I think I remember, too, saying a couple of times, on the blog, that I came away from writing

*Animals in Translation*thinking birds might just be as smart as people, or smarter. I didn't want to think that, particularly;

*How smart are birds, anyway?*wasn't the topic of the book, which was mostly about mammals, not birds. But every time Temple told me a bird story, I would think

*hmmmmm*.

The crow-and-the-rifle story, about a crow repeatedly taunting a rancher, has never left me.

So naturally, when I saw "Pigeons, Humans, and the Monty Hall Dilemma," I had to look.

The Monty Hall Dilemma is a probability puzzle that is notorious for eliciting suboptimal decisions from humans. A participant is given a choice from among three doors, one of which conceals a valuable prize. After an initial selection, one of the remaining, nonwinning doors is opened, and the participant is given a chance to switch to the other unopened door. The probability of winning is higher if the participant switches. Pigeons maximize their wins by switching on virtually all trials of a Monty Hall Dilemma analogue, whereas humans utilize a suboptimal strategy involving probability matching. Possible reasons for the difference between these two species’ performance are considered.No time to read or even skim --- but looking forward.

Pigeons, Humans, and the Monty Hall Dilemma by Walter T. Herbranson Current Directions in Psychological Science 2012 21: 297

## 38 comments:

Hi Cathy,

I checked in to see what’s new here and was surprised to see the Monty Hall problem reference. I think that problem stalks me. It was discussed extensively on Phys-L (a physics teacher email list). I just want to note that it is NOT necessarily the case that switching is the best strategy. It really depends critically on how the problem is presented.

If you are told IN ADVANCE that after you pick your door, Monty will reveal an empty door and offer you the chance to switch, then yes, you should definitely switch and in doing so double your chances of winning from 1 in 3 to 2 in 3. But that is not how the problem is usually presented.

Usually, they say something like: “You are on Lets Make A deal. You choose one of 3 doors, looking for a car. Then, Monty turns to the two doors you did not pick and reveals that one of them had spam behind it, offering you to keep your initial selection or choose to the still-unrevealed door. Do you keep or switch?” When I first heard this puzzle, I said I would not switch. I was mocked for being wrong…but was I?

Here is the key point: in this statement of the situation, it is in no way indicated that Monty’s offer is a standard part of the game. Monty is a friendly guy, but he is not YOUR friend. He is your opponent. He works for the network that has to pay for the cars. For all you know, the network has told him to make this offer to, say, 90% of people who have initially guessed right and only 10% of people who initially guessed wrong, as a way of reducing the number of cars they give away. Remember, Monty is not your friend and HE KNOWS WHERE THE CAR IS! If this is in fact the situation, switching vastly reduces your chances of winning. And unless you have been told the rules in advance, keeping your initial guess preserves the 1/3 chance you started with.

I haven’t read the article yet either (it’s behind a pay wall) but maybe pigeons switch because they are more trusting.

Hope all is well,

Phil Keller

Hi Phil,

You still have it wrong.

There are at least three key parts to the problem:

(1) Monty can only open one of the two doors you did not choose, and

(2) Monty cannot open a door with the car behind it.

(3) Monty has to open a door.

This is a conditional probability situation, so lets consider the three cases. I'll put the car behind door (A) for all of them, but it doesn't matter.

1) You pick A

2) You pick B

3) You pick C

1) Monty opens B or C, you switch to the other and you lose.

2) Monty can only open C, you switch to A and you win

3) Monty can only open B, you switch to A and you win.

By switching, you win 2 times out of three.

-Mark R.

Hi Mark,

I don't think I do have it wrong. Your argument is fine if your three assumptions are true. In fact, I said as much in my post. There is nothing about your reasoning that I disagree with. But what I am questioning is your third assumption. As the question is often posed, there is nothing that tells you that Monty HAS TO open a door. If he is happy to give away a car one third of the time, he can just move on to the revelation. If he wants to give away cars less often, he could make the reveal-and-offer to just those people who had in fact first picked correctly. After all, if even half of those people can be induced to switch (perhaps due to the prevalence of websites that tell them with confidence that switching is always better), then the rate of car winning drops to one out of six...

So you see it REALLY matters whether the offer to switch is announced in advance as a standard part of the game.

Phil

By the way, there is yet another possibility. Monty may have been told to give away MORE cars so that sponsors can have their new cars oohed and aaahed over. Perhaps he only makes that offer to people who initially guessed wrong. Well of course, then you should absolutely switch. The point is that without knowing more about how Monty decides to make the offer, switching can actually change your chances to a range from 0 to 100%. You just can't know without more info.

But again, if he always makes the offer, then the analysis that says that switching doubles you chances is correct, no issue.

Phil,

You're talking about a different situation. In *this* one, Monty does open a second door and never opens it on the car.

That's really it in a nutshell, and it's explained above.

You're arguing something more like this: math problem says, "John has two apples and gets three more, how many does he have now?" -- Well, I think he has twenty five, because I happen to know his mom works for an apple farm and has brought him more apples.

That's just not part of the problem.

Actually, it's more like: if we give John 3 apples and then 5 apples, how many does he have now?" and you are asking us to assume that we don't need to know anything about how many apples he started with!

I know he opened the door THIS time. But I really want to know if he ALWAYS opens the door or has he opened the door for some special reason. And I am saying that the typical statement of the problem does not address this. You are the one who is assuming something that isn't there!

The problem has a long history that you can read about on the web. The woman who calls herself the smartest woman in the world -- Marilyn something -- posted the puzzle and then the solution that you and Mark favor. But the original puzzle had the ambiguity that I am pointing out (and not claiming to be the first to notice, by any means). She later posted a correction, acknowledging the ambiguity and restating the problem so as to remove it.

I promise that I do in fact understand the analysis that leads to the answer Mark cites. But for that to be the answer, the problem has to be posed in a specific way -- a way that it is often not.

I'll ask you this way: suppose you were actually on Let's Make a Deal and you were put in exactly this situation. Would you really automatically switch? Wouldn't you be a little curious as to whether the offer was automatically given to every contestant or perhaps MAYBE used as a strategic way of affecting the frequency of payout?

But again, one more time: if you want to use the question as an exercise in conditional probability, just state it properly and all is good.

"I'll ask you this way: suppose you were actually on Let's Make a Deal and you were put in exactly this situation. Would you really automatically switch? Wouldn't you be a little curious as to whether the offer was automatically given to every contestant or perhaps MAYBE used as a strategic way of affecting the frequency of payout?"

Et voila -- you are now exactly back at the point of the article you commented on -- full circle!

The answer is that most humans would NOT automatically do so, even though it is in fact the better option. Interestingly, it seems birds do better at that than we do. So, it seems it's not so much Monty's motivations as those of the picker that matter!

(P.S. if you get your math problem on a test? I'd go with the answer, "I've given him 8 apples" -- and if it's multiple choice, it's highly unlikely there will be a choice that says "this problem cannot be solved.")

I don't know what else to say. It is only the better choice if you are sure the game is being played by consistent predetermined rules. And yes, on a multiple choice test, I'd pick 8 apples too. But that does not make the question defensible as well-posed.

It is just not irrational to suspect that Monty may be your opponent. Or he may be your guardian angel. Or a consistent neutral agent following the preset rules. You don't know unless the question is properly posed.

Btw, I am not actually commenting about the article -- I haven't read it :)

So we can stop arguing now: there is an extensive Wikipedia page about this problem. It says that the original intent was that the problem was to be interpreted as requiring the consistent offer. But it also acknowledges the ambiguity. And it has a nice chart at the end, showing the multiple interpretations and the resulting probabilities. I especially like the names they use: Monty-from-hell and angelic-Monty ...

For what it's worth, Phil, I've just read over this thread, and I agree with you completely. This was hashed out years ago, and it is an even better example of the non-intuitive nature of statistics than the properly qualified Monty Hall Problem alone. Few word problems are 100% explicit, with no reliance on any assumptions. In statistics, subtle differences in rules or assumptions or what is known or when it is known or by whom can change the answers.

Switching is the optimal strategy if everyone knows (in advance), believes, and follows the rules Mark listed---as Phil said. If conditions are other than this, the answer can change.

Thank you Glen. It IS an interesting problem on both levels. I may torment my physics students with it after the AP exam. But first I'll have to explain who Monty Hall was and what was "Let's Make a Deal".

Phil

Let's Make a Deal is still running on TV...in fact a friend of mine recently won a car on the show. No Monty Hall, though.

I once referred to "Let's Make a Deal" in an algebra class I was teaching. (No, I wasn't teaching the Monty Hall problem). No one had heard of the show, nor had they heard of Monty Hall. They were all familiar with the phrase "Door No. 1, Door No. 2 or Door No. 3", however.

Clearly unable to let this drop, I did want to point out that this posting clearly stated the problem:

"A participant is given a choice from among three doors, one of which conceals a valuable prize.

After an initial selection, one of the remaining, nonwinning doors is opened,and the participant is given a chance to switch to the other unopened door. The probability of winning is higher if the participant switches."So while Phil may want to discuss Monty Hall's motivations and sponsorships in some versions of this problem (although it would seem easier to just ask, "does he always open a door?") -- it was already stated in this one.

Right. But saying "a participant" sure leaves wiggle room: "each participant" would be stronger. But even then, that still misses the point. It doesn't matter that WE know what Monty is doing. For the participant to have enough information to make a rational choice, you have to tell HIM that you are going to be making him this offer before the game commences. That way, he knows that you are not out to get him (or help him).

It would not be hard to state this clearly. But you might not get results that are as fun to write about. If you tell a participant: "We have 3 doors. One has a car, the other two have canned squid. You can select a door. At that point, we will open one of the doors that you didn't select, revealing canned squid. At that time, you may keep your initial door choice or change your pick to the other unopened door."

If they said this, then there is no doubt that switching would be the optimal strategy and there would be no need to wonder about Monty's motives. And I bet more people would in fact switch -- if you have told them that the possibility of switching exists from the start, they might be less emotionally attached to their first guess. And I get it that the paradox is supposed to illustrate precisely that attachment. The problem is that if you don't give participants all of the information up front, your experiment can't distinguish the irrational from the wary.

Phil (who also can't let it go...)

Sorry, Jen, but I don't agree. "After an initial selection, one of the remaining, nonwinning doors is opened," describes a sequence of events, not a set of rules, much less whether the participant knew the rules. It just says, This is what happened.

It makes a difference. Consider two scenarios, one where Monty knows what is behind each door and MUST open a "goat door" in EVERY case, and one where Monty doesn't know what is behind the doors but must open a door, which turns out to be a goat in THIS case. "After an initial selection, one of the remaining, nonwinning doors is opened" could be describing either scenario.

I'll be the participant. There are three doors: car-door, goat1-door, and goat2-door. I don't know which is which, but I will always switch if given the chance.

SCENARIO 1: Monty knows which is which and must open a goat-door, which I know. I have an equal (1/3) chance of each of the following:

1) I pick car-door, Monty opens a goat-door, I switch and lose

2) I pick goat1-door, Monty opens goat2-door, I switch and win

3) I pick goat2-door, Monty opens goat1-door, I switch and win

If I switch, I win in 2 & 3. If I don't switch, I only win in 1. I'm twice as likely to win if I switch in this scenario.

CASE 2: Monty doesn't know, either, but he always randomly opens one of the doors I didn't pick, which I know. Each of the following is equally likely:

1) I pick car-door, Monty opens goat1-door, I switch and lose

2) I pick car-door, Monty opens goat2-door, I switch and lose

3) I pick goat1-door, Monty opens goat2-door, I switch and win

4) I pick goat1-door, Monty opens car-door, and I've lost

5) I pick goat2-door, Monty opens goat1-door, I switch and win

6) I pick goat2-door, Monty opens car-door, and I've lost

The description of what happened THIS TIME can only apply to 1, 2, 3, and 5 above (4 and 6 don't match the described sequence of events.) Since 1, 2, 3, and 5 are all equally likely, my chance of winning is two out of four if I switch and the same if I don't, so there is no advantage in switching.

There are other scenarios, too, but that should be enough to illustrate that if the problem merely describes a sequence of events, we don't have enough information to state with certainty that switching is always the best strategy. It depends on additional factors.

Here is an idea for an experiment related to this question:

Invite people to have a turn as a participant. Divide the participants into three categories:

1. Fully informed: Tell these people in advance that after they choose their door, a goat-door will be revealed and they will have the opportunity to change their selection. [This is the group that I think we all agree should rationally choose to switch].

2. Uninformed: Offer these people the chance to switch, but don't tell them in advance. [This is the group that does not have enough information to rationally decide].

3. Warned: Tell these people in advance that, BASED ON THEIR INITIAL SELECTION, they MAY be shown a goat-door and offered a chance to switch. And it is important to say it in just that neutral way: "based on...". Then, actually go ahead and offer all of these people the chance to switch as well.

I would be curious to see how these groups respond. And I think that the AP statistics students in my school do projects after the AP. I may see if any of them are interested in doing this. I would guess that the rate of switching would be highest in group 1, lowest in group 3... Not sure how to do this with pigeons.

Phil

It has been done enough with fully-informed people to demonstrate, to my satisfaction anyway, that very few people will switch when they ought to switch. If you did your three-group test, I'm not sure what your results would even mean. Would you be testing cog psych, or knowledge of probability, or knowledge of pop science (this problem was on Myth Busters), or even respect for authority?

I include that last due to an anecdote from one of Nassim Taleb's books. (Doing my best here from memory:) He asks a statistician, "If I take a fair coin, flip it 100 times and get heads every time, what are the chances of getting heads the 101st time?" The statistician grins and says, "Ah, you're trying to fool me, but the chances on the 101st flip are the same as always: 50%. The coin has no memory," he explains.

He asks "Fat Tony" the same question. "Ah, you're trying to fool me. If it came up heads 100 times in a row, the chances of heads on the 101st flip are basically 100%. It's obviously not a fair coin," he explains.

Taleb says that the statistician's brain is damaged after years in academia, but Fat Tony, a practitioner, understands the real world. He compares the likelihood of getting 100 heads in a row from a fair coin to the alternative, that the experimenter was lying about it being fair, and the choice is obvious--statistically.

Yeah, I see the problem. I ran it by one class, and though a few of them were interested, they did say that they saw it in the movie "21". I checked out a clip on youtube and sure enough, the movie professor says emphatically and unconditionally that switching is better, even though the scenario he sets up clearly does not fully inform the participant. In fact, he praises a student for NOT considering this issue...

As to what results may mean: if the informed group switches substantially more often, it may suggest that one the reason the uninformed group does not switch is that they have a fear that they are being manipulated. And the group to which it is suggested that their choice was the reason they got the offer? They would switch least of all.

But we'll see...

Please post your results. No matter what they end up being, I'm sure I'll find a way to use them to confirm my preconceptions.

Although I am not a student of a statistics problems, I do know that these figures can always be used to one's advantage, if I wished to manipulate same. The big hole in your argument of problems is that once the first box is seen to be empty, the contestant cannot exchange his box. So the problems still remain the same, don't they. . . one out of three. Oh, and incidentally, after one is seen to be empty, his chances are no longer 50/50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so. It was always two to one against him. And if you ever get on my show, the rules hold fast for you -- no trading boxes after the selection. ... Sincerely, Montyhttp://www.letsmakeadeal.com/problem.htmHeh. I love the image of researchers attempting to create "fully-informed" pigeons along the lines of the experiment outlined above.

It could be a life's work.

Then again, the pigeon's don't seem to need it.

The Monty Hall problem is one that I've never understood but had to memorize the correct answer. In fact, I have actually used it in my homeschool as an example of "sometimes the correct answer makes no sense so you just have to memorize it."

" pigeon's don't seem to need it."

Where oh where did that apostrophe come from? Heavens.

I thought that Mark R's explanation above was pretty clear -- one of the best that I've seen. I'm trying to commit *it* to memory!

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I vaguely remember that Monty didn't always open one of the other doors before asking whether the contestant wanted to switch. I think he started by asking whether the contestant wanted to switch before opening one of the remaining doors. If the contestant switched, then I don't think he always opened another door at that time. Often, the contestant had one bird in hand (refrigerator) and was offered a chance at a bigger prize in the bush (a car behind the door).

That's an interesting separate question; how much will people give up for what probability of how much reward? It's easy to bet two dollars on the lottery for the chance at 222 million (PowerBall today), but what if you had to pay for a full year at a time and then you looked at the probability? It's also easier to get lots of little money from people that one big chunk.

Maybe it's hard to think when you are dressed up in a chicken's costume and just got done yelling "PICK ME!" I don't think that pigeons have to do that.

"It's easy to bet two dollars on the lottery for the chance at 222 million (PowerBall today), but what if you had to pay for a full year at a time and then you looked at the probability? It's also easier to get lots of little money from people that one big chunk."My guess is that there are two things going on with lotteries (maybe three).

(1) The amount per week (say $2) is small. My guess is that some people who play at $2/week would not want to play at $104/year for 52 weeks worth of tickets. But lots still would. So small dollar amount each week is better than big dollar amount once per year.

(2) But ... even for those folks who would be fine paying $104/year in one go, I think they want 52 tickets. Because they want 52 chances to win, not just one chance.

(3) And ... they also don't want the 52 tickets all at once! A *large* part of the "value" of a lottery ticket is the weekly hope that *THIS* week will be the week that you win. That hope/excitement/anticipation might even be worth $2!

So ... of these two choices, which do you think would be more popular:

(A) $2/week and one big drawing at the end of the year, or

(b) $104 up front for one ticket per week for a year?

My guess is that (B) would be much more popular...

-Mark Roulo

"Let's Make a Deal" was an after-school staple along with "Truth or Consequences", "Queen for a Day", the "Art Linkletter Show", and for those in CT, the "Ranger Andy Show". How many remember Lippy the Lion and Hardy Har Har? How about Deputy Dawg? And "Concentration" was by far the most loved game show.

CW: The best way I've found to describe it is this: In the beginning, your choice has a 1/3rd chance of being right, but the other two combined have a 2/3rds chance. After opening one of the other two, all of that 2/3rds chance is concentrated on the one remaining unopened, unchosen door, and your choice is still stuck at 1/3rd.

It doesn't make sense to me. The original door started at 1/3 but once Monty opens the lousy prize door logically the odds should be 50/50 between the original door and the 3rd door. I have memorized the textbook answer but I do not understand why that is correct.

Just because there are two choices left does not make the odds 50-50. The two choices are not symmetrical. One card stands by itself (the one that you initially picked) and the other represents the survivor of a two-card elimination process that Monty performed for you. So in a sense it "carries the weight" of two doors. In that sense, the doors are not the same.

There is a variation that was invented to address exactly this point. It's the 100 door version. Suppose to win the car, you had to pick from 100 doors. Clearly, your chance of guessing right is 1/100. Now suppose of the 99 doors that you didn't pick, Monty opens 98 of them and reveals 98 goats [just as he said he would before the game started -- after all this thread's arguing, I can't believe I almost forgot to say that!]. Now, would you keep your initial guess? I bet you would switch. And would you think that just because there are two doors left, your odds of winning were 50-50? They are much better than that. They are 99/100. It was 99/100 that the car would be behind a door you did not initially pick. Then Monty kindly performed the winnowing out process so that the one that you switch to "carries the weight" of the 99 that it survived from.

Hope that helps --- Phil

P.S. My students say that they are going to try that study. So I hope to have results before the school year is over.

Crimson Wife,

Another way of thinking about this is, 1) you choose a door, then b) Monty asks you if you would like to switch to BOTH of the remaining doors.

-Mark Roulo

Anonymous: " My students say that they are going to try that study. So I hope to have results before the school year is over."

That's a lot of goats.

Switching to BOTH of the remaining doors is a very different situation. In that case, of course you should switch because it's 1/3 vs. 2/3.

But once you find out that one of the lousy prizes is behind door B, then door A or door C should logically have an equal chance. Why does opening door B suddenly change the chance of door C from 1/3 to 2/3? It should logically change it from 1/3 to 1/2, the same as Door A.

Flip a coin. It either lands flat or on an edge. Those are the only two possibilities, but they are not equivalent so the chances are not 50-50. I know that seems silly, but it is intended to emphasize that when choosing from two options, the probability is only 50-50 when the two options are equivalent. In your example, door C is NOT equivalent to Door A. Mark's way of saying it is really concise -- my last pulse takes the longer route. Since Monty will reveal the empty door, you really effectively ARE being given the option of switching to, effectively, BOTH doors.

Crimson Wife:

"Switching to BOTH of the remaining doors is a very different situation. In that case, of course you should switch because it's 1/3 vs. 2/3."Ah, we are making progress :-)

We can agree that opening a door does not move the car (or goat or whatever) :-)

So try this: There are three doors. You can put the doors into two groups: Group #1 has one door and Group #2 has two doors.

Then you can:

A) Pick group #1, or

B) Pick group #2.

After you select (A) or (B), Monty will open one of the doors from your group #2, which the rules guarantee will *not* contain the prize.

Do you select (A) or (B)?

-Mark R.

Crimson, you would be right if Monty opened his door randomly, which was the point I was making in my earlier explanation. If he didn't know what was behind any of the doors either but just opened one of the doors that you didn't open and---this time---it just happened to be a goat, the chances would be equal for each of the remaining doors, just as you are saying. Yes, his opening the door and finding a goat would give you (both of you) information, but it wouldn't tell you anything more about the door he didn't open than it told you about the door you had chosen. Both doors had originally had an equal 1/3 chance, but now that he has opened a goat-door, both unopened doors have an equal 1/2 chance. This is what your intuition is telling you, and it is CORRECT as long as Monty is opening doors randomly.

But if Monty's choice is not random but based on knowledge he has that you don't have plus rules he has to follow, and you know those rules, you can glean additional information from seeing what he chooses. That's an information theoretic way of looking at it, which is admittedly a bit abstract, yet it might help you obtain that gut feeling you are looking for--that maybe this makes some intuitive sense.

A more concrete view is that two out of three times, the car will be behind one of the two doors you didn't pick. EVERY time that happens Monty, using information you don't have, will be forced to eliminate the goat door and leave the car door for you. Since you know he's required to do that, you can take advantage of his extra knowledge by always choosing the door he leaves for you. You won't always win, but the two out of three times the car is behind one of the doors you didn't pick, the door he leaves for you will have the car.

This is one of the things that makes probability so challenging. Very subtle changes in information can change the answer in non-intuitive ways. Here is another classic example to illustrate that point:

The Johnsons have two children. You know that one is a boy. What is the probability that the Johnsons have two boys?

The Smiths also have two children. You know that the older one is a girl. What is the probability that the Smiths have two girls?

The questions are almost but not quite the same and the answers are different...

-Phil

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