Tuesday, 17 February 2015

The Reverse Monty Hall Problem and Conditional Probability

Please note that since I wrote this article, I have been persuaded that the argument it contains is wrong.  The correct answer for the scenario as it is worded is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).

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Originally, this was to be my last article on The Reverse Monty Hall Problem – or at least the last on why the solution is “it doesn’t matter whether you switch or stay, the likelihood of benefitting from either choice is 1/2".  It didn't work out that way ...

As the title suggests, we need to consider conditional probability.  The relevant equation is:

Pr(A | B) = Pr(A B) / Pr(B)

This means “The probability of A happening, given that B has happened, is equal to the (a priori) probability of both A and B happening, divided by the (a priori) probability of B happened”.  It may be phrased more briefly as “the probability that A given that B is equal to the probability that both A and B divided by the probability that B”.

In this case (please follow the links above if you don’t yet know the context), what we have is:

Pr(A) = the (a priori) likelihood that a car is hidden behind the door that you chose, but the philosopher didn’t open = 1/3

Pr(B) = the (a priori) likelihood that a goat is hidden behind a specific selected door = 2/3

Pr(A B) = the (a priori) likelihood that the car is hidden behind the unselected door AND a goat is hidden behind a specific selected door – noting that this is one of only three possible, equally likely permutations = 1/3

I have to stress, and I cannot stress this enough, that in the scenario I described, the philosopher had already opened the door before offering you the opportunity to switch.

I might need to ram this home, so please forgive me if you have already had a blinding flash of light and you now agree with me.  Let’s get more specific.  The doors don’t have any indications on them when you get there, but you are a paranoid sort and are convinced that there is some Derren Brown/Dynamo-type shenanigans going on, so you tell the philosopher you want to mark the doors just after he tells you which one he is going to open, but before he opens it.

  • You walk to the door that you selected that will remain closed, and you mark it as Door 1,
  • You walk to the door that the philosopher has told you will be opened, and you mark it as Door 2,
  • You walk to the unselected door, and you mark it as Door 3, and
  • Then you return to the philosopher to continue the show.   


The questions you must ask yourself are now made a little more simple:

Before I got here, what was the likelihood that the car would be behind Door 1?
There is only one car and three doors, therefore, Pr(A B) must be 1/3

Before I got here, what was the likelihood that Door 2 would have a goat behind it?
There are two goats and three doors, therefore, Pr(A B) must be 2/3.

Before I got here, what was the likelihood that the car would be behind Door 1 AND a goat would be behind Door 2?
This is precisely the same likelihood that there would be a car behind Door 2, because if there is a car behind Door 2, then there must be a goat behind Door 1, therefore Pr(A B) must be 1/3

Then it’s just a simple process of plugging in the numbers:

Pr(A | B) = Pr(A B) / Pr(B) = (1/3) / (2/3) = 1/2

It’s not rocket science, but it seems to have confused the heck out of a lot of people (including, to be totally honest, me).

Now that I have gone through this process, I am really hoping that the cognitive dissonance stays away!  However, there remains a feature that may still confuse people.

If you were asked, prior to entering the arena with the philosopher, whether it would be better, on average across multiple runs, to stay with your door (the one of two that remains to you) or to switch, the correct answer is it is better to stay – because, exactly as many have argued, this would be equivalent to the Monty Hall Problem.

This seems to be hugely counterintuitive.

However, you have to remember that at this point in time, the philosopher has not yet opened any doors.  Therefore, at this point in time, the door that the philosopher opens could be any of the three (depending on which doors you select and which doors have goats behind them).

This seems to be the default position taken by all armchair statisticians, despite being encouraged repeatedly to consider the scenario from just after the door has been opened.

Someone did ask to explain how my two images could be reconciled (from The Reverse Monty Hall Problem – The Solution):



And
 



Someone also complained about my graphical representation of the solution to the classic Monty Hall Problem:
 



This might be difficult to explain, but please bear with me.
 

First I need to re-jig the graphical representation of the classic Monty Hall Problem, not because what I did was wrong so much as it was not adequately explained.  I am assuming that, when given the opportunity to select between two goats, the philosopher chooses at random.  Therefore, the possibilities with the classic Monty Hall Problem are:



Now, imagine that we run the Reverse Monty Hall scenario over and over again, still assuming that when there is an option to select between two goats, the philosopher makes that selection at random.  This means that the possibilities are:






Then, removing invalid scenarios, we have: 



This can be compared with the results from the classic Monty Hall Problem.
  So, yes, if you run the scenario over and over again, you will get the 2/3 - 1/3 split.  But ... there was never any intention that Reverse Monty Hall Problem be performed repeatedly.

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Once again, please note that the argument above is in correct.  The correct answer is that the 2/3 - 1/3 split is a true (in this scenario) as it is for multiple iterations of it.

54 comments:

  1. "So, yes, if you run the scenario over and over again, you will get the 2/3 - 1/3 split. But ... there was never any intention that Reverse Monty Hall Problem be performed repeatedly."

    Wait, what?

    So, the odds are 2/3 if you stick and 1/3 if you switch. You admit this is empirically verifiable. Then you claim that, contrary to this, the actual probability is 50/50. How can you reconcile these two statements?

    ReplyDelete
    Replies
    1. Update: This is actually giving me cancer.

      Here are all the possible distributions of Mary/Ava/Car(men)

      AM C
      AC M
      MC A
      MA C
      CA M
      CM A

      Assuming your choice of doors is always the first two terms in this table (and you will agree that this represents all possible choices) you have a 1/3 chance of selecting both Ava and Mary. This is undeniable, right?

      Our philosopher will ALWAYS open one of our selected doors, and that door will ALWAYS reveal a goat. Given this new information, we now know where at least one of the goats in our initial selection is hidden.

      There is still here a 2/3 chance that our remaining door hides a car and a 1/3 chance that it hides another goat. This is undeniable.

      Whether the revealed goat is Ava or Mary is irrelevant: both represent undesirable results and are, logically, the same. They are not unique entities, as you claim.

      Whether our philosopher opens the leftmost or rightmost door is thus also irrelevant.

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    2. I hear you about the "this is actually giving me cancer". While experiencing it, the cognitive dissonance is quite uncomfortable. I apologise for foisting that on you.

      The problem is that you are generalising when, in the scenario, we are talking about a very specific situation, because the door is already opened. Because it's a specific situation, it is not so obvious that the question of "which goat?" is irrelevant (but I agree that in a sense it is), and the goat is certainly not a representation - it's an actual goat (in terms of the scenario).

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    3. It certainly is a representation. Whether we ultimately open the door that reveals Mary or Ava, the net result is the same: You lose.

      I think you need to be more clear and concise with what you mean by "the door is already open." It seems like every time you reply to somebody who points out the seemingly simple (and as you admit, empirically verified) result of this case (the 1/3-2/3 split), you say almost verbatim "the door is already open" and proceed to rather passive-aggressively insult their intelligence (see above: "While experiencing it, the cognitive dissonance is quite uncomfortable. I apologise for foisting that on you.")

      So, on to my interpretation of "the door is already open" -

      We've selected our two doors. Our resident philosopher reveals one of the goats we have chosen. The door, as you say, is open, and we now know:

      -Which one of our doors is a goat.
      -The order of the prizes behind the doors has not changed. It is the same problem.
      -Our remaining door and the one we have not chosen are still a mystery; they are, as you would say, closed.
      -Our initial choice gave us a 2/3 chance of our group containing the selection "car/goat"
      -Our initial choice gave us a 1/3 chance of our group containing the selection "goat/goat"

      Given all of the above, the informed decision would be to stay, because there is still a 2/3 chance that I initially selected a goat and a car, one of my initial choices was already revealed to be the incorrect choice (goat), and the 2/3 odds carry over to that last unknown.

      If I'm misunderstanding what you mean by "the door is already open," please try to explain with more clarity, as you've manged to present your problem in a way that has confused all of the commentors on reddit (most of who's logic is undeniably sound) and your blog here. The onus is on you to present your scenario properly.

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    4. There's no "net result". You are suggesting that this scenario will be re-run, over and over. This was never suggested.

      I've just suggested this to someone else on reddit, perhaps you can try it too:

      "I know it sounds silly, but can I suggest trying to crystallise the scenario by role playing it out. You don't even have to hide what's behind the doors. Just pick three objects, two of which are the same or similar. Put them in front of you and go through the options - just like you would once the door has already been opened. You might want to move one 'goat' forward to suggest that it was behind the door that was opened."

      Delete
    5. I suspected you were a troll; now I'm nearly positive of it.

      Are you truly suggesting that the probability of a given situation (given exactly the same circumstances) actually changes based on how many iterations of the circumstances are performed? That's not mathematics, that's insanity.

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    6. There's another possibility, as unlikely as you might think it is, and that is that I am right. I am willing to accept that there is a possibility that I am wrong. You seem unwilling to even contemplate the possibility that I am right. I can't really blame you, since if I had not gone through the process I did I probably would not believe me (in other words, the me that existed last week would not believe the me that exists now). You're sort of an incarnation of "me last week" - although I really hope that "me last week" would have entertained the very slight possibility that "me that exists now" might be right.

      This is not "an additional troll" as you may well suspect, it's just a reflection on how difficult it sometimes is to persuade people of things.

      Delete
    7. If you can't logically dismantle my argument (or present yours), your argument is illogical, and thus incorrect when we're dealing with a mathematical issue.

      Telling me "You're wrong, you just can't see it" does not do any justice for your argument. I have contemplated the possibility that you may be right. I've read your blog posts, pointed out the flaws in those posts, and given you ample chance to reply (rather than writing you off and moving on to something more interesting, which I would love to do at this point,) and every time you come back with some dismissive remark.

      So, last chance - if the odds are 50/50, why are the results of repeated experiments 66-33?

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    8. I do think that this might be THE most interesting aspect to the whole thing. You are right that I should explain this and I will put my mind to how I can best explain it. I have, via modelling, and role play, demonstrated to my entire satisfaction that it's right, but explaining why it is right is less simple.

      I understand that you are impatient with me, because I seem to be unable to give you a quick, succinct and convincing answer *right now*. Can I ask for your patience while I ponder a way to explain this to everyone's satisfaction? When I feel that I can clearly explain, I assure you that I will post it in an article here.

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    9. neopolitan,

      I would love to see a video of this model/role play of the situation you've developed.

      Every time you're backed into a logical corner, you assure us that you're able to demonstrate the situation to your satisfaction. So, if that's the last bastion of your argument, please provide video evidence of this so that it can be interpreted and we can all be done with this. I don't want to read through another one of your mad blog posts.

      The way I see it now, there's two likely outcomes to this whole thing:

      1) The demonstration you've set up that has you so convinced of your "solution" is flawed. I don't think I've heard you consider this notion yet. Or,
      2) You have not accurately explained the situation you are trying to present.

      That's the cool thing about math: It's demonstrable. So if you're right, let's see the demo.

      Delete
    10. Well, of course I don't think I've been backed into a logical corner. However, amusingly enough, a you-tube video of a demonstration was precisely what I was talking about last night with a couple of real live people and pondering how to implement at 4am this morning.

      The thing is, when I demonstrated it for these people (admittedly they were not maths PhDs), they found it so compelling that they are completely flummoxed that no-one can grasp what I am saying. But, of course, you could dismiss them as ignorant buffoons, stooges or people I simply made up. So, a video might just be the way to go.

      It's not going to happen overnight though.

      Delete
    11. Well, you have been backed into a corner.

      You have not been able to appropriately respond to any of the recent criticisms of your claim, which leads me to believe you simply don't have a response.

      Anyway, I'm awaiting your video proof. Until I've got that in my hands, or you manage to piece together a pretty astounding case for yourself (I would advise doing this from the ground up, as everything you've posted to date has been effectively vanquished), I remain unconvinced.

      You'll be on the cover of Scientific American if you can prove this one, champ.

      As a side challenge, go to your nearest university, find a math PhD, and have the good doctor verify to us - again, video testimonial with his academic reputation at stake - that you are in fact correct.

      Surely a doctor of mathematics will back you up, right?

      Delete
    12. Another good idea that I have also considered. It's actually part of the reason that I want to be so certain before going to the next step. If this is ground-breaking, if I am right, then I guess I should try to protect it a little. However, I think that if I am right, there is enough of an evidence trail by now that it would be rather difficult for anyone who has been arguing against me to suddenly turn around and claim ownership (or even anyone who has just been watching quietly.

      I suppose that I should be able to convince a maths PhD to give me half an hour of his or her time, given a rather simple cost benefit analysis. It might be very unlikely that I am right, but there are kudos galore for any PhD whose name is attached to verifying that vos Savant was wrong all this time (remember it was maths PhDs who argued most vehemently that vos Savant was wrong back in 1990). So half an hour for a possible academic gravy train? Might be worth it :)

      That said, I'd have to be pretty sure that I am right, huh?

      Would a published paper suffice instead of video testimony?

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    13. If you can get enough PhD mathematicians to agree with your theory that you get a peer-reviewed paper out of it, I'll eat my hat.

      And regarding vos Savant: Most of them (the mathematicians) misunderstood some part of the situation while it was being explained to them, it not being as popular a brain-teaser as it is nowadays. Presented with a clearer definition of what the Monty Hall problem was they changed their minds. And even the stragglers, presented with computer simulations and empirical proof, eventually came around.

      Don't romanticize yourself so much. You're not vos Savant. You're a dude with a blog and a decent mind for mathematics that's fooled himself into thinking he's smarter than everyone else.

      I won't be spending any more time here. Once you're published in an academic journal I'm sure I'll hear about you.

      Delete
    14. If there is a peer-reviewed paper on it, I will be wanting video evidence of you eating your hat.

      Delete
    15. I have a Phd in maths. I spent more than half an hour reading your blog and your reddit posts. My conclusion is that you lack mathematical training to see the tiny mistake you made.

      Just after giving the (correct) definition of conditional probability, you throw that sentence :

      > Pr(A) = the (a priori) likelihood that a car is hidden behind the door that you chose, but the philosopher didn’t open

      What does that mean ? The "but" between the two part of the sentence is VERY ambiguous. Is that :
      * the probability that both events happen (probability that a car is hidden behind the door AND the philosopher did not open it),
      * or is that the probability of one KNOWING the other one (probability that a car is hidden KNOWING that the philosopher did not open it).

      These are two very different things, that are not equal.

      By the way, what does "specific door" mean, and why do you use a different vocabulary than for the "unselected door". Aren't the two equivalent ? Could you not just say "selected door" and "unselected door" instead ? So this is a general advice : you should be more careful with the wording of your definitions. It's in the little details that the big mistakes are made ...

      So, there is this ambiguity of the definition of the event A. It seems that it makes you believe that you computed the correct probability for it (using one way to interpret the event). But it also makes you believe that you can use this probability in another formula (using the other way of interpreting the event). You don't see the global mistake because at each step, the thing that you are doing is correct given a possible interpretation of your event A.

      Your approach of the problem with conditional probability is good. But your lack of mathematical training is not helping you.

      You should think about the problem a little bit more, but this time, with 11 doors, 8 goats, 3 cars, you choose 7 doors at the beginning, and the host voluntarily open 4 of them that contain goats, and he also open one of the door that you didn't pick without a goat. And now you are asked to choose one door among the 6 doors left (3 of them were part of your initial pick, the other three were not). Which one do you choose, and what is the probability that you will win ?

      Your reasoning should apply in the exact same way. I'm not saying this to make you lose your time. By using these unique values for each parameter of the game, it will be easier for you to see what is the role of each thing.

      Delete
    16. (There was a minor cut and paste error in the original of this, right in precisely the wrong spot!)

      Thanks for your input. I agree that my wording obviously leaves a lot to be desired. I clearly wasn't expecting the interest of such people as yourself, so I didn't get a maths degree first! Seriously though, I should try to be more clear and I fully intend to so when I do my modelling. I don't want to confuse things at the moment with the Franken-Monty Hall Problem, but I suppose I could always come back to it.

      My intention is to do this:

      ---Colour the doors, left door Red, centre door White and right door Green
      ---Indicate whether a door is selected (S) or unselected (U)
      ---Name the goats (already done)
      ---Tag an opened door with O

      This will mean that the sample space will be populated by elements like this:

      (Red-S-C White-O-M Green-U-A)

      Which means the player selected the Red and the White doors and the host opened the White door to reveal Mary, while Ava was behind the unselected Green door.

      My question would then resolve down to: if you are playing the game, what is the likelihood, given that the host has already opened the White door to reveal the goat called Mary, that the car is behind the Green door?

      My answer is 1/2, based on the pretty obvious (to me) notion that the sample space consists of:

      (Red-S-C White-O-M Green-U-A) --- and --- (Red-S-A White-O-M Green-U-C)

      There is no other option. If you go back to the original spread of options, these two were equally likely right at the start and they are equally likely after the door is opened.

      There's no magic to it, as far as I can see.

      If this is more clear to you, and you now agree with me, then, as an anonymous mathematics PhD, could you please assist by using your authority to point it out to everyone else in mathematical terms that are clearly beyond me?

      If not, then I suppose I'll still be doing my modelling over the weekend :)

      Delete
    17. Not a mathematician, but it seems to me that the crucial problem here is that you're treating the two goats as separate events, when the standard formulation of the Monty Hall problem makes no such distinction.

      Delete
    18. Neopolitan, I think I can point out exactly where you’ve gone wrong, and I don’t think this is something that has been (specifically) pointed out yet. You’ve left one massive ambiguity in this problem you present that has been the cause of all of the debate to this point, and I think this has snowballed into an arduous trail of non-sequiturs that has taken this debate way into left field.

      In your first posting, you presented the problem with the following rule:

      >There are three doors, there is a goat behind two of the doors and behind the third is a car.

      Fair enough. This problem you’ve presented has (to this point) earned the title of Reverse Monty Hall, as these conditions (2 goats, 1 car) are the same as the original MHP, with the distinction being that the players choice is inverted (2 doors instead of one.)

      Your next posting is where things start to go south, when you “clarify” the conditions of the situation:

      >First things first. I did try to provide a couple of hints, one of which was that the goats are unique entities. To highlight this, I will give the goats names. There is the pretty, pleasant natured goat called Mary (M) and there is the unattractive, unpleasant goat called Ava (A). We might not care about the names of the goat, but I think we can agree that there are two goats and these two goats are not the same goat – meaning that they are not entirely interchangeable.

      Here, you’ve given your two goats names, but (you will admit) a goat by any other name is still a goat, no? A condition you didn’t set up is that the player can distinguish between the two goats! This is not an implied part of the problem you presented.

      So the issue here is that you’ve presented a different situation than the one that’s in your head.

      A better way to present this would look like this:

      There is a car, a goat, and an elephant, and each one hides behind one of three doors, randomly. The player picks two doors, and Monty opens one of those doors, revealing a goat or an elephant. What are the odds that the car is behind each of the remaining doors?

      Since the third term is always revealed, this problem can then be reduced to:

      There are two doors, behind one is a car, behind the other is either an elephant or a goat. What are the odds that the car is behind each door?

      I think we will all agree that this problem is neither difficult, groundbreaking, or interesting.

      This is NOT a reversal of the MHP. This is a problem with three unique entities behind the doors, while in the MHP there are only two.

      The title and the conditions from your first posting (see above) are the reason for all the debate, as both present an entirely different scenario than the one you’re attempting to solve.

      TL;DR – Everybody was right.


      Delete
    19. Then again, even this puts us in a precarious position, because you still have a 2/3 chance of the original set containing the car and a 1/3 chance of it not containing the car. So, perhaps whether or not the goats are distinguishable makes no difference, which would still leave you incorrect.

      Math PhD's, support me here.

      Delete
    20. Thanks Alex (both for commenting and also for not being another "Anonymous" (not that I don't also appreciate the engagement of those who are labelled "Anonymous", I just wish they'd give themselves a nom de plume).

      If I read correctly, you are saying that if it's a goat and an elephant, then the 1/2 answer is correct (per your TL;DR). The problem though, is that at least in the game on which the name of the classic problem is named, both the goat and the elephant would be "zonk" prizes (read all about here - http://en.wikipedia.org/wiki/Let%27s_Make_a_Deal). Therefore we could say that, in a sense, the term "goat" is interchangeable with the term "zonk". In terms of the scenario, the only really important thing about the goats and the elephant is that they are NOT the car.

      The question then is, if I am right when the goats are actually different (and they could be a goat and an elephant), but really goat=zonk=(NOT a car) and elephant=zonk=(NOT a car) ... where does it leave us? Am I still right, or am I now wrong? And if I am now wrong, why does defining the animals behind the doors in this way make any difference?

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    21. > There is no other option.

      You are right.

      >these two were equally likely right at the start

      I don't think so. Let me explain what I understood :

      You have three painted doors. Good.

      You are (rightfully) assuming that a priori each door has probability 1/3 of containing Ava, Mary or the Car.

      You are (also rightfully) assuming that a priori each door has probability 1/3 of being Open, Selected and Unopened.

      So, just to have the same vocabulary, you have three doors of different colors. Each door has two other variables which are the "object hiding behind" and the "status".

      If these two variables were independent, your answer would be completely correct. Because If these two variables were independent, then the probabilities of the event (Red-S-C White-O-M Green-U-A) and of the event (Red-S-A White-O-M Green-U-C) would be exactly the same in the set of all possible situation, which should be 1/36. And then you would rightfully conclude that knowing that "the white door is Open and contain Mary and the green door is unselected" (which is an event with probability 1/18), then the conditional probability of these two events is 1/2.

      So if we assume that the variables "object hiding behind" and "status" are independent, then your answer is perfectly logical. It shows a perfect use of conditional probabilities and everything.

      PROBLEM : The two variables are NOT independent. You can easily see that for the very simple reason that an Open door can never contain a Car. So this means that the probability of having a configuration with G-U-C has absolutely no reason to be equally likely to a situation with configuration G-U-A.

      So I'm not going to compute again everything in your scenario, but I urge you to reconsider this sentence :

      > these two were equally likely right at the start

      with what I just explain. I hope I was clear enough to make you understand where your mistake is. I reassure you, this is a very common mistake (or may be that's not what you wanted to hear).

      Delete
    22. Hi Mathematician,

      I think we might be getting somewhere. I think also that the clues to the solution are already available to anyone who wants to put the story together, spread across the articles here, the comments to those articles and my response to various exhortations to conform with the "staff answer" at reddit. So, perhaps I should keep my powder dry for a paper, but if there IS a paper in this, then I doubt that giving you the solution here here won't do much more harm. My idea is to put this into a more detailed article, but in short:

      We agree that there are six equally likely arrangements of goats and car at the beginning, before the game even starts. Using Red-White-Green and Ava (A), Car (C) and Mary (M), we have: ACM, AMC, CAM, CMA, MAC and MCA. All equally likely.

      My argument is that the Reverse Monty Hall Problem (RMHP), as with the Monty Hall Problem (MHP), consists of six possible games, or "mini-games", with each arrangement of goats and car being a different mini-game. When we start, we don't know which mini-game we are playing, and we have absolutely no information that would help us assign a likelihood of any of the mini-games we might be playing.

      Selecting two doors gives us no more information. However, when a door is opened, then we do get some useful information. Since all mini-games are equally likely and each randomly selected pair of doors is equally likely, let's say that the host opens the Red Door.

      Once the Red Door is opened, we get the information that we are not playing the CAM or CMA mini-game. But we could still possibly be playing any of the others, ACM, AMC, MAC or MCA - unless we can distinguish between Ava and Mary. It is my argument that these mini-games are still equally likely at 1/4 - if we can't distinguish between Ava and Mary - and 1/2 - if we can.

      This information about which door is opened and which mini-game we are playing is only available in a single iteration of the RMHP or MHP. The information as to which goat is revealed is only available in a single iteration of the RMHP or MHP. Once you generalise and start thinking about multiple iterations of the game, that information is no longer available and each of the mini-games is equally likely (which has been demonstrated via the standard simulations, as helpfully reported a few times here and at reddit).

      I really very very strongly encourage you to take three dice, one of which is different from the others (perhaps from a game of Risk). Put them in front of you. If the middle die is a "goat", pick it up and put it in your pocket. Otherwise, take the left die and put it in your pocket. Then, pick up the two remaining dice, one in each hand, and swap their positions.

      Then, try to convince me that, in a single iteration of the game - as the contestant, you could somehow tell that one of these two possibilities was more likely.

      (Slightly edited in last paragraph.)

      Delete
    23. Let's do this :

      > we have: ACM, AMC, CAM, CMA, MAC and MCA. All equally likely.
      > When we start, we don't know which mini-game we are playing
      > Selecting two doors gives us no more information.

      Absolutely.

      In your setting, the selection of two doors by the candidate is EXACTLY the selection of which of the three doors will be the Unselected one. This gives three possibilities for each of the 6 "mini-games" we are playing.

      This means that there are 18 possible situations after the choice of the two doors by the candidates, which are all equally likely.

      I'm going to denote these situation as A-CU-M (the situation is ACM and the player "unselected" the White door) or MU-A-C, and so on

      > However, when a door is opened, then we do get some useful information.

      First, let's focus on the different possibilities for the open door. As the player already selected its two doors, it means that there are only 2 possibilities left for Open door.

      So a priori, we get 36 possible configurations of the form AO-MS-CU (we are in the AMC situation, the player selected the Red and the White door, and the host opens the red door, reveling Ava). Or something of the form MO-CS-AU or even CO-AS-MU.

      As you can see, among those 36 configurations, some are impossible (the last one because the host cannot reveal the car). So these 36 configurations are NOT equally likely. The last choice of which door is going to be opened is NOT independent from the previous equally likely choices.


      Let's do two examples. Let's say that we are in the partial situation A-M-CU which has probability 1/18. This can lead to two final configurations :
      AO-MS-CU
      AS-MO-CU

      If we assume that the host opens a door randomly when he has a choice, these two configurations are equally likely and add up to 1/18. So obviously each of these situations is of probability 1/36.

      Now let's say that we are in the A-C-MU situation. This can lead to two configurations :
      AO-CS-MU
      AS-CO-MU

      One configuration is of probability 0 (the situation AS-CO-MU) because the host cannot open a car door. And the sum of the two probabilities must be 1/18. So the conf AO-CS-MU has probability 1/18.


      As you can see with this extremely explicit computation (I can hardly make it more explicit), the two situations AO-CS-MU and AO-MS-CU are not equally likely.

      > let's say that the host opens the Red Door.

      By the rules of the game, he reveals a goat. So there are 8 configurations corresponding to that :

      AO-CS-MU (1/18)
      AO-CU-MS (1/36)
      AO-MS-CU (1/36)
      AO-MU-CS (1/18)
      MO-CS-AU (1/18)
      MO-CU-AS (1/36)
      MO-AS-CU (1/36)
      MO-AU-CS (1/18)

      The total probability of these 8 situations : 4*(1/36) + 4*(1/18) = 1/3. This is exactly the a priori probability that the host will open the red door, which was the expected result.

      So now, what is the probability, KNOWING that the host opened the Red Door, that the car is behind the Unselected door ?Well, you just have to compute it using the formula for conditional probabilities. With obvious notations :

      P(CU | RO) = P(CU ∧ RO) / P(RO)

      We just said that P(RO) = 1/3.
      And for P(CU ∧ RO) we just add all the probabilities for the conf containing "CU". You should get 4*(1/36) = 1/9

      P(CU | RO) = (1/9) / (1/3) = 1/3

      So as you see, knowing that the host opened the Red door, the a priori probability that switching will give you the car is only 1/3.

      I cannot be more explicit than that, and I used your exact situation, with your notations. So I cannot imagine how you could disagree with this result.

      Delete
    24. > I cannot be more explicit than that, and I used your exact situation, with your notations. So I cannot imagine how you could disagree with this result.

      All I can tell you is, "prepare to be amazed!"

      Delete
  2. You say that after the host has opened either the leftmost or the rightmost door, then there are four possible scenarios (correct) and that they are equally likely (wrong).
    I agree that each of ACM, AMC, CAM, CMA, MAC and MCA is equally likely with probability 1/6. If it's MCA or ACM (a combined 1/3 of the scenarios), then the host has to open the leftmost door. If it's CMA or CAM (another 1/3 of the scenarios), then the host has to open the rightmost door.
    The remaining two scenarios, accounting for the remaining 1/3 chance, are AMC and MAC. These scenarios are split between leftmost and rightmost doors, since the host has to choose one or the other. Therefore, the chance of the host opening the leftmost door after AMC or MAC is half the chance of opening the leftmost door after MCA or ACM. Include that fact in your solution, and you get the obviously correct answer that staying wins 2/3 of the time.

    ReplyDelete
    Replies
    1. I think you have missed the point here, in the same way as pretty much everyone misses the point. You are considering the general while, in the scenario, you need to deal with a specific.

      You made a brilliant point on r/math:

      "Let me propose a scenario: I happen to look down and notice my shoe is untied, just as the host is opening a goat door. By the time I look back up, he's closed it again, and I have no way to know which of my doors he opened. Now he's asking me if I'd like to switch or stay.

      I claim that in this version, I have no information to go on, other than what I knew before the door was opened: the car is behind the door I didn't pick with 1/3 likelihood, and behind one of the doors I did pick with 2/3 likelihood. Since that's all I know, I choose to stay, and I'm 2/3 likely to win the car.

      Of course, I can force your scenario to become my scenario, by deliberately closing my eyes while the host opens the door. Do we conclude that closing my eyes makes the "always stay" strategy more effective than with eyes open?"

      This certainly does seem (but only it only seems) to put paid to my argument. I struggled with this as well: surely more information is going to **increase** the likelihood that you can make the right decision, rather than reduce it (as apparently seems to be the case, if I am right).

      The thing is that if you do not see which door (which specific door) the host opened to reveal a goat, then you are not able to eliminate two possibilities (out of the sample space of six), namely those in which the car sits behind the door that the host opened. I will have to work through this specific "eyes closed" scenario to work out why it resolves to 1/2, but in the short term, perhaps you might want to look at http://probability.ca/jeff/writing/montyfall.pdf (noting that this paper was not written by me).

      Delete
    2. I think that my argument shows that yours is wrong, by contradiction. But it's better to find the flaw in your specific argument, which I have done, and explained on r/math. Briefly, the four arrangements which are possible after seeing a door being opened are not equally probable, as you have assumed.

      Delete
    3. Maybe you should review that paper! His "proportionality principle" is exactly why you can't consider all the arrangements (MCA, CMA, ACM, etc...) equally likely after seeing a door open.

      Delete
    4. If (and only if) the likelihood of all the arrangements were not equal after the door is opened, then the "proportionality principle" may well be an excellent explanation as to why. However ... that wasn't the issue that I really wanted to draw your attention to.

      What I wanted to draw your attention to is that if Monty falls and knocks a door open, AND it just happens to have a goat and thus does not reveal a car, then the likelihood of benefiting from a switch becomes 1/2, rather than 2/3.

      Can I suggest a different situation? We can call it the "Monty Lies" variant of the Monty Hall Problem. After you have selected your door, the producer of the show tosses a fair coin with "Left" on one side and "Right" on the other and whispers the result into Monty's ear. Monty opens a door based entirely on the result of that coin toss, revealing a goat.

      Monty tells you that he selected the door based on the fact that there is a goat behind it, but he's lying. You don't know that he is lying.

      What is the likelihood of benefiting from a switch?

      Note that this scenario is directly analogous to my Reverse Monty Hall problem. You might not agree, but try assuming that it is.

      Delete
    5. In this post it seems like you just explained the correct answer to yourself, neopolitan, and then somehow in the last paragraph are fighting to figure out a reason that the answer is not correct.

      quoting from your description of the problem:
      "The philosopher will then open one of the doors you selected, revealing a goat."
      So having your eyes open or closed does not change anything about the possibilities "in which the car sits behind the door that the host opened"
      I read through that monty fall paper, and both of the variant scenarios that he describes do not relate to your problem, unless you are just somehow intending a situation other than the one you described.

      Delete
    6. The situation in the Monty Lies or Monty Falls games is a different situation from the one you have described.
      "The philosopher will then open one of the doors you selected, revealing a goat."
      This sentence is different from saying that monty opened a door completely at random and it happened to reveal a goat.

      Delete
    7. neopolitan said, "If (and only if) the likelihood of all the arrangements were not equal after the door is opened, then the "proportionality principle" may well be an excellent explanation as to why."

      No, there's no "if" about it: the proportionality principle tells you, unambiguously, that the probabilities are not all equal. Look at the "sister in the shower" example. Replace Alice and Betty with MCA and MAC, and replace "singing" with "the host opens the left door". Since the host *always* opens the left door for MCA, but only 1/2 the time for MAC, you must conclude that, given that the left door was opened, MCA is twice as likely as MAC.

      This is the paper that *you* cited, and it clearly does not support your conclusion.

      Delete
    8. Both versions of Anonymous and ChalkboardCowboy:

      I'm not really suggesting that any of the scenarios in the Monty Falls paper are a match for mine. What I am pointing out is that the paper indicates that, if Monty does not select a door deliberately, but rather falls against a door - a variant of which would be that he has a fit and randomly stabs a button which opens a random door - then the likelihood of benefiting from a switch is 1/2. What is being suggested here is that if you are not absolutely certain that the door opened for you is opened on purpose, or maybe even if you know that the door opened for you wasn't opened on purpose, then the probability can change - even if the door opened were the same door that Monty would open if he deliberately opened it. Many arguments against my position is that it is insane that probability might change on the basis of opening the door. However, we are told here that probability changes on the basis of the INTENT behind opening the door. I find this to be an even less reasonable suggestion. What makes far more sense to me is that the probability, in any single shot Monty Hall Problem (and in any single shot Reverse Monty Hall Problem), the likelihood of winning with either door is the same, at 1/2.

      I've suggested a way to test this out, yet no-one has indicated that they have tried it.

      Delete
    9. I'm not trying to talk about Monty Falls, Monty Crawls, Monty Lies, or whatever. I'm talking about *your* solution to the *Reverse Monty Hall* problem, as described on this page. First I gave you a reason to suspect you had a problem, since apparently by closing my eyes (to reject *reliable* information) I can improve my chance of winning the car.

      Then I took the time to find the exact flaw in your argument, explain it, and give you a reference in the very paper you cite (the "proportionality principle").

      All you've done is throw out different scenarios, and repeat that this is a "specific" event and so we can't talk about "general" probabilities, whatever that means. I asked you on r/math if that is an established concept, but you haven't responded. I looked in the paper you cited, to see if the proportionality principle was somehow contingent on repeating a situation multiple times, but it's not.

      Meanwhile, you've been tremendously condescending (here and especially in r/math), assuring us that you know the "cognitive dissonance" is painful, and that we're just not ready to accept the truth, and even that you'll have to "ELI5" for us (on reddit, that means "explain it like I'm five"). Look, you're obviously a bright guy. But you're *wrong*. And this isn't high school or your dorm where you can be the smartest guy in the room every time, and just bully and handwave your way out of a jam. Try that in academia and you may not ever recover the respect of your peers.

      I'm not going to engage any of the other scenarios you've tossed out. I've shown you the error in your RMH solution, and you haven't addressed it. If you choose to, you'll need to specifically justify treating all the arrangements as equally likely after a door has been opened (i.e., justify ignoring the proportionality principle).

      Delete
    10. Sadly, Chalkboard, I think there's no dealing with our friend here. In my discussion with him (above) and in his comments on /r/philosophy, his response to (valid) criticism is always the same: "You don't understand the situation, it's a specific situation, blah blah, and so on." Never does he refute any claim using numbers or logic, just dismissive arrogance.

      It's interesting to me that you came across this on /r/math. I think out friends repeated posting all over the internet, condescension, and argumentative nature are only a pathetic ruse to draw attention to his little blog.

      Delete
    11. I disagree that it's a ruse. I think he truly believes he's come across some new insight that has escaped everyone else up to now. It's ironic that he's telling everyone about their own "cognitive dissonance", because he's clearly the one who is so invested in having made a new discovery that he's blinding himself to the (multiple) clear explanations of his error.

      If he's planning to go into academia, I hope for his sake that he develops some better argumentative hygiene, or he's gonna have a bad time, as they say.

      Delete
    12. Truth. Held to any academic standards his argument holds little to no water, yet he insists we are the ones who need to think harder. I'm no mathematician (as stated, I found this on /r/philosophy), but I'm well-enough read in analytical traditions and logic to know bunk when I see it. I've been through the academic sieve, and I assume you have too. Telling somebody that a problem is beyond their comprehension and that they should just appeal to the (evidently) superior ethos of the author... yeah, doesn't fly. Sounds more like faith than empiricism, no?

      Delete
    13. Anon, I'm winding up a math PhD right now and, although my focus is not probability, I have more than enough experience to spot someone who is using words, rather than math, to make a point. His whole argument hinges (as you or another Anon pointed out above) on this distinction between a "single iteration" or "multiple iterations", which as far as I can tell is not an established notion in probability theory, and cannot be quantified mathematically.

      Delete
    14. Aye, that's me. If he's trying to make that point he'll need an accredited university and a good research team. And of course a stringent peer-review process. Hell, maybe he'll start a mathematical revolution of sorts if he can prove it.

      As it stands, he's a dude with a blogger account and a poor theory.

      Delete
    15. If he was *right*, then he wouldn't need any of those things. Unfortunately (or fortunately, for the consistency of probability theory), he's not.

      Delete
    16. ChalkboardCowboy,

      You asked at reddit "Do you have a reference for this distinction between "generalities" and "specific situations"? Is this standard terminology for an accepted concept?"

      No, I don't. It's just me scrabbling for terms to explain (I don't think I used the word 'generalities' though, perhaps I did, but at this moment it strikes me as wrong).

      As to quantifying mathematically, I did that. It's just conditional probability. Perhaps I could put it this way: if I were right that conditional probability is applicable, then would my application of conditional probability be mathematically correct?

      If so, then it's not a question of bad maths, or tossing words around, or mindless arrogance on my part, it's a question of whether or not conditional probability is applicable. I'm not going to use the standard ruse "prove to me that conditional probability isn't applicable". What I will try to do is present my modelling and show that the results are compatible with conditional probability.

      Delete
    17. neopolitan, I've shown you several times now exactly where the flaw is. Why do you not apply the proportionality principle to the four possible scenarios compatible with the leftmost door being opened?

      You've decided, with no justification offered, to disregard a principle that clearly applies to the situation. You then proceed to come up with an answer that everyone but you agrees is wrong (you call it "counterintuitive"). You need to have a very good reason, one you can explain, for disregarding the proportionality principle. But if you have it, you haven't shared it.

      Delete
    18. I should also add that, if your argument is amended to take account of the proportionality principle, then the answer (always stay for a 2/3 chance at the car) agrees with every other way of analyzing the problem. And that your "solution" without the proportionality principle implies at least one apparent paradox (that receiving new, true information can make the best strategy less effective than if you hadn't gotten the information).

      Is this not raising any warning flags for you?

      Delete
    19. ChalkboardCowboy,

      Sure, the apparent reduction in likelihood of winning the car if you keep your eyes shut is apparently paradoxical. From my perspective, it's a fascinating insight - I assume that, if I were right, it would also be a fascinating insight for you, would it not? Maybe worth a paper even?

      Delete
    20. It would be worth a dozen papers. But it's not right, I know it's not right, I know exactly how it's not right, and I've explained it to you multiple times. Why aren't you answering my direct question from two posts up?

      Delete
    21. Um, which one? I was addressing the "warning flags" question, I guess a little obliquely. Yes, it did raise warning flags for me, when I thought that it must be paradoxical. When I realised that it wasn't, the warning flags disappeared.

      The other question was "Why do you not apply the proportionality principle to the four possible scenarios compatible with the leftmost door being opened?" I think I am applying it. Perhaps you think I am applying it incorrectly and this would be consistent with you position on the applicability of conditional probability - the use of which brings us to the answer 1/2.

      I also asked you a direct question which you did not address: "if I were right that conditional probability is applicable, then would my application of conditional probability be mathematically correct?"

      By the way, I have written my stage by stage explanation of the scenario which should help either expose the errors in my thinking or help others to understand where I am coming from. It's being looked at by someone who helps me be less obscure (coy, evasive, confusing, pretentious, etc) than I might otherwise be. I can't push them to drop everything for me, so I have to wait.

      I know what ELI5 means, it's not intended to be an insult. If I am right, then I know that it's not easy to grasp and I suspect that it might be harder to grasp for someone who has been convinced ages ago that it cannot be right. If I am wrong, then I guess that I am just a pretentious prat. There have been situations in the past in which all the experts agree on something only to be shown to be wrong about it, so while it might (at least, a priori) be very very highly unlikely that I am right, it's not absolutely, totally impossible, is it? (Such a claim would surely be arrogance on someone else's part!)

      Delete
    22. Yes, the question about using the proportionality principle is the one I was referring to. You are not applying it, i.e. not reducing the probability of scenarios that could also go the other way (with the rightmost door being opened) relative to those which can only result in the leftmost door being opened. This step is not optional, or a matter of opinion--it's the way probabilities are calculated. The mathematical derivation is given in that paper you linked. If claim (as you have) that it should be ignored for this problem, then you need to give a very good reason. But so far, you've given no reason at all, beyond, "if I'm right, then it would be worth a paper".

      You're right, I didn't address your question, "if I were right that conditional probability is applicable, then would my application of conditional probability be mathematically correct?" It's a non sequitur. I know how to prove that 1=2 and that pi=4. Granted, each proof has a single problematic step, but IF the problematic steps could be justified, then my proofs would be mathematically correct (and worth no telling how many papers!). I hope you see the problem and the parallel here.

      Yes, I realize you know what ELI5 means; you're the one who used it. The parenthetical was for other readers who may not frequent reddit.

      Delete
    23. I removed your erroneous post (the one you deleted but to which information about you was still attached).

      The "worth a paper" isn't a reason why I might be right at all. That would be mere wishful thinking. I guess you my reasonably suspect that this is all just wishful thinking. All I was trying to encourage you to do was to go the extra step. Run the role play yourself. Or go get three dice, one of which is a different colour or style to the others (that'd be the "car"), and put them in front of you, remove the "goat", then pick up the other two and swap their positions. Then think about the probabilities.

      As you say, there are mathematical proofs for pi=4, so I might have to establish the fact of the matter before rolling up my sleeves (or begging a proper mathematician to roll up her sleeves) to do the proof.

      Delete
    24. Clarification: "As you say, there are erroneous mathematical proofs for pi=4"

      Delete
    25. Thanks for the delete. I think we're just talking past one another at this point. You don't seem interested in justifying your choice to ignore the proportionality principle, and honestly there's nothing else worth discussing.

      Delete
  3. I can point you to the error in your reasoning regarding this conditional probability approach.

    You say:

    "Pr(A) = the (a priori) likelihood that a car is hidden behind the door that you chose, but the philosopher didn’t open = 1/3"

    This is where error is introduced. The a priori likelihood that a car is hidden behind the one of the two doors that you choose, but that the philosopher didn't open, is 2/3. Given a random distribution of car, goat, goat, and a random selection of two of these, as you admit, 2 out of 3 times, a car will be in your selection. The philosopher will then open a door that does not contain the car, and it will reveal a goat. You know this going in to the situation (it is a priori).

    -B

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    Replies
    1. What you are talking about is not an a priori probability, B. You're (almost) forcing the solution by inserting your preferred answer there. Fortunately you are not actually forcing the solution, since Pr(A) doesn't factor into the equation. Sadly, some of the responses indicate that conditional probability cannot be used - this is annoying because if was universally agreed that conditional probabilities could be used, then my self-imposed task would be complete!

      Delete
    2. Fair enough. Admittedly, I am not familiar with conditional probability. I am about to make a post on one of our older threads, though.

      -B

      Delete

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