one makes initial choice one is right not to. If one played with someone on the street for money, they would only offer you the chance to switch if your inital choice was wrong. Vos Savant got it wrong. Most of her critics were also wrong, [their intuition was right, but their justifications were wrong].
http://rareformnewmedia.com/
Bleh, i meant they would only off you chance to switch if initial choice was right.
http://rareformnewmedia.com/
That shows that the mathematicians were correct, actually, as her statement of the problem is missing the critical component. Whether or not the host knew where the car was is irrelevant.
The key to the correct statement of the problem lies in knowing that the host will always eliminate a remaining door with a goat. If this is not known to be true, the action of the host does not provide you with any additional information.
I don't get how the first step affects the probability of winning the prize.
In this scenario, why are you not always left with two doors to choose from?
Does ANYONE really know math?
Aside: I'm a math major and I think I'm confident in saying that no one does.
If you were offended by anything I said... No, I'm not sorry. Please lighten up.
No, Monty doesn't pick the car. He just doesn't. HE knows where the car is and doesn't pick that door. This is the fact that lies at the heart of the paradox. The statement of the problem "Monty opens a door that reveals a goat" excludes that possibility and gives away information. This fact is what changes the odds into 1/3 v. 2/3 instead of 50-50.
- 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?
Think about it from Monty's point of view and not yours. The trick is that Monty is forced to pick a door with a goat, as opposed to picking a door at random. That is what skews the odds in favor of switching.
Here's a Perl script to run the simulation (enough with the conceptual thinking. =) ) It shows that switching gives you a 2/3rd chance of winning and staying gives you a 1/3rd chance.
[code]
use strict;
my $total = 10000;
my $switch_total_wins = 0;
my $switch_total_lose = 0;
my $noswitch_total_wins = 0;
my $noswitch_total_lose = 0;
for(my $i=0; $i$total; $i++) {
my $car = int(rand(3));
# Our first pick
my $pick = int(rand(3));
# Time for Monty to open a goat door
my $goat_door;
if ( $car == $pick ) {
# Two goat doors available, pick one
if ( rand() 0.5 ) {
$goat_door = ($car + 1) % 3;
} else {
#$goat_door = ($car + 2) % 3;
$goat_door = abs(($car - 1) % 3);
}
} else {
# player picked a goat door, open other goat door
if ( $car != 0 && $pick != 0) {
$goat_door = 0;
} elsif ( $car != 1 && $pick != 1) {
$goat_door = 1;
} else {
$goat_door = 2;
}
}
# Now switch from original pick to new pick
my $new_pick;
if ( $pick != 0 && $goat_door != 0) {
$new_pick = 0;
} elsif ( $pick != 1 && $goat_door != 1) {
$new_pick = 1;
} else {
$new_pick = 2;
}
print "car: $car pick: $pick goat: $goat_door final: $new_pick\n";
## make sure logic is good
die if $pick == $goat_door; # cannot open door we picked
die if $car == $goat_door; # cannot open door with car
if ( $new_pick == $car ) {
$switch_total_wins++;
} else {
$switch_total_lose++;
}
if ( $pick == $car ) {
$noswitch_total_wins++;
} else {
$noswitch_total_lose++;
}
}
print sprintf " switch wins: %5.2f%% losses: %5.2f%%\n", 100.0 * $switch_total_wins / $total, 100.0 * $switch_total_lose / $total;
print sprintf "noswitch wins: %5.2f%% losses: %5.2f%%\n", 100.0 * $noswitch_total_wins / $total, 100.0 * $noswitch_total_lose / $total;
[code]
This is what I thought, but the Wikipedia entry for the problem states otherwise:
http://en.wikipedia.org/wiki/Monty_Hall_problem#Aids_to_understanding
I'm not quite clear on why Monty's knowledge matters, even after reading this.
"Now gluttony and exploitation serves eight!" - TV's Frank
That would be true if and only if the host did not know what was behind the door that was opened. If the host randomly chose a door, your reasoning would be correct. However, the host knows where the car is and chooses to open a door that has nothing behind it.
*** Where are we going? And what's with this handbasket?
According to my tests, all odd numbers greater than 1 are prime. 3 tested as prime, 5 is prime, 7 is prime. Now I admit that 9 tested as non-prime, but considering 11 and 13 did test as prime, I'm considering 9 an erroneous data point. The general trend is clear.
Related to this story is a follow-up: http://tierneylab.blogs.nytimes.com/2008/04/10/the-psychology-of-getting-suckered/#more-264 In this you read "In fact, when we presented probability puzzles including a version of Monty Hall and the Mr. Smith problem to 327 MBA students who had all been trained to use such computations, they were almost seven times as likely to rely on some form of the partition-edit-count strategy than an explicit computation. Strikingly, none of the MBA students who tried to compute the answer to these puzzles arrived at a correct solution." Anyone surprised?
GP is probably a PnP gamer and confused him with the "Monty Haul" campaign type. Understandable if you've never seen the show, since it is also a play on Hall's "name"
You might win twice, duh.
If you can't see the value in jet powered ants you should turn in your nerd card. - Dunbal (464142)
Who the fuck remembers Monty Hall? I'm pushing 40 and even I barely remember him. Might want to update that to "Howie Mandel" at least.
SJW: Someone who has run out of real oppression, and has to fake it.
Unfortunately false. Note that there is guaranteed to be an empty door that you didn't select. As your chance of getting it right first time was obviously 1/3rd, then the chance that it is behind the other two doors must be 2/3rds. We already know that it is not behind at least one of them, and so opening one of them gives us a 2/3rds chance it is behind the other.
For a more extreme example, try thinking about it with a pack of cards.
"Lisp
If Monty does know, and you know what conditions he is revealing the door to you, you have gained some information about the two unopened doors. The classic form of the problem he always reveals a goat door - so you should switch to the door that he did not pick.
It's not a matter of probability. If a number N is not prime then it must have one or more pairs of integer divisors A and B such that (1 < A <= B < N) and (A * B = N). If (B >= sqrt(N)) then ((N / B) < sqrt(N)). (A = (N / B)), so (A <= sqrt(N)). Thus the lesser divisor in the pair can be no greater than sqrt(N); if there is no integer divisor between two and floor(sqrt(N)), inclusive, then N must be prime.
"The state is that great fiction by which everyone tries to live at the expense of everyone else." - Bastiat
No, Marilyn was wrong. Under the Buddhist (and therefore correct) interpretation, the answer is to choose what's behind the door that Monty opens and reveals to be empty. Material goods only promote greater attachment to the random, impermanent dream-world our consciousness inhabits, whereas recognizing that enlightenment, the capacity for which is intrinsic to all things, is the only true prize means you begin to demonstrate the attitudes necessary to cast off the shroud of this false existence.
If you think this is a joke, ask yourself: would a car *really* prevent you being unhappy, or it is simply another thing you have to worry about?
Ok, now stop talking about your wife's box and GBTW.
Part of the problem with your thinking (and there are many) is that you're stuck on this arbitrary number of doors, 3. Let's rewrite the problem, and apply your reasoning.
There are 100 doors now. After you make your initial choice, Monty will open up 98 doors with goats behind them, and offer you the choice to switch your initial choice to the last remaining unlocked door.
Now would you do it? The choice is clear, yes, for the same reasons as the choice is clear for n =3. According to your reasoning, when I initially picked the first door out of 100, my probability was 1/2 of guessing correctly. Your reasoning leaves much to be desired.
She wasn't wrong, but she left out a crucial piece of information in how she presented the problem. Even I realized this when I read her article, and my IQ is below 140.
The movie "21" (I happened to see it last night) also botched the presentation of the Monty Hall Problem in the same way. Nothing was said to indicate whether the game show host's action (showing the goat) was independent of the contestant's choice. It's a subtle point and people presenting the problem often leave the ambiguity.
--- Often in error; never in doubt!
By symmetry, the expected return on the initial switch MUST be exactly 1.0x, yet the simple math says 1.25x.
Your expected return for switch was never 1.25X. The issue is the mixing of methods. You either look at the magnitude of the expected gain, or the factor relative to your current expectation, but not both. I.E., if switching results in getting the larger check your return increases by a factor of 2. If switching results in getting the smaller check your return increases by a factor of 1/2. But you don't add 2 + 1/2 to get the average.
With equal probability, switching will either change your return by a factor of 2 or 1/2. That is, starting with X, you switch to either 2*X or (1/2)*X. If you start multiplying, you must finish multiplying. The expected gain of the switch is a factor of 2*(1/2) aka 1 aka no advantaged is gained by switching.
If you want to add/subtract the different, with equal probability, switching will either increase your return from X to 2X--a change of X--or decrease your return from 2X to X--a change of -X. The expected gain of the switch is X-X aka 0 aka no advantage is gained by switching.
I, for one, am really glad it's the psychologists who don't know Math. Just image what would happen if it was engineers or computer scientists who didn't!
http://erichsieht.wordpress.com/category/english/
If you want a goat, and Monty opens the door with the car, should you switch doors to get a better goat?
"I only speak the truth"
Karma: null(Mostly affected by an unassigned variable)
Occasionally there would be a "booby prize" that was actually valuable. At the end of the show, Monty would give a few contestants the option to trade their winnings for a shot at the big end-of-show prize. Sometimes he would offer this to a booby prize winner, and on at least one occasion, she gleefully accepted, only to discover her prize was fairly valuable.
Sorry I can't remember the specifics, it was decades ago.
In any case, booby prize winners would have had to be given the choice to keep their prize, or a monetary equivalent. Game show laws in the U.S. have been very strict about this since the 1950's quiz-show scandals.
If he showed you the car it wouldn't matter if you changed or not. There's only goats left.
Still, I agree with you. It's very important to understand he doesn't pick a door at random.
No, that only brings about faster rebirth. Haven't you read the Lotus Sutra, or did you skip class that day?
http://en.wikipedia.org/wiki/Lotus_Sutra
"The tradition in Mahyna states that the Lotus Sutra was written down at the time of the Buddha and stored for five hundred years in the realm of the dragons (or Ngas). After this, they were re-introduced into the human realm at the time of the Fourth Buddhist Council in Kashmir. The tradition further claims that the teachings of the Lotus Sutra are higher than the teachings contained in the gamas and the Sutta Pitaka (the Sutra itself also claims this), and that humankind was unable to understand the Lotus Sutra at the time of the Buddha (500 BCE). This is the reason given for the need to store the Lotus Sutra in the realm of the dragons for 500 years, after which humankind was able to understand the Lotus Sutra."
But it doesn't say that the host always opens another door and offers the switch. Those are crucial. If the host only opens and/or offers when you've picked correctly, for example, then you should never switch.
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Thank you. That is the *only* way to correctly calculate the odds. The commonly-used intuition-based argument works, but is misleading. I've had some very frustrating discussions with people who tried to apply the same logic to the "Deal or No Deal" game, where if you get down to just two cases you get the opportunity to switch. Of course, in that game, the host doesn't have any extra information to give you, so there is no statistical advantage to switching, but just TRY to convince people who've had the Monty Hall problem pounded into them, thinking they understand it, but really having no clue.
Walking through the calculations properly will always give you the right result, regardless of whether you can come up with a nice intuitive explanation.
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I'm sorry but I still think that the accepted answer to the Monty Hall problem is just flat out wrong. The reason why is that because as the great Mathematicians of Rush once said in their song "Freewill", "If you choose not to decide you still have made a choice."
Essentially thus-- irregardless of the initial odds, your choice after the first door has been opened is not to keep your original choice or not, but a choice between one of two doors (bringing the odds to 50%, rather then 33.3%). "Staying" with your original door is as much a choice as "switching", so you are still in the effect of the new, 50/50 odds.
And thus, once again, Rush has saved the world from faulty math.
(Note I use fix() not int() with the random numbers, as fix() doesn't do any decimal rounding which would distort the results of the rnd() function which is supposed to output 0 <= rnd() < 1.)
My point was that you make two choices with the first one being essentially useless since the host always opens one of the doors that has no prize behind it. And then you get to choose between two doors, one with a prize and one without a prize. Thus, a 1/2 chance of winning regardless if you once again pick the door you picked first or choose the other door.
/Mikael
Greylisting is to SMTP as NAT is to IPv4
But we know the host always opens a door with nothing behind it, which leaves two doors and a second choice between two doors and we know there's a prize behind one of these doors. Basically, the first choice you make is useless, you might as well just have two doors with a prize behind one of them.
/Mikael
Greylisting is to SMTP as NAT is to IPv4
i\hbar\dot{\psi}=\hat{H}\psi
No, as there is a crucial extra piece of information. After our first selection, we know that the chance of the prize being behind the other two doors is 2/3rds, and after the host opens an empty one, we /know that it cannot be one of these/.
/do not know which one/. That's the crucial extra bit of information.
Look at it like this. Let P(1) denote the chance of it being behind door 1. P(2v3) is the chance of it being behind door 2 or 3.
Presume we select door 1 to start.
P(1)=1/3 (3 doors, pick one)
P(2v3)=2/3 (if not behind 1, must be behind one of others.
P(1v2v3)=1 (must be behind one door)
Note that the chance of it being behind door 2 or 3 is 2/3.
Now, presume that the host shows us it is not behind door 2. Our original selection cannot suddenly get more likely, as we selected from three doors.
P(1)=1/3
P(2)=0 (not behind door 2)
But wait! P(2v3)=2/3. Therefore,
P(3)=2/3
Basically, the first choice does matter. Obviously if we were to select from a hundred doors, our chance of being right at first is 1/100. This will not suddenly rise if 98 others are eliminated, so the chance of it being behind the unselected door becomes 99/100.
Yes, you always know that one of the remaining doors will be eliminated, but until it is, you
"Lisp
But the rules of the game are pretty simple, no matter what your first choice is you always end up with a choice between two doors, one with a prize and one without a prize. Thus no matter what your first choice you will always have a 50% chance of winning with your second choice.
Basically, the first choice and the removal of one of the doors is just for dramatic effect, your choice is really always between two doors with an equal chance of winning.
/Mikael
Greylisting is to SMTP as NAT is to IPv4
Not sure anyone will see this, but after mapping out an algorithm to test the theory, I realized the error of my ways. The first choice is indeed 1/3 (each door has a value of 1/3). The second choice has two doors left with assymetrical odds; the initially selected door remains 1/3 and the other remaining door is indeed 2/3 (as it is basically a collapsed choice identical to choosing two of the three original doors). If anyone still doesn't believe, I would recommend the same learning process. T.
No it isn't. The first choice and removal gives you a better chance of winning.
If you're right first time, which is a chance of 1/3, then changing for your second choice will make you lose. That's 1/3rd chance of losing if you switch.
If you're wrong first time, which has a chance of 2/3, then changing for your second choice will make you win. That's a 2/3rd chance of winning if you switch.
The reverse applies if you stay with your original choice.
The important thing is that the first selection was picked from 3 doors, and restricts the possibilities for the next door to be opened.
"Lisp
Whoaah havent run the code yet (netsplit if you catch my drift) but hey:
http://xkcd.com/386/
This time around its me being wrong! Statisticly speaking.
Hivemind harvest in progress..