## How to explain Monty Hall problem when they just don't get it

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Talking to some friends, I was asked to explain the answer to the Monty Hall problem (see also here;) .... they were having some trouble because whoever explained it to them didn't do a very good job.

#Humblebrag I was able to explain it in under 30 seconds to most people (winning on staying is 1/3; winning on switching is 2/3 because it's based on losing the first pick, which is a 2/3); but some people just didn't understand....is there any sure-fire method of teaching the answer to people?

Note: This question is not being asked about a math class per se, but is probably equally applicable to the classroom setting as it is to a group of friends.

EDIT: I notice that a lot of the new answers don't understand what this question is asking. Pay attention to the bold text above, and realize that I understand the solution to the problem, but had some trouble explaining it to some friends....the focus of this question is how to explain the answer, not what is the answer.
A question focusing just on the answer would probably fit in a lot better on Mathematics or MathOverflow; possibly Cross Validated (statistics).
Thanks! :)

After posting where to find other questions regarding directly to the problem itself, I found this question on Cross Validated -- it appears to be very, very related to my question.
(The letter copied in this answer to that question is hilarious, but very telling....)

What worked for me is something akin to ChrisA's rendition, but in slower-motion. Since the revealing of a losing door and the offering of the choice complicate things, cut them out and reintroduce them within a thought experiment; to wit, consider the following 4 (or 5) versions of the game:Vandermonde 2015-11-20T05:29:50.633

(1) Choose a door and it is immediately opened, and you win or lose based on that one door; (2) Normal version of the game, except that the host fast-talks you and opens your door before you can protest or exercise your option to switch; (3) Normal version of the game, except that due to, e.g., mind control, you never switch; and (4) Normal version of the game, except that you are determined (by free will or otherwise; the point is that the reason you don't switch is irrelevant) never to switch.Vandermonde 2015-11-20T05:30:35.430

If it helps as a stepping stone, you could include (1.5) Normal version of the game, except that you are given the opportunity to switch without any door being revealed. It should be clear that each change of rules leaves the probability of winning unchanged from the last version, that the chance under (1) is 1/3, and that the chance under (4) is the desired answer.Vandermonde 2015-11-20T05:35:02.293

(Of course, the 1/3 figure is (as it must be) under the caveat that luck never enters as in Valentin's or my other comment, but that's merely an inevitable occupational hazard in applying probability theory to real situations. The issue of whether or not the probabilities you use reflect reality or are accurate is exogenous to the validity of the calculation.)

Vandermonde 2015-11-20T05:37:44.323

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Tutor 2014-06-15T03:19:05.783

5If time allows, you could always do an experiment. If that will not convince someone, then I don't know what will. (Of course, it doesn't explain reasoning, i.e. why.)dtldarek 2014-06-15T07:58:39.777

@dtldarek No, it won't explain reasoning, but it probably would help break people out of a strong misconception, so that they could allow themselves to understand something they now know to workTutor 2014-06-15T13:30:54.797

3It would be interesting to know if people who don't "get" the MHP also do not get the right answer to "Mary has exactly two children; at least one is a girl; what is the probability that the older child is a girl?" Again we have a situation where intuition says that it's 50% when it is in fact 67%.Eric Lippert 2014-06-15T14:53:37.517

@EricLippert You're right; they probably wouldn't....just for laughs, after explaining MHP to the guys, I gave them Schrodinger's cat to think about.... really threw them for a loop ;)Tutor 2014-06-15T15:55:28.100

I resisted the claim for decades, including writing a simulation which seemed to confirm my belief. I honestly don't remember how I was finally convinced. I need to reconstruct that since I'm on the verge of talking myself out of it again.keshlam 2014-06-15T16:50:23.693

@Eric I was verging on using that "children" puzzle as a variant with a more everyday feel until I realized that it tricked me, and figured it wouldn't help anyone else.Ryan Reich 2014-06-15T20:41:43.547

Meanwhile there are plenty of good ideas around, and I feel that some significant overlap starts to emerge. Thus, I protect the question for now. Let me know if you want it unprotected later. (Protection means only that very new users cannot answer the question; users with at least 10 points gained on this site can still answer.)quid 2014-06-15T22:57:34.463

I don't have reputation enough to add an answer, so I'll put it in a comment instead. If one out of three doors contains the prize, the only time you'll gain anything from not switching is if you initially picked the correct door, the odds of which are one in three. If you chose a dud door (2/3 chance), Monty will helpfully eliminate the only other dud and leave the winning door as the only one left.Andreas Eriksson 2014-06-16T09:50:51.643

Viktor Mellgren 2014-06-16T13:32:06.480

As out of curiosity, I came to this exchange site. Now I'm like .... "Wat"Hugo Rocha 2014-06-16T17:57:57.513

@HugoRocha What's confusing you?Tutor 2014-06-16T18:28:34.433

@HugoRocha ....I understand the correct answer to the Monty Hall problem, but had some trouble conferring that knowledge to some friends, so I asked this community (of "Mathematics Educators") how they thought the answer should be explainedTutor 2014-06-16T18:29:00.423

I can't post an answer, so here's one that hasn't come up yet: you are a prisoner, kept with two others in a cell. One day, the guard announces two of you are to be executed, but doesn't say who. You're terrified, because that means the chance of being executed is 2/3! You plead with the guard, but he tells you he won't do anything to change the probability that you are executed. Finally, you ask him: "Well, could you at least tell me a prisoner who isn't me who will be killed?" The guard says sure, and names one of the other prisoners. You smile, knowing that since there's onlyBen Aaronson 2014-06-16T19:01:50.933

two prisoners left, and you're one of them, you've improved your chances of survival from 1/3 to 1/3. Most people will realise this story is counterintuitive, and it's a direct analogy of Monty Hall. Realising why your chance can't possibly be 1/2 explains why the chance of picking the car can't be 1/2 either.Ben Aaronson 2014-06-16T19:02:18.220

@Tutor I am confused myself, but I read the wikipedia article and now it's clear. English is not my native language, so sometimes I'm a bit slower to get.Hugo Rocha 2014-06-16T19:19:51.737

@BenAaronson Your scenario is incorrect. If you think about your analogy, you, the prisoner, are the door that was picked by the contestant. The other two prisoners are the doors the contestant did not pick. The other prisoner that was named by the guard is the door that got eliminated. The other prisoner that was not named is therefore the one whose chances of "being the right choice" just got increased. Not you.Kai 2014-06-16T22:31:55.120

I think the wording "increased the chances" of the other door being the right one is a little misleading. What really happened is by eliminating doors as being "wrong" you are revealing information about the doors that were not chosen. Obviously, once you have information revealed, you end up recalculating the odds based on this new information. But the new information doesn't really say anything about the door that was chosen, so the way you calculate its odds doesn't change.Kai 2014-06-16T22:38:23.877

@Kai Think of the scenario in terms of the misconception about Monty Hall, not the truth. In the misconception, once Monty reveals his goat, both other doors change from having a chance of 1/3 of being the car to 1/2. By analogy, both other prisoners would change from a chance of being spared from 1/3 to 1/2. The point is that this is obviously counterintuitive in the prisoner case, which should shine a light on why it's also wrong in the monty hall case.Ben Aaronson 2014-06-16T22:41:38.993

@BenAaronson Gotcha. I was thrown off by the fact that you were using the "you" prisoner as an example of a "flawed thinker." But I think most people don't think of it as both of the un-eliminated doors are "increasing their odds to 1/2." I think the misconception comes from that people are trained that if you flip a coin 4 times, and get 4 heads, it's still only 1/2 chance for heads next flip. In other words, hyper-correction for another common fallacy. Applied to Monty Hall, I believe they're thinking the uneliminated doors both remain at 1/3 odds.Kai 2014-06-16T22:53:49.103

@Kai Hm, you might be right, but I don't think so. If they say both remaining doors are 1/3, you can say "Where's the other 1/3?" If they say it's the opened door, you can point out that it's clearly impossible that it's in the opened door, so the chance of that is 0. If they dont agree with that point, their understanding is so lacking they probably need to start with something simpler than Monty Hall.Ben Aaronson 2014-06-16T23:30:03.897

I believe that to truly understand the solution to MHP, the audience should know probability . BUT if the audience dont, Then this particular problem may be explained like this perhaps; IT IS NOT ABOUT WHAT YOU(PLAYER) PICK, IT IS ABOUT WHAT THE HOST DOESN'T PICK, HE KNOWS WHERE THE CAR IS, HE WONT SHOW IT TO YOU, SO GRAB THE ONE HE DOESN'T, COZ OUT OF THE TWO REMAINING CLOSED DOORS THE HOST WILL ALWAYS PICK THE ONE WITH THE GOAT NOT THE CAR!!!! THERE IS A HIGH CHANCE WHAT THE HOST LEFT UNPICKED HAS CAR IN ITrps 2014-06-17T06:39:17.987

If you can write software, try writing a little simulation that plays the game multiple times and makes random choices each time. I tried doing that, and it clicked for me when I was coding Monty's "choice" and realised I only actually had to invoke the random number generator if the contestant picked the correct door in the first place.BenM 2014-06-18T11:11:59.940

Play the game. Get two students: One switches, and one doesn't switch. You choose a door to put the car behind. Let the student guess which door, and reveal one of the "goat" doors. The students employ their strategy, then see who gets a goat and who gets the car. In a few turns, the class will realize that switching is way better. Once they realize that switching works, they'll be better able to understand why. You can also show the nine possible combinations for each turn, (Car behind Door #1 - Picked #1, Car behind Door #2 - Picked #2). They'll get it.David W. 2014-06-18T13:52:33.593

The very simple and direct way to explain this is as follows: when selecting randomly at the beginning, you are more likely to be wrong. Consider an extreme case where you are extremely unlucky and you select the wrong door(one of the goats) 99% of the time. Now that the other door is opened, and you know you're so unlucky, swapping almost guarantees the car. Swapping at the end takes advantage of the fact that you will probably get the wrong door.Valentin 2014-06-18T19:48:51.697

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I haven't seen it mentioned here but if anyone has ever seen the movie 21 (With Kevin Spacey) they actually explain the problem in one of the scenes. They call it "The Game show host problem" If you haven't seen it it's posted on youtube https://www.youtube.com/watch?v=8mSGNRfHSWQ pretty horrible quality, but still watchable. The movie itself is also pretty good. - AE

A.E 2014-06-15T21:56:05.030

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Your explanation, by the way, is very elegant. As an experienced mathematician, I see immediately that it cuts right to the heart of the matter and admits no ambiguity. Unfortunately, this is precisely the quality that makes it unconvincing to others; the main confounding aspect of Monty Hall is that it ruthlessly exploits an intuitive misunderstanding of probability, and that misunderstanding remains even when one hears the correct explanation. People don't learn by being told what's true if that truth is in conflict with a previously more strongly believed falsehood; you have to directly confront and discredit their preconceptions before they will be able to appreciate the slick answer.

One possible misconception is that, after one door is eliminated and two remain, many people will feel that as an "either/or" choice, each door has probability of success 1/2, whereas in fact the distribution is not uniform. You might test the possibility of this being the problem by posing other, simpler questions about non-uniform probabilities, for instance: play a game where I flip a coin and give it to you if it comes up heads. What's the probability after two flips that you have 0, 1, or 2 coins? If they know the right answer, ask them to explain why, and if not, work it out with them.

(This can become Monty-Hall-like if you change the game once they understand. I flip both in secret and announce that there were "more than zero" heads, without saying exactly how many. Now what are the probabilities? This is very close to one of the branches of the actual Monty Hall problem)

Because of the coins setup where the "doors" are dynamically generated rather than static and the implication that the game is repeatable is more obvious, it may conform better to the intuition. The language a problem is stated in is very important for it being intuitive: people have different opinions about different but isomorphic versions of a single problem. I don't claim that everyone will like coins either.

I would avoid an explanation that rests upon the other person successfully adopting a mathematical mindset. For instance, the "100 doors" version immediately displays the problem but the concept of specializing from a generalization, or solving a related problem, especially one with extreme behavior, is not always natural to people.

@hildred: Well, she is onto the intuition that in order to conclude anything about probabilities or optimal strategy, you need assumptions about probabilities -- one cannot get something for nothing (in reasoning deductively, all of the content of any conclusions you make is already contained in your hypotheses) -- and those assumptions can well be wrong, resulting in garbage-in/garbage-out, TANSTAAFL, whatever you like to call it....Vandermonde 2015-11-19T23:56:00.610

... If you believe in luck being an inherent quality of a player (and I do know some nearly consistent winners over quite a long term) or in lack of true randomness, the model assumed here no longer applies.Vandermonde 2015-11-19T23:56:14.367

1Great answer....thanks for the compliment, and for explaining that "you have to directly confront and discredit their preconceptions before they will be able to appreciate the slick answer."Tutor 2014-06-15T13:25:59.250

1Think I should put "my" method in as an answer of its own? [not really my own...it belongs to whoever explained it to me properly (though not first -- whoever that was did not do a good job;)]Tutor 2014-06-15T13:33:04.123

1No, it's in the question already. This question shouldn't be about "all methods to explain Monty Hall" anywayRyan Reich 2014-06-15T13:37:05.050

True....but it's still a good method when the other party is not too proud to look at things a different wayTutor 2014-06-15T13:53:17.800

"This question shouldn't be about "all methods to explain Monty Hall" anyway" .....too late :(Tutor 2014-06-17T02:30:32.553

1@Tutor It was timely when it was written :)Ryan Reich 2014-06-17T04:05:22.120

3When I tried to explain the Monty Hall problem to may Grandpa's CNA, she insisted you should always hold because there was a reason you picked it in the first place. I was unable to successfully counter this belief.hildred 2014-06-18T16:56:30.987

@hildred That is totally hilarious! Did you point out that, whatever you think the reason is, you might have been tricked?Ryan Reich 2014-06-18T17:04:48.210

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For some reason, the 'extend it to 100 doors and eliminate 98' explanation doesn't make it any clearer for me.

Rather than talk about probabilities as fractions, I explain it this way:

"If you picked the car (without knowing it) on the first choice, you'll lose it by switching, whereas if you didn't pick the car, you'll gain it by switching."

(stop here and make sure they get that. Then...)

"Suppose you do the experiment 100 times*, and always stick with your initial choice. About 33 of those times, you picked the car on the first choice. So if you always stick, you'll end up with the car 33 times out of 100.

However, about 66 of the times, you didn't pick the car on the first choice, but if you switch, you'll switch to the car.

Therefore if you always switch, you'll get the car about 66 times out of 100, whereas if you always stick, you'll get the car about 33 times.

So it's twice as likely to get the car if you switch, as if you stick.

*or 99 if they're picky

Easiest explanation I have ever read! KudosDexters 2014-12-21T10:21:32.120

7Very nice. The only thing to spell out is the fact that Monty knows to remove a non-winner, so in effect the '2/3' is packed into that one remaining door. But your explanation is great, and nicely avoids the 100 door extension.JoeTaxpayer 2014-06-15T13:48:27.127

2I have always prefered the 100 door extension and don't understand how it would be difficult to explain in that way.David G 2014-06-15T17:24:49.380

5@JoeTaxpayer: I think the thing which makes this explanation so good is that you don’t have to say anything slippery like “the 2/3 is packed into the one remaining door” to make it rigorous. Monty always picking a non-car door is encapsulated in the step “If you picked the car, you’ll lose it by switching,” etc. — which is a deterministic step. It avoids using conditional probability (which for many people is unintuitive) by breaking the argument into a conditional-but-deterministic part, plus a non-conditional probability part.Peter LeFanu Lumsdaine 2014-06-15T21:16:57.123

1The 100 door extension is perfect as long as you make the person who doubts you actually play a few rounds using it. :-)R.. 2014-06-16T17:15:04.223

1@DavidG The reason I don't like the 100 door version is because the intuition it gives you- for me at least- is not based on Monty knowing which door the car is behind. If Monty picked randomly (with the possibility of him unknowingly opening the door with the car) then switching would bring no benefits. So any explanation that gives you the right intuition in the case where he knows which door it's behind should also gives you the right intuition in the case where he doesn't. Otherwise you're just replacing one misleading intuition with another.Ben Aaronson 2014-06-16T18:52:53.943

1Here's my problem with the $n$-door version - it's not at all obvious that the correct generalization of the Monty Hall problem is that the host opens $n-2$ doors rather than still only opening $1$ door. Both of these reduce to the same thing in the $n=3$ case, but only the opens-$n-2$-doors version gives you good intuition about how the problem works.senshin 2014-06-16T23:40:56.897

5Both of the previous two comments perfectly demonstrate my objection that to someone who doesn't know what's going on, a generalization of a tricky problem may not convincingly illustrate the original problem at all. You have to know how it generalizes, and which features are relevant.Ryan Reich 2014-06-17T04:08:13.130

@DavidG The numbers are too big to hold in your head and sort through. I could not understand the 100-door version until I had a better explanation of the 3-door version, and even now it's difficult to visualize.Izkata 2014-06-18T04:38:53.807

I prefer to use 52 doors. Represent each door with identical paper cards, with a different pattern on the front. Choose a single pattern to represent the car, and call it the "lady".Aron 2014-06-19T06:31:28.283

For me I convinced myself with n-door generalization, and with opening all doors, except one. For n = 1 million, I have a 1 in a million chance, and all the others together have 999,999 in a million chance. After opening 999,9998 doors, I am drawn to choose the other closed door. Whether we open 1 door or n-2 doors has the same result, you still have more chance changing your door with one of the non opened doors. The larger n is, the smaller the increase in probability. Opening n-2 doors makes it much more obvious the choice to change door.devMomentum 2014-06-19T12:37:37.060

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There isn't any sure-fire method of explaining anything, and especially in math. But specifically in the case of the Monty Hall problem it has been proven by extensive experience that many individuals with otherwise above average intellectual capacities exhibit an exceptional tenacity in refusing to accept the (otherwise) widely agreed upon solution; don't waste your time on repeating this experience, unless you should find pleasure in doing so.

It should also be noted that for experts in probability theory there is a lot of nitpicking to be done about the exact hypotheses that need to be formulated in order to get a completely well defined probabilistic problem with the purported solution as correct answer (the Wikipedia page on the subject illustrates this). So if the resistance you get is of the type: "it is not so clear cut, it might depend", then very possibly they could actually be right, unless you took all the required precautions in explaining the problem.

2I think you raise a good point, and I always felt that some stories on some mathematicans having a hard time to understand it, is actually based on them not getting a complete description (for example glossing over the fact that the host would never open the door with the prize in the first round.) However, perhaps you could add something to make your post more of an answer to the question, like stressing that explaining the initial set-up again can be helpful in some cases.quid 2014-06-15T09:07:54.167

Thanks for expanding the answer!quid 2014-06-15T10:58:45.630

3+1 ...."There isn't any sure-fire method of explaining anything, and especially in math."Tutor 2014-06-15T13:15:13.360

3...."Just make sure you agree about the method and its validity before applying it." ....that very well may have been my issue....Tutor 2014-06-15T13:15:41.877

4I upvoted this answer because I've ran into people who blindly assert the 2/3 solution regardless of the game that is being played. It is absolutely imperative to explain that the host always opens one non-winning door regardless of if your choice is correct or not, and there are some other nit-picky things to be considered as well. If you fail to explain this, Monty could play by different rules - for example, only open a door if you chose the right door initially.Kikanaide 2014-06-16T16:17:36.877

3@Kikanaide: The real Monty Hall would sometimes open the player's door immediately (instant win or loss), would sometimes open a door with the prize (instant loss), and would sometimes open an empty door. If the host--unlike the real Monty Hall--always opens an empty door and allows a switch, then switching will be a winning strategy 2/3 of the time, but if the host doesn't always allow the choice, the probability that the original guess was wrong in those cases where the host does allow a switch may be anywhere from 0% to 100%, depending upon the host's strategy.supercat 2014-06-17T23:13:46.663

@Kikanaide: Suppose one chooses door #1 and the host always picks the lowest available door. Then there will be a 1/3 probability of the host showing door #3 (in which case the prize must be behind #2) or a 2/3 probability of the host choosing door #2 (in which case, the prize would be just as likely to be behind #1 as #3). That wouldn't change the fact that if the prize was uniformly randomly placed, and the host always shows an empty door and allows a switch, the probability of winning by switching if one doesn't know the hosts reasons for picking the door he did would be 2/3.supercat 2014-06-18T19:32:27.663

Simul-posting fail: @supercat, sorry I deleted my initial post (and then my next one). My claim is that while the odds of that strategy succeeding are 2/3 in the expectation, the odds are no longer 2/3 once a door is opened. And I don't claim that there exists a bias by which switching becomes unfavorable, or that alters the 2/3 in the expectation. Simply that without that constraint, the odds in a particular game may not be 2/3.Kikanaide 2014-06-18T19:32:54.103

@Kikanaide: Suppose that not switching and winning would pay 3.0, but switching and winning would pay only 1.8. Assuming door choices are uniformly random, the expected value from not switching would be 1.0, while the value from always blindly switching would be 1.2. If the host always shows the lowest available door, then the value if one switches only upon seeing the host show the higher door would be (1/3 3.0) + (1/3 1.8), or 1.6--a clear improvement. No matter what means the host chooses for picking a door, however, the optimal strategy must yield an expected value of at least 1.2.supercat 2014-06-18T19:47:59.493

@Kikanaide: Note that in the proposed scenario, blindly switching would cause one to have an expected value of 1.8 in that one case out of three where the host would open door #3, and an expected value of 0.9 those two cases out of three where the host opens door #2. Both cases contribute 0.6 to the player's expected value.supercat 2014-06-18T19:51:56.693

@supercat: thank you for the examples (yes, cost-for-switch changes the problem tremendously). These examples show that one must be particularly careful when structuring this problem. Never assume the full set of assumptions is so "natural" that the person you'll talking with has already grasped them. The answer can change if the game changes. Period.Kikanaide 2014-06-19T20:41:51.020

@Kikanaide: I find it somewhat odd that people describing the problem expect everyone to assume the hypothetical game-show host will behave in a way no real game show host I know of ever did. Nobody should be expected to accept the claimed probabilities in the absence of a complete explanation of the game.supercat 2014-06-19T21:30:19.743

13

Perhaps it's not the explanation that's the problem. I suggest you have them explain to you their understanding of the problem. Listening to their justification might reveal why your explanation is not gaining traction.

Even in the cases where people are saying they now agree with you, you don't necessarily know that they understand the problem. It's possible that all you did was get them to agree with you. People often just agree with another person who seems to have a better grasp of math.

However, if they explain it to you, you'll have a much better idea of their understanding. No matter whether they agreed or disagreed with your conclusions.

5This is usually very helpful. Alas, some people just don't know why they think what they do.Ryan Reich 2014-06-15T13:34:17.590

4That's why every chance to help them reflect on that fact is a chance worth taking. Otherwise they may never realize the value of their own efforts to explain and justify. If we can't establish justification as a basic mathematical practice, we're not really teaching mathematics.JPBurke 2014-06-15T16:53:10.897

1Usually if they don't believe that it's better to switch, they see the problem as a choice between two doors. The difficulty is then to show them that it's not a choice between two equivalent options, in the sense that a coin flip is such a choice.ChrisA 2014-06-18T06:22:06.237

1@ChrisA Yes, essentially, the disagreement is about equivalence vs. non-equivalence. I think what we, as math educators, have to help students do, though, is see that they have to go beyond saying that the choice is equivalent for their explanation to be considered a justification. How do they show you their basis for believing it is equivalent? If it is similar to a coin flip, why is it similar? Whether or not they can explain it, it certainly helps to establish what counts as justification, and the norm of requiring justification.JPBurke 2014-06-18T14:21:28.293

12

I've had best luck by simplifying the problem to asking whether you want to pick one door or two doors. Everyone understands they'd be better off picking two doors. At that point I tell them to just ignore the fact that Monty showed them what was behind one of the doors they didn't pick. They're still picking either one door or two doors.

Edit: The more sophisticated, but still confused listened will object that this ignores the additional information gained from Monty opening one of the doors. But that's a red herring. There's always a losing door in the two you didn't pick, so Monty is not revealing any new information.

1Perhaps I am one of the confused, but how is switching equivalent to picking two doors? Yes, you make two choices, but you don't benefit from actually having your fingers in both jars, so to speak. You still only get the car if it's in your final choice, however much more probable that turns out to be.Ryan Reich 2014-06-17T04:44:18.740

1@RyanReich Assuming you switch: you will always switch to the door that has the car (if either of the two unchosen doors has the car) because the other one was eliminated. So you are effectively choosing both (both of the initially unchosen doors)Richard Tingle 2014-06-17T12:22:34.690

1Yeah, but the other door doesn't always have the car.Ryan Reich 2014-06-17T13:04:50.583

1@RyanReich if you imagine the doors as sets, two empty one containing a car, then with the switching strategy you get the union ("both") of the two doors you did not pick originally.quid 2014-06-17T21:32:27.227

1@quid Oh, that makes sense. I was misled by the fact that, in a way, neither of the two doors you pick is actually your choice: Monty chooses one, and you are stuck with the other (if you desire). Nonetheless, two are better than one, as long as Monty doesn't conspire against you.Ryan Reich 2014-06-17T22:13:46.427

@RyanReich Indeed Monty promises not to remove a door that contains the car. If he could then the whole thing breaks down and returns to a 1/3, 1/3, 1/3 probabilityRichard Tingle 2014-06-18T14:41:44.073

furthermore, the analysis only works if Monty always offers the swap, or at least offers it independent of whether you actually picked the car or not.Llaves 2014-06-19T15:16:34.010

11

I found the most helpful way to think about the problem is to expand it to a larger number of doors. For example, if you have them select from 100 doors, where 99 are losers and 1 is a winner. Then after the initial selection is made, eliminate 98 doors.

The crux of this explanation is from one of Polya's heuristics

**EDIT**

The initial answer remains above, as I feel it is a quick answer to the question, thought it might not satisfy everyone. I answered almost immediately after the question was posed, and in the time since this edit, many more thorough answers exist on the board that should satisfy such readers.

I do wish to provide an answer to the comment by @Tutor who asked "what Polya's heuristics are". George Polya is a well-renowned 20th century mathematician whose seminal work is How to Solve it. In it, he discusses his heuristics (techniques) for solving (perhaps difficult) problems. Under the principle of "Devise a Plan" he discusses the idea of solving a different problem if you are having difficulty solving the one posed. Sub-heuristics of this are: (1) solve a simpler problem, (2) solve a special problem, (3) solve a more general problem. The statement of a new problem with 100 doors could fall under any one of these categories.

4I have never understood how this is even supposed to begin to be helpful. Nothing about it triggers any sort of intuition for me; you might as well be talking about the color of Monty's tie for all the relevance I've been able to see. What is this supposed to trigger in the learner's head?user2357112 2014-06-15T08:07:42.093

7@user2357112: This argument does not intend to transmit any intuition. It specifically targets at destroying the blind conviction that two possibilities that are equally possible are therefore equally likely; one this fallacy is abandoned, maybe the door to a proper analysis will be opened (and no goat will be found there ;-). The destruction of conviction is intended to happen by making the conclusion drawn from the fallacy more "evidently absurd". I have no idea about its effectiveness, but this is clearly the aim.Marc van Leeuwen 2014-06-15T12:06:56.013

@MarcvanLeeuwen Two reasons I upvoted your comment: explaining 100 doors, and "maybe the door to a proper analysis will be opened (and no goat will be found there ;-)."Tutor 2014-06-15T13:04:58.517

4This is precisely how I first made intuitive sense of this problem. The essential realization that worked for me is how absurdly unlikely it would be to choose the correct door from among 100 doors. I realized that "switching" was the same as being able to flip the probability (instead of the number of doors being against you, now they were working in your favor). Essentially: let's say the door is chosen for you at random. Would you rather have that door, or the 99 other doors? I'm not saying this is a good explanation for others, but it was my own thought process.JPBurke 2014-06-15T17:00:43.880

In connection to expanding the number of choices, there is a strategy in contract bridge that depends on what is known as the "law of restricted choice". The Monty Hall problem is perhaps the simplest case of the law of restricted choice. There are a large number of beginning bridge players who simply do not get, and cannot apply the law of restricted choice.Walter Mitty 2014-06-17T10:49:45.483

@user2357112 To me it's pretty obvious why this is helpful. People often claim that it's a 50:50 choice once one of the doors is opened, and some of them are completely resistant to any suggestion that this is not the case. Expanding the choice to 1,000,000 doors blows that argument away completely, so they're prompted to think a little more (and come up with another objection, perhaps, but some progress will have been made).Dr Eval 2014-06-17T12:36:46.007

Thank you for clarifying :)Tutor 2014-06-22T02:21:37.430

I may now have to track down a copy of How to Solve It ;)Tutor 2014-06-22T02:22:03.610

The reason why the 100 doors version is not necessarily helpful is because of where people's idea of 50% chance is coming from. If it's coming from the fact that there are two options for what is behind the doors, then it may help. But if it's coming from the fact that after all doors are open, you have two doors to choose, then it is unlikely to help. I find it interesting that people stick to their own preferred explanation despite the obvious evidence that it doesn't work for everyone.DavidButlerUofA 2014-07-23T19:44:46.717

8

The best sure-fire method for teaching this to people who don't want to learn it is to set up a Monty Hall style game for small stakes of real money. When they start losing 2/3 of the time, they will become more receptive to your explanation.

1This seems somewhat related to the set up an experiment/simulation already mentioned. But then to add a more noticeable consequence for some could actually be important. So, it seems like an actual variation of the answer.quid 2014-06-15T22:52:19.647

2Yeah, the money is important. It changes some people's perceptions.Walter Mitty 2014-06-16T12:14:14.500

1

There's an online demo at http://onlinestatbook.com/2/probability/monty_hall_demo.html which helpfully keeps a tally of your successes as well.

Superbest 2014-06-17T07:55:14.393

1This works pretty well if you're explaining it to a programmer who can quickly simulate the gambling - and losing - of large sums of money very quickly!Dr Eval 2014-06-17T12:38:15.747

1The people who can't be persuaded by logic aren't going to be persuaded by a simulation either. If the logician can lie to them, why can't the author of a simulation?Walter Mitty 2014-06-18T11:05:19.920

7

I never understood it until my Maths lecturer explained it to me. Unfortunately I can't remember the exact way he explained it, but I'll try my best to remember. I really like your own explanation, so there is a chance that the following won't work. In that case, you may have to resort to empirical experiment. Another observation from my own experience is that sometimes I get too "mathsy" too quickly, and people don't follow my reasoning. Perhaps you need just need to slow down a bit!

Try this explanation. Get three cards to represent the doors. Use an ace as the winning door. Now, take two of the cards, and give one to your friend. There is a 2:3 chance that you have the ace, because you have two of the three cards, and there is a 1:3 chance that your friend has the ace, because he has one of the three cards. Now, peek under your cards, and discard one. Ask your friend if they want to swap for the card that you kept.

This is essentially identical to what happens in the problem, but it feels different because you are choosing two of the cards. You can highlight that you had a 2:3 chance of having the ace, because you had 2 of the cards. This may appeal to their intuition.

2By bringing physical aids such as cards into it, you add a visual component. This increases the chance of visually confirming that, by switching, you are in effect switching to both of the non-chosen cards. And we are guaranteed that (1) at least one of those is a loser and (2) will be discarded by Monte. Now, ask if they'd be willing to switch their one card for the two remaining cards before Monte looks at and throws away a loser.user2338816 2014-06-17T05:49:48.027

1Yes! That is a very good point.daviewales 2014-06-17T16:58:08.333

7

Speaking from my own experience, a stage I had to go through before understanding any explanations of the logic and math involved was indignation at being baffled by a cheap fairground trick that probably goes back centuries. Also the eventual recognition that what put me, the punter, at a disadvantage to the operator was the fact that s/he is working to a strategy and is therefore prepared to lose a game every so often (1 out of 3 in this case), whereas I'm limiting myself to only thinking I've got to win the present game each time.

Another motivational prepping step I'd recommend is a highly entertaining presentation (available on YouTube) by the great tele-mathematician Marcus du Sautoy playing the game with someone who sportingly agrees to be the fall guy who never switches and so loses heavily. This brings out the compelling practicalities.

A couple of points about the explanation stage. A clear diagram with your three doors and a list of the three possibilities of what's behind them: Car Goat Goat; Goat Car Goat; Goat Goat Car. If my first choice of door happens to be the correct one, and I switch, then sure as hell I'm going to lose on that occasion. Don't mince words about that, but use it to show how it means that if instead my first choice happens to be the wrong one, which is going to be 2 times in 3, then switching will reverse those odds into my favour.

(I agree with others that I've never understood how considering lots and lots of doors is helpful)

1

Found the Sautoy video https://www.youtube.com/watch?v=o_djTy3G0pg ....didn't watch it yet, but will....showing a bunch of experiments to the stragglers among my friends probably would help them

Tutor 2014-06-15T13:12:26.887

1

And see if this comment helps you to understand 100 doors [it sure helped me;)]

Tutor 2014-06-15T13:13:24.313

1I hope you don't mind that I edited your answer to include a link to the Marcus du Sautoy video....just to let you knowTutor 2014-06-15T19:54:03.483

2That's fine. I just watched it again, and its great strength is it vividly dramatizes the consequences of thinking it doesn't matter if you switch or not. But in my opinion the downside is that to organize the actual explanation round this fallacy complicates things, which is why the video isn't so effective at this stage. Your approach starting with the consequences of wrong and right picks is the clearer and more elegant. But maybe both are needed when dealing with what's essentially a trick.Wanderlust 2014-06-16T09:33:40.323

1@Tutor Sure, but if your explanation requires further explanation in order to understand why it helps, then is it really useful as an explanation?DavidButlerUofA 2014-07-23T19:41:29.330

5

While your explanation is correct, and easily understood by people like us, it's a bit too terse (a quality we like) for a lay person to understand. You can simply expand on it. And most importantly ask the audience questions along the way. Every explanatory sentence should have a question that goes a long with it. The key is for you to identify where the hangup is. I like to think about the problem in terms of the chosen door rather than the switching door. So this is how I might go about asking the questions.

1. Ask them before removing a door "what is the probability that the door they chose has the car?"

2. Then say, "I remove a door. Since I am not evil, I am not going to remove the door with the car." Then ask "Why does the probability that the car is behind the door you initially chose not change?" If they don't get this right, work out the cases, as others have mentioned (continuing to ask questions along the way).

3. Then say, "Since there is 1/3 probability that the car is behind your door, what must be the probability the car is behind the other door?"

Explanation at each stage may be required, but this socratic method seems to work well. I deployed something similar to a bunch of small children at the NY county fair. By making the argument composed only of leading questions, its a way of making sure they are following at every step. Of course you can ask them different questions. The key is to make it socratic.

1"And most importantly ask the audience questions along the way." "The key is to make it socratic." When there are problems understanding a basic idea, a lot of times there is room to improve the presentation.....+1Tutor 2014-06-15T18:06:21.997

I think things can be made much clearer if one makes the probability of the host showing an empty door and allowing the player to switch contingent upon whether the player initially guessed correctly (the real Monty Hall would sometimes show the door with a car without offering the player a chance to switch). If the probability of being allowed to switch is affected by the correctness of the original choice, then the fact that one is offered a car should affect one's estimated probability if the original choice having been correct. Note that a "neutral" host would pick a door at random...supercat 2014-06-16T21:15:58.830

...and only allow the player to switch if the randomly-picked door didn't hold a car. In that case, a player whose initial choice was right would always get a chance to switch, while one whose initial choice was wrong would only be offered a chance to switch half the time. Switching under such circumstances would be a 50-50- proposition. An evil host would always show the car if it wasn't behind the player's door, and a massively benevolent one would only show it if it was behind the player's door. To always show an empty door and allow switching is mildly benevolent, not neutral.supercat 2014-06-16T21:19:06.997

5

I find that most people who THINK they understand the Monty Hall Problem, actually don't. For about 5 years I was one of them, until a further insight made me understand it better. More of that at the end.

However, I came up with a super-exaggerated version that seems to give people pause for thought at least.

In the UK, our lottery has a probability of 14,000,00:1 for the jackpot. Suppose I buy all 14,000,000 tickets one Saturday and let you choose 1 ticket with your favourite numbers, the rest I put in my shed. I lock you in a room until the draw with no access to the lottery result and I sit in my shed and watch the lottery draw. After the draw, I walk out of my shed with 1 ticket in my hand and set light to the entire shed and the rest of the tickets. I unlock the door of your room and announce, “Good news! One of us has the winning ticket! Would you like to swap with me?” Why would you think that your ticket has suddenly become a 50/50 chance of being the winner? If you believe that, imagine we did the very same thing NEXT week and the week after. Why has your lottery ticket suddenly acquired a 50/50 chance of winning every week – yet never wins it?

The crucial thing I alluded to at the top is that Monty HAS TO KNOW where the car is before any of the normal interpretations become valid. The “100 doors” scenario illustrates the point very well here. You pick door one and Monty opens door 2, then door 3; all the way to door 36. He bypasses door 37 and moves to 38, 39; all the way to 100. He then asks if you want to swap door one for door 37. Clearly the car is behind door 37 (well, 99 times out of a 100).

However, imagine you watch the program for 3 months before you are a participant and every week Monty keeps opening doors to find that the car is behind: door 17, door 81, door 53 – each week he keeps stumbling upon the car BEFORE the point he would give the contestant the chance to swap. When you go on the show, you choose door 1, then Monty proceeds to open each door from 2 to 99 and does not find the car. When he offers to swap door 100 with you, there really is a 50/50 chance that you have the car already and there is no advantage in swapping.

Interesting....except in your last paragraph, I think that there still is 99:1 chance of winning by switching.....since there are two doors left, the first door you chose is either the car or isn't; and door 100 [in your last paragraph] MUST be the other....since in your first choosing, you had 99% chance losing, switching now is 99% chance winningTutor 2014-06-15T20:27:39.437

Good example with the lottery; very well may use it next time ;)Tutor 2014-06-15T20:28:01.930

The real world kinda intrudes on the lottery example. Unless you're a game show host, you wouldn't want to give up a winning ticket. :) So if you come to me after, wanting to swap tickets with me, and you already know the outcome...heh. Unless i know you're so rich that that much money means nothing to you, i'd suspect i have the winning ticket, and wouldn't swap.cHao 2014-06-15T20:31:21.580

1@Tutor the last paragraph has to do with how he framed the problem. In this case the host opens doors in order until he reaches the car. If he reaches the last door its a 50:50 chance. This is not how the host behaves in the monty hall formulation (in an n door monty hall formulation, he'd open all the doors in one shot, seemingly at random - so that the position of the door does not give added information), its an illustrative example.MHH 2014-06-15T21:11:09.973

@Tutor: Saying "Door A is the car" is equivalent to saying "Every door but A is a goat". If Monty knows where the car is, and isn't allowed to reveal it, then all 98 of the doors he opens must be goats, and he has a 99% chance of ending up with the car. (The other 1% is yours.) On the other hand, if he doesn't know, then by eliminating all the other doors, he's effectively choosing one at random -- just like you did. His chances at that point are as good as yours.cHao 2014-06-15T21:24:55.290

+1 for explicitly confronting the 50/50 misconception with an intuitively plausible scenario. As you say, it may not convince them, but it should make them think about what they believe, which is of course where they are wrong.Ryan Reich 2014-06-17T04:49:55.173

4

The big problem with the "Monty Hall" problem is that there are many problems that sound superficially the same, but have different solutions. The terms of the game have to be stated very precisely.

As an example, Marily vos Savant's statement of the problem as it is quoted in the Wikipedia article is imprecise. It doesn't state whether the game host must or may open another door, whether he must open a door with a goat or whether he is allowed to open a door with the car, and if he has some freedom of choice, whether he is trying to help you or not. Or whether he is even allowed to "cheat" and swap the contents of the two unopened doors sometimes.

So you listened to the explanation why changing doors is so good, but they just changed the rules - after you choose the first door, if you chose a goat the game host will open the door and say "tough luck, you lost". If you chose the door with a car, the game host will open another door with a goat, so if you swap, you lose. So you see that without stating the rules precisely, you can't give the correct answer.

The "100 doors" explanation doesn't help. So you are the lucky guy that by a one in hundred chance picked the car, and you are shown 98 doors with goats. If you swap, you lose. On the other hand, people who picked a goat will be shown fewer than 98 doors (once in a while, someone will be shown 98 doors, or the game would be given away).

As commented elsewhere I agree that imprecission of the statement can be an issue. But I think the 100 doors explanation in fact can help, as it has as a side effect that the terms are made more precise.quid 2014-06-15T17:58:02.393

Still, the distinction between "may open" and "must open" is one that has to be kept clear in the definition.Walter Mitty 2014-06-17T10:52:30.410

4

Here's my take on explaining it. (And it usually involves a drawing for me :-) )

Assuming the host opens a door the odds of making the right choice without switching the door is 1/3 because it was made before having the knowledge the host provided.

Once you decide to make the switch here is a breakdown of the probability of losing VS winning:

this is very nice because it actually is tersely clear about the host's nature. I think many people who don't understand the problem actually do understand probability. The problem for many is merely the deliberate muddying of the waters as the question is asked. I have similar grumblings about probabilities of children gender. Like, "what is the probability you have 3 girls in a row?" Verses, "you have two girls, what is the probability you have a third girl?" Usually, when people comment I have three girls, wow, what are the odds? They're not talking about the conditional hypothetical.The Chef 2014-06-17T14:05:34.560

they're talking about traveling back in time and asking, if I have three children, what are the odds they're all girls? That is what makes having 3 girls in a row interesting.The Chef 2014-06-17T14:07:27.047

3

The problem is most of the time something in the line of that a switch isn't believed to be helpful, because nothing has changed in the eyes of the person you are explaining the problem to: they consider the chances to be the same as before.

Explaining that something has changed can be the best step to take first: make sure they understand that they are getting more information about the system.

Ofcourse, then you go on, that because of this information, the chances have been changed. Then you can explain that the information is different based on the case you were right (1/3, no real new information) and the case you were wrong (2/3, new information is he tells you were the winner is).

3

The best way I came up with is to draw all 9 possible outcomes.
Prize is in A, you pick A;
Prize is in B, you pick A;
Prize is in C, you pick A;
Prize is in A, you pick B;
and so on.
Each outcome is equally probable and 6 of those will make you win.
I don't think it get's any easier than this.

1What if the person doesn't understand probability?Ryan Reich 2014-06-17T04:52:37.007

3

Short and sweet summation:

It's more likely that your first choice will be wrong than it will be right. Therefore, if you are given the opportunity to switch, it's more likely that switching will be right than it will be wrong.

Any ideas on how to make it more accessible to those who just don't get it from this?Tutor 2014-06-15T19:19:33.010

1This is literally the same as the method explained in the question.Ryan Reich 2014-06-17T04:45:25.093

That's absolutely incorrect. It is also more likely that your second choice will be wrong than it will be right, so by the exact same (wrong) argument it is more likely that staying with the first decision will be right than it will be wrong. Any explanation needs to take into account the additional information.gnasher729 2014-06-18T17:30:52.083

@RyanReich Yeah, I was just trying to reword the original method so it was possibly more approachable. It helped me to think about it in such terms.SFlagg 2014-07-08T20:22:39.963

2

Switching turns a winning guess into a losing guess, and a losing guess into a winning one. Since your guess is twice as likely to be losing as it is to be winning, you do better to switch.

The question was not how to explain the problem; the question was how to best explain it to people who don't understand....I know this is true (see question!); but I wanted to know what professional Math Educators had to say about how to explain itTutor 2014-06-16T03:13:41.940

@Tutor I had a hard time with this problem. This is the encapsulation of the argument that convinced me. Your mileage may vary. I suggest you try it.Thumbnail 2014-06-16T05:04:27.093

I know -- that's pretty much how I got it also, and how I still think of the answer (see question) ....however, I found myself in a situation where that did not work....Tutor 2014-06-18T04:31:18.733

2

Once I set up a bet, using a bunch of (future) lotto drawing. Lotto in Italy uses 90 numbers, and traditionally those from 1 to 30 are marked 1, those from 31 to 60 are marked 2, nd those from 61 to 90 are marked X. I would have taken the role of Monty Hall, and I explained in advance the algorithm which I will use to show a result (among 1, 2, X) which did not appear.

Exposing the algorithm makes people think that Monty Hall cannot usually choose a random door (or a random value, in my example). At least for me, in this way is easier to understand that for the opponent too the choices are not casual.

1

Originally, the chances are 1/3 that the car is behind Door 1, 1/3 that the car is behind Door 2, and 1/3 that the car is behind Door three. If you choose Door 1, your chances of winning are now 1/3. More to the point, the chances are 2/3 that the car is behind Doors 2-and-3 together. This fact will hold true for the rest of the problem.

Monte Hall opens Door 3 and shows a goat, meaning that there are ZERO chances of the car behind Door 3. But the chances are still 2/3 that the car is behind Doors 2-and-3 together. What's left of this "together?" Door 2. ALL of these chances are now behind Door 2, which "inherited" Door 3's chances. So the increase in your chances of winning by "switching," to Door 2, are 2/3 to 1/3, or 2 to 1.

Another example: Suppose there was an election with Candidates A, B, and C, each with 10 votes. Suppose candidate C (Door 3) drops out and says, don't vote for me, vote for B over A (and his supporters follow him faithfully so he gets zero votes). What happens next? Candidate B gets candidate C's ten votes plus his own ten for 20, versus ten votes for candidate A.

1I know it's a lost cause by now, but this question, on this forum, really can't be a repository for everyone's favorite explanation of this problem from the mathematical point of view. It's well known how to solve it; the problem is how to teach it. Just presenting an allegedly correct computation will not convince anyone who doesn't understand all the things that are confusing about this puzzle.Ryan Reich 2014-06-18T00:49:56.630

@Ryan Reich: This was a "self-taught" exposition. Frankly, I had trouble understanding it myself, until I came up with the explanation above. The key trick was that the extra "chances" were "transferred" (I used the term "inherited") from Door 3 to Door 2 when Monte Hall opened Door 3. The reference to Bayesian, were just to tie to the literature. You can't teach anything to anyone until you can first teach it to yourself.Tom Au 2014-06-18T00:54:02.473

I don't want to criticize your solution or you. But I'm not convinced by your answer that its style is good for teaching others, or that it consciously addresses something in particular that needs teaching. It's really just a presentation of your favorite explanation, as-is.Ryan Reich 2014-06-18T01:02:08.240

@RyanReich: I am a historian "dabbling" in math. As such, I am a better "proxy" for a layman than the average mathematician. The idea of a "windfall" is one that is probably appealing to a layman. It's actually a simple, layman construct: You and I together have 100 dollars. I have 0 dollars. How much money do you have? 100 dollars." I learned to use such explanations working as a math tutor in high school,Tom Au 2014-06-18T01:15:45.333

-1

TL;DR There are still 3 doors even though one of them has been opened.

Fact 1. After your initial choice there is a 2/3 chance that the prize is behind one of the other two doors. Your chosen door has a 1/3 chance.

Fact 2. When the empty door is shown to you, there is still a 2/3 chance that the prize is behind one of those two doors.

Fact 3. Except now you know that one of those two doors is empty, so you choose the door that was not shown. That one door has a 2/3 chance that the prize is behind it.

Conclusion: The unknown door has a 2/3 chance of having the prize, but your chosen door only has a 1/3 chance. So always switch.

The mistake that most people make is thinking that it is a now a two door problem and therefore has a 50-50 chance. It is still a 3 door problem and can be easily solved with basic logic.

7If you think it can be "easily solved" and involves "basic logic" then you are insulting your students. Smart people still have a hard time with this problem, and acknowledging this is the first step to successfully educating them.Chris Cunningham 2014-06-16T02:15:42.107