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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

withoutany 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 – 2015-11-20T05:37:44.3231

See here also....didn't have access to it at the time ;)

– Tutor – 2014-06-15T03:19:05.7835If 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 work – Tutor – 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....

reallythrew them for a loop ;) – Tutor – 2014-06-15T15:55:28.100I 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 – 2014-06-15T20:41:43.547Meanwhile 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

Numberphile has some good videos about this https://www.youtube.com/watch?v=7u6kFlWZOWg https://www.youtube.com/watch?v=4Lb-6rxZxx0

– Viktor Mellgren – 2014-06-16T13:32:06.480As 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 explained – Tutor – 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'tme who will be killed?" The guard says sure, and names one of the other prisoners. You smile, knowing that since there's only – Ben Aaronson – 2014-06-16T19:01:50.933two 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,

bothother doors change from having a chance of 1/3 of being the car to 1/2. By analogy,bothother 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 IT– rps – 2014-06-17T06:39:17.987If 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

2

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