And Behind Door No. 1, a Fatal Flaw

[Brownian Motion]

The New York Times

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April 8, 2008

Findings

And Behind Door No. 1, a Fatal Flaw

By JOHN TIERNEY

The Monty Hall Problem has struck again, and this time it’s not merely embarrassing mathematicians. If the calculations of a Yale economist are correct, there’s a sneaky logical fallacy in some of the most famous experiments in psychology.

The economist, M. Keith Chen, has challenged research into cognitive dissonance, including the 1956 experiment that first identified a remarkable ability of people to rationalize their choices. Dr. Chen says that choice rationalization could still turn out to be a real phenomenon, but he maintains that there’s a fatal flaw in the classic 1956 experiment and hundreds of similar ones. He says researchers have fallen for a version of what mathematicians call the Monty Hall Problem, in honor of the host of the old television show, “Let’s Make a Deal.”

Here’s how Monty’s deal works, in the math problem, anyway. (On the real show it was a bit messier.) He shows you three closed doors, with a car behind one and a goat behind each of the others. If you open the one with the car, you win it. You start by picking a door, but before it’s opened Monty will always open another door to reveal a goat. Then he’ll let you open either remaining door.

Suppose you start by picking Door 1, and Monty opens Door 3 to reveal a goat. Now what should you do? Stick with Door 1 or switch to Door 2?

Before I tell you the answer, I have a request. No matter how convinced you are of my idiocy, do not immediately fire off an angry letter. In 1991, when some mathematicians got publicly tripped up by this problem, I investigated it by playing the game with Monty Hall himself at his home in Beverly Hills, but even that evidence wasn’t enough to prevent a deluge of letters demanding a correction.

Before you write, at least try a few rounds of the game, which you can do by playing an online version of the game. Play enough rounds and the best strategy will become clear: You should switch doors.

This answer goes against our intuition that, with two unopened doors left, the odds are 50-50 that the car is behind one of them. But when you stick with Door 1, you’ll win only if your original choice was correct, which happens only 1 in 3 times on average. If you switch, you’ll win whenever your original choice was wrong, which happens 2 out of 3 times.

Now, for anyone still reading instead of playing the Monty Hall game, let me try to explain what this has to do with cognitive dissonance.

For half a century, experimenters have been using what’s called the free-choice paradigm to test our tendency to rationalize decisions. This tendency has been reported hundreds of times and detected even in animals. Last year I wrote a column about an experiment at Yale involving monkeys and M&Ms.

The Yale psychologists first measured monkeys’ preferences by observing how quickly each monkey sought out different colors of M&Ms. After identifying three colors preferred about equally by a monkey — say, red, blue and green — the researchers gave the monkey a choice between two of them.

If the monkey chose, say, red over blue, it was next given a choice between blue and green. Nearly two-thirds of the time it rejected blue in favor of green, which seemed to jibe with the theory of choice rationalization: Once we reject something, we tell ourselves we never liked it anyway (and thereby spare ourselves the painfully dissonant thought that we made the wrong choice).

But Dr. Chen says that the monkey’s distaste for blue can be completely explained with statistics alone. He says the psychologists wrongly assumed that the monkey began by valuing all three colors equally.

Its relative preferences might have been so slight that they were indiscernible during the preliminary phase of the experiment, Dr. Chen says, but there must have been some tiny differences among its tastes for red, blue and green — some hierarchy of preferences.

If so, then the monkey’s choice of red over blue wasn’t arbitrary. 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. If you work out the permutations (see illustration), you find that when a monkey favors red over blue, there’s a two-thirds chance that it also started off with a preference for green over blue — which would explain why the monkeys chose green two-thirds of the time in the Yale experiment, Dr. Chen says.

Does his critique make sense? Some psychologists who have seen his working paper answer with a qualified yes. “I worked out the math myself and was surprised to find that he was absolutely right,” says Daniel Gilbert, a psychologist at Harvard. “He has essentially applied the Monty Hall Problem to an experimental procedure in psychology, and the result is both instructive and counter-intuitive.”

Dr. Gilbert, however, says that he has yet to be persuaded that this same flaw exists in all experiments using the free-choice paradigm, and he remains confident that the overall theory of cognitive dissonance is solid. That view is shared by Laurie R. Santos, one of the Yale psychologists who did the monkey experiment.

“Keith nicely points out an important problem with the baseline that we’ve used in our first study of cognitive dissonance, but it doesn’t apply to several new methods we’ve used that reveal the same level of dissonance in both monkeys and children,” Dr. Santos says. “I doubt that his critique will be all that influential for the field of cognitive dissonance more broadly.”

Dr. Chen remains convinced it’s a broad problem. He acknowledges that other forms of cognitive-dissonance effects have been demonstrated in different kinds of experiments, but he says the hundreds of choice-rationalization experiments since 1956 are flawed.

Even when the experimenters use more elaborate methods of measuring preferences — like asking a subject to rate items on a scale before choosing between two similarly-ranked items — Dr. Chen says the results are still suspect because researchers haven’t recognized that the choice during the experiment changes the odds. (For more of Dr. Chen’s explanation, see TierneyLab.)

“I don’t know that there’s clean evidence that merely being asked to choose between two objects will make you devalue what you didn’t choose,” Dr. Chen says. “I wouldn’t be completely surprised if this effect exists, but I’ve never seen it measured correctly. The whole literature suffers from this basic problem of acting as if Monty’s choice means nothing.”

[John L]

Strange. I am kind of vague on how the monkey M&M and Monty Hall problem are related.

All the critique seems to be saying is that a reason that the same monkeys that prefer red over blue also usually prefer green over blue could be they don't like blue and never liked blue. The orignial experiment presumed that the monkeys were indifferent between red and blue before being forced to make a choice, and that choice then influenced subsequent choices.

But the Monty Hall problem is something a little different.

[AllenLowe]

I think they meant Monty Python -

[Tom Storer]

All the critique seems to be saying is that a reason that the same monkeys that prefer red over blue also usually prefer green over blue could be they don't like blue and never liked blue.

I think the point is that it's more precise than "could be"--in fact, it's "have a 2-in-3 chance to not like blue." If I understand correctly, it goes like this...

Monkeys who prefer red over blue then go on, when given the choice, to prefer green over blue, by about a 2/3 margin.

When you look at the odds, there are six possibilities for order of preference:

Red, blue, green

Red, green, blue

Blue, red, green

Blue, green, red

Green, blue, red

Green, red, blue

If we take only those where red is preferred to blue, we have:

Red, blue, green

Red, green, blue

Green, red, blue

In 2/3 of these cases, green is preferred to blue. So the fact that 2/3 of monkeys who preferred red to blue also preferred green to blue can be explained by simple statistics rather than by choice rationalization. Hence the choice rationalization explanation is less convincing than it would otherwise be.

But you knew that. I think the comparison to Monty Hall is just that in both cases you have to think about the odds.

Edited April 9, 2008 by Tom Storer

[DukeCity]

When you look at the odds, there are six possibilities for order of preference:

Red, blue, green

Red, green, blue

Blue, red, green

Blue, green, red

Green, blue, red

Green, red, blue

If we take only those where red is preferred to blue, we have:

Red, blue, green

Red, green, blue

Green, red, blue

In 2/3 of these cases, green is preferred to blue. So the fact that 2/3 of monkeys who preferred red to blue also preferred green to blue can be explained by simple statistics rather than by choice rationalization. Hence the choice rationalization explanation is less convincing than it would otherwise be.

But you knew that. I think the comparison to Monty Hall is just that in both cases you have to think about the odds.

Excellent explanation! Thanks, Tom.

[JSngry]

Unless Carol Merrill is present, none of this speculation has merit.

[BruceH]

My brain hurts.

[Aggie87]

I don't follow the logic in changing doors.

Initially you have a 1 in 3 chance of picking the door with the car behind it. You pick a door. Then Monty opens one of the other doors to show a goat.

Now it's a new ball game, and what happened before is basically irrelevant. There are two doors left, one of which has a car behind it.

Monty is giving you a choice between the two doors. You can either keep the door you initially picked, or go with the other.

Doesn't that therefore mean there is a 50/50 chance that the car is behind the door you selected - regardless when you picked that door? And there's also a 50/50 chance it's behind the other door.

If that's the case, it should make no difference in odds whether you keep your original door or pick the other one.

Edited April 9, 2008 by Aggie87

[Dan Gould]

I don't follow the logic in changing doors.

Initially you have a 1 in 3 chance of picking the door with the car behind it. You pick a door. Then Monty opens one of the other doors to show a goat.

Now it's a new ball game, and what happened before is basically irrelevant. There are two doors left, one of which has a car behind it.

Monty is giving you a choice between the two doors. You can either keep the door you initially picked, or go with the other.

Doesn't that therefore mean there is a 50/50 chance that the car is behind the door you selected - regardless when you picked that door? And there's also a 50/50 chance it's behind the other door.

If that's the case, it should make no difference in odds whether you keep your original door or pick the other one.

Aggie,

I thought the same way before I followed this link to an online version of the Monty Hall game. Within, there is an explanation for why the odds do change with the new information given. The critical element is that Monty Hall always opens a door with a goat behind it.

Let's say you picked the car initially. The odds are 1/3 that you've picked right. Switching away is a bad idea - always.

But let's say you picked a goat first. Monty is going to reveal the door with the other goat, and that means that switching will mean that you've won the car. And most importantly, that applies whether you've picked goat 1 or goat 2 - the other goat gets revealed, and switching is the winning play.

Basically, 1/3 of the time you'll switch away from the car. The rest of the time you'll switch to the car, and win.

You need to have an NY Times account to access the link, but it explains it better than I did. And you can play the game as many times as you want. I went 20 times clicking on door 1 and then staying - I did win a higher percentage than expected - 67%. But then I clicked on door 1 and always chose "switch doors" - and I won 80%!

It may be counterintuitive, but its accurate.

[catesta]

Unless Carol Merrill is present, none of this speculation has merit.

I'll give merit to the first person that can show me a purple marble or a matchbook from the Excelsior Hotel.

[Dan Gould]

Aggie,

thinking about it some more, I think this is the clearest way to explain:

When you make your choice, the odds are 1/3 that you picked the car, 2/3 that you picked a donkey.

That means that after the second donkey is eliminated, there is a 2/3 chance that by switching, you will pick the car (the same 2/3 chance you were wrong in the first place).

I probably screwed that up, too. Just follow the link, play the game ten or twenty times under each strategy, and read the explanation section. All will become clear.

[Swinging Swede]

Initially you have a 1 in 3 chance of picking the door with the car behind it. You pick a door. Then Monty opens one of the other doors to show a goat.

Now it's a new ball game, and what happened before is basically irrelevant.

The thing is that it isn't a new ball game. You already knew that there would be a goat behind at least one of the two other doors, so Monty opening one and showing a goat doesn't actually reveal any new information and therefore you still have a 1 in 3 chance if you keep the door you first picked.

Here's another way of looking at it. Suppose that you always pick door 1 (in bold). Now there are three possibilities.

door1 door2 door3

car goat goat

goat car goat

goat goat car

Now Monty will open one of the other doors and show a goat (which is crossed over below). We will then have:

door1 door2 door3

car goat goat [Monty could also open door 3; it doesn't matter.]

goat car goat

goat goat car

As you clearly see here, in 2 of 3 cases you will win the car if you switch to the other remaining door, and in 1 case of 3 you will win the car by keeping your originally chosen door.

[Dan Gould]

i found dan goulds explanation very good actually

Well, thanks - I thought Swede's description nailed it much better. His explanation told us that if you always make a bad decision under pressure and wish you could change it - you should.

I think it all boils down to

There is a 2/3 chance of being wrong initially.

But in those 2/3rds of the time, changing after the first goat is revealed means you win.

There's a 1/3 chance of being right initially.

Not changing means those odds stay the same.

[Aggie87]

I understand what both of you are saying, and appreciate the descriptions.

But if you KNOW up front that Monty is going to open a door with a goat behind it, then you know that one (goat) door will be eliminated no matter what you pick.

Which means that you are really picking one of two doors in the first place, right? Because you KNOW that your choice is going to come down to the door you pick and one other - not two others.

And if you are picking one of two, don't you have a 50/50 chance from the outset?

Hmmm.

[Dan Gould]

I understand what both of you are saying, and appreciate the descriptions.

But if you KNOW up front that Monty is going to open a door with a goat behind it, then you know that one (goat) door will be eliminated no matter what you pick.

Which means that you are really picking one of two doors in the first place, right? Because you KNOW that your choice is going to come down to the door you pick and one other - not two others.

And if you are picking one of two, don't you have a 50/50 chance from the outset?

Hmmm.

I think you've missed a key issue: the question asked what is the best strategy if you are given the option to change doors. The Times, Swede, and myself have given you different-yet-similar explanations of why your odds improve by switching.

On the other hand, it seems as though you are simply looking at it as a straightforward binary set of outcomes - goat or car. In a sense that is absolutely correct - in a given situation, there is one car and one goat and you've chosen one or the other. It is true that when you chose the door, your odds were 1 in 3, and when Monty eliminated one door, your odds went up. The question is whether your odds improve by switching at that point, and I think its pretty clear from each of our different modes of explanation that switching doors will improve your chances. Did you run the simulation under the two different strategies? If nothing else, that should have convinced you.

[Swinging Swede]

Picking one of two is not the same as having a 50/50 chance, if you have additional information about the two choices.

Suppose that there were 100 doors with 99 goats and 1 car behind them. You pick one door randomly. You should now have a 1 % chance of having picked the door with the car behind it. In other words it's very unlikely that you have picked the right one. Then Monty opens 98 of the other 99 doors and reveals 98 goats behind them. Once again you are left with two doors. Would you still think that you have a 50 % chance if you keep the initially chosen door?

[Aggie87]

I haven't run the simulation, Dan - can't access it from this computer. I also understand the arguments that have been shared here, and they make sense.

I'm still saying that Monty's opening of a door with a goat behind it leaves 2 closed doors, one with a car behind it. It seems immaterial which door you chose first, KNOWING that you will again have to choose one of the two doors that are left. The data at that point in time is that you have two closed doors, and a car behind one of them - and that's it. That's what you have to go by AT THAT POINT IN TIME.

I'm not arguing that you're wrong. I just am not convinced by the logic, whether it's true or not.

Picking one of two is not the same as having a 50/50 chance, if you have additional information about the two choices.

Suppose that there were 100 doors with 99 goats and 1 car behind them. You pick one door randomly. You should now have a 1 % chance of having picked the door with the car behind it. In other words it's very unlikely that you have picked the right one. Then Monty opens 98 of the other 99 doors and reveals 98 goats behind them. Once again you are left with two doors. Would you still think that you have a 50 % chance if you keep the initially chosen door?

Well, if the result is that there are two remaining closed doors, one of which you've picked and one you didn't, there's one with a car behind it and one with a goat. At that point in time there's a 50/50 chance the car is behind your door, is there not?

[rostasi]

Erik is a Zen Master who lives in the moment.

[rostasi]

Wikipedia has everything

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

[rostasi]

Next, we will have an explanation of Brownian Motion

by the originator of this thread -

[Aggie87]

Wikipedia has everything

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

Thanks for posting that link, Rod! LOL

I think this quote is where I'm tripping up:

The reasoning above applies to all players on average without regard to which specific door the host opens and all individual players at the start of the game, but not to a specific player at the point the player is asked whether to switch given which door the host has opened

I keep putting myself into that point of time where the player is asked whether to switch, which isn't the correct logic to use, I guess.

But try this: Initially I pick Door A. Monty opens Door C with a goat behind it, leaving Door A or Door B. He lets me choose which one I want. I should switch to Door B, because it gives me a 67% chance of winning.

Another person is brought into the studio at that point (without any prior knowledge of this game), and given a choice of Door A or Door B as well. Are his odds different than mine? Why?

[Swinging Swede]

Picking one of two is not the same as having a 50/50 chance, if you have additional information about the two choices.

Suppose that there were 100 doors with 99 goats and 1 car behind them. You pick one door randomly. You should now have a 1 % chance of having picked the door with the car behind it. In other words it's very unlikely that you have picked the right one. Then Monty opens 98 of the other 99 doors and reveals 98 goats behind them. Once again you are left with two doors. Would you still think that you have a 50 % chance if you keep the initially chosen door?

Well, if the result is that there are two remaining closed doors, one of which you've picked and one you didn't, there's one with a car behind it and one with a goat. At that point in time there's a 50/50 chance the car is behind your door, is there not?

No, in this example there is a 1 % chance that the car is behind your door, and a 99 % chance that it is behind the other door.

Just one question at this point: Are you aware that Monty isn't randomly opening another door, but is purposely avoiding opening the door with the car behind it? In other words, if the car is behind 1 of the other 99 doors (which it has a 99 % chance of being), Monty will purposely avoid the door with the car behind it and open the other 98 with goats behind them. If you are aware of that, and still think there is a 50/50 chance in the end, then I/we have to think hard about some other way of convincing you.

[zen archer]

I just played the game and i picked door #2 ....then they showed me the goat behind door #3.

i stuck with door #2 and WON the Car , do i have to pay the sales tax on it ?

[rockefeller center]

Because I've too much time on my hands I wrote a quick and dirty Monty Hall simulation (Linux and Mac users need Mono, Windows users need to have .NET installed to run it) for those who only believe what they can see . See attachment.

Here's the output for 10000000 iterations:

10000000 iterations

Wins without changing doors: 3334363, 1/2.99907358616923

Wins with changing doors: 6665637, 1/1.50023171078773

MontyHall.zip

Edited April 9, 2008 by rockefeller center

[Swinging Swede]

But try this: Initially I pick Door A. Monty opens Door C with a goat behind it, leaving Door A or Door B. He lets me choose which one I want. I should switch to Door B, because it gives me a 67% chance of winning.

Another person is brought into the studio at that point (without any prior knowledge of this game), and given a choice of Door A or Door B as well. Are his odds different than mine? Why?

The chances are always evaluated in relation to prior knowledge. If Monty himself were to step in at this point, he would have a 100 % chance of winning the car if he chooses the door he already knows that the car is behind, and a 0 % chance if he chooses the other. You would have a 67/33 chance because of your prior knowledge. The third person who has no prior knowledge and just knows that there are two doors with a car behind one would correctly evaluate his chances as 50/50. In the other example with 100 initial doors, you would have a 99 % chance if you switched, but the third guy would once again have a 50/50 chance.