# Psychologists Don't Know Math

#### Zonk posted more than 6 years ago | from the one-plus-one-equals-your-mother dept.

566
stupefaction writes *"The New York Times reports that an economist has exposed a mathematical fallacy at the heart of the experimental backing for the psychological theory of cognitive dissonance. The mistake is the same one that mathematicians both amateur and professional have made over the Monty Hall problem. From the article: "Like Monty Hall's choice of which door to open to reveal a goat, the monkey's choice of red over blue discloses information that changes the odds." The reporter John Tierney invites readers to comment on the goats-and-car paradox as well as on three other probabilistic brain-teasers."*

## Nice try! (5, Funny)

## geekoid (135745) | more than 6 years ago | (#23029966)

## Re:Nice try! (5, Funny)

## Ilan Volow (539597) | more than 6 years ago | (#23030092)

## Re:Nice try! (4, Funny)

## mrbluze (1034940) | more than 6 years ago | (#23030110)

goats-x## Re:Nice try! (1)

## geekoid (135745) | more than 6 years ago | (#23030162)

## They don't know math? (2, Insightful)

## hkgroove (791170) | more than 6 years ago | (#23029978)

## The Monty Hall Problem (0)

## Anonymous Coward | more than 6 years ago | (#23029984)

"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*

## Re:The Monty Hall Problem (1)

## beckerist (985855) | more than 6 years ago | (#23030086)

## How much is gas this week? (0)

## StefanJ (88986) | more than 6 years ago | (#23029986)

## Hmmm.... (5, Insightful)

## Otter (3800) | more than 6 years ago | (#23029998)

2) The issue seems easy enough to settle empirically, given a few monkeys and a bag of M&Ms, besides the fact that it seems to have been empirically settled decades ago anyway.

3) This is, though, a good opportunity to ridicule "21" for completely botching the Monty Hall problem, along with pretty much everything else relating to math, gambling and Boston-area geography.

## Re: movie 21 / Monty Hall stuff (0)

## Anonymous Coward | more than 6 years ago | (#23030394)

## Re:Hmmm.... (0, Redundant)

## B3ryllium (571199) | more than 6 years ago | (#23030444)

## Re:Hmmm.... (5, Funny)

## Gat0r30y (957941) | more than 6 years ago | (#23030460)

## Re:Hmmm.... (1)

## TheRaven64 (641858) | more than 6 years ago | (#23030502)

[1] I know it's bad form, but I don't do it very often...

## Re:Hmmm.... (0)

## Anonymous Coward | more than 6 years ago | (#23030518)

But you can write it off as a ploy by filmmakers to try to prop up the main character as some genius capable of counting cards even though the +/- counting schemes are simple enough for pretty much anyone to master if they put in the effort. Really, two other scenes are much more indicative of high intelligence than that one...the scene where they have him calculate the price of a customers entire purchase, complete with differing discounts per item and tax, and the scene where he figures out how to exchange $250k worth of chips without arousing suspicion. And even those two scenes are more faux-intelligence than actual intelligence.

## Re:Hmmm.... (1)

## coolhaus (186994) | more than 6 years ago | (#23030548)

Seriously. It's a bit like saying that paleontologists don't know chemistry. Not that a good grounding in formal chemistry would hurt a paleontologist, but all they need is a basic grasp of it.

## Seems to make sense (4, Interesting)

## 26199 (577806) | more than 6 years ago | (#23030026)

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 :)

## Re:Seems to make sense (1)

## b4dc0d3r (1268512) | more than 6 years ago | (#23030482)

## I dislike things that "seem". (4, Interesting)

## jd (1658) | more than 6 years ago | (#23030492)

(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.

## Re:I dislike things that "seem". (1, Insightful)

## Anonymous Coward | more than 6 years ago | (#23030594)

farm animalsfor radioactive contamination a mere nine years after the first MRI scan of a human body and only three years after the first commercial MRI installation in Europe. If you have sources to the contrary I'd be really curious to see them.## Why should we be surprised? (1)

## actionbastard (1206160) | more than 6 years ago | (#23030028)

## To be fair, mathemeticians didn't know math either (5, Interesting)

## ZombieRoboNinja (905329) | more than 6 years ago | (#23030032)

## Re:To be fair, mathemeticians didn't know math eit (1)

## feijai (898706) | more than 6 years ago | (#23030056)

## Re:To be fair, mathemeticians didn't know math eit (0)

## Anonymous Coward | more than 6 years ago | (#23030388)

## Re:To be fair, mathemeticians didn't know math eit (1)

## Trespass (225077) | more than 6 years ago | (#23030512)

## Re:To be fair, mathemeticians didn't know math eit (0)

## Anonymous Coward | more than 6 years ago | (#23030064)

## Don't worry. (3, Funny)

## ZombieRoboNinja (905329) | more than 6 years ago | (#23030140)

http://en.wikipedia.org/wiki/Monty_hall_problem [wikipedia.org]

## Re:To be fair, mathemeticians didn't know math eit (5, Insightful)

## wurp (51446) | more than 6 years ago | (#23030136)

She does seem to be brilliant, but everyone makes mistakes, and calling them on them will educate them if they were wrong, and educate you otherwise.

## Re:To be fair, mathemeticians didn't know math eit (1)

## 1729 (581437) | more than 6 years ago | (#23030164)

## If she's so smart... (1)

## spiffmastercow (1001386) | more than 6 years ago | (#23030494)

** 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?

## Re:To be fair, mathemeticians didn't know math eit (4, Informative)

## melikamp (631205) | more than 6 years ago | (#23030522)

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.

## Re:To be fair, mathemeticians didn't know math eit (1)

## Planesdragon (210349) | more than 6 years ago | (#23030572)

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.

## We're being played (4, Informative)

## Naughty Bob (1004174) | more than 6 years ago | (#23030038)

In truth, the 1956 experiment

mayhave had flaws (though Chen's paper doesn't prove this), but many subsequent ones have upheld the original findings, and are not subject to the alleged problems.## Re:We're being played (4, Interesting)

## yuna49 (905461) | more than 6 years ago | (#23030294)

## Dude (3, Funny)

## bogie (31020) | more than 6 years ago | (#23030044)

## Re:Dude (3, Funny)

## u8i9o0 (1057154) | more than 6 years ago | (#23030438)

The "Monty Hall problem" link in the summary informed me that I need Flash to understand the problem.

However, on that page they then offer "Need to know more? 50% off home delivery of The Times."

This confuses me terribly - if I now pick the home delivery choice, does the probability of learning about the Monty Hall problem go down 50%?

Damn - I should have picked the Flash answer from the start.

## What? (1)

## ArchieBunker (132337) | more than 6 years ago | (#23030060)

## Ummm, I don't get it. (2)

## Ralph Spoilsport (673134) | more than 6 years ago | (#23030062)

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

## Re:Ummm, I don't get it. (1, Informative)

## bunratty (545641) | more than 6 years ago | (#23030098)

## Re:Ummm, I don't get it. (4, Insightful)

## wurp (51446) | more than 6 years ago | (#23030208)

When you choose one door out of three, and one of those three was pre-chosen randomly to be "the winner", your chance of having picked the right door is 1/3. At least one of the other two doors is not the winner, so the fact that Monty can show you that one is not the winner doesn't change your chance of having chosen the winner.

HOWEVER, now your chance is the same (1/3), but the chance of either the door you chose or the remaining door closed door being the winner is 100%. Therefore the chance that the remaining door is the winner is 2/3. Switch doors to double your chances.

I have a BS in math (not statistically oriented, but I had the normal discrete math sequence) and I still had to think about this a lot before I switched answers from the wrong one to the right one

## Re:Ummm, I don't get it. (1)

## bunratty (545641) | more than 6 years ago | (#23030254)

## Re:Ummm, I don't get it. (1)

## wurp (51446) | more than 6 years ago | (#23030336)

## Re:Ummm, I don't get it. (5, Informative)

## Jeremy Erwin (2054) | more than 6 years ago | (#23030146)

Suppose the car is behind door number one.

If you pick door number one, then Monty has a choice of picking door number two, or three. If you switch, you lose.

If you pick door number two, then Monty must open door number three. If you switch, you win.

If you pick door number three, then Monty must open door number two. If you switch, you win.

Monty's choice of which door to open is constrained in two out of three choices. Pick the door he didn't open, and you'll win two out of three times.

But the problem assumes that Monty has to offer you that choice. On the game show, he didn't.

## Re:Ummm, I don't get it. (1)

## sysrammer (446839) | more than 6 years ago | (#23030566)

## Re:Ummm, I don't get it. (0)

## Anonymous Coward | more than 6 years ago | (#23030148)

## Re:Ummm, I don't get it. (1)

## RedWizzard (192002) | more than 6 years ago | (#23030516)

## Re:Ummm, I don't get it. (2, Informative)

## zulater (635326) | more than 6 years ago | (#23030176)

Your original odds were 1/3. Monty has a 2/3 chance of having the right one. Monty's odds of having the right one is greater than your odds of having the right one so statistically you should switch.

Look at it by way of cards (in the article).

You need to pick the ace of hearts. Monty will then go through the deck and pick the ace of hearts or a random card. He will then show you the other 50 "goats" and ask if you want to trade. You have a 1/52 chance of picking it. Monty then has a 51/52 chance of picking it. Obviously, statistically speaking you should switch.

## Re:Ummm, I don't get it. (3, Informative)

## rmcd (53236) | more than 6 years ago | (#23030182)

If 1 has the car, he can pick either door. If you switch, you lose. Prob 1/3

If 2 has the car, Monty *has* to open 3. If you switch, you get the car, Prob 1/3

If 3 has the car, Monty *has* to open 2. If you switch, you get the car, Prob 1/3

Thus, there's a 2/3 chance of getting the car when you switch.

The other way to think about this is that Monty is revealing no information about *your* door when he opens one of the other two. Thus, the probability that your door has the car must be 1/3 both before and after Monty opens one of the other doors. Since there's only one closed door left, the car is behind it with prob = 2/3.

## Re:Ummm, I don't get it. (1)

## stevelinton (4044) | more than 6 years ago | (#23030194)

## Re:Ummm, I don't get it. (1)

## SeekerDarksteel (896422) | more than 6 years ago | (#23030200)

You have a 1/3rd probability of choosing the car initially and a 2/3rds probability of choosing the goat. If you do not switch, you have a 1/3rds probability of having the car. After one of the other doors has been revealed to be a goat, however, the following is true: If you originally picked a car, you will get a goat. If you originally picked a goat you will get a car. Since you had a 1/3rd chance of originally picking the car and a 2/3rds chance of originally picking a goat, if you switch you end up with a 1/3rds chance of getting a goat and a 2/3rds chance of getting a car.

If you still don't understand, do this: Write out every possible combination of two goats and one car. CGG, GCG, and GGC. Now say you pick the first door. Elimiate one of the goats other than door 1. You are left with CG, GC, and GC. Notice that two of the three possible combinations have the car behind door 2, not door 1.

## Re:Ummm, I don't get it. (1)

## Todd Knarr (15451) | more than 6 years ago | (#23030212)

Because your initial probability of picking the car isn't 50/50, it's 2:1 against the car. You choose from 3 doors, remember, not 2. So initially the probability is 1/3rd that you've chosen the car, 2/3rds that the car is behind one of the doors you haven't chosen. Then Monty opens one of the doors you haven't chosen. He's constrained to open a door with a goat behind it, but the fact that he's opened a door doesn't change the initial probabilities. So the probabilities remain 1/3rd that you've chosen the car, 2/3rds that the car's behind one of the other doors. Monty's helpfully told you which door the car isn't behind, so the 2/3rds-chance door must be the one neither you nor Monty has chosen.

## Re:Ummm, I don't get it. (1)

## WAG24601G (719991) | more than 6 years ago | (#23030214)

Imagine there are 100 doors. You still pick only one (say, door 15). You have a 1/100 chance of being correct. Monty Hall throws open every door except Door 15 (your choice) and Door 72, and all open doors have goats. Will you stick with your choice or switch? This example makes it a bit more intuitive to see that while you still only had a 1/100 chance of being right, you had a 99/100 chance of being wrong, *and* (most importantly) if you are wrong then the car must be behind Door 72. Therefore Door 72 now has a 99/100 chance of being the car, while Door 15 only has 1/100.

Try that whole exercise again now, but with just three doors. You find that while your choice has only 1/3 probability of being correct, the remaining door will have 2/3 probability.

## Re:Ummm, I don't get it. (2, Insightful)

## Nos. (179609) | more than 6 years ago | (#23030222)

## Re:Ummm, I don't get it. (1)

## spun (1352) | more than 6 years ago | (#23030232)

It has nothing to do with cognitive dissonance. The cognitive dissonance experiment has been show to contain a similar type of error, that is all. I don't think you really read the article.

## Re:Ummm, I don't get it. (2, Insightful)

## Sorcha Payne (1047874) | more than 6 years ago | (#23030234)

## Re:Ummm, I don't get it. (1)

## EMeta (860558) | more than 6 years ago | (#23030244)

If you pick car first (1/3) and don't switch, you win. If you pick goat first (2/3) and don't switch, you lose.

Better yet, imagine that there was 2,001 doors, one car and 2,000 goats, and then when you picked a door 1,999 other goats were revealed. Now you know almost for certain that you picked a goat, so you switch to the other remaining door. The Monty Hall problem doesn't give you those odds, but they make it a bit clearer for me.

## Re:Ummm, I don't get it. (0)

## Anonymous Coward | more than 6 years ago | (#23030256)

After you pick, I remove 50 cards that are NOT the ace of the spades.

What are the odds you picked the ace of spades versus the odds that the remaining card is the ace of spades?

Now, change the set size from 52 down to 3 and you're back to the original problem.

## Re:Ummm, I don't get it. (1)

## Peter Mork (951443) | more than 6 years ago | (#23030260)

If you initially choose the door with the car, are shown a goat, then swap: you lose!

If you initially choose a door with a goat, are shown a goat, then swap: you win!

Obvious, you say. However, these are the only two possible scenarios. What are the odds that you're in the first scenario? It's not 1/2, it's 1/3. Thus, you should always swap because your odds of winning are greatly increased.

Clearly, showing you a goat doesn't change your initial odds of having chosen the car. But, it does give you more information about the initial scenario. Hence the reason to switch.

## Re:Ummm, I don't get it. (1)

## geekoid (135745) | more than 6 years ago | (#23030292)

this explains it best.

http://en.wikipedia.org/wiki/Monty_Hall_problem [wikipedia.org]

## Re:Ummm, I don't get it. (1)

## thatseattleguy (897282) | more than 6 years ago | (#23030304)

Another way to look at it that works for some skeptics:

Expand the problem to 100 doors, behind which can be found 99 goats and one car. You pick one of those 100 doors at random.

Now Monty opens 98 of the remaining doors, shows you 98 goats, and asks if you want to keep your door or switch to the other remaining closed door the (one you didn't pick).

Are you better off switching? Nearly everyone understands here the answer's "yes". Monty's stacked the deck greatly in your favor - you have a 1/100 chance of your original pick being right and a 99/100 chance of the other door being right.

The same principle works when you reduce the problem to just three doors, even though the odds go down to 2/3 in your favor instead of 99/100.

## Re:Ummm, I don't get it. (0)

## Anonymous Coward | more than 6 years ago | (#23030306)

I agonized over this in much the same way you did until I realized that making a change does give you stronger odds. The way I finally reconciled it was to use the case of N=1000...i.e. there are 1000 doors instead of 3 doors.

The host tells you to pick a door. You pick door 978 (for example). The host then opens up

all the remaining doorsto show you gag prizes except for door number 322. Now, what are the odds that your first guess (door 978) was right...or is the prize more likely behind door 322?Again, there are only two doors at the end even though you started with 1000, but the probability is most definitely

not50/50 between them. The same thing happens in the case of three doors (N=3), so therefore you are better off changing your selection, as you will have a better chance of winning.## Re:Ummm, I don't get it. (1)

## adamofgreyskull (640712) | more than 6 years ago | (#23030364)

Door 1: Goat

Door 2: Goat

Door 3: Car

Scenario 1: You pick door 1 (Goat), Monty reveals the other Goat, behind door two. You switch to the other unopened door, door 3. You get the Car.

Scenario 2: You pick door 2 (Goat), Monty reveals the other Goat, behind door one. You switch to the other unopened door, door 3. You get the Car.

Scenario 3: You pick door 3 (Car), Monty reveals the Goat behind door one. You switch to the other unopened door, door 2. You am teh fail.

It seems counter-intuitive at first, but search your feelings, you know it to be true..

## Re:Ummm, I don't get it. (1)

## jmorris42 (1458) | more than 6 years ago | (#23030524)

Work the problem backwards. On your initial pick there is a one in three chance you picked the car and a two in three you picked a goat. Then Monty opens a door. If your original pick was a goat the remaining door has the car. So there is still a one in three chance your original door has a car and a two in three chance the remaining door is the right one. The key is realizing the odds are still x in three and not 50-50.

.

## Re:Ummm, I don't get it. (1)

## SpinyNorman (33776) | more than 6 years ago | (#23030534)

In other words, there's a 2/3 chance the car is behind one of the other two doors, and this probability doesn't depend on whether you actually open any of the doors.

Now, when Monty opens a door he's revealing additional information! It doesn't change the 2/3 odds of one of those other two doors being the right one, but it does rather limit which one!

So, do you want to stick with your original 1/3 chance of being right, or switch and have a 2/3 chance of being right?

## Re:Ummm, I don't get it. (1)

## jonadab (583620) | more than 6 years ago | (#23030550)

The truth, of course, is that he doesn't care as much about what effect showing the goat behind door number three will have on *you*, as he does about the effect it will have on the audience. What effect will it have on you, the participant? Who can say? From his perspective that's not very predictable, and it's not the point anyhow. But the effect it has on the audience matters very much. He's a showman, building dramatic tension. That's his job. He'd do the same thing whether the car was behind the door you picked, or the other one. So we're back at 50/50.

But not everyone understands this, so if they try to second-guess his motivation, that _can_ have an impact on their choice.

## Re:Ummm, I don't get it. (1)

## the_humeister (922869) | more than 6 years ago | (#23030584)

## Re:Ummm, I don't get it. (1)

## superwiz (655733) | more than 6 years ago | (#23030590)

## Put it into more physical/visual terms (5, Insightful)

## davidpfarrell (562876) | more than 6 years ago | (#23030592)

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.

## scientology? (1, Informative)

## Anonymous Coward | more than 6 years ago | (#23030074)

The protests against scientology is this Saturday in every major city around the world! Sunday for Philadelphia.

## Article title misleading? (3, Insightful)

## Prien715 (251944) | more than 6 years ago | (#23030082)

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?

## Re:Article title misleading? (1)

## Toonol (1057698) | more than 6 years ago | (#23030338)

hasto be true. I guess possessing a mathematics degree is no guarantee that you're still capable of learning. Weird that so many educated people can't bother themselves to do that.Getting off topic, but sometimes I run into people who don't want to learn any more. They want things to be the way they think, because changing the way they think to reflect the way things are is too troublesome. How boring! Those people must be dead inside.

## A the social sciences (0, Flamebait)

## Shadow Wrought (586631) | more than 6 years ago | (#23030094)

Not that I ever bought that a small sample, however truly random, really does prove what the larger whole would do.

## As long as (1)

## sadgoblin (1269500) | more than 6 years ago | (#23030142)

## The problem is a fallacy (-1, Troll)

## daveime (1253762) | more than 6 years ago | (#23030188)

The fallacy lies in stating that before Monty opens the door and shows the goat, your chance of picking the car is 1/3.

It is NOT, because as Monty will always pick a door with a goat behind it, your choices are always going to be two

Thus your chance of picking the door with the car are 1/2, they were 1/2 at the start, and they are STILL 1/2 after Monty opened his door. The odds do not change "in your favour", because they simply do not change AT ALL. Ergo, there is no advantage or disadvantage in changing doors.

## Re:The problem is a fallacy (1)

## Nos. (179609) | more than 6 years ago | (#23030246)

## Re:The problem is a fallacy (2, Insightful)

## 1729 (581437) | more than 6 years ago | (#23030266)

## Re:The problem is a fallacy (1)

## bunratty (545641) | more than 6 years ago | (#23030274)

## Re:The problem is a fallacy (1)

## hidannik (1085061) | more than 6 years ago | (#23030314)

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.

## Re:The problem is a fallacy (1)

## BengalsUF (145009) | more than 6 years ago | (#23030322)

Only after you pick your one of three doors does Monty reveal one of the remaining two doors which contains a goat.

## Re:The problem is a fallacy (5, Insightful)

## 1729 (581437) | more than 6 years ago | (#23030344)

## Re:The problem is a fallacy (4, Insightful)

## ThreeGigs (239452) | more than 6 years ago | (#23030368)

"It is NOT, because as Monty will always pick a door with a goat behind it, your choices are always going to be two"Your argument *only* works if Monty opens a door *before* you pick. *And*, you get to pick *twice*. First time from three doors, second time from two doors.

You pick, from a choice of three, giving Monty a choice of two.

Your argument is based on the reverse, Monty being able to pick from three doors, and you only get two.

Do you see it now? You 'lock' a door, precluding Monty from choosing it.

Remember, since you have first pick, your chances of getting a goat are 2/3. Meaning you most likely picked a goat. Meaning when Monty reveals a goat, the remaining door is most likely a car.

## Re:The problem is a fallacy (1)

## hidannik (1085061) | more than 6 years ago | (#23030400)

http://c2.com/cgi/wiki?MontyHallSimulation [c2.com]

## Re:The problem is a fallacy (0, Troll)

## daveime (1253762) | more than 6 years ago | (#23030452)

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

## Re:The problem is a fallacy (3, Informative)

## thatseattleguy (897282) | more than 6 years ago | (#23030602)

Re:The problem is a fallacySorry, 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.

## Indeed (5, Interesting)

## sustik (90111) | more than 6 years ago | (#23030204)

Some researchers involved in pchycology (social behaviour etc.) came to high schools and drew up the friendship graph of the class. (Maybe school works differently where you live, we had a class of size 30-40 students attending exactly the same lectures.)

They assumed friendship to be mutual (if not, than it was not considered friendship). One clever cookie made the observation that almost always there is a group of 6 students who all friends to each other (a clique), or alternatively a group of 4 students, who do not like each other.

There were excited discussions among the researchers what social forces are the reason that one of the above situations always seemed to occur.

They were somewhat disillusioned when our math teacher explained them Ramsey's theorem. Since R(6, 4) is between 35 and 41, indeed one can expect either a frienship or hateship clique to appear with quite high probability... (This does not mean that properties of the frienship graph worth not examining, but one needs to know the math to do it properly.)

## Pot, Kettle, Black (5, Funny)

## ryu1232 (792127) | more than 6 years ago | (#23030288)

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.

## Real World & Monty Hall Problem (1)

## Kotukunui (410332) | more than 6 years ago | (#23030290)

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.

## Re:Real World & Monty Hall Problem (1)

## robo_mojo (997193) | more than 6 years ago | (#23030448)

Yes, and in that one attempt, you'd be better of switching.

You don't NEED to play the game numerous times to get a conclusion that switching doors is better, you can reason that switching doors is beneficial in any given trial.

There is a 1/3 chance that your original door is the prize, and a 2/3 chance that the other door is the prize. A person using reason will switch doors.

The author only suggested that you play the game in case you didn't understand the reasoning. If you understand the reasoning then you don't need to play the game.

Nice anecdote. Nobody said that you must win, of course in either option there is still the chance to lose.

Then playing game shows isn't for you. A reasonable person might feel unhappy at losing if he switched doors, but he will not have regretted his decision (he WOULD regret losing by keeping his original door, however, because that decision would have been unreasonable).

## You know who can't do math? (5, Insightful)

## geekoid (135745) | more than 6 years ago | (#23030318)

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.

## mod parent up (0)

## Anonymous Coward | more than 6 years ago | (#23030484)

## Could someone please correct the title (1, Flamebait)

## Anonimouse (934959) | more than 6 years ago | (#23030328)

## Pots and kettles (0)

## Anonymous Coward | more than 6 years ago | (#23030362)

## Cognitive Dissonance (5, Funny)

## jdbolick (804666) | more than 6 years ago | (#23030404)

## Sadly, not as wrong as shown (4, Interesting)

## DynaSoar (714234) | more than 6 years ago | (#23030408)

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.

## Way to Go (MORON)! (1)

## redlaceparasol (1241732) | more than 6 years ago | (#23030412)

## Monty Hall in TI-BASIC (1)

## Digi-John (692918) | more than 6 years ago | (#23030414)

## A Simple Explanation of the Monty Hall Problem (3, Informative)

## ThinkFr33ly (902481) | more than 6 years ago | (#23030432)

Anyway, here is the simple explanation that I've found helps people realize their error in thinking:

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.

## Re:A Simple Explanation of the Monty Hall Problem (1)

## daveime (1253762) | more than 6 years ago | (#23030598)

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.

Except that you didn't complete the truth table !!!

If you have already chosen the door with the car behind it, Host can reveal door 2 OR 3, leading to one extra entry in each of your outcome sets.

[car] [goat] [goat]

Choose door 1. Host reveals door 3. Switch to door 2. NO CAR.

Choose door 1. Host reveals door 2. Switch to door 3. 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.

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 1. Host reveals door 2. 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.

Therefore whether you switch or not, you still end up with 50:50 chance of winning.

## Tricky Monty (1)

## Sloppy (14984) | more than 6 years ago | (#23030504)

I love this one; it's

sodevious. If Monty doesn't show you which door he opened, then you gain nothing.## Some slashdoter's don't know math either (1)

## Jeff1946 (944062) | more than 6 years ago | (#23030528)

## I *KNEW* it! (1)

## Trogre (513942) | more than 6 years ago | (#23030542)