Tierney on Keith Chen, Monty Hall, and Psychology Experiments

John Tierney hits a home run with this fantastic column about a recent paper by Keith Chen (whose work on capuchin monkeys has previously caught our attention).

The Monty Hall problem is as follows:

You are chosen to compete on Let’s Make a Deal. There are three curtains. Behind one of the curtains is something wonderful like a new car. Behind the other two curtains is something worthless (at least to an American) like a goat.

You choose curtain #1. Then Monty Hall pulls open curtain #2 revealing something terrible behind it. That leaves curtain #1 and curtain #3. One has a car behind it, the other has the goat. The question: should you stick with curtain #1 or switch to curtain #3? To find out the answer, read Tierney’s column.

Keith Chen’s paper is not directly about Let’s Make a Deal, but rather a subtle application of the same logic to a common method of measuring how choices are made in psychology lab experiments. I have to confess that I had to read Chen’s paper a few times before I got what he was talking about, just as it took me a long time to understand the Monty Hall problem the first time I heard it.

Congratulations to Keith on writing an extremely insightful piece, even if it will probably never be published in a peer reviewed journal because of the inherent biases in the process!

The Dread Pirate Robert

Post #38
An n (nearly but not infinit) amount of time can pass, as long as no re-shuffling is allowed


I understand the solution, but my friends who propose the 50/50 answer are saying that after Monty opens door 3 to reveal the goat that it is a brand new problem. This brings up the question:

How much time must transpire before one probability problem becomes another?

Jason W

Just tried doing the little game that came with the article 100 times with switching every time, just to see how it would turn out. Got 67 out of a 100. Love counterintuitive things like that.


#12 had it right but did not go far enough.. Assume 1,000,000 doors. You pick one and Monte eliminates 999,998 doors. Anybody think you should stay with your original pick? The 3 door example confuses people since you get down to 2 doors and folks think 50-50


#1 Adam - Thank you! With your simple example, it FINALLY makes sense to me.


This is for Adam, the fellow who had the bleg request about doing his masters in the US. I don't know anything about doing your masters in a particular field, but I know that when people do a masters in order to have a better chance at getting into an economics PhD program, they usually do that in the UK or in Canada, with Canada usually being the cheaper choice. Popular programs for this include Toronto, UBC, and Queen's.


That website is VERY helpful.


If Monty asked you in the beginning of the scenario would you rather pick two doors, or just pick one door. Everyone would choose to pick two doors.

That is equal to the Monty situation described above and is the easiest way I've found to explain it to others.


Tierny links to another article that has the most concise, digestible explanation I have seen:

"Mr. Hall said he realized the contestants were wrong, because the odds on Door 1 were still only 1 in 3 even after he opened another door. Since the only other place the car could be was behind Door 2, the odds on that door must now be 2 in 3."


I think that one of the appeals of the wrong answer to Monty hall is that you feel as if you have worked out the trick of the problem (and you know there has to be one) by treating the two as independent. Usually when people make reasoning errors regarding probability it is because they fail to treat independent events as independent. The wrong answer is more appealing because you feel smart when you are giving it (i.e. smarter than people who think that lottery numbers that haven't come up recently are more likely to come up next week).

Regarding Adam's bleg- I am an "international" student doing my PhD in the US, but I am not from the UK and it is not in economics so this info might not be the most specifically helpful. One thing I think it will help to know is that often fees and tuition (make sure you keep track of that distinction that isn't always made in other countries too)- are often purely notional amounts, that are not paid by any students in the program, so it all comes down to how the specific programs you are looking at fund their students, and how many of them they fund. In the US professors/supervisors are much more involved in getting funding (i.e. money for tuition and/or living expenses) than they are in countries with the UK system. Hope this helps! good luck! (PS- I suspect that familiarity with British English is an advantage on the GRE verbal- hope it works out that way for you)



My first instinct was that this had to be wrong, but after I read it a couple time like you suggested, it indeed made sense. Amazing how intuition can steer not only us layman wrong, but trained researchers.


I think this question was in the Times a while ago.

My probability professor gave this to us as homework. We all knew that it wasn't the apparently 50/50 answer, but we all had a hard time trying to figure out how to prove that you should switch. The best way is to use the method illustrated by Adam in #1 (I made it even more ridiculous by saying that there were 1 million curtains)

It's kinda funny how after explaining it to people, they still vehemently defend their claim that it doesn't matter because now its 50/50.


Kit (32), saying that {BB} is a different permutation to {BB} is essentially saying that is is possible to have 'the older brother' born first, the same goes for {GG}. Families can have two kids in just four ways, {a girl, then a boy}, {a boy, then a girl}, {a boy followed by another boy} or {a girl followed by another girl}. All four of these outcome are equally likely, if we know one child is a boy, then there is only a 1 in 3 chance the other child is a boy. I think where you are having trouble is that a family who already have a boy, have a 50-50 chance with their NEXT baby.

The Monty Hall problem just needs someone to actually get an empirical result... then the doubters would fade away... anyone seen it done anywhere?



I have a bit of an issue with that answer, because it implies that order matters. So wouldn't the full set of possible children be more like {GG}, {GG}, {BB}, {BB}, {BG}, {GB} to allow for the permutations of the order of the children, and hence make the probabilities 50/50?


If you haven't seen the film 21 already, this is presented at the start of the film as the "Game show host dilemma", or something along those lines. I remember learning it in Quantitative Methods last year, is great fun.

I have a bleg request also:

I am looking at doing my Masters in the US.
1) Are there any students who read this who are doing their Masters in the US, however did their BA/ BSc in the UK?
2) Is there a large difference in cost? LSE is apparently ?14k/$28k ish.

My MA would be in Economics, most probably Behavioural Economics.

Thanks in advance for any input,



The Monty Hall Problem is fun and interesting (see comments above), but I really question the usefulness of Dr. Chen's discovery. He critiques past psych experiments in which subjects choose between equally attractive choices (e.g., different colored M&Ms), arguing that the subjects may have had a logical reason for choosing one option over another. But that reason materializes only if the subject had an undetected preference for one option. But lack of preference is one of the fundamental *premises* of the experiments, a premise that Dr. Chen appears to just assume away. His critique boils down to saying that *if* a subject secretly preferred green M&Ms over blue ones, the subject would more likely chose green M&Ms over blue ones. Well, duh.

Phil Steinmeyer

The linked-to column repeats the often made mistake that the Monty Hall problem has a simple, clear solution - that switching produces a 2 in 3 chance of winning and therefore, you should always switch.

In fact, the correct solution depends on how the problem is worded. Usually, the problem is not worded clearly enough to resolve ambiguities, and the correct answer depends on what guesses the subject makes about these ambiguities.

The key issues is if the switch offer will ALWAYS be presented.


1) Switch offer ALWAYS presented: In this case, switching gives you a 2 in 3 chance of coming out ahead.

2) Switch offer only presented when the contestant has already chosen the winning prize: In this case, switching is a sure loser.

3) Switch offer only presented when the contestant has already chosen a losing curtain: In this case, switching is a sure winner.

In a game show situation, to heighten dramatic tension, it's not unreasonable to think that case 1 is likely. If the contestant has viewed previous episodes of the game show, they can perhaps make a more informed decision.

But in the real world, the analogy that comes to mind for me is a customer and a salesperson. The customer indicates a preference to buy item A. The salesperson tries to persuade the customer to buy item B. It's not an exact analogy, as there may not be an item C, but in this situation, I think the customer is wise to question the salesperson's motives, and quite possibly, to stick with item A.



ok I've just deleted a huge post to say


(my God that thing really does know everything...)

nice diagrams help ALOT, as well as knowing the entire problem - I didnt realise a goat must be shown.


In regards to the Monty Hall problem, I'm still amazed at how many people are stuck on the idea that the odds should be 50/50. I've talked with some of my friends about this and the easiest way I've found to explain the reasoning is to just apply the same logic but to a larger data set, such as a deck of cards.

If you pick the ace of spades and I tell you that these 50 cards are wrong (leaving only the ace of spades, and the five of clubs, for example), the other person quickly sees that the odds are not 50/50 and that the probability of them being correct (if they do not switch) is only 1/52.


I was confused about this for awhile too. What convinced me thinking about how the other choices are being eliminated.
To expand on Adam's helpful post above, suppose we're looking for the ace of spades. So if we start with a deck of 52 cards and set one aside thinking it's the ace of spades, we're left with 51 cards. After eliminating 50 that we know aren't the ace of spades, we're left with two cards: one that we just randomly set aside at the start, and another that has survived this process of elimination out of a pool of 51.
There was a 1/52 chance that we set aside the right card from the start.
In the remaining 51/52 scenarios, there will be a 100% chance that the card that wasn't eliminated is the correct one.


@ #27 Josh

That made me laugh.