Imagine that you’re in a game show and your host shows you three doors. Behind one of them is a shiny car and behind the others are far less lustrous goats. You pick one of the doors and get whatever lies within. After making your choice, your host opens one of the other two doors, which inevitably reveals a goat. He then asks you if you want to stick with your original pick, or swap to the other remaining door. What do you do?
Most people think that it doesn’t make a difference and they tend to stick with their first pick. With two doors left, you should have even odds of selecting the one with the car. If you agree with this reasoning, then you have just fallen foul of one of the most infamous of mathematical problems – the Monty Hall Dilemma. In reality, you should actually swap every time – doing so means double the odds of getting the car. I will explain why shortly but if you’re currently confused, you are not alone. Over the years, the problem has ensnared countless people, including professional mathematicians. But not, it seems, pigeons.
Walter Hebranson and Julia Schroder showed that, after some training, the humble pigeon can learn the best tactic for the Monty Hall Problem, switching from their initial choice almost every time. Amazingly, humans who get similar extensive practice never develop the optimal strategies that the pigeons pick up. This doesn’t mean that pigeons are “smarter than humans” as some news stories have claimed, but it does mean that the two species approach probability problems in different ways. We suffer because we overthink the problem.
The Monty Hall Dilemma takes its name from Monty Hall, the presenter of a show called Let’s Make a Deal, which involved similar choices. The dilemma became truly legendary when it featured in a column called Ask Marilyn in Parade magazine. When columnist Marilyn vos Savant, the then holder of the Guinness World Record for highest IQ, wrote the correct solution, she was inundated with complaints.
Around 10,000 readers disagreed with her, and wrote in to say as much (these were the days when trolls had to actually pay for postage; read the last letter in particular). Many of them had PhDs and many were mathematicians. Even Paul Erdos, the most prolific mathematician in history, refused to believe this explanation until computer simulations proved beyond all doubt that always switching was the best strategy.
The problem is that most people assume that with two doors left, the odds of a car lying behind each one are 50/50. But that’s not the case – the actions of the host beforehand have shifted the odds, and engineered it so that the chosen door is half as likely to hide the car.
At the very start, the contestant has a one in three chance of picking the right door. If that’s the case, they should stick. They also have a two in three chance of picking a goat door. In these situations, the host, not wanting to reveal the car, will always pick the other goat door. The final door hides the car, so the contestant should swap. This means that there are two trials when the contestant should swap for every one trial when they should stick. The best strategy is to always swap – that way they have a two in three chance of driving off, happy and goatless.
All over the world, people are spectacularly bad at this. We almost always stick or, at best, show indifference. Hebranson and Schroder wanted to see if other species would be similarly vexed. They worked with six Silver King pigeons and altered the game show format to suit their beaks.
Each pigeon was faced with three lit keys, one of which could be pecked for food. At the first peck, all three keys switched off and after a second, two came back on including the bird’s first choice. The computer, playing the part of Monty Hall, had selected one of the unpecked keys to deactivate. If the pigeon pecked the right key of the remaining two, it earned some grain. On the first day of testing, the pigeons switched on just a third of the trials. But after a month, all six birds switched almost every time, earning virtually the maximum grainy reward.
Every tasty reward would reinforce the pigeon’s behaviour, so if it got a meal twice as often when it switched, you’d expect it to soon learn to switch. Hebranson and Schroder demonstrated this with a cunning variant of the Monty Hall Dilemma, where the best strategy would be to stick every time. With these altered probabilities, the pigeons eventually learned the topsy-turvy tactic.
It may seem obvious that one should choose the strategy that would yield the most frequent rewards and even the dimmest pigeon should pick up the right tactic after a month of training. But try telling that to students. Hebranson and Schroder presented 13 students with a similar set-up to the pigeons. There were limited instructions and no framing storyline – just three lit keys and a goal to earn as many points as possible. They had to work out what was going on through trial and error and they had 200 goes at guessing the right key over the course of a month.
At first, they were equally likely to switch or stay. By the final trial, they were still only switching on two thirds of the trials. They had edged towards the right strategy but they were a long way from the ideal approach of the pigeons. And by the end of the study, they were showing no signs of further improvement.
Why is the Monty Hall Dilemma so perplexing to humans, when mere pigeons seem to cope with it? Hebranson and Schroder think this is a case of our own vaunted intelligence working against us. When faced with a problem like this, we try to think it through, working out the best solution before we do anything. This would be fine, except we’re really quite bad at problems involving conditional probability (such as “if this happens, what are the odds of that happening?”). Despite our best attempts at reasoning, most of us arrive at the wrong answer.
Pigeons, on the other hand, rely on experience to work out probabilities. They have a go, and they choose the strategy that seems to be paying off best. They also seem immune to a quirk of ours called “probability matching”. If the odds of winning by switching are two in three, we’ll switch on two out of three occasions, even though that’s a worse strategy than always switching. This is, of course, exactly what the students in Hebranson and Schroder’s experiments did. The pigeons, on the other hand, always switched – no probability matching for them.
In short, pigeons succeed because they don’t over-think the problem. It’s telling that among humans, it’s the youngest students who do best at this puzzle. Eighth graders are actually more likely to work out the benefits of switching than older and supposedly wiser university students. Education, it seems, actually worsens our performance at the Monty Hall Dilemma.
Reference: Herbranson, W., & Schroeder, J. (2010). Are birds smarter than mathematicians? Pigeons (Columba livia) perform optimally on a version of the Monty Hall Dilemma. Journal of Comparative Psychology, 124 (1), 1-13 DOI: 10.1037/a0017703
More on bird brains:
- Confirming Aesop – rooks use stones to raise the level of water in a pitcher
- Clever New Caledonian crows use one tool to acquire another
- Bird-brained jays can plan for the future
- Sparrows solve problems more quickly in larger groups
- City mockingbirds can tell the difference between individual people
April 2nd, 2010 at 10:20 am
It’s not really a problem in logic, nor in conditional probabilities. It is a problem in mathematical modelling, as you say here: “The problem is that most people assume that with two doors left, the odds of a car lying behind each one are 50/50. But that’s not the case – the actions of the host beforehand have shifted the odds, and engineered it so that the chosen door is half as likely to hide the car.” One either models the actions of the host correctly, or one does not. The logic in each case is then a short and easy argument. As so often in life, it’s getting your axioms accurate enough that matters.
April 2nd, 2010 at 11:20 am
@1: au contraire, it’s pure, simple logic. 1/3 of the time, you have picked a winner. 2/3 of the time, you have picked a loser, and the game show host then shows you which of the doors is the winner. You can either stick, and win 1/3 of the time, or switch, and win 2/3 of the time. Mathematical modelling is not required.
April 2nd, 2010 at 11:37 am
I read this paper a little while back, and I think it is a great example to add to the growing literature on non-human animal intelligence. Unfortunately, it’s a bit of an unfair comparison. The pigeons ran the task something like 300 times per day, everyday, for the full 30 days. The college students only ran the simulation 200 times total over the 30 day period. Seems a bit of an unfair comparison to say that pigeons are more capable of figuring this problem out when they have had 9000 attempts to solve it. After that much repetition, I wonder if the college students wouldn’t have gotten the idea as well. Maybe they still wouldn’t have. Who knows.
April 2nd, 2010 at 12:16 pm
@davem: you miss the point; you are taking too narrow a view of what a mathematical model is. The assumption you make about the host’s motive, and therefore action, is the key. That’s your model. The rest then is trivial. If you assume something different about his motive and therefore action, you’ll come to a different conclusion.
April 2nd, 2010 at 1:00 pm
I think such a study shows the futility of trying to give a universal definition of ‘intelligence’
April 2nd, 2010 at 1:05 pm
I aced statistics…except for the probability section. I’ve never gotten it and I still don’t, not even after reading this. However, as I remember this type of show, the options aren’t 2 equal goats and a car. There are the equivalent of a live goat, a dead goat and a car. So sure you’d much rather get the car, and you have a 2 out of 3 (I take your word for it) chance of getting the car. But if you look at what you have, a live goat, vs the downside of switching, a dead goat, it makes sense to stick. At least you’ve got a goat. And you don’t get to go on the show more than once. It’s like that in life too. All those real life choices don’t occur exactly the same way many times a day. A bird in the hand is worth two in the bush so to speak. It creates a blind spot in human choice making, but wisdom in the grand scheme.
April 2nd, 2010 at 1:39 pm
[...] computer simulations proved beyond all doubt that always switching was the best strategy.
Excuse me for nitpicking, but I’m not sure the term “proof” is appropriate here, since this a mathematical, not a science problem. You prove a mathematical statement by deriving the statement from known theorems (and ultimately from the basic axioms). Examples – because that’s what we’re really talking about here – may serve as a hint towards the truth value (and of course counterexamples can demolish any universal claim), but don’t constitute proof themselves. Here in particular a computer simulation necessarily runs only through a finite number of examples. You could modify the problem so that the best strategy is to switch only in x% of all cases (where x is sufficiently large), and computer simulations would likely mislead you.
Admittedly, the last scenario may seem disconnected from reality, but that’s the difference between science and math.
I’m curious as to how the students “overthought” their task. Did they keep track of which buttons they pressed, which lit up and so on? I don’t know how many iterations you’d need – now that’s a problem where computer simulations would come in real handy – but at some point I’d think just looking at your previous successes and failures you should be able to figure out that switching is the better strategy (although I don’t know if 200 runs are enough for that). Did they formulate a theory on how to best play the game? Basically I’m wondering if they actually approached the game from a thinking, analytical perspective or whether they relied on their intuition.
Also, I thought Monty Hall was a topic in school math. So, had the student never learned (or forgotten) about the Monty Hall problem? Or did they know it, but didn’t recognise it in the button pressing variant?
Did Hebranson and Schroder ask the students about any of this?
April 2nd, 2010 at 1:49 pm
Just imagine there are two candidates, A and B. A and B both choose the same door. After the moderator picked one door A always stays with his first choice, B always changes his choice to the remaining third door. Now imagine you run this experiment 999 times. What will happen? Because A always stays with his initial choice, he will win 333 cars. But where are the remaining 666 cars? Of course B won them!
Another illustration:
Let us conduct the experiment with 100 doors! Now let’s say the candidate picks door 8. By rule of the game the moderator now has to open 98 of the remaining 99 doors behind which there is no car. Afterwards there is only one door left besides door 8 that the candidate has chosen. Obviously you would change your decision now! The same should be the case with only 3 doors!
April 2nd, 2010 at 3:23 pm
Brad – At the end of the experiment, the students weren’t showing any further signs of improvement though. I think their performance had hit a plateau.
April 2nd, 2010 at 4:16 pm
For those interested, Jason Rosenhouse has written a book about this, *The Monty Hall Problem* (Oxford University Press, 2009). Here’s the Amazon page for it: http://www.amazon.com/dp/0195367898/
April 2nd, 2010 at 7:30 pm
In regards to ‘overthinking’. The idea here is that the humans had come to the conclusion that the choice did not matter. They had convinced themselves that there was a 50/50 chance of getting the grain, and therefore stopped really looking at new data. Or, (in regards to probability matching) they had come up with a better feel, but were trying to game a random situation. That is, they maybe thought that if they choose the less optimal strategy 33% of the time, it would somehow make them more likely to succeed.
Lastly: You can prove monty hall using binary logic (Brute force all possible outcomes of the binary choices the contestant makes) and you can determine a strategy to win 2 in 3. The reason why a computer model is used to ‘prove’ this to mathematicians, is because they were no doubt a little leery of accepting the proof that contradicted their intuition so strongly. But having a simulation repeat the test millions of times becomes increasingly hard to deny.
April 2nd, 2010 at 8:50 pm
Most decriptions of the Monty Hall Dilemma I have seen rely on an unstated assumption. That assumption is that Monty either will always show you the goat, or at least Monty showing you the goat is random and not related to whether you selected the door with the car. But what if that isn’t the case. After all the original premise is pick the door you think has the car, not pick the door you think has the car, then I’ll show you one of the goats and give you chance to switch.
So, what if Monty is short on cash and wants to avoid giving away cars. If you pick a goat, he declares the game over, and you’ve got a goat, or maybe he’ll give you a chance to trade your door for a curtain that has a sheep behind it. But if you pick a car, then Monty will show you a goat and try to get you to switch. In this scenario, switching is a guaranteed loss, and not switching a guaranteed win.
In statistics terms the 2 variables, whether you picked the car, and whether you get to make a second choice are assumed to be independent, while that is never never stated in the scenario. If they are not independent then the statistical reasoning behind it is incorrect.
April 2nd, 2010 at 9:57 pm
It’s funny: for me, the Monty Hall problem is like one of those figure/ground illusions, like the two faces and the vase, where sometimes I totally understand the solution, and other times I completely lose the insight. Maybe it’s because I’ve had a few drinks.
April 3rd, 2010 at 12:44 am
Just a point of possible confusion if one is not familiar with the show. You wrote: “You pick one of the doors and get whatever lies within.” To me that sounds like you get – immediately – what was behind that door. That leaves two doors unopened. The host opens one of them and, behold! a goat. If you don’t see a car behind the door you’ve chosen then of course, you switch.
That’s not how the game works, of course, but that’s how it seems to be described in the article (at least to me).
BTW, I would have stayed put…Oh well! Now I know better. It took me some time to work out the reasons, but now I understand. Thanks!
–Michael
April 3rd, 2010 at 3:16 am
In other words, pigeons are frequentists, humans are Bayesians (and bad Bayesians at that).
April 3rd, 2010 at 6:34 am
@Jerry: “Most decriptions .. rely on an unstated assumption. That assumption is that Monty either will always show you the goat, or at least Monty showing you the goat is random and not related to whether you selected the door with the car.” That’s exactly what I meant by saying it all depends on your model of the host’s intentions and therefore action. When I first tried to explain that point to a mathematician, he missed it entirely, cried “No, it’s just mathematics”, and damned near tried to hit me.
@Lilian: “I aced statistics…except for the probability section.” My memory of being taught introductory probability was that it was littered with unstated assumptions of the sort Jerry mentioned. That, and illogical use of English, managed to make the intellectually trivial contents needlessly demanding – but with the demand being placed on your ability to guess what your teacher had in mind, rather than on the theory.
April 3rd, 2010 at 9:38 am
I suspect the difficulty arises from faulty questions. The car never moves. I think that the chances were always 50/50. There are not really three choices in the beginning because Monty knows where the car is and manipulates the outcome. Each of the two goat doors is equivalent and so can really only be considered flip sides of the same choice- the non-car choice. So from the outset the contestant only has two possibilities: to pick the car or the non-car flip side choice of a goat. 50/50. It reminds me a little of the “missing dollar riddle’ where the wrong question is being asked, resulting in great confusion.
What if we ask the correct question? “What are the chances of something there is never going to happen, happening?” One in three? Or zero? It will never occur that the contestant picks one door and then learns the result of that choice. It never really is a one in three probability because the choosing is prolonged through another step where the appearance of 50/50 is more obvious. But it was 50/50 from the start.
If Monty was in the dark, as well, and could not manipulate things we would see a very different outcome, were he to blindly open one of the three original doors and then ask if the contestant wanted to switch.
April 3rd, 2010 at 10:50 pm
Actually the pigeons learned to switch because their first pick never garnered any reward!!!
April 3rd, 2010 at 10:52 pm
What would happen if the pideons actually won a reward one third of their first choices as they should?
April 4th, 2010 at 5:49 am
Love the Family Guy reference on the graph.
April 4th, 2010 at 8:22 am
To write a whole book about it. That’s where they lost me… I’m with what XiXiDu says, as I am with what Charles is saying. It’s not that complicated. It’s logic. The car stays in the race, always. Pigeons aren’t that complicated and apparently solely interested in the food (reward). Students might be not that interested in the ‘reward’ at all. Comparing totally different factors and non-competing interests isn’t great science / mathematics /o.w.n.
April 4th, 2010 at 9:32 am
The Monty Hall Problem is never explicit enough: is the prize moved (to the un-chosen door) when a booby-prize is revealed or is nothing is moved and the choice is still truly random?
The remaining door should (if nothing is switcheroo-ed behind the scenes) be just as likely as the initial choice to have the prize behind it.
April 4th, 2010 at 11:48 am
It sounds to me like they did not offer the students enough trials or a good enough reward. If they are really there because it gets them bonus points in their psych class (my default assumption), why would they give a crap about how well they perform? I don’t get the feeling that they were being offered something that was inherently valuable to them the way that food is valuable to a pigeon. Possibly if they run this test again with money or chocolate bars for the people I’ll be less skeptical.
April 4th, 2010 at 4:38 pm
Flip the point of view:
First pick:
your odds are 1 in 3 for the car, so host’s odds are 1 in 3 for 2 goats;
your odds are 2 out of 3 for a goat, so host’s odds are 2 out of 3 for 1 car, 1 goat(!).
Thus, 2 out of 3 times the host will choose the goat over the car and leave the car behind the remaining unchosen door.
(Note, that if the host gives you a second chance only 2 out of 3 times, then all bets are off.)
(also, note tc @ 18.)
April 6th, 2010 at 12:15 pm
There really is no problem here. There is nothing to doubt either. It’s logic. You don’t need to simulate this, it’s a self-contained truth. Your chance of picking the car first time is 1/3 but your chance of choosing a door with a goat behind it, at the beginning, is 2/3. Thus on average, 2/3 of times that you are playing this game you’ll pick a goat at first go. That also means that 2/3 of times that you are playing this game, and by definition pick a goat, the moderator will have to pick the only remaining goat. Because given the laws of the game the moderator knows where the car is and is only allowed to open a door with a goat in it. What does that mean? That on average, at first go, you pick a goat 2/3 of the time and hence the moderator is forced to pick the remaining goat 2/3 of the time. That means 2/3 of the time there is no goat left, only the car is left behind the remaining door. Therefore 2/3 of the time the remaining door has the car.
May 9th, 2010 at 5:28 pm
Preposterous. Unless the car is being switched, it is one door and one door only that has the car behind it. You either pick it or you don’t, odds and chances are constructs of an imperfect species.
June 22nd, 2010 at 4:49 pm
I think the family guy graph is labeled wrong (or the text describing it is wrong). The text says:
By the final trial, they were still only switching on two thirds of the trials.
But the graph shows switching at only about 1/3.
April 2nd, 2011 at 6:20 pm
Just came across this and noticed that the second graph you’ve shown is incorrect – it is incorrectly labelled as the human response, but actually shows data from the pigeons response on a second trial (where ‘staying’ was optimal). The real results for humans show a shift toward the optimal strategy but only switching 2/3rds of the time, as you describe in the text.