Talk:Monty Hall problem/Arguments/Archive 12

Distinguishable goats
Consider the two scenarios below. in both you know in advance that there are 3 doors, you can see all three doors. You know that behind 2 doors are goats and behind 1 door is a car. You know that you will be allowed to choose a door, that the host will open a different door to reveal a goat. The only difference is that in (A) You know that one goat is Black and the other is White, in B you know that at least one is black and in C you know nothing about the goats except that they are goats.

(A) The host reveals a goat, you can therefore see it's colour, you therefore have the knowledge of which coloured goat is "the revealed goat" and which is "the hidden goat" you have new infomation.

(B) If the host reveals a white goat, you know "the revealed goat" is white and "the hidden goat" is black. you have new information. If the host reveals a black goat you know "the revealed goat" is black and that there is some possibility "the hidden goat" is not black. You have a very small amount of information.

(C) The host reveals a goat of any colour. You gain information about "the revealed goat" (height, girth, number of legs, ears, horns etc. and fur color) but you knew you would gain these details about the goat revealed. The specifics of them cannot be contrasted with another goat and therefore while you have gained specific information about the revealed goat, you have gained no information about the problem.

To highlight this further consider (d) where there are empty doors instead of goats.

(D) The host reveals an empty space. You now know all detectable details about "the revealed empty space". You have gained no new relevant information.

If the goats are distinguishable, from the perspective of the contestant, then what is different about (C) than (D)? SPACKlick (talk) 12:09, 29 July 2014 (UTC)


 * Good, just approaching the point of the matter. Which door is opened by the host (#3 or #2) and which goat is shown (#A or #B) is irrelevant "if" we assume the unknown host to be indifferent in case he has two goats, given the contestant's initial choice was the prize. No need to condition on door number or goat's colour. Of course we CAN condition on door numbers and on goat's names or colour. But this would be an unnecessary laborious task, then. In case that there are no goats at all, but just two empty doors, we cannot condition on "emptiness" but would be limited to condition on doors (resp. their numbers, colours or functions). But fruitless if we assume the host to be indifferent which door he opens if the contestant by chance selected the prize. As to the paradox, we have to assume an unbiased host, in this respect. All we know is that his role is extremely biased, as he will never show the car, and by that giving birth to the paradox. Henze says the MHP is quite simple and, technically, has no bearing on "conditional probability". Gerhardvalentin (talk) 14:05, 29 July 2014 (UTC)


 * You've missed my point. From a contestant view perspective we CAN NOT condition on goat (any more than we can condition on the host existing), the contestant has no knowledge about goats on which to condition outside that which is preordained as background to the problem. Without some information distinguishing goats, merely seeing A goat doesn't let you condition on the goat. No matter how biased a host we presume, we gain no information relevant to the problem by a single goat being revealed. It transfers no information. I don't know how else to put it. So I'll ask the question again. In what way, would the problem, be different, at all, ever with empty doors rather than doors hiding goats? [All above still being asked from the position of the contestant] SPACKlick (talk) 14:54, 29 July 2014 (UTC)
 * I agree that it makes no sense to condition on either the goats or the doors because the player has no knowledge about the significance of either but some people insist that we must condition on the door revealed by the host, even if we then assign equal probability to the options the host has and get the same answer as if we had not conditioned. We can either ask the question 'Does it make any difference?', to which the answer is clearly 'No', or, 'Could it ever make any difference?', to which the answer is 'Yes', if for example we knew the host's door preference.


 * If we accept the argument that, because it could in other circumstances make a difference if we condition on the door number opened by the host, then we can, and indeed must, condition on the goat revealed. The fact that we only see one of the goats is irrelevant.  The question we must ask ourselves is whether we could tell, in a future game, whether the same goat or a different goat had been revealed.  In practice it might be easier to distinguish between doors than goats but goats are not bosons and thus are always, in principle, distinguishable.  Just as in the case of the doors, if we knew the hosts goat preference, which goat was revealed would be inportant. Martin Hogbin (talk) 15:58, 29 July 2014 (UTC)


 * To answer your question, if by empty doors we mean the doors had identically nothing behind them then there would clearly be a difference, in that case we could not distinguish between the objects revealed by the host, and Morgan's solution would be the complete one. There are no circumstances in which it would matter which nothing was revealed (not which door was opened).  The host could never, even in principle, have a preference for one nothing over the other.  In the stated problem there are two goats though.  One is revealed and one is not. In the case where the player knew that the host had a dispreference for the goat that was revealed he would know that winning by swapping was a sure thing.  You cannot do that with identical nothings. Martin Hogbin (talk) 16:08, 29 July 2014 (UTC)


 * Close but no cigar Martin. You have had to add an extra bit of information to make the statement you've made. "In the case where the player knew that the host had a dispreference for the goat that was revealed" In order to do this the contestant has to be able to identify the goat that was revealed as distinct from the goat that was not revealed. The player however cannot do this because he does not have ANY information about the goat not revealed. To know that this was the disprefered goat it would have to have some identifying feature (sya being white) that the player knew was unique to the dispreferred goat. Without the additional piece of information that is unique to this goat and revealed the revelation of this specific goat is informationally identical to the reveal of any goat.


 * This is distinct from the doors where he has information (such as location, colour, state of closedness etc.) about all 3 simultaneously. That being said, I do not subscribe to the argument that because it affects the probability from the hosts POV if we condition we should therefore condtion. but there are different reasons as to why we shouldn't condition based on Goats (contestant cannot possibly indentify or distinguish them) and the doors (pure symmetry by lack of knowledge of host strategy) SPACKlick (talk) 16:57, 29 July 2014 (UTC)
 * On the contrary, I suggest that I visit to Cuba on my behalf may be in order. As explained above, the question is not whether the goat revealed will matter in all circumstances but whether it might matter in any circumstances.  After the goat is revealed, the player could, for example, hear a member of the audience say, 'That is unusual, he hardly ever shows that goat', (how that audience member knows that fact is immaterial); that could not happen with identical nothings.  Goat identity could matter in at least one case so we need to deal with that fact.  To put it another way, goat 1 being revealed is a different event from goat 2 being revealed, however the goats are labelled.  That means that, even if I cannot think of it, there could be a case where it would matter.


 * Even if we are told (as in the original question)that the doors were numbered 1, 2, and 3 this information does not help us on its own. The only way we have or relating this information to probability calculations is by their role in this example of the game; we have also have such information on the two goats. Martin Hogbin (talk) 18:32, 29 July 2014 (UTC)


 * Again, if the player hears "wow he hardly ever reveals that goat" there is some unique feature of that goat, known to the audience member that allows the audience member to distinguish that goat from other goats. The audience member is using that piece of knowledge to pass on a reasoned conclusion. But in the question that piece of information doesn't exist. There is no circumstance where the specific goat revealed can matter (from the POV of the contestant) because the specific goat reveals no information other than it is a goat. Once you add any other piece of information (like the goats are painted 1 and 2) then you can condition on it. It'll then be pointless for the same reason as the doors, the number doesn't link to any other information. But without some way of identifying the goat uniquely, some way of knowing "aha it is this goat and not that goat" then there is no information to condition on from the contestants POV. It's "a goat" and even once it's revealed it's still "a goat" not distinguished from any of the millions of ur-goats that existed in potentia prior to the door opening.
 * Not so. Hearing "wow he hardly ever reveals that goat" is sufficient for the player to know that he has a better chance of winning by switching than he had before he heard that remark.  Hoe has no need to know anything about the hidden goat at all.   On the other hand, the revealing of one of ywo identical nothings can never tell you anything, under any circumstances. Martin Hogbin (talk) 10:55, 30 July 2014 (UTC)
 * It is sufficient BECAUSE it is dependent on the audience member having sufficient additional information to identify the goat. The Audience member must know some trait this goat has is unique to goats in the set or must know some trait this goat has which he knows the other goat does not in order to identify it. That is the piece of information missing that I've been talking about above. SPACKlick (talk) 11:23, 30 July 2014 (UTC)
 * I have agreed that it is harder in practice to distinguish between doors than goats but that is irrelevant.
 * To stress. With the doors, if they were un-numbered and a-locational from the perspective of the player. Like they're in a different studio. The player picks a door by pressing a button which stops a flashing light over one of the doors (he can't see this, it's in another studio) and then the host opens a door to reveal a goat and tells the contestant that he's opened another door with a goat in, the contestant cannot possibly condition on the doors, he has no way of identifying them. If the host shows the contestant an live video feed of the door he opened with the goat in it, but not of either of the other doors, the contestant still can't distinguish this door from the other two, because it is the only door about which there is information. In order to condition on something it has to be different from I, the background information. That there is "a goat" and "an open door" is in the background detail. Even when you've seen the goat, it doesn't add anything unless you have some way of making it distinct from the unrevealed goat whether or not this distinction will make a difference to the probabilities. SPACKlick (talk) 20:15, 29 July 2014 (UTC)
 * Sight is not the only form of knowledge. When a goat is revealed we know that it is not the same goat as the one that is still hidden.  We do not need to know anything about how it is different from the other goat, only that it is a different goat.  With the doors we might indeed see that the opened door is painted red and the others are not but this information tells us nothing of use beyond the important fact that it is the door that the host opened (and the obvious fact that all the doors are not the same door).
 * There can be a way the revealed goat can add something to your state of knowledge without knowing what distinguishes it from the other goat, as I have described above. There is no way that revealing one of two identical nothings can tell you anything.  The event 'goat A is revealed behind door 3' is a different event from 'goat B is revealed behind door 3', that is a plain and simple fact.  If we are going to analyse the problem thoroughly we cannot treat these two events as the same, even if it is not immediately apparant why it might matter if we do this.  We have no option but to treat '"nothing" A is revealed behind door 3' as the same event as 'nothing B is revealed behind door 3', where the "nothing"s are identical and indistinguishable by definition.


 * I am very puzzled by people's resistance to accepting that there are two different goats in this problem. Martin Hogbin (talk) 10:55, 30 July 2014 (UTC)


 * Your way a revealed goat can add something to knowledge required another actor with additional information. The goat alone provided nothing. I agree that "Goat A is revealed behind door 3" is distinct from "Goat B is revealed behind door 3" My argument is that in principle, from the contestant POV t"Goat A is revealed behind door 3" is indistinguishable from "Goat B is revealed behind door 3". There is nothing in the act to indicate it as one over the other. The player has no way of knowing, with any level of probability that "a goat" is or is more likely "Goat A" or "Goat B" without information unique to one goat or distinct between both goats. In order to refute it all you would have to do is show how the contestant, could in principle, make a distinction without use of this additional information, the reason you cant is that the two distinct acts are indistinguishable. SPACKlick (talk) 11:30, 30 July 2014 (UTC)


 * Ther is nothing in probabilty theory that says that I must 'show how the contestant, could in principle, make a distinction without use of this additional information'. The contestant cannot revise his probability estimate without additional information but neither can he use the door identities revise his probability estimate without additional information.  Seeing the three doors (if indeed he does) does not make a scrap if difference. Martin Hogbin (talk) 18:08, 30 July 2014 (UTC)


 * SPACKlick, if there are two goats behind three doors then there's exactly the same reason to condition on goats, as on doors. Because of course we can identify and distinguish even goats with the same colour and size. They ARE different: One of them IS SHOWN, and the other one was NOT shown. I am with you, what counts is only the host's POV. What do we know about the host's POV, his secret goat preference? I agree with you: Nothing. But who knows, PERHAPS the unknown host has a strict preference to show just only one of them, the one that he LIKES to show, that he ALWAYS shows if ever possible. For us the goats may look identical, but the host may know them by name, who knows! And it COULD be that the goat actually displayed is exactly the one that he uses to NEVER show, if ever possible. You are right, we have no information whatsoever on the host's goat preference, but he COULD just have shown the goat that he does not like to show, the one that he avoids to show if ever possible. Then it is obvious for us that switching is most likely to get the car, probability to win by switching almost 1. So why not condition on goats, likewise? Imho exactly the same reason as to condition on doors, imho "no different reasons" at all. Gerhardvalentin (talk) 18:58, 29 July 2014 (UTC)
 * No Gerhard, you've misread me. The only thing that matters is the contestants POV. The puzzle is about what the contestant should do, this can only be conditioned on what the contestant does or should know. From the contestant's POV there aren't 2 distinguishable goats. At the beginning there are two potential goats, which are entirely indistinguishable and I'm going to ask you again, what information does seeing the revealed goat pass on to the contestant that would allow her to distinguish it from the potential goat behind one of the other two doors? What information does seeing the goat give other than a door, say door 3, has a goat behind it? I'm not asking about information about host preferences, or strategies. The issue at hand is that information about one object, of which you know there are two, does not allow you to distinguish the objects. You need some information about both, even if that information is that some trait of the seen goat is unique. SPACKlick (talk) 20:15, 29 July 2014 (UTC)

Swap the goats for balls
(1)Here is a simple problem. There are three different coloured balls in an urn. One of the balls is red. The rules of the game are that we take a ball without looking at its colour and put it into a bag. The host must then look in the urn and remove a non-red ball from the urn. He must then offer you the choice between keeeping the ball that you originally chose or taking the ball left in the urn.

At this stage you can chose whether to play and bet. You can bet £100. If you end up with the red ball you get £50 (and your £100 back of course) but if you do not get the red ball you lose your £100.

(2)At this stage there is not much to gain by playing. You know you will have a 2/3 chance of winning by swapping and thus should be able to break even in the long run.

(3)Now the host shows you the ball that he removed. It is black. Do you want to play now?

(4)Finally someone truthful and knowledgeable says 'He only shows the black ball when he absolutely has to'. Now I will play, as I will win every time if I swap. I still have no idea what colour the ball that I originally chose was, but I do know that it is not the same one as the ball revealed by the host.

(5)Now try that with one red ball and two identical black balls. The game is never worth playing. Martin Hogbin (talk) 10:55, 30 July 2014 (UTC)
 * There is some similarity with the MHP, but no equivalence. The setup is such, that the first choice and final decision of swapping comes down to a Bernoulli trial with 2/3 chance of succes. Introducing the behavior of the host as dependent of the first choice, may be described by a variable H, with values 0 in the case the host does not show the black ball, and 1 if he does. In the situation (4) the event {H=1} is the same as swapping gets the red ball, and knowing you originally have chosen a black ball. I do not quite get what situation (5) means. In the foregoing the black balls also may be identical, i.e. not to be distinguished by the host. Nijdam (talk) 09:22, 31 July 2014 (UTC)
 * Thanks for your response. For the moment can we ignore the question of how this problem relates to the MHP and just discuss the problem as posed.  There are three different, and easiliy distinguished colours of ball in the urn.  We are told at the start that one ball is red but we are told nothing about the other two balls except that they are of different colours.


 * Do you agree that, in (4) above (game A) I would gain by swapping?


 * Do you agree that a different game (game B) were played with one red ball and two identical black balls there is would be no advantage in switching if you were told 'He only shows the black ball when he absolutely has to' (which would be somewhat nonsensicalin the circumstances.


 * What would you consider to be the sample space for game A? Martin Hogbin (talk) 10:58, 31 July 2014 (UTC)
 * Sorry, Martin, I read too hastely, and thought at the start there were one red and two black balls. Now I see you start with a red and two balls of different colour, different from red. I introduce variables X and H; X being the colour of the ball you initially choose, and H the colour of the ball removed by the host. Values (r=red, b=black, w=not red and not black):


 * XH
 * rb 1/3 p
 * rw 1/3(1-p)
 * bw 1/3
 * wb 1/3
 * In case (3) you may calculate 1-P(X=r|H=b) as the probability you'll win when you swap. P(X=r|H=b)=p/(1+p), hence tou win when swapping with 1/(1+p). In (4) I understand what you say as "the host only removes the black ball, if he's forced to do so, i.e. in the case the other not chosen ball is the red one, otherwise he'll remove the w-ball." This comes down to p=0. Now about (5): w=b, the black balls being identical, better: indistinguishable.


 * XH
 * rb 1/3
 * bb 2/3


 * conditional probability on the red when swapping: 2/3. Nijdam (talk) 13:01, 31 July 2014 (UTC)


 * Good, you seem to agree with me completely on this puzzle. We can, and should, condition on the fact that a black ball was revealed, even though we do not know the colour of the third ball. We need to calculate P(X=r|H=b).
 * Now suppose that, after the player has taken his ball, a fly settles on one of the two remaining balls inside the urn. The host must still never show the red ball.
 * The host removes a ball, which proves to be black and to have a fly on it.
 * How would you analyse this? Would you agree that we need to calculate P(X=r|H=b.H=f)? Martin Hogbin (talk) 17:47, 31 July 2014 (UTC)


 * Would you agree that the player wins by swapping with probability 1/(1+p) where p=1-P(X=r|H=b.H=f) and that we could calculate the probabilty of winning by switching if we were told for example, 'That is odd, the host does not like flies or the colour black so he would only reveal a black ball with a fly on it one time in ten if there were any choice in the matter.'? Martin Hogbin (talk) 17:47, 31 July 2014 (UTC)
 * Now what about the fly. Firstly a question: When the game is repeated, and again a fly lands on a black ball, is it the same fly as before? Nijdam (talk) 20:05, 2 August 2014 (UTC)
 * I am sure you knew where I was headed with the fly so let me go on to this version. There are three different-coloured balls randomly placed into three closed boxes numbered 1,2, and 3. One of the balls is red. The rules of the game are that you choose a box but do not look inside it.  The host must then look in the boxes and open a box to remove and show a non-red ball. He must then offer you the choice between keeeping the ball that you originally chose or taking the ball in the unopened, unchosen box.
 * You intially choose box 1 and the host shows you the black ball that he has removed from box 3. Should we now calculate (X=r|H=b,HB=3)?


 * Is this the same as the MHP? Martin Hogbin (talk) 11:52, 3 August 2014 (UTC)
 * No, it isn't. In the (standard) MHP the goats are indistinguishable. In this problem at the moment of decision, the player knows there is not only a red ball, but also a black one, which is revealed to him, and a third ball differently coloured. Nijdam (talk) 17:47, 4 August 2014 (UTC)
 * I what way are the goats indistinguishable? I that because we see only one of the goats? Martin Hogbin (talk) 22:36, 4 August 2014 (UTC)
 * Just as you introduced, in your problem with the three balls, 2 identical balls, in the standard MHP the goats are "identical", meaning not to be distinguished, as there is no refence to a specific goat. Of course we may formulate a version of the MHP with distinct goats, and see where this leads; fine with me.Nijdam (talk) 13:30, 5 August 2014 (UTC)
 * The goats are necessarily distinct, one is behind one door and one is behind another. The host knows there are two goats and knows which one they are revealing, they are distinct.
 * Whether or not they are distinguishable (other than by location) is another matter and one in two parts, because they could be distinguished by the host and not by the contestant or by both or neither. I see no reason to assume the physicists spherical goat in a vacuum here, I don't think it changes MHP significantly whether you do treat them as Distinguishable to all, or just to the host or to neither but I don't see a good reason for assuming one over the other 2. SPACKlick (talk) 13:44, 5 August 2014 (UTC)
 * Whether two goats are distinguishable is not a matter of opinion or conjecture it is a simple matter of fact. All real goats are distinguishable.  The only things that are indistinguishable are bosons and (hypthetical) objects which have been defined, either explicitly or by convention, to be indistinguishable.  For example, if the MHP had said, 'behind the others, indistinguishable goats', or, 'behind the others, identical goats', they could be treated as indistinguishable, but it does not say that.  This not some crazy notion of mine but the generally accepted  scientific and mathematical opinion.
 * Of course, it may be hard for the average person to distinguish between two goats but that is anothe matter altogether. Martin Hogbin (talk) 18:35, 5 August 2014 (UTC)
 * Martin, for ease of reference I've added some numbers into your above comment, I apologise if you feel that's intrusive.
 * (1) I agree this sets up an identical puzzle.
 * (2) I agree that your expected return is even
 * (3) No, I don't want to play at that point because I don't have any new relevant information since (2)
 * (4) Note, this knowledgeable, truthful person has to know both (A) the hosts preference for the black ball and (B)that there is only one black ball. (B) is equivalent to saying "can identify the revealed ball uniquely". And yes, these two pieces of information being added to the universe change the probabilities in the problem.
 * (5) True the game is not worth playing with 2 identical balls. Just as it wasn't at (2) and just as it wasn't at (3)
 * Additional (6) Note the game is not worth playing at (5) for the same reason it was at (3) and (2), because information to identify the ball didn't exist within the possible space of the contestant (And that host preference information didn't exist in that space either). At point (3) and at point (5) the two problems are not just "not distinguished" from one another they are "indistinguishable" from one another. The information to do so simply doesn't exist in the sample space. SPACKlick (talk) 11:38, 30 July 2014 (UTC)


 * The numbers are helpful and I agree with everything that you wrote up to (5).


 * I am not quite sure what you mean when you say that the two problems are "indistinguishable" from one another. In the first case we know that all three of the balls are distinguishable, in the second two are, by definition, indistinguishable, so the two problems themselves are clearly different.


 * The question being asked is whether we should play or not, there is no requirement to identify the balls (whatever that means). The host revealing a black ball is clearly a different event from the host revealing a ball that is neither red nor black and therefore separate outcomes including these two events must exist in our sample space. How we calculate various probabilities from our sample space and what additional information we later use is another matter.


 * I agree, of course, that we need extra information to make use of the identity of the ball that was revealed but that is just as true of the door identities in the standard MHP. The argument for including door numbers in the MHP exactly the same as I have given above for the balls. Opening door 2 is a different event from opening door 3 so outcomes including both these events must be in our sample space as a matter of principle, regardless of the fact that we need additional information to revise our probability calculation depending on which door was opened. Martin Hogbin (talk) 17:45, 30 July 2014 (UTC)


 * I would be interested in Nijdam's thoughts on the correct sample space for this game. It seems a straightforward probability calculation with three balls of different colours to me.  The fact that we do not know the colour of the third ball is not important, only that it is not the same colour as the other two. Martin Hogbin (talk) 18:08, 30 July 2014 (UTC)


 * The below was written below your last edit, I think it is still relevant but if I missed something, apologies.


 * When I say the two problems are indistinguishable I am still talking from the contestants POV. The contestant's relevant information at (2) and at (3) and at (5) is identical. The contestant can never distinguish whether she is at (3) or at (5) unless more information is added. When you talk about needing extra information to make use of the identity of the ball, you're assuming that we already have that information, but merely seeing the ball doesn't provide its identity unless we can distinguish it from the other balls it could potentially have been.


 * The argument for including doors is this. When the contestant picks door one, she knows what it means to have picked door 1 as opposed to door 2 or door 3, because she can see all three doors. Compare that to a situation using the contestant in another studio picking a door at random and then being shown the live feed of only the door they picked. They can't identify the door they picked, they can't distinguish it as an entity, unless they also have some knowledge about the set of doors which limits which doors it is. The same logic applies to the hosts door. The contestant can see the two possibilities and see how thye are different and distinguish which path on the probability tree she is on.


 * In short, observing an object, even in supreme detail, is insufficient to distinguish its identity unless you also have some additional information about the set of objects that could have fulfilled its function/role/position, therefore the contestant cannot identify the goat they've seen, cannot distinguish it from the other goat, or from the vast set of potential goats, therefore they do not gain any information by seeing it. Therefore it is meaningless, when talking about the probability from the contestants POV, to condition on the revealed goat, because "there will be a revealed goat, I will label it revealed goat" is identically true whatever goat is revealed and whatever goat is not revealed. SPACKlick (talk) 18:38, 30 July 2014 (UTC)


 * Additional post edits. Opening Door 3 and Opening Door 2 are only different events because door 2 and door 3 can be distinguished from each other. If the doors were bosons they would be identical events. From the contestant POV, the doors can be distinguished/identified [I use the two almost synonymously]. The goats however cannot due to lack of information. SPACKlick (talk) 18:38, 30 July 2014 (UTC)

Suppose that you are told at the start that the third ball is white?
Would that make any difference? Martin Hogbin (talk) 18:15, 30 July 2014 (UTC)
 * In short yes. The black ball is now distinguishable from the set of balls. You now have the additional information required to distinguish and identify the balls. As I said above when I set this up, if he knows at least one of the goats/balls is white and you reveal a non-white ball, he has additional information. He now has a probability tree with two branches, white revealed, Black revealed.


 * With two balls of identified colour (say white and black), the contestant has a set of possible events {Black revealed&White hidden, White revealed&Black hidden}. The information that the revealed ball was black narrows this possbility space. AS it turns out irrelevantly because the possibility space is symmetric.
 * When the two balls are both of unidentified colour to start with, the contestant has no set of possibilities to whittle down. OR rather a set with no specifics {There will be a revealed ball, it will have a colour&There will be an unrevealed ball, it will have a different colour}. This set is unchanged by the revelation of the colour of the first ball.SPACKlick (talk) 18:38, 30 July 2014 (UTC)


 * What if you were told that the third ball was less red than the red one?


 * What if you were told that it was a colour that was not red or black? Martin Hogbin (talk) 19:14, 30 July 2014 (UTC)


 * We need to be specific with that first question. What exactly are you told? There is 1 red, 1 black and 1 red but less red than red? Or that there is at least one ball with a lower R value in its RGB colour? The specifics of that information tells you if you can identify which ball has been revealed. In the first case, you can distinguish you have black not red but less red than red. In the second case you can't distinguish. The information is Red(255R), Lower than 255R, Some color Seeing a black ball doesn't tell you which this is. Seeing a bright yellow ball would tell you you had Some color and not Lower than 255R.


 * If you know the three balls are Red, Not Red and Not red or black. Then seeing a black ball does let you ditinguish, you know you've see the Not Red ball and not the Not red or black ball. But you are given information that can distinguish the two balls to start with. In the original scenario you have 3 pieces of information about the balls:


 * One ball is [Red]
 * Two balls aren't [Not Red 1, Not Red 2]
 * Not Red 1 is a different color from Not Red 2.
 * When a black ball appears you gain "Either Not Red 1 or Not Red 2 is black". You cannot distinguish which of Not Red 1 and Not Red 2 is the revealed black ball.


 * I think that method may be the best to explain it. At the beginning of the game we have:


 * Car,
 * Two goats with some properties Goat1{A1,B1,C1...}, Goat2{A2,B2,C2...}
 * (from background intelligence about the world) {A1,B1,C1...} =/= {A2,B2,C2...}. (Although A1 may equal A2 etc.)


 * When the goat is then revealed we can detect some or all of its properties. {An,Bn,Cn...}. We don't have enough information to narrow down whether our revealed goat is Goat1 or Goat2 so the two events are indistinguishable.


 * If we had some information, say at least one of the goats is black {Cn=RGB(0,0,0)} then our initial set up is changed by that information to


 * Car,
 * A Goat which is black Goat1{A1,B1,RGB(0,0,0)...}
 * A Goat with some properties Goat2{A2,B2,C2...}


 * The reveal of a not black goat, {Cn =/= RGB(0,0,0)}, would then let us distinguish that we were on the Goat2 path. (Note: Goat1 and Goat2 are arbitrary labels and can be switched without changing the argument). I'm not sure whether the reveal of a black goat in this scenario gives you any information. I will think about it further


 * This argument doesn't apply to the doors, because we can detect differences in the properties of the doors from the beginning. for instance


 * Door 1 {On the left}
 * Door 2 {in the middle}
 * Door 3 {On the right}.


 * Now the distinctions I picked there were arbitrary but when we can see all 3 doors we can distinguish them. In fact, even if we can only see two we can distinguish them


 * Door 1 {On the left}
 * Door 2 {Out of sight}
 * Door 3 {On the right}.


 * In the scenario I described above, where you are in a different studio, pick your door at random and then are shown the revealed door. Your starting information is


 * There are 3 doors [Door1{A1,B1,C1...}, Door2{A2,B2,C2...} and Door3{A3,B3,C3...}]


 * A door is revealed and you learn that the revealed door has properties {An,Bn,Cn...} that doesn't narrow it down to any of the three doors, so the three doors are still indistinguishable. This is the difference between the doors and the goats. Is that any clearer?SPACKlick (talk) 20:10, 30 July 2014 (UTC)

Sorry, I did not follow that. You say above that if you were told at the start of the game that the third ball was white it would make a difference. Would it make a difference if you were told at the start that the third ball was not red and not black? Martin Hogbin (talk) 21:51, 30 July 2014 (UTC)


 * Sorry, I went off on a thought there. The relevant section to that question is
 * "If you know the three balls are Red, Not Red and Not red or black. Then seeing a black ball does let you distinguish, you know you've see the Not Red ball and not the Not red or black ball. But you are given information that can distinguish the two balls to start with."
 * To extend that slightly. By being told One of the balls is neither red nor black at the beginning, that extra piece of information lets you non-arbitrarily label all 3 balls as


 * Red
 * Not Red or Black
 * Not Red


 * When Black comes out in the hosts hand that tells you that the host is holding Not Red and you know he is not holding Not Red or Black.
 * If any other color came out, say blue, You can't identify whether the ball the host is holding is either of the non-red balls.
 * As I said above with reference to the Black Goat I'm unsure but I think you can't then condition on balls because the contestant cannot distinguish which of the two balls was revealed. You can condition on "the revealed ball being blue" but that doesn't seem to mean much. I'm going to run the information states on that to see if it means anything at all.SPACKlick (talk) 22:06, 30 July 2014 (UTC)
 * Ok, having run the information states. You can condition (with the additional information that one ball is additionally not black) on the revealed ball being black and the revealed ball not being black. When the host reveals a non-Black ball You can't condition on NotRed and NotRedorBlack being revealed because the contestant can't distinguish them.SPACKlick (talk) 22:27, 30 July 2014 (UTC)
 * It was all my fault. The  reason that I could not make sense of your reply was that I asked the wrong question.  This made your answer much more complicated that I intended.  To avoid confusing existing threads I will ask my intended question in a section at the end. Martin Hogbin (talk) 09:26, 31 July 2014 (UTC)

The Initial Information
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
 * There exists a door [Door1{On the Left, Closed, Unchosen}]
 * There exists a door [Door2{In the Middle, Closed, Unchosen}]
 * There exists a door [Door3{On the Right, Closed, Unchosen}]
 * Behind One Door is a car [Car{Behind door n}]
 * Behind the other doors are goats [Goat1{A1, Behind door m}, Goat2{A2, Behind door l}]
 * You may pick a door [Add property {Chosen} to one door and remove property{Unchosen}]
 * Then the host will open an {Unchosen} door [add the property {Open} remove the property {Closed} from one of the 2 {Unchosen} doors] Revealing a Goat [Goaty{Ay, Behind door x}, where x = the open door]
 * Then you will be allowed to open a {Closed} door and win the prize {Behind door z} where z is the door you opened.

At the point where you make the choice (here assuming contestant chose 1 and host opened 3
 * There exists a door [Door1{On the Left, Closed, Chosen}]
 * There exists a door [Door2{In the Middle, Closed, Unchosen}]
 * There exists a door [Door3{On the Right, Open, Unchosen}]
 * Behind One Door is a car [Car{Behind door n}]
 * Behind the other doors are goats [Goat1{A1, Behind door m}, Goat2{A2, Behind door l}]
 * Behind door 3 is a Goat [Goaty{Ay, Behind door 3}]
 * Now you may open Door1{On the Left, Closed, Chosen} or Door2{In the Middle, Closed, Unchosen} and win the prize {Behind door z} where z is the door you opened [z=1 or z=2].

With that information you cannot distinguish Goat1 from Goat2 nor can you identify Goaty. Conditioning on Goaty{Ay, Behind door 3} is identical to conditioning on Door3{On the Right, Open, Unchosen}, because they are dependent.

You can't condition on Goat1{Ay, Behind door 3} because the contestant cannot determine from the available information whether or not that's the case and the same goes for Goat2{Ay, Behind door 3}.SPACKlick (talk) 20:31, 30 July 2014 (UTC)

What if the host tells us that the third ball is white?
I meant to ask whether you thought that we should condition on the revealing of a black ball if the host, as he reveals the black ball, tells us that the third ball (wherever it might be) is white. Martin Hogbin (talk) 09:26, 31 July 2014 (UTC)


 * I'll answer in the above style Information state notation. Please note I use {x,y}={w,z} to mean (x=w&y=z)OR(x=z&y=w)OR(x=y=z=w) ie. the sets contain the same elements, represented order is irrelevant eg. {1,2,3}={3,1,2}]I've added an apparently unnecessary ballz, to stand in for "the unrevealed, Not Red, ball" because it makes things clearer later.


 * Initial
 * 1 Red Ball [Ball1{Red, Not Chosen, Concealed}]
 * 2 Not Red Balls [Ball2{NotRed2, Not Chosen, Concealed} & Ball3{NotRed3, Not Chosen, Concealed}]
 * I will choose a ball and change it to chosen Balln{Colorn, Chosen, Concealed}
 * The host will then choose a Not Chosen ball and reveal it Ballx{NotRedx, Not Chosen, Revealed}
 * I will then choose between sticking with Balln{Colorn, Chosen, Concealed} and switching to Bally{Colory, Not Chosen, Concealed}
 * {n,x,y} = {1,2,3}
 * Ballz{NotRedz, ?chosen?, Concealed}
 * {z,x} = {2,3}


 * Final Choice
 * 1 Red Ball [Ball1{Red, ?chosen?, Concealed}]
 * 2 Not Red Balls [Ball2{NotRed2, ?chosen?, Concealed} & Ball3{NotRed3, ?chosen?, Concealed}]
 * Balln{Colorn, Chosen, Concealed}
 * Ballx{Black, Not Chosen, Revealed}
 * I can stick with Balln{Colorn, Chosen, Concealed} or switch to Bally{Colory, Not Chosen, Concealed}
 * {n,x,y} = {1,2,3}
 * Ballz{White, ?chosen?, Concealed}
 * {z,x} = {2,3}


 * All probability is conditioned on the initial I. So you can only additionally condition on differences from I. Here are things we can condition on
 * NotRedx = Black ["given the revealed losing ball is black"]
 * NotRedz = White ["given the unrevealed losing ball is white"]
 * {Not Red2, Not Red3} = {Black, White} ["given the two not red balls are 1 black and 1 white"


 * Note we cannot condition at all on Ball2 or Ball3. We can also see that all three things we can condition on have no effect on the probability calculations P(n=1) or p(y=1). We have no probability distribution of possible Colorx, Colorz, Not Red2 or Not Red3's. No known rule depends upon them. No other part of the problem can be written in terms of them.


 * To summarise, There are three things it is meaningful to condition on, it is not useful to condition on any of them.
 * We are interested in P(n=1) [and P(y=1)].
 * It makes sense to ask P(n=1| Colourx=Black), P(n=1| Colourz = White) and P(n=1| {Not Red1,Not Red2} = {Black,White})
 * All three of the above conditional probabilities trivially equal P(n=1)   — Preceding unsigned comment added by SPACKlick (talk • contribs) 11:45, 31 Jul 2014)


 * It will take me a while to read and understand all that. Are you using any standard probability approach?


 * In the meantime is that a 'yes' or a 'no'? Or do I take your last remarks to mean 'yes we can' but it is pointless? Martin Hogbin (talk) 13:02, 31 July 2014 (UTC)


 * Sorry, I wrote that in several stages and seem to have edited out the plaintext summary.
 * It is meaningful to do ask the Probability of the Red Ball being the Chosen ball Given the ball the host revealed was black and we are told the ball the host did not reveal is white. It is then trivial to show conditioning on these two events doesn't alter the probability of the red ball being the chosen ball. SPACKlick (talk) 13:35, 31 July 2014 (UTC)
 * I agree that, if we have no information about the host's colour preference, conditioning makes no difference to the probability of the red ball being the originally chosen ball. Presumably you agree that if we are told that the host never reveals the black ball, except when he has to, the probability of the red ball being the originally chosen ball is affected.


 * What if, instead if being told that the third ball is white, the host tells us that it is either white, yellow, or blue? Martin Hogbin (talk) 16:37, 31 July 2014 (UTC)


 * As long as the host tells us that we can determine of the revealed NotRed ball is different in a manner we can determine for the remaining NotRed ball we can condition on that specific property.SPACKlick (talk) 17:15, 31 July 2014 (UTC)


 * So if you were told that the third ball was not red or black that would be OK?


 * The problem does say that the balls are all of different colour, does that not amount to the same thing?


 * Can I just check? This argument is about whether, in the case where you can only see one of the things, two balls (of dfferent colour) can be treated in exactly the same way as two 'nothing's? Martin Hogbin (talk) 10:22, 1 August 2014 (UTC)


 * In answer to your first question, If you know that the balls are {Red, Black, Not Red Or Black} that is sufficient If you know they are {Red, Not Red, Not Red Or Black} that is only sufficient once you see a black ball because it becomes the first case.
 * We are talking about the point in the game where the host reveals a black ball. We already know that the balls are all different colours so you do know for sure that the balls are {Red, Black, Not Red Or Black}.
 * Ah, I mistook when you were being told. If you were told it at the beginning it makes a difference. If you are told it when the black ball is revealed, it makes no difference and adds no information.
 * What I said above is it depends on what you knew initially, if you knew that the one of the balls couldn't be black initially, then the reveal of a black ball identifies the ball and you can condition on that ball being revealed. If, however, you didn't know to begin with that either ball couldn't be black initially, you simply know that one ball cannot be black now, but that is after the balls identities have already been set "objectively", you cannot narrow the revealed ball down to either of the existing balls so you cannot condition on its identity.SPACKlick (talk) 13:50, 1 August 2014 (UTC)


 * In answer to your second question, no. It isn't the same thing. {Red, NotRed1, NotRed2} & NotRed1 =/= NotRed2 doesn't resolve when you learn that NotRedn = Black. Because you can't identify or distinguish the two balls in that case. Which means when you see the black ball it could still be either ball1 or ball2
 * I cannot understand what you mean. What does  NotRed1 mean?  How would we ever know which ball was NotRed1?
 * Ok, I may not have made my notation clear. To begin with you know there is 1 Red ball and 2 that are not red. We know the 2 Not Red balls are different colours I denote Ball{Colour of Ball} meaning the information we have is;
 * Ball1{Red}
 * Ball2{NotRed1}
 * Ball3{NotRed2}
 * NotRed1 =/= NotRed2
 * If we denote the revealed ball as Ballh{NotRedh} then what I'm saying is we can condition on "the revealed ball being black" that is P(C=X | NotRedh=Black} but we cannot condition on the revealed ball being either Ball1 or Ball2 that is P(C=X | h=1)SPACKlick (talk) 13:50, 1 August 2014 (UTC)


 * At the point when the host reveals a black ball, all the balls are distinguished. We do not know where they all are but we know exactly what they are, we know the balls are {Red, Black, Not Red Or Black}.  We know the host has the black ball and we have to calulate the conditional probability that the red ball remains in the urn, given that the host has removed the black ball.  You are inventing a problem which does not exist. Martin Hogbin (talk) 13:13, 1 August 2014 (UTC)


 * At that point we know we have a complete set of properties for the balls{Red, Black, Not Red Or Black}. However when the balls first appeared we were given information about them that made them distinct entities (that they had different colours) but not enough to distinguish them. If we ignore the red ball then it works as per below
 * We know we have Ball1 which is some colour that's not red. We know we have Ball2 which is another color, also not red
 * We then see a Black ball. So we know one of those colours is Black and one is not Black.
 * We don't know, and couldn't know, whether we are in the situation where Ball1 was revealed or where Ball2 was revealed
 * Therefore we Cannot condition on which ball was revealed.

Does that make it a little clearer?SPACKlick (talk) 13:50, 1 August 2014 (UTC)
 * Essay,_Monty_Hall_Spacklick_Information_Approach.pdf I'm going to try to be specific in response to your third question because our definitions here matter. I'm not so much arguing that the balls are the same as nothing (because the balls have properties) I asked about nothings to work out a starting point to discuss from, the understanding of why they're different and why it matters. The point I'm trying to make is that you cannot condition on "The revealed goat is goat1" or "The revealed goat is goat2" any more than you could condition on "the revealed emptiness is emptiness1". The only goat related condition, with any meaning, is "the revealed goat has {set of observable traits}" AND this condition, while it has meaning, is contentless. I'm drafting a document laying out the argument but it's taking some time to perfect. Please feel free to read the draft version. SPACKlick (talk) 11:04, 1 August 2014 (UTC)
 * I will read it but can we make sure that we are talking about the same thing. Martin Hogbin (talk) 13:13, 1 August 2014 (UTC)

Three scenarios
Here are three scenarios. The question I am asking is whether there is any significant difference between the way we would calculate the required probability in each case.

First
There are three different coloured balls in an urn. One of the balls is red. The rules of the game are that we take a ball without looking at its colour and put it into a bag. The host must then look in the urn and remove a non-red ball from the urn. He must then offer you the choice between keeeping the ball that you originally chose or taking the ball left in the urn.

Now the host shows you the ball that he removed. It is black.

We have to calculate the probability that the red ball is still in the urn.

Second
There are three different coloured balls in an urn. One of the balls is red. The rules of the game are that we take a ball without looking at its colour and put it into a bag. The host must then look in the urn and remove a non-red ball from the urn. He must then offer you the choice between keeeping the ball that you originally chose or taking the ball left in the urn.

Now the host shows you the ball that he removed. It is black. He tells you that the third ball is white

We have to calculate the probability that the red ball is still in the urn.

Third
There are three different coloured balls in an urn. One of the balls is red. The rules of the game are that we take a ball without looking at its colour and put it into a bag. The host must then look in the urn and remove a non-red ball from the urn. He must then offer you the choice between keeeping the ball that you originally chose or taking the ball left in the urn.

Now the host shows you the ball that he removed. It is black. He tells you that the third ball is neither red nor black

We have to calculate the probability that the red ball is still in the urn.


 * In short, no, there is no difference on how we can meaningfully calculate the probability (with a minor caveat). In all three cases it makes sense to calculate the probability of the ball in the urn being red given the revealed ball was black. In all three cases it makes sense to calculate the probability given we know the ball in the urn is not black. And in all three cases it makes no sense to calculate the probability given revealed ball being a specific ball. SPACKlick (talk) 14:03, 1 August 2014 (UTC)
 * Forgot to add the minor caveat, in the second scenario, you can condition on knowing the ball in the urn is white, rather than just not black. Trivial but included for completeness sake.


 * I realised I was quite short there on what "given the revealed ball was a specific ball means" it will be a little in depth but here goes. Everything I say is about the start of the problem until I say otherwise, because the initial information is the key here.


 * At the start of the problem we know we have 3 distinct and distinguishable balls.
 * We know each of these balls differs in color
 * We can therefore label each of the balls arbitrarily
 * Ball1{Color1}, Ball2{Color2}, Ball3{Color3},
 * We know one of the balls is Red, as all the balls are currently identically labeled(bare arbitrary suffix) we can declare one of those labels to the red ball.
 * Ball1{Red}, Colour1=Red
 * We know we will choose a ball, but not which ball it is. We will add the property {Bag} to the ball we place in the bag
 * We can arbitrarily label this ball Ballc{Colorc, Bag}, we won't know what color it is or what number
 * We know we will be shown a ball by the host. We will learn what color it is but not what number.
 * Why can we not assign it an arbitrary one of the remaining numbers, as we did with the red ball? Martin Hogbin (talk) 15:14, 1 August 2014 (UTC)
 * We can assign it any label we want that is previously unused or previously undistinguished. If we call Ball2{Color2, Host} then that excludes that ball from ever having been assigned {Bag}. All three balls must be available to be picked by the contestant. However, even if you complicate the maths by making Ball2 the ball revealed by the host, you still can't condition on Ball 2 being the ball the host will reveal, because you've just defined it as such within I. SPACKlick (talk) 19:22, 1 August 2014 (UTC)
 * We will add the property {Host} to that ball and label it. Ballh{Colorh, Host}
 * We know that ball will not have color red, h=/=1 and will not have property bag h=/=c
 * We know that we will then get to choose between the remaining 2 balls
 * Ballc{Colorc, Bag} and Ballr{Colorr}
 * We know that Ballc, Ballh & Ballr are the three balls Ball1, Ball2, Ball3 but not which is which


 * Now we move to After the host has revealed a ball to us.
 * We have learned is Colorh = Black. and by extension Colourr =/= Black.


 * Now there are 6 Arbitrary labels of balls but we can condition on none of them because [P(r=1) is probability red ball is in the urn)
 * P(r=1|h=1) - h=/=1 this can never be true.
 * P(r=1|h=2) - we can never determine h=2, conditioning on null information is meaningless
 * P(r=1|h=3) - we can never determine h=2, conditioning on null information is meaningless
 * P(r=1|h=h) - Tautology, h=h is part of I
 * P(r=1|h=c) - h=/=c this can never be true.
 * P(r=1|h=r) - h=/=r this can never be true.


 * Therefore we can never condition on any specific ball, only on the property of the revealed ball.SPACKlick (talk) 14:29, 1 August 2014 (UTC)
 * But the properties are fixed to the balls. The same ball is always the black one.
 * No, not from the contestants POV. There is no black ball in the contestant's universe until one is revealed, until then there are only Red and NotRed balls. SPACKlick (talk) 19:26, 1 August 2014 (UTC)

Fourth
There are three different coloured balls in an urn. One of the balls is red, one is black and one is white The rules of the game are that we take a ball without looking at its colour and put it into a bag. The host must then look in the urn and remove a non-red ball from the urn. He must then offer you the choice between keeeping the ball that you originally chose or taking the ball left in the urn.

Now the host shows you the ball that he removed. It is black.

We have to calculate the probability that the red ball is still in the urn.

Is any significant difference in the way we would calculate the required probability in this case? Martin Hogbin (talk) 15:11, 1 August 2014 (UTC)


 * In the case where the three balls are identifiable prior to the contestant choosing a ball it is meaningful to condition on the host or the contestant drawing a specific ball, so in this case it would be meaningful to calculate P(R=1|H=3) (assuming you labeled the black ball 3).
 * To answer the question literally though, I wouldn't say it was required, showing that white and black are trivially symmetric is a better solution. SPACKlick (talk) 20:10, 1 August 2014 (UTC)
 * Of course, I agree with your last statement but that applies to doors also in the MHP. My aim, as you know, is to show that, if you condition on doors, you must also condition on goats.


 * Are you saying that this problem is fundamentally different from the other three in some way? Martin Hogbin (talk) 09:37, 2 August 2014 (UTC)


 * Yes, The fourth problem is fundamentally different from the others. But without looking at the full analyses above I find it really hard to put in english.
 * In all 4 cases You know there are 3 distinct and distinguishable balls. In the first three cases you can't actually distinguish 2 of the balls except functionally [Ball I chose, Ball the host drew, Ball Remaining]. In the fourth case you can additionally identify them, prior to their assigned functions, as [Red, Black, white] and what you can condition on is how those two sets map to eachother. SPACKlick (talk) 10:41, 2 August 2014 (UTC)
 * I think that the reason that you are finding the difference difficult to put into English is that it does not exist.


 * I do not see the logic of arbitrarily numbering the balls than puzzling about which arbitrary (and completely unknown to the player) number is assigned to which colour. In all four cases, by the time that we (the player) have to make a choice we know that there are three balls, a red one, one a black one, and one that is neither red nor black. That seems a reasonable basis on which to distinguish the balls.  Martin Hogbin (talk) 11:19, 2 August 2014 (UTC)


 * It truly does exist. We really are going to have to work this from an "Information available to the contestant angle". I'll try once more in English but I've tried to start an information discussion below, working from the ground up so we agree at each step.
 * In scenarios where you know no traits of the Goats/unred balls first, you have two dstinct items you cannot distinguish. When you then pick a ball/chose a door one of those two items could be the item you protected from being revealed by the host, but you don't know which one. The host then reveals a ball/goat and all you know about it is it's colour and that it's the ball/goat. that's been revealed. Those are the only two things you know about that ball/goat. One of them you knew would be true from the beginning. The host will always reveal "the ball that's been revealed" because that's how it gets that identity. What you didn't know is that it would be black, and as I've said numerous times it is meaningful to condition on "the ball the host reveals" being "black". SPACKlick (talk) 12:15, 2 August 2014 (UTC)
 * Only in this instance of the game do we know that the ball that he reveals is the black one. In future instances we it could be the black or the not-black-not-red one.  To condition on the revealed ball being black we just consider games wher the black ball is revealed.  That is what conditioning is. Martin Hogbin (talk) 19:51, 2 August 2014 (UTC)


 * Where you say "In all four cases, by the time that we (the player) have to make a choice we know that there are three balls, a red one, one a black one, and one that is neither red nor black." What you're missing is that in the fourth case, we KNEW we were going to know that, we've known it all along and it will be true in EVERY case. In the first three cases we didn't. We didn't know what the color would be, so we only identify the ball functionally as "the revealed ball".SPACKlick (talk) 12:18, 2 August 2014 (UTC)


 * The only thing that matters is what the player knows at the point he has to decide what to do. At that point he knows that the revealed ball is the black one and that the other one is neither red nor black.
 * That's NOT the only thing that matters. Unless you define an initial, a background information against which you are adding conditions, then conditioning on something is meaningless. Conditioning means adding a condition in addition to the background. SPACKlick (talk) 20:51, 2 August 2014 (UTC)


 * This is the point on which we disagree in several places on this page. I am not sure how to resolve this.  Have you seen Nijdam's solution to the original problem?  He is quite happy to calculate the conditional probability P(X=r|H=b). Martin Hogbin (talk) 19:51, 2 August 2014 (UTC)
 * And the reponse to P(X=r|H=b) Depends on how you define B. At one point, Nijdam was talking about b as "The goat that is revealed" but we can't condition on that because |H=b means Given that the host reveals the goat that is revealed and we ALREADY know that to be the case.
 * If you define b as "the second goat" then the contestant can't condition on it because he can never distinguish the second goat from the first goat. He only sees one goat, he sees it after the goats have been differentiated and so he can't condition on it.
 * Read the sections on the information. Read them without trying to answer any questions other than, does the contestant know all of this to begin with and does the contestant know anything else to begin with. SPACKlick (talk) 20:51, 2 August 2014 (UTC)


 * The frustration is of course two way and it seems to be the norm for the MHP. Everyone seems to have their own opinion and I have never yet seen anyone change their mind.  One reason, which Richard Gill agrees with, is that in the end the MHP takes you back to the meaning of the word 'probability'.  I think it was Richard who said that in the last 300 years we have not made any fundamental progress in the subject.  Mathematicians have refined the manipulation of probabilities but still cannot tell us exactly what it means.


 * As I am sure you know there are two models of what 'probability' means; the frequentist model and the Bayesian model. I think it is generally accepted that, if the situation is modeled correctly (which is often the problem) the two approches should give the same answer to any question.  The frequentist approach has the advantage that it does give a clear answer to the question of what the probability of an event is.  Its is the fraction of times that the event would occur in a large number of repititions of the experiment; that difficulty is in deciding exactly what is to be repeated. and in the MHP this would be somewhere between the two, utterely unrealistic, extremes of: the same people set up the same show in exactly the same way and the same doors are chosen every time; and the entire game is reconstructed from scratch with a different venue, host, doors, car, goats and maybe even rules, every time that it is played.


 * For the MHP the most usual assumptions are that the rules, host, doors, car, goats are the same every time and the car and goats are distributed uniformly (at random) at the start. It is often taken that the player chooses a door randomly at the start and, althogh I thing that this is the correct way to do things, let us only consider cases wher the  =====A thought=====
 * Actually I have just had a thought. We (the player) do not need to be able to distinguish between the two non-red balls at all.  We just need to know that the host can. It is his choice that matters.  Do you agree? Martin Hogbin (talk) 18:49, 31 July 2014 (UTC)
 * No I wouldn't agree with that. Because if you cannot distinguish the balls you cannot probablistically determine, let alone determine for certain, which eventspace you are in, you have no given to condition on.SPACKlick (talk) 19:46, 31 July 2014 (UTC)
 * I understand the standard procedure to be that you set up a sample space containing all the possible outcomes. On that basis elements which are distiguishable (in principle) can be included even if they are not distiguishable or of interest in a specific case.  To some degree what is included is a matter of convention and taste and how much 'preprocessing' you want to do.


 * The argument being used by some people is that you should always set up the most comprehensive and complete sample space and then process this formally to get your solution. I can certainly see the merit is this method, especialy when the problem may have counterintuitive aspects, but there also is a case for spotting obvious and correct symmetries at the start of a probelm in order to make the solution simpler, expecially when the problem was intended to be a simple brain teaser and the complications were, in fact, unintended by the problem setter.


 * My argument is that if you take the former approach of setting up the most complete (using some reasonable criterion) sample space then you must include the two goats. I am not sure if you are convinced of this yet so I will continue the argument above. Martin Hogbin (talk) 10:07, 1 August 2014 (UTC)


 * I agree, that if we are setting up a sample space of all possible games we need to include the goat's locations as distinct. But that is an objective sample space. When calculating the probability in the question, we are not using an objective sample space because in any SPECIFIC game the sample space is either P(C=X)=1 or P(C=X)=0 because the car is in a fixed location.
 * Probability is dependent on information so we need to set up the sample space in terms of the information available to the Contestant. If you picture your sample space as a branching tree then this tree is the same as the objective tree with some branches superimposed over each other, where information is unavailable to the contestant. One of the differences between the objective tree of all games and the contestants tree of possible games is that any fork dependent on Goat1 and Goat2 is closed.SPACKlick (talk) 10:20, 1 August 2014 (UTC)
 * You have got a point but I would divide that information into three different classes:


 * 1 Information which has an effect on the probability of interest. This would exclude both goat and door identity in the MHP.  Both require additional information to affect the probabilty of winning by switching.


 * 2 Information which, combined with other information which we may later get could affect the probability of interest. Both door and goat edientity fall into this class in the MHP.  If we are told, for example, that the host never reveals the particular goat shown unless he has to, we know swapping is a sure thing.


 * 3 Information which could never matter. In an urn version of the MHP with one black ball and two indistinguishable (by definition) white balls, which white ball was revealed would fall into this category. Martin Hogbin (talk) 14:22, 1 August 2014 (UTC)


 * Except that because we are calculating the probability at the point where the contestant is making a choice, we know there is no more information to come. So there is no category 2 at that point. If the information isn't in the model at the point of the decision it's never going to be there, so if more information is required to make conditioning on something meaningful and purposeful then it could never matter because no more information will arrive.
 * It's the issue of perspective I've been talking about all along. This probability has to be calculated using the information the player knows. I, the background information, is what the player can know at the start of the game. Any conditions to that will be the additional information the player could know at the moment they make the decision. Any other information is irrelevant.SPACKlick (talk) 20:14, 1 August 2014 (UTC)
 * There is a category 2 because I defined it as, 'Information which, combined with other information which we may later get...'. You are using the argument that I, Gerhard, and many others have been using over the years to support the simple solutions; we have no information that allows us to distinguish in any useful way between doors (or goats) so we can, at the start of the problem and before formulating it mathematically, apply an obvious and intuitive symmetry.
 * No, I'm very much not using that argument, I'm using the argument that while you can identify and condition on doors with the available information, even though it's useless, you cannot condition on goats because you know nothing about the identity of the goats which you didn't know initially. There are only 4 things you know at the point of making the decision that were not in the initial information.
 * Which door you chose
 * Which door the host opened
 * Which door the other door is
 * The observable properties of the revealed goat (such as, it is Black)
 * These are the only things which CAN be conditioned on and it is a seperate argument that it is pointless to condition on any of them.


 * However, all those who preferred the Morgan solution still do. They argue that that it is better not to trust to intuition (as this problem demonstrates) but to proceed in systematic way covering all the options and then to allow for the assumed state of knowledge of the problem solver is the subsequent maths. If you approach the problem this way then you should include anything in my category 2, even if it later proves to be irrelevant.


 * The point that I am currently trying to make is that if you adopt a logical and consistent approach to the problem then you should either ignore the doors or initially include both the doors and the goats in your formulation. As we all know, with the information available to the contestant, neither makes any difference but could do if additional information were available.  Martin Hogbin (talk) 10:39, 2 August 2014 (UTC)


 * And I am diasgreeing with that point Martin. Have you read the informational approach I put above as yet which demonstrates the difference between the goats and the doors? The only way to formulate a solution to a probablistic problem in the whole is to identify the I, the background information, from the relevant point of view, and then look at every possible set of information at the point of decision. Because the contestant can know which door was opened, and that can very between games, you can condition on which door opened. Because the contestant can never know which goat was revealed, even though that can vary between games, you cannot condition on which goat was revealed.SPACKlick (talk) 11:00, 2 August 2014 (UTC)
 * The contestant does know what goat was revealed, he sees it. He does not know what the hidden goat looks like but he does know that it is not the same as the revealed goat.  If the game were repeated many times we could only count the times that the same goat was revealed.  We would not count the times what the other goat, whatever that looked like, was revealed.  That is what conditioning is, only counting events in which the given condition is met.  We have seen one goat revealed.  If we wish to condition on that event we only count times when the same goat is revealed.  The only requirement is that the two goats are distinguishable, in other words we would know if the other goat were revealed. Martin Hogbin (talk) 17:04, 2 August 2014 (UTC)
 * I honestly don't know how to progress further with this Martin, you have closed your mind. The contestant doesn't know that he can distinguish the revealed goat from the hidden goat. The contestant doesn't know which of the two goats he knew existed to begin with has been revealed. You say we would not count the times with the other goat, but we're not running it several times, across which we get to see both goats, conditioning is about "considering all the times where it was this goat revealed" and the contestant cannot distinguish those times from other times without gaining information about the other goat. The contestant lacks the information. IT's as simple as that and I'm sorry you can't see it.SPACKlick (talk) 20:47, 2 August 2014 (UTC)

How to progress
The frustration is of course two way and it seems to be the norm for the MHP. Everyone seems to have their own opinion and I have never yet seen anyone change their mind. One reason, which Richard Gill agrees with, is that in the end the MHP takes you back to the meaning of the word 'probability'. I think it was Richard who said that in the last 300 years we have not made any fundamental progress in the subject. Mathematicians have refined the manipulation of probabilities but still cannot tell us exactly what it means.

As I am sure you know there are two models of what 'probability' means; the frequentist model and the Bayesian model. I think it is generally accepted that, if the situation is modeled correctly (which is often the problem) the two approches should give the same answer to any question. The frequentist approach has the advantage that it does give a clear answer to the question of what the probability of an event is. Its is the fraction of times that the event would occur in a large number of repititions of the experiment; the difficulty is in deciding exactly what is to be repeated. In the MHP this would be somewhere between the two, utterely unrealistic, extremes of: the same people set up the same show in exactly the same way and the same doors are chosen every time; and the entire game is reconstructed from scratch with a different venue, host, doors, car, goats and maybe even rules, every time that it is played.

For the MHP the most usual assumptions are that the rules, host, doors, car, goats are the same every time and the car and goats are distributed uniformly (at random) at the start. It is often taken that the player chooses a door randomly at the start and, although I think that this is the correct way to do things, let us only consider cases where the player chooses door 1, just for simplicity. The host's decision should, in my opinion, be treated the same way and we should take it that the host chooses uniformly between legal doors. Morgan do not do this in their paper. Also those who wish to demonstrate the relevance of the 'conditional' solutions often use the example of a host who has a fixed door preference known to the person answering the problem.

Why am I saying all this? Because, although I am a Bayesian at heart, I think a frequentist approach is the best way to resolve our disagreement. Let us make some resonably realistic assumptions, that the rules, host, doors, car, goats are the same every time and the car and goats are distributed uniformly (at random) at the start of each game. The game is going to be played many times and we are going to attend one game. There were games before and will be games after the game that we see. We will take it that the doors have actual numbers on them and that the goats are, unknown to the player, called Alfie (who is white) and Bertie (who is black). Let us also assume for the moment, that the host does not open a door uniformly but might have some preference based either on door or goat. We can change that later if you like.

We (the player) pick door number 1. We can see that the door is number 1 and that we might have chosen doors 2 or 3. The host then reveals a goat (Alfie) behind door 3. We can still see door 2 and appreciate that the host might have opened that door instead, We can see the goat, it happens to be white, but we has no idea whether it is Alfie or Bertie. What is the probaility that you will win by swapping. From a frequentist perspective, it is the fraction of times that the player wins over all the games (past, present, and future) in which the host opens door to reveal the same goat that you saw. This is a simple mathematical fact, you can condition the whole set of games on the door opened and the goat revealed, that simply means that you only count the games in which those conditions hold. You do not need to know the name (Bertie) or colour of the other goat. To get any value from the conditioning you do, of course, need to know the host's policy. If the revealed goat were the one that he tries not to reveral then swapping would win every time that goat was revealed.

I think, though, that I can see your point. Suppose Alfie and Bertie are not available and two alternative goats have been used for your game. They are a black one (Charlie) and a White one (Dave) and the host does not care about colour at all but actually tries not to reveal the goat that comes last on his alphabetical list. In your case you will still see a white goat but you will win by swapping half the time. The point is that the property of the goat that we see must be the same property as that which the host uses to decide which goat to reveal. Does this make any sense to you and does it relate to what you are saying? Martin Hogbin (talk) 11:12, 3 August 2014 (UTC)


 * I think what you've said does relate and helps me narrow down the exact point of disagreement. Where you've said "We can see the goat, it happens to be white, but we has no idea whether it is Alfie or Bertie. What is the probaility that you will win by swapping.  From a frequentist perspective, it is the fraction of times that the player wins over all the games (past, present, and future) in which the host opens door to reveal the same goat that you saw.  This is a simple mathematical fact, you can condition the whole set of games on the door opened and the goat revealed, that simply means that you only count the games in which those conditions hold." This is where I disagree. When you run across all games (past, present and future) You're structuring those games with the objective or hosts perspective. The contestant has a wider set of possible games with hundreds of goats. Sometimes these goats are the same colour, sometimes they are not. Sometimes neither of them is a white goat. So it's this universe that he's conditioning in. As it turns out, with the additional knowledge of the host, the majority of games in that universe will never happen, but the contestant doesn't know that.
 * If we make a simplifying assumption that goats only have 2 properties, Colour and Name. This is to represent all those properties the contestant can detect when the goat is revealed and all those the contestant can't detect but the Host knows. When the contestant sees a white goat, he doesn't know it's Bertie. It could be any white goat. So while he can't narrow down his set of possible games to those where Bertie is revealed, he can narrow it down to the set of games where a White goat is revealed (including those games where both goats are white).
 * Essentially what we're disagreeing on is the universe of games which the Contestant is working in initially. You're taking a frequentist approach in the universe of possible games with the Hosts knowledge and I am arguing that you should be taking a frequentist approach in the universe of possible games with the contestant's knowledge and this includes some games we know as an objective observer to be impossible. SPACKlick (talk) 15:01, 3 August 2014 (UTC)
 * My set of resonable assumptions included the assumption that the same host, car, goats, and doors were used every time the game is played. It would be a rather odd version of the problem if these were changed every time.


 * You seem to be muddling up two different models of probability. Using the frequentist model we set up a sample space of all possible outcomes based on the information given in the problem statement.  It does not change as the game progresses it is a fixed space that we process mathematically.  We have agreed to consider only events which are described in the problem statement so we have no need to include the infinite range of possible outcomes that our imaginations can contrive.


 * This is not some madcap scheme that I have just made up, it is one of the standard methods of dealing with probability, as any probabilist can confirm. To try to convince you I have asked a really simple problem below. Martin Hogbin (talk) 10:08, 4 August 2014 (UTC)


 * Two disagreements here but we're getting closer which is good. "Using the frequentist model we set up a sample space of all possible outcomes based on the information given in the problem statement." We should be using that information in the problem statement available to the person whose persepective we're being asked the calculation in, here the contestant. I don't believe there's a difference here.
 * The sample space does not need to be restricted to information known to the player. What the player knows can be dealt with via the assumed distributions. Martin Hogbin (talk) 13:49, 4 August 2014 (UTC)
 * I fundamentally disagree with you here. Consider; In the actual game the car IS in a location. Therefore the probability of the car being behind any door is either 0 or 1, the only reason we calculate the probability as 1/3 or 2/3 is because the contestant doesn't know where the door is. It is this lack of information that makes our answer probablistic rather than certain. We can only make a probability judgement based on limited collection of information. The only collection of information that makes sense is the collection of information available to the contestant. It is from this information that we can set up "the prior", "The Initial", "The universe", "the sample space" whatever you want to call it. SPACKlick (talk) 16:53, 4 August 2014 (UTC)
 * I would accept that we do not include in our sample space anything that the player does not know is possible. But we must include what the player know could happen.  The player one goat but he knows there are two different goats both in elements of our sample space.
 * Maybe I have seen your point though.


 * Suppose the game is repeated and the player gets to the same point in the game. When the goat is revealed the player is asked, 'Is that the same goat you saw last time?'.  The player see what looks to him like the same goat, but can he be sure that it is, knowing that he may not have seen both goats?  Is that your point? Martin Hogbin (talk) 19:18, 4 August 2014 (UTC)
 * Exactly, that's how I should have said it days ago. Sorry for going round the houses on that one. He can't know, in a second run of the game if he's seen the same goat twice or two similar goats. My next point would be that, because of this he can never limit his sample space to an individual goat.SPACKlick (talk) 19:31, 4 August 2014 (UTC)
 * You would know if they were different though. Martin Hogbin (talk) 22:39, 4 August 2014 (UTC)
 * Yes, If you saw multiple runs of the game and If you knew there were only two goats and if the goats were easily distinguishable and If on at least 1 run of the game you saw each of the two goats then you would be able to distinguish them. But that's a lot of ifs the first of which isn't true in the MHP.SPACKlick (talk) 06:06, 5 August 2014 (UTC)
 * Having thought about it overnight, I sort of see a way round my objection. Given that the contestant knows there are two goats, and given that the contestant knows that the revealed goat is the revealed goat and not the other goat you can sort of model a condition with two assumed distinct goats, given that you've seen this one. However, since you knew whatever goat you saw you would model in the same fashion even if that condition was meaningful, and I'm still not sure it is, it certainly collapses under symmetry. That needs more thought.SPACKlick (talk) 09:31, 5 August 2014 (UTC)


 * And "My set of resonable assumptions included the assumption that the same host, car, goats, and doors were used every time the game is played. It would be a rather odd version of the problem if these were changed every time." I disagree about assuming the same goats every time because that isn't in the problem statement, isn't true of game shows I have seen with live animal "zoink" style prizes and massively restricts the sample space.
 * OK. Can we just consider the case where the goats are not swapped. It does not make any fundamental difference but just complicates things. Martin Hogbin (talk) 13:49, 4 August 2014 (UTC)
 * If we want to limit it to that I will need to know. Does the contestant know that it is always the same 2 goats? Does the contestant know anything about the 2 goats? I'm willing to run with just considering this subset of cases but I want us to agree on eactly which subset because it does make a fundamental difference to the sample space. SPACKlick (talk) 16:53, 4 August 2014 (UTC)
 * Yes the contestant does know? He only knows that they are different goats and that they are the same ones that are always used.  We can assume that he has never seen either before.  Martin Hogbin (talk) 19:18, 4 August 2014 (UTC)


 * I've answered your simple problem below but it doesn't contain any informational asymmetry, so it's not really relevant. SPACKlick (talk) 10:25, 4 August 2014 (UTC)

Suppose it were a matter of national security
As you know, I am attempting to show that the Morgan solution can be considered as complete. Here is a rather contrived scenario that illustrates the point that I am trying to make.

Suppose that the outcome of a particular game is a matter of national security. We not only need to know whether to swap or not but, as accurately as possible, how beneficial it is to us to do so and to maximise that benefit. We know the game rules, that the host always opens an unchosen door to reveal a goat and always offers the swap. We have read the Morgan paper but have somehow lost all out historical data on the show, so we do not know in advance what the doors or goats look like. We do know that the car and goats are placed uniformly at random but we are not sure how the host chooses which legal door to open when he has a choice (I think this matches Morgan's tacit assumptions quite well). For this reason, some of our spies are looking into the host's general background and preferences, his favourite numbers, colours and animals etc but we have not yet received a report from them.

Luckily we have got our man as the contestant on the show and he is secretly in contact with us so that we can give him advice. We have decided in advance that our man should initially choose door 1 (we think the doors are actually numbered). Our player chooses door 1 and the host opens a door to reveal a goat. Having read the Morgan paper, we ask our player for details of the door that the host opened so that we can correlate with the reports from our spies. I would also ask the player to dscribe, as fully as he could, the goat that was revealed in case that helped in any way. Would you? Martin Hogbin (talk) 13:07, 5 August 2014 (UTC)


 * Yes, if there is the possibility of more information existing at the time the decision is made then I'd ask him to describe the goat, where Monty was standing, what hand he opened the door with, any words he said to me or to others. I would ask for every detail I could get from my man inside. Because if I also get the information "he always opens with his left hand if he's opening door 2 and the contestant chose the car", his handedness is relevant. If I get the information "when he says switch and not choose in his patter you chose the car" then his wording is relevant. When there is more information to come all information could be relevant.
 * But, that's not the case here, there's no more information to come in, the system is closed at the point of decision with a finite, describable information state. SPACKlick (talk) 13:29, 5 August 2014 (UTC)


 * Yes I agree. In the problem as stated we have no knowledge of the things that you talk about so if we are to solve it we cannot make use of this information.  This, of course, applies to the door that the host opens.  We are given no information as to what the host's door preference might be or as to what significance the door numbers might have, apart from representing the roles of the doors in this instance of the game, so conditioning on the door number opened is completely pointless. Can we do it?  Yes, we can, you can condition on any given event.  Should we do it? Not in my opinion because it is obviously pointless.


 * I would split solutions to the MHP into three categories.


 * 1 Solutions of the problem as it was intended to be (essentially an urn problem with one black and two indistinguishable {by convention} white balls). It is, in my opinion, traditional in the case of simple brain teasers not to invent complicating factors that spoil the problem. As Gerhard points out, it is a natural assumption of the problem that the host does not give us any clue as to where the car is. People who get the answer wrong often go on to defend their wrong answer by giving other categories of solution as discussed below.


 * 2 Real-world solutions. Imagine you are a consulting probabilist and someone comes into your office and asks you a question along the lines of the MHP.  You would want to ask a million questions along the lines that you have mentioned above.  Interesting to talk about but the most important point here would be to ask yourself what exactly is it that your client wants to know.


 * 3 Intermediate solutions. These, in my opinion, are neither fish nor fowl and ultimately just end up as pointless arguments as to how far from the intended simple problem in the direction of the real-world problems you wish to go.  There is, in my opinion, one logical midway position, and that is, as advovated by Morgan, to use only events and choices mentioned in the problem statement and to parameterise unknown facts for posssible future use.  On that basis you sould condition on doors and goats if you want to be really fussy and comprehensive.  If you want to be downright perverse you might also consider the words spoken by the host. Martin Hogbin (talk) 09:12, 6 August 2014 (UTC)

A really simple example
There is an urn containing two balls of different colour. One of the balls is black. You pick a ball from the urn. What is the probability that you will pick the black ball? What is your sample space? Martin Hogbin (talk) 10:08, 4 August 2014 (UTC)


 * The sample space is those times where I pick a ball and it is black and those times where I pick a ball and it is not black. For completeness sake, the space where a ball is not black is made of tiny spaces where it is each of the set of possible colours but we can ignore that information and consider only the two events Black/NotBlack. That however doesn't contain the informational asymmetry of the Monty Hall Problem and therefore isn't particularly useful. SPACKlick (talk) 10:25, 4 August 2014 (UTC)


 * I understand what you mean about the asymmetry of the MHP but do not think it is relevant here. However. I will not persue this further as it causes unnecssary disagreement. Martin Hogbin (talk) 13:49, 4 August 2014 (UTC)

Back to the roots
The solution presented by in Parade shows the three possible arrangements of one car and two goats behind three doors and the result of staying or switching after initially picking door 1 in each case:

After she first selected a door, the contestant will remain within her scenario that she is in. The contestant knows that with 1/3 chance she selected the car and did arrive in the lucky guess scenario, and with 2/3 chance the car is behind one of the two unchosen doors, having arrived in the wrong guess scenario. She never knows where the car is, nor in which scenario she actually is in, but she remains there, even after the host's opening of another door. No way out. She remains there, irrespective of the door opened, and irrespective of the goat shown. Which one of the two unchosen doors was opened by the host (whose extremely biased role is to never show the car), or which goat was shown, doesn't matter at all, because the rules of the intended paradox firmly say that the host will never give further advice as to the actual location of the car, nor as to the scenario she actually is in. Ergo after the host showed a goat behind one of his two doors, her information regarding the scenario she actually is in did not change (Henze). Even if some mathematicians ignore the host's role regarding the intended paradox (Mlodinow, a physicist and mathematician, but no petty one: "the role of the host goes unappreciated". Any aberrant rules of other "game shows" do address variants, but never address the intended paradox. The host's role has to be shown clearly and without ambiguity. Gerhardvalentin (talk) 14:10, 1 August 2014 (UTC)
 * See my last post in the section 'A thought' above. Both the goat and door identities fall into my class 2; information which makes no difference unless we get some further information (which we do not).   Martin Hogbin (talk) 14:26, 1 August 2014 (UTC)
 * Martin you are right, as to the intended paradox, we impossibly can get any further information. And the article should show the host's role without any ambiguity. He cannot open all three doors at once nor arrange for transparent doors, as Morgan et al. (1991) sneakily did. See Henze (1997) and Mlodinow (2008): "... the role of the host goes unappreciated." Gerhardvalentin (talk) 17:55, 1 August 2014 (UTC)

Do we still agree it's meaningless? Comment
I didn't want to derail the discussion above but Martin Hogbin, Nijdam, do we all agree that ultimately this conversation will have no effect on the final answer? The answer will be 1/(1+p) and p/(1+p) where p is the hosts preference for the door that was opened given the exact nature of the current situation, in as far as we consider it potentially including goats, days of the week, tie colors and breakfast choices. And none of the discussion above will give the contestant any more idea as to the value of p.SPACKlick (talk) 10:20, 14 August 2014 (UTC)


 * Yes I do agree but add that, lacking any information we have to take p to be 1/2, so the final answer is 2/3. I also agree that p, like any probability, can only take a value from 0 to 1, so the probability of winning by switching can only vary from 1/2 to 1, subject to the normal game rules.


 * I would also point out that the answer is also 1/(1+z), where z is the is the host's preference for the goat that was not revealed given the exact nature of the current situation, in as far as we consider it potentially including the door which hid the goat that was revealed, days of the week, tie colors and breakfast choices. There is no reason why we must express tha answer in terms of doors.  Martin Hogbin (talk) 20:31, 14 August 2014 (UTC)


 * Agree on both points (although I'd have worded it differently). We don't take p to be a half, we take the average p over a universal distribution and that average is 1/2 by symmetry. And your right, we shouldn't focus on doors, or goats. p is the preference for the door/goat combo given all other factors. SPACKlick (talk) 20:41, 14 August 2014 (UTC)


 * I certainly agree that we should use the preference for the door/goat combo; that has been my point all along. To consider the door opened but to ignore the goat revealed is an incomplete solution.  Note also that the presence of two goats is mentioned in the problem statement; days of the week, tie colors and breakfast choices are not.


 * We can take p (the probability that the host will choose door 3 given the car is behind door 1) to be 1/2 without averaging. It is a fundamental (Bayesian) principle that we must take probability to be evenly distributed over all the possible options, if we have no information to the contrary.  This is probably why many get the answer to the MHP wrong, they distibute the probability of the car being behind each of the two unopened doors evenly, because they do not realise that we have been given some information by the host. Martin Hogbin (talk) 20:54, 14 August 2014 (UTC)


 * I have just realised that my last statement puts a but of a spanner in the works of the argument just above that the host must not give any clue as to where the car is. He actually has to do this to make the problem work. Martin Hogbin (talk) 20:56, 14 August 2014 (UTC)


 * My point above (and we'll finish that discussion above) is that the contestant cannot differentiate Door3withGoatA and Door3withGoatB and so he can only consider preference for Door3 but I'm glad we agree that ultimately whether we consider doors or goats or tie colors or whatever, ultimately every piece of preference we consider has to be resolved to 1/2.


 * On the host not giving a clue argument, that reason is why I'm certain Gerhard is misreading or at least overstressing the Henze passage and wish I could read it in English rather than my very broken German where I lose all the subtleties of stress and context SPACKlick (talk) 21:02, 14 August 2014 (UTC).


 * I cannot read German at all so I cannot comment on Henze.

As to the intended paradox, we have to assume that the host chooses randomly when he has a choice: "then he is equally likely to open either non-selected door." (MH revisted). In conformity to Henze, who explicitly says "in effect, the MHP has nothing to do with conditional probability theory". Gerhardvalentin (talk) 17:07, 15 August 2014 (UTC)
 * Yes, I do agree that it is all a bit pointless because, from the likely state of knowledge of the player, the answer is 2/3, even Morgan et al agree that now. I will respond to your comment about the goats in a new section above. Martin Hogbin (talk) 08:56, 15 August 2014 (UTC)
 * Not pointless at all, because it IS necessary to distinguish the intended paradox from other variants.
 * Not even slightly, do don't have to assume anything about the hosts protocol because even if the host was 100% biased against a specific goat/door combo the contestant can never know this and either by the Bayesian rule or by averaging protocols using symmetry the contestant can only act as if the hosts choice is random. No other strategy will be more accurate. SPACKlick (talk) 17:47, 15 August 2014 (UTC)
 * So you finally approach the intended paradox? Regards, Gerhardvalentin (talk) 18:10, 15 August 2014 (UTC)
 * I am where I have always been, we can condition on doors, we can condition on goat properties and doing either changes nothing about the expected return of 1/3 vs 2/3 because we don't have information. SPACKlick (talk) 21:22, 15 August 2014 (UTC)
 * Splendid. As to the host's secrecy regarding the paradox: Actually, the contestant irrevocably is either in the lucky guess scenario (1/3) or in the wrong guess scenario (2/3), and the actual scenario she actually is in, is unalterable. By switching, she'll obviously get the car with probability 2/3 *whatever* strategy is used by the host. This fact is unchangeable. Henze: But quite another issue is her actual KNOWLEDGE about in which of both scenarios she actually is in, after the host did show a goat. Henze enunciates regarding the paradox, that her ACTUAL KNOWLEDGE regarding the specific actual scenario that she is in, cannot be "better" or "closer" after the host's opening of a door in order to show a goat. Otherwise the host would have failed in keeping secrecy regarding the actual location of the car, and this is what we have to understand in distinguishing the paradox from other variants that are used in textbooks to teach conditional probability theory. Imo the article should clearly show that difference, and it must keep this clearly apart, that's all what matters. Gerhardvalentin (talk) 05:58, 16 August 2014 (UTC)
 * Sorry Gerhard as I've said a dozen or so times now, I believe you're overstressing Henze's point here. But none of this maths changes any of that, the point is to show accurately the 1/3, 2/3 split and until you have mathematically accounted for the lack of knowledge about the hosts preferences you haven't fully answered the question. And in fact as the question is "should you switch" Showing that whatever tactic the host is using you are always at least as well off, and usually better off, to switch is a more complete answer to the question. When you start assuming a specific preference on the host's part you're beyond the standard MHP but accounting for the fact that the host might have a preference and working out what effect that fact has on the contestants chances (as it turns out, no difference because of lack of information about which preference) isn't breaking the MHP it's answering it more fully. SPACKlick (talk) 08:10, 16 August 2014 (UTC)
 * Sorry, but I hear some editors say "you only can tell the overall probability" to win by staying or switching, but never the probability for the actual game. You can read this argument here not only one dozen or so times. Without Henze and others who know to distinguish the intended paradox and other variants, this argument will survive. And survive. Sorry, no way out. Gerhardvalentin (talk) 10:57, 16 August 2014 (UTC)
 * Spacklick, there is no rule in mathematics that everything must be done the hard way. It is posssible to observe at the start of the problem that it is symmetrical with respect to door number.  Changing the door numbers round at the start of the problem can make no difference to the answer.  Spotting symmetries and using them to solve problems, which might otherwise be very difficult, is something mathematicians do.  It is not considered a bad or sloppy approach but a skill to be commended.  In the case of the MHP the symmetry with respect to door number is quite obvious to most people, who probably apply it without thinking too much.  There is, of course, a false symmetry between the originally chosen door and the remaining door after the host has shown a goat and this is what leads most people to get the wrong answer, but the overall door symmetry at the start of the question is natural, obvious, and correct.


 * The problem can be made into a useful educational tool by hinting at some knowledge of a host door preference. The removes the door-number symmetry and requires a conditional solution.  Martin Hogbin (talk) 20:49, 16 August 2014 (UTC)


 * By the way, I do not know what Henze said but what he should have said is that the host must give no clue to the location of the car, beyond that required by the rules of the game. Martin Hogbin (talk) 20:51, 16 August 2014 (UTC)

Doors
We have to end the discussion about the doors. It is commonly accepted, no, I better say it is a given fact, that the contestant, and we, see three doors. I mistrust all the authors who try to get accepted that the doors only are distinguished by their role. Many of them use this argument to solve (to hide their original mistake) the problem with the simple explanation. Of course you may label the chosen door number 1, but that does not mean the door, chosen in further instances of the game, will also be number 1. Or if you want to insist in calling the chosen door number 1, this number it's merely a rephrasing of "chosen door", and does not refer to a physial door. Let's end this nonsensical way of reasoning. Nijdam (talk) 18:08, 4 August 2014 (UTC) For good understanding I'll add here that refering to the function of a door is no more than introducing random variables, like C,H,X as I did to describe the sample space. They do not label a specific physical door. Nijdam (talk) 13:23, 5 August 2014 (UTC)
 * I think that's a key point to realise. The doors have identities, we can label those doors by their identities, Traditionally {1,2,3} or I have seen {leftmost, middle, rightmost}. These labels are equivalent.
 * The doors also have functional roles based on who chooses them. {Contestant, Host, Remaining} or however you choose to label them. In each specific run of the game, when the contestant makes the choice, you can identify how these two sets of labels map, and the mapping will be distinct each time.
 * The doors further have functional roles relating to what they hide {Car, Goat1, Goat2} when you comes to make the choice you only know what one door, say {3}, doesn't map to one of these labels {Car}, the rest of the mapping isn't known to you. The host knows that mapping from the beginning.
 * Similarly the Goats have two sets of identities. {Goat1, Goat2}, their actual identities, and {Revealed, Concealed}, their functional roles. Within a single game you don't know how those two sets of identities map, however once a goat is revealed, the host does. SPACKlick (talk) 09:26, 5 August 2014 (UTC)


 * @Nijdam, I don't say this in order to hide "their" original mistake, but in effect there's no difference between door numbers (1, 2, 3), door names (leftmost, middle, rightmost), or door functions (the one originally chosen, the one having been opened and the second still closed one). They all are "distinct", and it is possible to condition on ANY of them, given you are persuaded that the host just opened his preferred door resp. he just opened his avoided door. And there are two distinct goats, so you can condition on them as well. You can condition on the goat shown, given you are persuaded that the host just showed his preferred goat resp. he just showed his avoided goat. And you also can condition on the goat that was not shown, given you are persuaded that he did not show his preferred goat or did not show his avoided goat. – But quite another question, regarding the intended paradox, is: what are we ALLOWED to assume? The question is, whether the host MAY BE assumed to provide for transparent doors (contradictory to the relevant literature), or MAY NOT be assumed by us to provide for transparent doors (see Henze and others). As to the paradox, the host's bias is reduced to never show the car, and – as to the literature – we "may never assume" any host's bias regarding doors or goats. Gerhardvalentin (talk) 11:11, 5 August 2014 (UTC)
 * Gerhard, while it is trivial to show that there is no difference between {Door1, Door2, Door3} and {Left, Middle, Right}, one being a numeric label and the other being a label by property but both assigned by the identifiable properties of the door. It is false to say that there is no difference between that style of label and a functional label. You cannot condition on "The host opening the door the host opens" because that is necessarily true and part of the initial information. You can only condition on the host opening a door identified in some other way, say rightmost.
 * As for your comments about not allowing host bias, nothing in the problem suggests we should ignore host bias. The argument you are making does smack of an excuse for the simple solution. It is trivial to make the case that with no information about bias or selection strategy the contestant can only calculate even chance for either door when he has selected the car. SPACKlick (talk) 11:38, 5 August 2014 (UTC)
 * Thank you. May be trivial, yes. But as to the literature however, the host of the intened paradox has to be assumed to choose evenly (flipping a coin,or so). This is constitutive to the intended paradox. But, in departing from the paradox, we offhand can condition on his preference for the door opened versus the door not opened in case he had a choice (not 1/2 e.g., but 1/4 or 7/9). Gerhardvalentin (talk) 12:28, 5 August 2014 (UTC)
 * It is not constitutive at all. It should be calculated as part of the answer. Both ways of doing it are valid, 1) to show that on average across all possible strategies the host acts as if choosing at random or 2) to show that in the extreme cases of host strategy it is still of advantage to the contestant to switch. You could of course make it an explicit assumption (as is properly done with the unstated parameters that the host always opens a door containing a goat that the contestant did not choose and always offers the choice) but it is not in the MHP itself. SPACKlick (talk) 12:49, 5 August 2014 (UTC)

"In all these considerations, of course, it is crucial that the car-hiding door must be kept secret by the host, who is also obliged to open a goat-hiding door." Please note the words "of course, without any question...". And you know that any host's door preference provides for transparent doors. Henze knows to distinguish the world famous paradox from the matter of teaching probability calculus, and is using any "host's door preference" only in later chapters, in teaching probability calculus. Gerhardvalentin (talk) 16:06, 5 August 2014 (UTC)
 * So you contradict the literature. Norbert Henze, an eminent professor of probability theory and stochastics (btw, he knows in person both Richard D. Gill as well W. Nijdam) says in his book "Stochastics for Beginners.  An Introduction to the fascinating world of randomness (chance), 9th Edition":


 * There is no contradiction Of course the car hiding door must be kept secret, but that merely means the host doesn't say "the car is behind door 1" or open the car door, it doesn't mean the host can do nothing to indicate the location of the car door, because he indicates the location of the car door in any formulation of the game, at twice as likely behind the remaining door as the closed door. You're over-extending the meaning of that passage. SPACKlick (talk) 21:13, 6 August 2014 (UTC)


 * It is a pity that no English translation of Henze is available. Ithought that I had seen a translation of a passage somewhere that made his thoughts clearer. Whether or not Henze says it though, it must be a fundamental principle of the puzzle that the host is not allowed to give any clue as to where the car is, either in an obvious manner, like opening the door which hides it, or in a more subtle manner, like choosing a specific door when the player has originally chosen the car.  Once that principle is breached all sorts of other, obviously unintended, possibilities arise, such as the host giving a clue to the player by the words that he uses.  Martin Hogbin (talk) 09:50, 7 August 2014 (UTC)


 * Regarding Nijdam's original comment, there is nothing special about doors. Doors, goats, like all real objects (except bosons), are distinguishable.  I use the word 'distinguishable' in the standard sense used in physics and combinatorics to mean that they are, in principle, different objects and that they can, in at least one circumstance, be distinguished. Indistinguishable objects can never be distinguished.  An action involving one object is a different event from the same action involving a different object.


 * In a practical sense, I fully understand that doors with numbers on are easily distinguished and that most goats look alike to most people but that is not the point. In principle, they are the same; real distinguishable objects.


 * The distinction has be made my SPACKlick between cases where we can see all the available objects at the start, for example the doors, and cases where, at the point of decision making, we can only see one of the objects, for example the goats. I thought we had reached agreement that in at least some cases we would still be able to distinguish between the two goats, they are therefore not indistinguishable.  In terms of conditioning I have yet to see a valid reason for conditioning on the door opened but not on the goat revealed. Martin Hogbin (talk) 09:50, 7 August 2014 (UTC)
 * It is not possible to condition on the revealed goat, as we do not know which one it is. We may as well treat the goats as "identical", "indistinguishable", just as we can use identical balls in an urn, which we cannot tell apart. The alternative is, as I already discussed above, to average over the goats:
 * P(C=2|X=1, H=3)=P(C=2|X=1, H=3, G=2)P(G=2|X=1,H=3)+P(C=2|X=1, H=3,G=3)P(G=3|X=1, H=3)


 * If we consider the goats as identical,
 * P(C=2|X=1, H=3)=P(C=2|X=1, H=3, G=2)=P(C=2|X=1, H=3,G=3)


 * Nijdam (talk) 11:06, 7 August 2014 (UTC)


 * I do not follow the basis for your distinction between doors and goats. Is that fact that we see only one of the two goats? Martin Hogbin (talk) 11:11, 7 August 2014 (UTC)


 * Right, we only see one and we have no info to tell them apart.Nijdam (talk) 19:15, 7 August 2014 (UTC)


 * You would accept that revealing goat 1 (the goat that is revealed) behind door 3 is a different event from revealing goat 2 behind door 3, would you not? Martin Hogbin (talk) 20:50, 7 August 2014 (UTC)
 * You will remember your answer to my question which started, 'There are three different coloured balls in an urn. One of the balls is red'. Note that we are not told the colour of either of the non-red balls and that we only see one of them, exactly the same as with the goats. Martin Hogbin (talk) 08:44, 8 August 2014 (UTC)
 * No, in your example I was told there were 3 differently coloured balls. In the MHP we do not know the goats have different charateristics. Nijdam (talk) 16:20, 8 August 2014 (UTC)
 * Is that the basis of your argument for not conditioni9ng on the goat revealed? Martin Hogbin (talk) 14:05, 9 August 2014 (UTC)
 * I really think there is some sloppiness here between "Identical" and "Indistinguishable". "Identical" objects cannot be distinguished in theory, they hold all the same properties. "Indistinguishable" objects cannot be distinguished in practice, possibly due to availability of information or detection techniques. Two goats, where one is seen are indistinguishable. Two goats of the same size, colour, smell etc with different genetics, say, are indistinguishable in a situation where genetic testing is not possible. Two identical goats will be indistinguishable in all circumstances. I think it is more than reasonable to suggest that the two goats are not identical. One goat is definitely to the left of the other wrt some fixed point. Whether the goats, if both revealed would be indistinguishable or not is not told within the question and the goats are indistinguishable when only one is seen. SPACKlick (talk) 14:01, 8 August 2014 (UTC)
 * The word 'indistinguishable' is commonly used with the sense with which I use it. It is the word usually used in particle physics with reference to bosons and the term is also used in combinatorics, having the sense with which I use it, see for example: ,,.  I only ever use it to refer to thingw which can never, in principle, be distinguished.  The word 'identical' seems to be used with the same meaning sometimes as well.  Martin Hogbin (talk) 14:31, 8 August 2014 (UTC)
 * I had not appreciated that usage from physics. I think it would be useful to distinguish the two concepts via different labels in this conversation but I am not tied to any specific label for either but "quintessentially indistinguishable" is a ludicrous suggestion wrt goats, as you say only bosons exist with that property. "Indistinguishable with available information" is relevant here. SPACKlick (talk) 14:48, 8 August 2014 (UTC)
 * I agree that we should distinguish between the two concepts. My preference would be to use the words 'indistinguishable' and 'identical' to refer to the fundamental properties of objects and to use phrases to refer to other meanings.
 * You refer to 'Indistinguishable with available information' but can you describe exactly what that means and can you describe how your definition would apply differently to the goats from how it would apply to the doors? Martin Hogbin (talk) 15:59, 8 August 2014 (UTC)


 * My thoughts about this are more or less in agreement with SPACKlick. Two identical object, meaning they share all (intrinsical) properties, are indistinguishable. We may put them in front of us, and call the left one Peter, and the right one Paul, but once they have been out of sight, there is no way for us to tell which is Peter and which is Paul. Two object may be indistinguishable for us, due to restriction in our observational possibillties. If we are only allowed to look at two plates, placed on a table, and with no visible differences, they are indistinguishable for us. However they may be numbered on their bottom, not visible to us: they are not identical. Of course two real world goats are not identical. However they may be indistinguishable to us, in the sense that we are not be able to tell them apart by observing them. The goats in the MHP however do not need to be real world goats, they may be identical. Even if they are not identical, they may well be indistinguishable to us. And then of course they may be striking different, for instance black and white. The point is, do we know the difference? If yes, we can distinguish between them. Now what, when we know they are different, but we do not know the difference? In the MHP we are not told whether the goats are different, nor whether a specific goat is shown. So, when we consider the goats as indistinguishable, we need not bother about which goat is shown. When the goats are different, we are not told which one is shown, hence we have to average over the goats., as I showed above. Nijdam (talk) 16:55, 8 August 2014 (UTC)
 * Standard usage of the word 'indistinguishable' seems to me to be that it properly is only used to refer to objects that are fundamentally indistinguishable, that is to say they cannot ever be distinguished by any means at all. Other meanings should be shown by a descriptive phrase.
 * You seem to be talking about a different meaning of the word from both myself and SPACKlick. You are seem to be talking about the possibility that a person could distinguish between two different goats in practice.  I agree that it is more likely that a typical host and player could distinguish between two doors that between two goats.  We could assign probabilities to these two possibilities.  I would say tht probability that both the host and the player could distinguish between the doors is close to (but not identically) 1, and the posssibility that the host and player could distinguish between two goats (where they could see both of them) is about 0.4.  Two similar goat would be hard for the average person to tell apart but the goats could be different colours for example.  You may quibble about my figures but you must agree that the probability that a player could distinguish between two goats is not zero.
 * Are you basing your argument for not conditioning on the goat that was revealed on the fact that goats are harder to distinguish between than doors or are you basing it on the fact that the player only sees one goat? These are two completely different arguments. Martin Hogbin (talk) 18:13, 8 August 2014 (UTC)
 * The MHP is treated as a mathematical problem, so don't bother about practical restrictions. And then, I repeat myself, a goat is shown, but no indication is given about which one it is: I cannot condition on an it. Nijdam (talk) 10:01, 11 August 2014 (UTC)
 * Good, let us not concern ourselves with the practical difficulties in distinguishing between goats. Can we take it that two goats are distinguishable in principle if you can see both of them?
 * You did condition on the ball revealed in by urn problem. You only saw one of the balls but you were told that the balls were all of different colour.  How then does this not apply to two goats (which we agree are different and in principle distinguishable) when we see one of them? Martin Hogbin (talk) 11:16, 11 August 2014 (UTC)
 * No, even if we see both goats, they might be indistinguisable. We may however make the assumption they're distinguishable. This then constitutes a different form of the problem. In your urn problem I was told there were three different colours, and also hat the revealed ball was black. Nijdam (talk) 08:35, 12 August 2014 (UTC)
 * In the MHP you know that there are two different goats. They will, for sure, be differnt in some way, they are distinguishable, and you see one of them, so you would know if you saw the other one.  I thought that we had agreed not to bother about practical difficulties of distinguishing between goats.  Of course it might be very easy; one goat could be white and the other black, the problem statement does not rule this out in any way.  If we are to solve the problem prperly, why then should we ignore this possibility? Martin Hogbin (talk) 18:02, 13 August 2014 (UTC)
 * I take it that you would agree that, if we were told that the two goats were of different colours and that the revealed goat was black you agree that we could condition on the revealing of a black goat. Martin Hogbin (talk) 18:04, 13 August 2014 (UTC)

Distinguishing between distinguishing between balls and distinguishing between goats
You seem to be using 'indistinguishable' in a somewhat unconventional sense to mean that the player might not be able to distinguish between two goats if he could see them both. This is, of course, correct. Some players might not be able to distinguish between two similar looking goats but that does not make the goats indistinguishable as the word is generally understood in physics and combinatorics, it is more of a practical restriction, as you called it, which we do not ned to bother about.

You must surely agree that some players could distinguish between some goats (say a balck and a white one) so, to solve the problem properly we must treat the revealing of one goat behind door 3 as a different event in our sample space from the revealing of a different goat behind door 3, even if not everybody could tell the difference between every pair of goats. Do you not agree? Martin Hogbin (talk) 10:14, 12 August 2014 (UTC)


 * No, I don't agree. Unless there is a reason to believe that the events CAN be distinguished, on a practical level, by the contestant, then they cannot treat them as different events.


 * On Seeing a black goat, the contestant knows that they goat they've seen is black, they don't know which goat it is and they don't know if they could tell seeing this goat from seeing the other goat, or any other black goat. Therefore the information the contestant received from seeing a black goat is not "I've seen this black goat" but "The revealed goat that I've seen is black". So we can't condition our sample space on some goat being revealed as opposed to another, but on the revealed goat being some colour as opposed to another and those are quite different. Again I'd point you towards writing out what information the player has at the start and could have at the point of decision.SPACKlick (talk) 10:24, 12 August 2014 (UTC)


 * The question that I was asking Nijdam was whether two goats could be regarded as distinguishable, if the player could see both of them.


 * You both seem to disagree with me still but for two different reasons. You argue that we should not distinguish between the goats because the player only sees one of them and does not know what the other goat looks like.  On the other hand, in my question about three balls of different colour, one of which was red, Nijdam agreed that we could condition on the ball that was revealed (which proved to be black), even though we could not see the other ball and did not know its colour.  Your response to the same question was that we could not condition on the revealed ball, because we could not see the other ball and did not know its colour.  These are two different arguments which we must take care to distinguish between.  11:05, 12 August 2014 (UTC)


 * By your definition of Distinguishable, the goats are clearly distinguishable. But the inherent non indentical nature of the goats is neither here nor there. Whether the goats would be practically distinguishable by a contestant who saw both is something which may or may not be the case. The fact that the contestant can only see one goat means that the contestant cannot distinguish the seen goat from the myriad potential goats they could have seen. But you seem to be assuming that the contestant knows that the two goats are practically distinguishable. Could you confirm if that is the case?
 * The player obviously knows that goats are distinguishable by what you call my definition. I do not assume that the player knows for sure that he will be able tell if the other goat were revealed next time but I do assume that he might be able to tell.


 * I do not think it actually matters whether the player knows that he will be able to distinguish between the goats. I agree that the question is to be answered from the player's state of knowledge but whether the player can distinguish in practice between the goats is a question of the problem setup.  Do we envisage that the player might be able to distinguish between the goats.  Clearly, it was not vos Savant's intention that we should do so, but neither was it that we should distinguish between the doors that the host might have opened.  Once you go beyond the known intent of vos Savant I think you are free to make the assuption that the player (and the host) might distinguish between the goats. Martin Hogbin (talk) 16:47, 12 August 2014 (UTC)


 * I haven't seen where Nidjam said we could condition when we are simply told that two balls exist, only where it was clear that there was "Red", "Not Red" and "Not Red or Black", and even there he conditioned only on the colour of the revealed ball. In the MHP we don't have that situation, we have "Car" "Not Car" and "Not Car", There's no way of knowing whether the two goats are practically distinguishable from eachother. SPACKlick (talk) 11:42, 12 August 2014 (UTC)
 * Yes, that is what I said above, there were three balls of different colour, one of which was red. I think Nijdam and I both took this to mean that the balls were all of clearly different colours, such as red black, and white.  As written the balls could be red, black and dark grey though.


 * My point is that these are all differences of degree not of principle. It is possible that the player might easily distinguish between the goats but not be able to distinguish between the two doors that the host might have opened.  Unlikely I admit, but still possible. Martin Hogbin (talk) 15:47, 12 August 2014 (UTC)


 * No, it's not possible that the contestant can't distinguish between doors, the contestant selected one out of the three doors and therefore distinguished that door from the three and was free to do so with any of the three doors. This is a difference in principle. The doors are distinguishable, the goats are not. SPACKlick (talk) 21:10, 12 August 2014 (UTC)


 * I can understand your argument that the player must be able to distinguish between the doors, although it is possible to contrive situations where the player cannot distinguish between the door the the host did open and the one that he might have done but these are unlikely scenarios.


 * I still cannot understand your argument that two different goats (possibly obviously different) cannot be distinguished by the player (they are distinguishable by the normal definition). Have a look at the goat article, I could easily distinguish between every single goat on that page.


 * Is your argument based solely on the fact that the player only sees one goat? Martin Hogbin (talk) 23:36, 12 August 2014 (UTC)


 * First off, it's not the normal definition, it's the definition in physics.


 * As the references I gave above showed, it is also the normal definition in combinatorics. Can I suggest that we use some agree terminology anyway to avoid taliking to cross purposes.  I suggest that we use the word 'indistinguishable' on its own to mean that two things are fundamentally indistinguishable and can never be distinguished between under any circumstances.  We can use terms like, 'indistinguishable by the player' to mean that a specific person is unable to distinguish  between the things in principle in given circumstances.  We could use 'indistinguishable in practice' to mean, for example, goats may look very similar and although they might be distinguishable in principle it might be impossible for the player to tell the difference in practice.


 * I do not claim that the above terminology is the only standard but I do belive that it is not at odds with general usage in statistics, physics, combinatorics, probability, and maths in general. If we agree to use it we can avoid going round in circles.


 * I did not see this comment Martin as it was in the middle of one of my comments, please try and avoid splitting comments. I agree that it's not at odds with usage in physics, I haven't seen enough of the other 3 categories to comment however even some physicists make the distinction (to avoid having to perform calculations for every electron in the universe as where the waveforms don't overlap identical particles are distinguishable). If we want to avoid confusion we should stop using the term indistiguishable at all. Identical can be used for the essential relation of two objects. non-identifiable could be used for the situation where the identity of an object cannot be ascertained from the given context (the term requires a context to be meaningful). So while the two goats are non-identical and likely the two goats are identifiable to the host (although not certainly), the two goats are non-identifiable for the contestant in a single run of the MHP and may be non-identifiable when stood side by side. SPACKlick (talk) 15:29, 14 August 2014 (UTC)


 * It's not solely based on seeing a single goat but that's the fundament of the argument. SPACKlick (talk) 07:18, 13 August


 * So the fundament of your argument is that, because in the single instance of the game described the player only sees one goat, the goats are indistiguishable to the player. You would agree that the host must be able to see both goats and therefore they are distinguishable to him?


 * Your argument is therefore different from Nijdam's. In my urn problem with three balls of different colour, one of which was red, Nijdam was happy to condition on the revealing of a black ball, even though we have no idea what colour the third ball was.  You maintained that such conditioning was not possible because the player only saw one ball, which proved to be black.   Do you agree? Martin Hogbin (talk) 08:02, 13 August 2014 (UTC)


 * I don't know how to make it any clearer. It depends on exactly what information you have. (Red, 2xnot red) you canonly condition on "Revealed ball was [Color]". (Red, Not Red, Neither Black nor Red) and revealed was black, can only condition on "Revealed ball was Black". (Red, Black, White), can condition on "Black Ball Revealed", "White Ball Revealed". My argument isn't that in the single instance you have only seen a single one, it's because at the moment where you are determining the probability you do not have enough information to distinguish between events where the same goat is revealed and events where a differing goat was revealed, this is in part caused by seeing only one goat, and in part caused by not knowing how distinguishable the goats would be if you were shown both in the context of the game. Once again I ask you to look at the information available to the contestant at the time of the decision.SPACKlick (talk) 10:39, 13 August 2014 (UTC)
 * I understand what you are saying but I think that you are creating a problem from two half-problems, so to speak. To try and make progress let us remove one of the half-problems.  Let us say that at the start of the game you are told that there are two obviously different coloured goats.  The host opens a door to reveal a white goat.  You do not see the other goat but you do know for sure that if you saw it you would be able to distinguish it from the one that you have just seen.  Can we condition on the fact that a white goat has been revealed now? Martin Hogbin (talk) 16:59, 13 August 2014 (UTC)
 * As I have said, for every scenario so far, we can condition on the revealed goat being any particular colour or other obvious feature. If you add the additional information that;
 * The same goats are always used in the game
 * The goats are practically distinguishable by some specified feature (or "one or more" of a set of specified features)
 * then the player has enough information to identify the goats and thus condition on goat identity. I don't believe either are reasonable assumptions for the MHP. In the MHP the only information the player has is there are on this stage two non-identical goats, I have seen one of those goats and it was (list of obviously discernible properties say "black"). SPACKlick (talk) 19:41, 13 August 2014 (UTC)
 * We are trying to calculate the probability for a single instance of the game. In the frequentist approach to probability we must envisage a large number of repetitions of the same game so your condition 1 is irrelevant.  The only things that change are those that are agreed to be random.
 * Your second condition is clearly a matter of degree. In a real game with a real player and real goats some players would be able to distinguish between some pairs of goats.  The probability of this is not zero so we must accout for it in our calculations in some way. Martin Hogbin (talk) 22:32, 13 August 2014 (UTC)
 * The first condition is absolutely relevant. In order to run a frequentist approach you must first define the universe you are within. A two goats only universe is significantly smaller than a many goats universe and contains different subsets for the same conditions.
 * The second condition, you miss the point of. If we are not dealing only with the set of games where we have the information that we can distinguish the goats, then we cannot identify the revealed goat. Period. It's not a matter of degree it's binary. Either we can identify the goat, or we cannot and if we cannot identify the goat we cannot condition on its identity, only on the revealed goat having those properties. SPACKlick (talk) 22:51, 13 August 2014 (UTC)


 * If you see a door opened and happen to notice that it is door 3, that is information that you could use (if for example you knew the host prefered to open door 3) to revise the probability that you originally chose the car. The chance that you will notice that the door opened is door 3 is quite high, let us say .98.


 * If you see a white goat revealed, you know that goats come in various colours and that the other goat might not be white. That is information that you could use (if for example you knew the host prefered to reveal white goats) to revise the probability that you originally chose the car.  The probability that the other goat is not white is lower, say 0.2, but you cannot ignore this information. Martin Hogbin (talk) 22:44, 13 August 2014 (UTC)

And I have at no point ignored the information that the revealed goat is (for instance) white. You can condition on any detectable property of the revealed goat, but that's not the same as conditioning on the revealed goat being that goat.SPACKlick (talk) 06:28, 14 August 2014 (UTC)


 * It may be useful here for you to express in words what you are wanting to condition on. As I think we may be at cross purposes here. The various conditions which differ from the unconditioned in principle (even if not in result) I have found in the original MHP are probability of the car being in a location
 * given the contestant picked the (left/middle/right) door
 * given the host opened the (left/middle/right) door
 * given the (left/middle/right) door is the remaining door
 * given the revealed goat was (black/white/three-legged/male....etc.)
 * and any combination of them. The conditions I am objecting to include (but are not limited to)
 * given the revealed goat is the goat that happened (to/not to) be revealed this time
 * given the revealed goat is Goat (1/2/A/B...etc)
 * given the (black/white/three-legged/male...etc.) goat was revealed
 * are any of these the condition you would like to be including? SPACKlick (talk) 10:09, 14 August 2014 (UTC)

Comment by Gerhard

 * In 1/3: lucky guess scenario, in 2/3: wrong guess scenario. Immutable, from the beginning to the end, after the host has shown a goat Immutable. Back to the intended unintuitive paradox: Three doors, behind them two goats and only one car. After the contestant had first selected her door, the host – who, in this famous paradox, is slave to his extremely biased role – has to open another door in order to show a goat BUT NEVER THE CAR and HAS to offer a switch to the second still closed door. What if he just showed his specific goat that he normally never uses to show? You need not be a mathematician to see what follows. And what if he just has opened the specific door that he normally strictly avoids to open? It's trivial, in both cases likewise you KNOW that the prize actually is behind his other unopened door. In both cases likewise you exactly know the actual location of the car. No maths required, no conditioning required, no sample space required, no probability calculus required. Allowedly it was better to use three balls in an urn, but that's not the problem. You simply have to understand the intended paradox that surely does never permit a host who "can" disclose or give an additional hint in which scenario the contestant actually is in, who "can" disclose or give an additional hint regarding the actual location of the door, passing such variant as the intended paradox. Ridiculous. As to the famous paradox, the chance is and remains 1/3 to 2/3 and no, the paradox may not be corrupted and perverted. Henze therefore says: "In all these considerations, of course, it is crucial that the car-hiding door must be kept secret by the host, who is also obliged to open a goat-hiding door." Henze knows to clearly distinguish between the intended paradox and those sneaky variants that he uses to teach probability calculus. Gerhardvalentin (talk) 22:41, 7 August 2014 (UTC)


 * Gerhard, do you accept that if the host gives you no indication of his preferences then him acting with preferences doesn't tell you where the car is? That is for a contestant ignorant of the hosts favourite door, the host opening his dispreferred door still keeps the location of the car secret? 213.106.233.97 (talk) 08:16, 8 August 2014 (UTC)

Thank you for this question, SPACKlick. Not only Henze knows the core of the intended world famous paradox, the literature overall says the host "has" to be assumed to choose evenly, if he has two goats to choose from (imagine he flips a coin). It is constitutional that we never may assume that any information regarding any host's bias in choosing from two goats could have infused. This is an important constitutive of the paradox. Thus the host maintains secrecy regarding the actual scenario the contestant is in, so the actual location of the car remains secret (we should avoid any unclear wording). Gerhardvalentin (talk) 08:54, 8 August 2014 (UTC)
 * SPACKlick, maybe Gerhard should have said, 'anything which might give a hint...'. The host does not know what the player might know and should therefore act with the greatest caution in whatever he does to avoid giving any possible hints.  Martin Hogbin (talk) 09:05, 8 August 2014 (UTC)
 * I agree that it's a sensible way to set up the paradox, and I agree it is the natural assumption for the paradox but I don't think it is required for the paradox to exist and I don't think it is an inherent part of the paradox. I'm not keen on adding unnecessary excess to the interpretation of the question and since restricting the hosts preference makes no practical difference to the underlying mathematics for the contestant it seems an unnecessary restriction. SPACKlick (talk) 13:55, 8 August 2014 (UTC)

Alfie and Bertie
I am still trying to understand why Spacklick and Nijdam think that you cannot condition on the revealing of a goat.

To try and get some kind of agreement let us assume that the goats are called Alfie and Bertie by the host but no one else knows this. Let us also assume that Alfie is white Bertie is mid grey. The host knows this but the player does not. The host prefers to reveal Alfie where possible.

The argument against conditioning on the goat revealed, as I understand it, is that, if the player sees a white goat, he has no way of knowing whether it is Alfie or Bertie, and he does not know what colour the other goat is or even that it is, in fact, a different colour.

I think it is a fact that you can condition a sample space on any event, although the sample space may remain unchanged after the conditioning. We also must agree that the player does know that there are two goats and he has seeen only one of them.

I argue that in the player's mind there is, conceptually, the goat he has seen and the other goat. As he has never seen the other goat and knows nothing about it he cannot be sure that he would know if the other goat were revealed in a replay of the game. On the other hand he knows that it is quite posssible that he would know if he saw the other goat. It is therefore possible for the player to set up, in his mind, a sample space in which elements involving both the goat that was revealed and the other goat are included. It may be that he proves unable to distinguish between the two goats but it may be that he is able to do so.

Now let us look at the information available to the player. We all agree that, in the absence of any information about the host's preferences, the player can only assess the probability of winning by switching as 2/3. In order to revise this estimate he needs some information about the host's preferences. Let us now imagine that he has heard a few whispers about the host; he has heard that the host prefers to open/reveal higher numbered doors and lighter coloured goats. The player now has the same kind of information about doors as he does about goats.

On the day, the player sees the host open door 3 to reveal a brilliant white goat. He can use the door number information to lower his estimate of winning by switching because he knows that the host is likely to have opened 3 rather than door 2, because he prefers higher numbers.

He can also use the information about the goat. He does not know what colour the other goat is so he cannot be sure that the other goat is not of a lighter colour however, knowing a little about goats and their general appearance, he can estimate that it is more likely that the other goat is darker. He then calls upon his conceptual sample space in which there are two goats, conditions it on seeng the hosts preferred goat, and revises his estimate of winning by switching down a bit more. Now I agree that, if the mid grey goat had been revealed, the player could not easily have done this, and I accept that I chose an extreme goat to reveal, but all I am trying to show is that revealing just one goat could affect the players estimate of the probability that he will win by switching. Of course, it is a highly contrived situation, but so it is with the door preference when you think about it. Martin Hogbin (talk) 09:45, 15 August 2014 (UTC)


 * First let me highlight the assumptions you make that are not inherent to the MHP. The same goats are used every time. It makes a difference whether the contestant knows this.


 * You say "I think it is a fact that you can condition a sample space on any event, although the sample space may remain unchanged after the conditioning." This is simply untrue. You can only condition on, events which are possible in the sample space, are different from the initial and can be distinguished with available information.


 * As for your conditioning on the goat in the final paragraph, you miss the point that you are not there conditioning on the goat but conditioning on the colour [See my comment of 10:09 yesterday at the end of this section], which we have all agreed you can condition on from the start. The information you gave there about doors and goats wasn't analogous. Imagine you'd been given the information that the host preferred Alfie to Bertie and would avoid revealing Alfie wherever possible. You couldn't ever use this information because you can never know if Alfie is revealed.


 * I have however from this realised an inaccuracy in what I've been saying. You can condition on a goat named Alfie being revealed, however as you never know if you are in that situation that conditional probability cannot inform your decision of whether to stick or switch irrespective of the value of that conditional probability. SPACKlick (talk) 10:09, 15 August 2014 (UTC)

Summary
Let me try to summarize the result this far. NB Concerning point 5: I have to give it some more thoughts.Nijdam (talk) 16:00, 15 August 2014 (UTC)
 * 1) The 'standard' MHP may be formulated with one door with a car behind it and two empty doors.
 * 2) Just for fun goats are introduced instead of emptyness. The goats are, just as the empty doors, assumed to be indistinguishable.
 * 3) The goats may also be considered to be different.
 * 4) They are named Alfie and Bertie. However the contestant is not told which goat is shown.
 * 5) One goat is black, the other white. This is a strange situation, as we do not know which goat is revealed, although for sure the contestant knows which one.
 * 6) The contestant knows which goat is shown
 * 7) If we know which goat is revealed, we should also condition on the revealed goat.
 * 8) If we do not know which of the two different goats is shown, we do not have to condition on the goat, but may as well average over both.


 * We know that there are two goats. Only one of them is shown (goatREVEALED), the other one was not shown (goatHIDDEN). If we suppose that the host has a bias just to show his preferred goat but not the other one, and if we suppose that he just has shown his avoided goat, then we know that the contestant actually is in the wrong guess scenario – strictly speaking in that half of the wrong guess scenario where the contestant has first selected the host's preferred goat. So switching will win the car. But if we CANNOT know about preferred goat and avoided goat, then similarly we CANNOT know about preferred door and avoided door. So what's all that "assumed door bias" good for? It's no little bit better than any "assumed goat bias". No difference at all, meaningless mathematical exercises without reference to the famous intended paradox. But if you use any preferred resp. avoided door, then you HAVE to use preferred resp. avoided goat, also; otherwise inonsistent. Gerhardvalentin (talk) 18:05, 15 August 2014 (UTC)
 * @Gerhard: It is one hour before the game is played. In a stable behind the stage are the two goats. Show me the goatREVEALED. Nijdam (talk) 16:52, 16 August 2014 (UTC)
 * Nijdam, it's the question of the goat revealed. Given that the host knows his two goats even by their name. If you "assume" that the goat that shortly will be revealed – resp. that an hour later already had been revealed – will be resp. was the one that the host usually avoids to show, if any possible, then the contestant is very likely to have initially selected the host's preferred goat and therefore switching will win the car – given these "assumptions" were correct. No matter whether this is shown by calculus, or not. That's a fact. Gerhardvalentin (talk) 23:05, 16 August 2014 (UTC)
 * I really do hope you understand this yourself, I don't. Nijdam (talk) 08:29, 17 August 2014 (UTC)
 * Gerhard, the issue here is you are labelling the goats by their function in the game. The Door opened will always bee (DoorOPENED), the goat revealed will always be (GoatREVEALED) however the door opened can be (DoorLEFT), (DoorMIDDLE) or (DoorRIGHT) there is no equivalent labelling for the goats. The contestant can always tell which door was opened and can never tell which goat was revealed so while the contestant can condition on the opened door they cannot condition meaningfully on revealed goat, because they cannot distinguish which goat was revealed. The goats are distinct from the doors in this sense. SPACKlick (talk) 07:57, 16 August 2014 (UTC)
 * What you say is not so. We se a single instance of the game and wish to calculate the probability of winning by switching for that one game.  To do this we envisage a number of repititions of the same game.  That is to say the same doors, goats, host, and even player.   In these reptitions the only things that change are the random elements of the game, the positioning of the car and goats, the player's original choice, and the host's choice.  The door originally chosen is a fixed door but in repitions of the same game the player may choose a different door.   The goat revealed is a real goat, we do not know what it looks like but the player does, he has seen it.  In repitions of the same game the host may reveal a different goat, which the player would notice.


 * I am going to set up a frequentist model of the game below which we can all agree on. This should help decide the issues that we disagree about. Nijdam, your opinion on whether I have set up the model correctly would be welcome. Martin Hogbin (talk) 16:06, 16 August 2014 (UTC)
 * Why not condition on the doors function or on the goats function? In case the host has the choice between his preferred goat X and his avoided goat Y to show, then we can assume that he will show X with probability $$q$$ and he will show Y with probability $$1-q$$. Then the car will be behind the door offered ("O") to switch on:

\begin{align} &P(Car_1) = P(Car_2) = P(Car_3) = \tfrac{1}{3}  \\ &P(M_Y|O) = 1-q \\ &P(M_X|O) = 1 \\ \end{align} $$
 * Gerhardvalentin (talk) 10:57, 16 August 2014 (UTC)


 * Response to Nijdam.Regarding your point 2, there is nothing in the problem statement to say that the goats are to be taken to be indistinguishable. Of course, we know that Whitaker and vos Savant did not intend us to distiguish between the goats, but they also did not intend ust to distinguish between the two doors that the host might open.


 * We are told to take the role of the contestant. This is an argument used by those who claim the doors are important.  They argue that the contestant could see the doors so we must distinguish between them.


 * I am going to start a section below in which we set up an agreed frequentist model of the MHP. From there we should be able to resolve all issues. Martin Hogbin (talk) 16:06, 16 August 2014 (UTC)
 * You're mistaken; it cannot be avoided to distinguish between the doors. I'm quite surprised you keep on saying so. And just as you said, I' m also pretty sure it was not intended to distinguish between the goats. Nijdam (talk) 16:46, 16 August 2014 (UTC)
 * Obviously the doors in the problem as stated are distinguishable but it was not the intention that we should care which door the host opens to reveal a goat.


 * A question for you. Do you agree that the MHP (as intended by Whitaker and vos Savant) can be expressed as this urn problem with one red ball and two (indistinguishable by convention) white balls?  The player takes a ball from the urn and without looking at it puts it in a bag.  The host then looks in the urn and removes a white ball.  The player, who wishes to win the red ball, is given the option of having the ball that is left in the urn or the ball that he put in the bag.  Do you agree that that was the problem that vos Savant intended or do you think that she intended that the doors should form an integral part of the problem? Martin Hogbin (talk) 20:33, 16 August 2014 (UTC)
 * I do not understand what you mean by "we should care", but we definitely do know which door is opened by the host.
 * Your urn problem is not equivalent to the MHP, nor is it the intended form. Nijdam (talk) 08:35, 17 August 2014 (UTC)



A frequentist model of the problem.
To try and resolve some of the issues under discussion I would like to set up a frequentist model of the problem. We assume the standard game rules; that the host always opens an unchosen door to reveal a goat and always offers the swap. We will also assume that the question asks us for a probability of winning by switching for a specific instance of the game in which we are the player.

From a frequentist point of view, the probability of winning by switching for this one game is given by envisaging a large number of repetitions of the same game. In other words, although throughout the game series the host, goats, and doors may have changed from time to time, we must imagine repeating the game with the same doors, host, goats, and even player every time. The only things that change are the random events, namely the original placement of the car and goats, the player's original choice, and the host's choice.

Do we all agree that this is the correct setup?
 * Okay, there is hardly anything to be disputed here. Nijdam (talk) 20:03, 18 August 2014 (UTC)


 * I am going to change the setup slightly. As we only see one goat, when we repeat the experiment, one of the goats must be the one that we saw.  The other goat may change. Martin Hogbin (talk) 08:32, 17 August 2014 (UTC)
 * This, however, is tricky. In a frequentistic model, we don't consider "real" things, but abstractions, i.e. if you like the model to be a descriptipon of a game with a changing goat, ethe changing goat is just considered to be "the other goat".
 * I agree this is tricky, which is wy I overlooked it initially. I think you will agree that the Bayesian and frequentist models of probability should always give the same answer, provided that the problem is modelled correctly in both cases.
 * Making this change is actually a move towards the position of you and Spacklick. As the player never sees the other goat we should not model it literally as being the same goat each time but as the population of posssible goats.  If we literally modelled the problem with the same goat every time the player would be able see the secod goat in our model, which he cannot do in reality.  I think we do agree on this, the second goat is an abstract entity, representing what the other goat could be.  Martin Hogbin (talk) 08:39, 19 August 2014 (UTC)

Do we all agree that, if the random events are taken to be also uniform, the probability of winning by switching (however we choose to calculate it) is 2/3? Martin Hogbin (talk) 21:02, 16 August 2014 (UTC)


 * No, for instance if you did calculate the probability as the fraction of the sum of the angles at the base of a equilateral triangle and te complete circle. Also other calculations won't do.Nijdam (talk) 20:03, 18 August 2014 (UTC)


 * I think you musunderstood what I was saying. All I was asking agreement for was the numerical answer of 2/3.  I was not expecing you to agree that any method of calculating this was acceptable. Martin Hogbin (talk) 15:19, 19 August 2014 (UTC)
 * Then I only say: the appropriate probability has the numerical value 2/3.

Agreement?
On thinking about the above model, I think we may all be able to come to some agreement on the subject of the goats. Here are things that we may all be able to agree on:

1 With the normal Bayesian assumptions (that the original car and goat placement, the player's original choice, and the host's choice are all uniform) it does not matter which goat is revealed.
 * Goats indistinguishable? In what way doesn't it matter? Nijdam (talk) 09:21, 23 August 2014 (UTC)
 * Only in the sense that it does not affect the numerical value of the probability of wining by switching if the host's choice is uniform.
 * I would agree that, if we are given information at the point of deciding on the host's door preference, for example if we are told that the host prefers to open the highest numbered door available under the rules, then we must condition on the door opened or we will get the wrong answer. We calculate the conditional probability that the player will win by switching given the host's door preference and that a specific door has been opened. We can agree to differ, for the moment at least, on whether we must condition on the door opened by the host in the uniform-host-choice case.

2 If we (somewhat perversely) allow the host's choice to be non-uniform then there is information that we can be given at the point of deciding that would mean that we should condition on the goat revealed. For example if we are told, 'He always reveals that goat when he can' or, after have seen a white goat, we are told told 'The other goat is black'.
 * Okay. Nijdam (talk) 09:21, 23 August 2014 (UTC)

3 Given less useful information about the host's goat preference, for example, 'The host prefers to reveal the lighter coloured goat', we can, upon seeing, for example, a brilliant white goat, revise our probabilities on the basis that the other goat is more likely to be darker than the one that we have seen. Whether we call this process conditioning on the goat, conditioning on a property of the goat, or something else, we need not worry about too much.
 * Bayesian approach with prior probabilities? Nijdam (talk) 09:21, 23 August 2014 (UTC)
 * Yes, if you want to take a Bayesian approach. We calculate the conditional probability that the player will win by switching given the host's goat preference and that we have seen the features of goat revealed.


 * There is a difference between conditioning on the goat revealed and conditioning on the door opened. We cannot revise our original estimate (based on a uniform-host-choice)  of numerical value of the answer (the probability that the player will win by switching) unless we are given some additional information about the host's door preference. We can, of course, plan ahead, by including the conditioning in our calculation, just in case we do get some information on the hos's door preference.


 * We cannot revise our original estimate (based on a uniform-host-choice) of numerical value of the answer (the probability that the player will win by switching) unless we are given some additional information about the host's goat preference and some information about the goat that is revealed.  We can, of course, plan ahead, by including the conditioning in our calculation, just in case we do get that information.  We must bear in mind, though, that the player would be able to see the revealed goat, so some information about the goat would be available to the player.  Martin Hogbin (talk) 10:18, 23 August 2014 (UTC)

Disagreement
I think that the things we disagree on turn out to be matters of taste and context.

1 Do we need to carry out a conditional calculation in the standard (all uniform) MHP just because the doors can be distinguished at the start, or can we decide that there is a symmetry between door numbers so we can ignore the door number that the host opens?
 * Both methods are correct, as long as we use the symmetry in our deductions. Nijdam (talk) 09:21, 23 August 2014 (UTC)
 * OK, so we can agree that the simple solutions are correct provided that they are preceeded by an argument that the problem is symmentrical with respect to door number? Martin Hogbin (talk) 11:05, 23 August 2014 (UTC)
 * I prefer to say that the simple way of reasoning is not correct. There is however a solution mthoid using symmetry. Nijdam (talk) 21:35, 23 August 2014 (UTC)
 * In what way is it wrong to point out at the start that we have no knowledge of the car and goat placement or the host's door preference so we can ignore door numbers and use a simple solution? Martin Hogbin (talk) 10:00, 24 August 2014 (UTC)
 * It's okay to point this out, but you also have to show how this effects your way of reasoning. Anyway, this is not the simple explanation. Nijdam (talk) 06:44, 25 August 2014 (UTC)
 * I would argue that we do not need to carry out a conditional calculation and can merely resort to symmetry. I would also say there are only four pieces of information which the contestant has at the point of decision which vary across games, "Which door the contestant opened", "Which door the host opened" "Which door is the remaining door" and "What easily detectable properties the revealed goat has". These are the only four pieces of information the contestant has they he can sensibly condition on. There is no need to condition on any of them because they all collapse under symmetry. SPACKlick (talk) 13:49, 26 August 2014 (UTC)
 * I agree. My example below from geometry shows how the simple, obvious, and intuitive application of symmetry at the start of a problem can greatly simplify the solution.  There is no rule on maths that you must do things the hard way. Martin Hogbin (talk) 22:34, 27 August 2014 (UTC)

2 Should we include in our sample spaces things which we are not able to distinguish between in order to allow for the case that something is revealed which means that we can distinguish between them?
 * Why should we? Nijdam (talk) 09:21, 23 August 2014 (UTC)
 * I am not sure, but you seemed to think that we should condition on the door opened by the host in the case where the doors are not numbered. In that case the only way we can distinguish between the two doors that the host is allowed to open, when the player has originally chosen the care, is by their role on the problem.  We can call them, 'the door the host opens' and 'the other door'.  Martin Hogbin (talk) 11:05, 23 August 2014 (UTC)
 * Aren't you somehow confused? The doors are always "numbered", i.e. distinct. We cannot identify them by their role. Nijdam (talk) 21:35, 23 August 2014 (UTC)
 * I am indeed confused by your use of the word 'numbered'. When I use that word I mean that the doors have actual numbers written on them and that these numbers are referred to in the question.  Of course, I agree that it is very reasonable to assume that the player can distinguish between the two doors that the host might open.


 * I see no reason why the doors and the goats cannot be identified by their roles in the specific instance of the game described in the problem statement. Suppose at the end of the game the player wins the car and is told that he can also have a goat to take home with him; he has still only seen one goat.  The host asks him which goat he would like.  He can either say that he would like the goat that he saw or he could say that he would like the other goat.  The only difference between the way he distinguishes between the doors and the goats is that when he gets his goat he might not be able to tell which goat he actually got. Martin Hogbin (talk) 09:56, 24 August 2014 (UTC)


 * Well, numbered or labelled or distinct, for a model of the problem it is all the same. (BTW: MvS mentioned numbers for the doors.) As for the identification by the role, you may label the items by their role in this instance of the game, but that is something different than labelling them acording to their role in the game. And there is no need whatsoever to label them by their role in this instance of the game, other than introducing a source of confusion. Nijdam (talk) 23:41, 24 August 2014 (UTC)


 * I agree that the doors are distinct, as are the goats; the only difference is that the player sees all the doors but only one goat. I am talking about a version of the problem in which the doors are not given numbers in the problem statement.    I do not quite understand what you are saying about labelling items by their roles in the game.


 * When we see a specific instance of the game we can use that to uniquely identify items that are revealed. When a door is opened we know that it is a specific physical door and that, in future instances of the game, that door may or may not be opened.  We must ask ourselves the question as to whether we could actually tell if it was the same door or not and, in the case of numbered doors, it would be exteremely likely that we could tell.  In the case of goats, one goat is revealed.  We know that it is a specific real goat and that, in future instances of the game, that goat may or may not be revealed. We must again ask ourselves the question as to whether we could actually tell if it was the same goat or not and, as we only see one goat, we would have to assign some realistic probability to this ability.


 * One way to solve the problem in a more general sense would be first to solve the unconditional problem than to apply Bayes' rule to get the conditional probability given any new knowledge that we gain. If we were told, for example that the host opened door 3 but he would always open door 2 where posssible, we could use Bayes' rule to get Morgan's solution for this case.  If we were told that a brilliant white goat were revealed but the host always shows the darker of the two goats when he can, we could also apply Bayes' rule to revise our unconditional probability, based on how likely we thought it was that the brilliant white goat was the darker of the two goats. Martin Hogbin (talk) 09:51, 25 August 2014 (UTC)


 * Martin, here is where I started to get concerned about conflation between the differing concepts I had labelled distinct and distinguishable. The player can always identify the doors independent of their role in the game, whether by numbers on them, or colour or position, the very fact that the contestant can choose between them and continuously identify them means they are distinguishable in practice throughout. When, at the initial, the contestant mentally labels the doors 1, 2 & 3 they maintain these identities throughout this problem irrespective of their role within it. With the Goats no such labeling can be maintained. Functional labeling, by the role the item plays in the game, is very different from independent distinguished labeling in how it interacts with existing information.


 * I have used consisten terminology throughout this discussion. I use 'indistinguishable' to mean that they cannot ever, under any circumstances, be distinguished.  This is consistent with the meaning generally used in physics, combinatorics and probability.  For othe meanings it is better to use phrases like, 'cannot be distinguished by the player at the point of deciding', or, 'may be treated as indistinguishable'.


 * Using my definition the goats themselves are certainly distinguishable. They are also distinguishable to the host, who can see both of them.  The only area where we may disagree is whether the goats are distinguishable to the player at the point he makes his decision.  I principle I assert that they still are; one he can see, and one is s behind a closed door.  The player knows that they are two different goats, in two different places.  Your argument is essentially that we should treat them as indistinguishable to the player because he can only see one of them.  Even this is not true if we allow some real world knowledge, as I have shown above.  The Bayes calculation that I propose is a standard probability calculation based on the player's knowledge of the host's general goat preference and the player's estimate of the distribution og goat colurs in the show. Martin Hogbin (talk) 20:45, 26 August 2014 (UTC)


 * Point missed again. You don't seem to get the other concept being discussed irrespective of its label. We all agree two goats are distinct/distinguishable/identifiable in the trivial sense of being two different goats. What you're missing is that they cannot be identified from Contestant POV at all. It's not because we only see one of them, they could be identified only seeing one of them if for example they were numbered one and two and it's painted on them. You would already have a non-functional identity for the goat. This can then be mapped later to a functional identity. It would make sense to say "in this instance goat 1 was revealed" and that statement adds information. It doesn't add information to say "in this instance the goat that was revealed was revealed" because that's a tautology true of every game. Without some form of identity which can map variably and is not tied to another identity the goat cannot be identified from the contestant POV SPACKlick (talk) 21:07, 26 August 2014 (UTC)


 * From the rules, you know the goat the host reveals will be labeled "The goat the host reveals". Nothing that happens in the game can change that. Similarly you know from the rules that the door the host opens will be "The door the host opens". However what you also know once the door has been opened is that "The door on the left" maps to "The door the host opens" (or right or middle or whatever). With the goats you never learn anything about their identity so you can't map a non functional label to a functional one.
 * Door Labels
 * Identity/Position eg. [Left, Right, Middle] or [1,2,3]
 * Function-Selection eg. [Contestant choice, Host opens, Remaining]
 * Function-Concealment [Car, Goat1, Goat2]
 * Goat Labels
 * Identity [Goat1, Goat2]
 * Function-Revelation [Revealed, not revealed]
 * Function-Concealment [Behind left, Behind middle, Behind right]

Do you see that there are two different types of label and that with the doors you can map independent labels to functional labels whereas with the goats you can't? We are currently disagreeing on either whether this difference exists OR where it matters.SPACKlick (talk) 13:49, 26 August 2014 (UTC)


 * No. The doors and the goats may be labelled by their role in the one instance of the show that is described.  We know that the the doors and goats could be different in othe instances.  I have aready shown this.  If we are told that the host always prefers to reveal the goat that he did in this instance then we can update our probability of winning by switching using Bayes' rule.  You may say that this is a different kind of information but so what?  In principle the goats are distinguishable, in practice whether we treat them as indistinguishable depends on what information we get. As we do not know what information we may get (in the actual question there is none on doors or goats) we cannot say whether we should treat the doors or the goatrs as indistinguishable.


 * Mapping arbitrary labels (such a door numbers) to door function is meaningless.Martin Hogbin (talk) 20:45, 26 August 2014


 * And that's what you're missing. The labels for the doors are NOT arbitrary. They are the identifying features of the doors that persist throughout the experiment. The doors are identifiable irrespective of their function and therefore it makes sense to talk about which door plays the role of which functionSPACKlick (talk) 21:07, 26 August 2014 (UTC)
 * Of course the door numbers are arbitrary; they could be anything.  What can you tell me about door 1, except that it is the door that the player originally chose?   The labels serve only the purpose of letting us know if the same door is referred to later in the problem.  We know that host cannot open door 1, because we labelled the door that the player originally chose as door 1.
 * The label is arbitrary but the method of identification is not. The labels do indeed serve the purpose of letting us know which door is which throughout the ENTIRE problem whereas the goats do not have that property. And Door 1 is not the same as the door that the player opened. Door 1 is the door that the player opened in some cases but it's door 1 in all cases (if actually numbered) or The leftmost door in all cases if identified by position but the identity holds across ALL cases and across the entire problem.SPACKlick (talk) 11:02, 27 August 2014 (UTC)
 * The goat that was revealed in thge instance of the game described in the problem is a real goat. It remains the same goat throughout the problem whether it is revealed or not. The player could, in principle, go up on stage and hang a number 1 round its neck. Martin Hogbin (talk) 22:29, 27 August 2014 (UTC)


 * In the end I think the only way that we can make progress might be to discuss actual questions. For example (with the usual assumptions), if we were told that a brilliant white goat was revealed but the host always shows the darker of the two goats when he can, what, in your opinion is the probability that the player will win by switching?


 * Is it exactly 2/3, Less than 2/3 or More than 2/3. Martin Hogbin (talk) 20:45, 26 August 2014 (UTC)


 * I have two responses to that before I answer. First, that's not a useful question A more useful question is the one I asked above. Do you deny that a difference exists between the functional and non-functional identifiability of the doors and goats or do you deny that the difference is mathematically relevant? Secondly the question is irrelevant because as I have already stated time and time again conditioning on the revealed goat having some set of properties is viable "The goat is brilliant white" would be an example of that. Third, The answer is that your odds are over two thirds. But please answer one of these two questions relevant to the disagreement.
 * 1 Do you agree or disagree that there is a fundamental difference between the characteristics by which the doors and the goats can be identified. Specifically that while both the doors and the goats can be identified by their functions (including relations to one another) the doors can also be identified separately to their functions?
 * You have not defined what you mean by, 'functional and non-functional identifiability of the doors and goats'. After the player has chosen a door but before the host has opened a specific door but knowing that he must do so to reveal a goat, the player can calculate that the probability of winning by switching is exactly 2/3.  By 'functional identifiability' do you mean:


 * After the host has indicated the door he intends to open, can the player use the number of the door opened to revise his probability? Answer: No.
 * After the host has revealed one of the two goats, can the player use the indentity of the goat revealed to revise his probability? Answer: No.
 * Is there any information the player might be given after the host has indicated the door he intends to open, that the player could, together with the number of the door to beopened, use to revise his probability? Answer: Yes.
 * Is there any information the player might be given after the host has revealed one of the two goats, the player could, together with the indentity of the goat revealed, use to revise his probability? Answer: Yes.
 * Or do you many something else by 'functional identifiability'? If so exactly what do you mean?
 * None of the above. A functional identity is an identity by function. As I listed above. "The door that was opened", "The door that was chosen". These are identities that will vary from game to game because something is being identified solely by its function in the game. A non-functional identity will exist throughout the whole game and can persist between games like "The goat named Alfie" and with a fixed stage "The door on the left".
 * Sorry, a complete misunderstanding.
 * What you call a functional identity also persists thgroughout the whole series of games. Suppose that there were no numbers.  On seeing the game the player could mark the door that he chose (say with a number 1).  It retains that number throughout future games.  He could also mark the goat that he saw with a number too.  That goat retains its number also, whether it is revealed or not. Martin Hogbin (talk) 22:29, 27 August 2014 (UTC)
 * No, the functional identity doesn't persist. The revealed goat is only a meaningful identity once a goat has been revealed. Once you restart the game there is no revealed goat, neither goat has that identity and neither goat had that identity at the beginning of the game. With the doors there is necessarily a persistent identity as 1) the player can see all three of them at all times and 2) the player is choosing between them (including a choice which relies on past identity). With your marking or hanging a sign on. The player can, in effect, mark the doors before the problem begins and keep track of them throughout the problem. With the goat, the contestant cannot mark it at the beginning they can only mark it once it is "the revealed goat" So whatever goat was revealed would be marked (identified) in the same fashion irrespective of which goat it is.
 * The persistence you claimed for the identity of the revealed goat only existed once the goat was marked because the two goats would then be distinct from the beginning of the next problem (marked and unmarked). I also want to apologise here for the diversion to persistence between games the key persistence is throughout the game. The door on the left is the door on the left from before the run begins to the point of decision. The goat named Albie is the goat named Albie from before the game begins to the point of decision, but the player doesn't know it. The chosen door, the revealed goat, the opened door. These functional identities are only acquired mid-game.SPACKlick (talk) 23:35, 27 August 2014 (UTC)


 * Would the question be different if it had said, 'the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat, say goat number 1, goat number 2 remaining hidden'? Martin Hogbin (talk) 08:33, 27 August 2014 (UTC)
 * We would then have to discuss whether that implied that the contestant KNEW it was goat1 as opposed to goat2. I'd interpret it as being an identity of the goats not available to the contestant but available to the host but it would be a different discussion.


 * 2 What is the condition you think we are discussing the viability of? "Given albie is revealed", "Given a white goat is revealed", "Given the goat that was revealed was revealed" or something else?SPACKlick (talk) 21:07, 26 August 2014 (UTC)
 * The condition is that one of the goats, complete with all its individual characteristics is revealed. The important thing actually is whatever characteristic the host may use to decide which goat to reveal. Martin Hogbin (talk) 08:33, 27 August 2014 (UTC)
 * Please specify a condition Martin so we're not talking across thin air and can bottom out this discussion. It's been taken off on tangents too many times now. From the above are you happy for the condition to be "Given that the goat that was revealed this time was/is revealed"SPACKlick (talk) 11:02, 27 August 2014 (UTC)
 * Yes, "Given that the goat that was revealed this time was/is revealed" is what I want to use to revise my estimate of the probability of winning by switching. There is nothing tautological or circular about this. If the other goat had been revealed the probability of winning by switching might have been different. Martin Hogbin (talk) 22:29, 27 August 2014 (UTC)
 * Right, that condition is, from the contestant pov, not just uninformative (like conditioning on the door there's no way of reducing the distribution to affect the probability) but also meaningless. The contestant has no more information about the identity of this revealed goat after it is revealed than he had initially. He knew a goat would be revealed and knew that it would be a particular goat. He doesn't know which particular goat it is now. "this goat" is effectively meaningless because the player cannot distinguish "this goat" from any of an infinite subset of the infinite goats in potentia that could have theoretically been revealed.SPACKlick (talk) 23:35, 27 August 2014 (UTC)

To tie together the two posts I just made. The doors have three sets of identifiers, one set fixed from the start of the game and two sets of functional (what they conceal[goat, goat, car] and how they are chosen [contestant, host, remain]) Throughout the game the third set is fully mapped to the first and the second set partly so. The goats also have three sets of identifiers. Their names/identities are fixed. What door they are behind and whether or not they were revealed by the host are functional. The player at the beginning of the game knows that "Doorn" will be the door the host reveals and that that door will conceal a goat and that that goat will be behind "Doorn" and that it will be the revealed goat. When the goat is revealed all he has learned is the value of n, he hasn't learned the name/identity of the goat. Whichever goat was revealed the contestant has learned nothing about it he didn't already know bar a subset of its properties. none of this makes "this goat" have any informational content.

By analogy. Consider this version of the monty hall problem. It starts as normal with the specification that the doors obvious properties are identical and the contestant can only distinguish them by position. The contestant picks a door (say leftmost). Then a curtain closes in front of the doors with the host behind it. The host picks which door to open as normal and that door is moved out in front of the curtain and opened to reveal a goat. The contestant is then asked if they'd like the prize from their chosen door or from the other closed door which are both still behind the curtain.

First off I assume we both agree that in this problem the probability is still 1:2 Stick:Switch. However in this problem I would contend that it is meaningless to condition on which door the host opened. The contestant cannot distinguish runs in which the host opened the middle door or one's in which the host opened the right door. In every sense the host does pick either the right or middle door, it's just the contestant doesn't have persistent access to that information throughout the puzzle. SPACKlick (talk) 23:35, 27 August 2014 (UTC)

Because we only see one goat
Your example above shows that your argument against conditioning on the goat revealed is based on the fact that we only see one goat. I agree that this makes a difference but it does not prevent us from trying to revise our probability on seeing a particular goat.
 * So you claim but you have yet to demonstrate it. You have yet to specify the information gained by the contestant that can be used to condition SPACKlick (talk) 10:17, 28 August 2014 (UTC)

Firstly, I agree that in the actual question asked, both the door opened and the goat revealed are non-informative; we have no information on the host's preferences. Once we got some extra information we may be able to revise our probability that we originally chose the car. We might be told where the car is, for example. However, let us stick to information that does not breach the basic rules of the game except in the sense that Gerhard talks about; information on the host's preferences that does, in my opinion, breach the spirit of the game. This could be things like, 'the host always opens door 3 when he can', or 'the host always opens the highest door number available under the rules', or, 'the host always reveals the darker of the two goats', or the 'host always reveals the goat with horns'. Now I agree that these bits of information may have different qualities but they all fall into the category of host preferences that are within the standard game rules. Upon receiving information like this, the player may revise his probability of winning by switching, as you have agreed above.
 * Note the information you included about the goats here were both "The host interacts with a goat with property x" not "The host interacts with goat x". This is entirely my point, the information "the host tries not to reveal Albie" doesn't make any difference at this point. You need to be given an identity of the goat in order to condition on it. If you were given the information "The host prefers not to reveal this goat" your condition is "Given the goat the host prefers not to reveal was revealed" another solely functional identity relation. SPACKlick (talk) 10:17, 28 August 2014 (UTC)

Regarding enduring properties of objects, what is important is the property or properties on which the host will make his decision. If the player sees a brilliant white goat but the host always reveals the heaviest goat then the whiteness is of no value.
 * Absolutely disagree what is important is the properties about which information is available to the contestant. The host can prefer the goat that ate the most hay before the show but that information does nothing for the contestant because he doesn't have access to that information. SPACKlick (talk) 10:17, 28 August 2014 (UTC)
 * Of course, that goes without saying. We were talking about properties that the player might discern.  The player can revise his probabilty estimate if he sees a brilliant white goat and knows that the host makes his choice on the basis of how dark the goat is. Martin Hogbin (talk) 10:41, 28 August 2014 (UTC)
 * No it doesn't go without saying because you've been ignoring it throughout. The host can choose based on many factors we CANNOT under any circumstances meaningfully condition on as the contestant. You said and I quote what is important is the property or properties on which the host will make his decision. Do you retract that in favour of what is important is the property or properties which the contestant can discern whether or not the host uses them to make a decision?SPACKlick (talk) 11:15, 28 August 2014 (UTC)
 * I thought it was understood throughout that we were calculating from the player's state of knowledge. What matters therefore is things that the player can discern and also knows will correlate with the the host's choice.  Martin Hogbin (talk) 13:58, 28 August 2014 (UTC)
 * No what matters therefore is things that the player can discern irrespective of what he knows about their correlation with host choice. SPACKlick (talk) 14:06, 28 August 2014 (UTC)

Regarding the fact that the player only sees one object, I agree that this gives him less information than seeing both but it may not give him none. In your example above, suppose there are 5 numbered doors. The player picks door 1. He also knows that the host always opens the highest numbered door available. Door 4 is wheeled out and a brilliant goat is revealed behind it. The player knows that 4 is nearer the high end of the door numbers available to the host so he can revise his probability on that fact, despite only seeing one door. On the other hand, if the player had known that the host always reveals the lightest coloured goat, the player could hace revised his probability on seeing just one brilliant white goat.
 * Again you resort to conditioning on a property of the goat and again i reply properties of the goat are information gained by the contestant and can thus be meaningfully conditioned on. With your Door 4 being wheeled out, you've assumed the contestant can know it's door 4, because you marked it door 4, but as specified in the example the doors were un-numbered and not distinctive being of similar outward appearance. By painting the 4 on, you've given the player access to information about the identity of the door. If the 4 wasn't there, as was my point above, the player would have no way of knowing whether this door was high or low numbered. It's just a door. Sure it is a particular door from before, could be middle, could be right and it is definitely one of the two but the contestant has no access to which and so cannot condition meaningfully on which. To break it down again below is the change of information in the problem detailing the only information gained between the initial and decision point.


 * At the start of the problem the contestant knows about the doors
 * Their locations (left, middle, right) [for convenience 1,2,3]
 * That he will choose one of them Doorz
 * That the host will open one of them Doorx that contains a goat and isn't Doorz
 * There will remain a Doory, The other door.
 * At the point of decision they also know
 * The values of x, y and z [Say x=3, y=2, z=1]


 * At the start of the problem the contestant knows about goats
 * There are two goats Goat1 and Goat2
 * One Goat will be the Revealed Goatn
 * The revealed goat will have some detectable properties
 * At the point of decision he also knows
 * The detectable properties of Goatn


 * What the contestant most emphatically does not know is the value of n. This is effectively what you are trying to condition on with the condition "Given that the goat that was revealed this time was/is revealed" it's "Given n = what n equaled this time which I do not know and could never know".


 * Consider the following, there is a 4 sided die which is unfair such that a higher number is more likely than a lower one. The host rolls the die where the player cannot see it. The player must now bet whether the die will come up with the same number on the next roll. The odds of getting a double vary from 0.25 (basically fair) to 1 (die which almost always lands on the highest side). The actual numbers are irrelevant because my point is in this game while it makes sense to condition on a specific value of the die it makes no sense to condition on "the value it landed on this time" because you have no access to that information and so it effectively resolves to "given that it landed on a value" which was already in the inital information and is thus a meaningless condition. SPACKlick (talk) 10:17, 28 August 2014 (UTC)


 * I do not understand what we are now arguing about. Is it just whether we condition on the revealing if a specific goat or that we condition in the property that the goat is white.  As I understand it, you condition on an event, and the event that I wish to condition is the event that the host reveals a brilliant white goat.  You have already agreed that, on seeing this event, and knowing that the host prefers to reveal the lighter goat, you would revise the probability that you had originally chosen the car. Martin Hogbin (talk) 10:41, 28 August 2014 (UTC)
 * This is why i made sure we specified the wording of the condition. Do you see the difference between the conditions "Given that the revealed goat is the goat that was revealed this time" and "Given that the revealed goat has the properties of the goat revealed this time"? The first is about the identity of the goat and conditions on a specific goat. The second is about the properties of the goat and can therefore apply to an infinite subset of the infinite goats in potentia. SPACKlick (talk) 11:01, 28 August 2014 (UTC)
 * Fine. If you think that there is a difference between a goat and the properties unique to that goat I am not going to argue.  My point is that when the event that the host reveals one of the goats occurs, the player, who will be able to determine at least some of the features of that goat, may be able to revise his estimate of his probability of winning by switching, provided that he has some knowledge of the host's goat-feature-preferences.  Do we agree on that? Martin Hogbin (talk) 13:27, 28 August 2014 (UTC)
 * Note the additional words you said If you think that there is a difference between a goat and the properties unique to that goat [emphasis mine]. You have no way of knowing what properties are unique to that goat, or if any of the detectable properties are unique to that goat. As i have said before yes I agree that the contestant can condition on the detectable properties of the revealed goat. I also agree that he could revise his probability if he also had information regarding the hosts preference relating to that category of property.SPACKlick (talk) 13:38, 28 August 2014 (UTC)
 * That is my point. The player, on having certain informtion, can revise his chances of winning by switching after he has seen tha goat revealed.  Whether you call that conditioning on the properties of the goat or the goat I do not care. Martin Hogbin (talk) 14:01, 28 August 2014 (UTC)
 * I, however, do care because a) they are different things and b) the second one cannot be done. As s simple example of how it cannot be done consider the half balls problem below. SPACKlick (talk) 14:08, 28 August 2014 (UTC)

All of that is well, but an uncompleted angular field. As to the intended paradox, literature says that we must assume that the host, having a choice, made a coin toss. But in the parody of a host who knows both goats by their names and prefers to show just only one of them if he has a choice, but usually avoids to show the other one if any possible, and if the contestant knows (had been told?) that the host just revealed the latter one, the avoided one, then the contestant doesn't need mathematical games to know that she is very likely to have initially selected the host's preferred goat, and therefore switching doors will give her the car. Conditional probability addresses a parody, no more the intended paradox. Leonard Mlodinow and Norbert Henze say that no conditional probability theory is needed to solve the paradox that actually has nothing to do with "conditional" probability. And Mlodinow exposes the only known bias: it is in the role of the host to never show the car, by that generating the paradox. Gerhardvalentin (talk) 22:47, 28 August 2014 (UTC)
 * 1) If you read this section, you'll notice the agreement at the top that conditional probability is unnecessary to solve MHP because everything you can condition on is either symmetrical or irrelevant to all other information. The only disagreements were about a) what could meaningfully be conditioned on in the original MHP (now resolved I believe although there may still be a small amount of disagreement between Martin and I) and Nijdam's disagreement that the solution rooted in symmetry is the solution in general referred to as the simple solution.
 * 2) Far more literature addresses the MHP with conditional formatting than without.
 * 3) As has repeatedly been explained to you Gerhard, nothing in the puzzle expresses or means that the host cannot have preferences. The puzzle does indicate that the contestant doesn't know about those premises, so even if the host had extreme preferences it wouldn't affect the probability calculation because that information isn't available to the contestant. Your return to Henze is almost monomaniacal and I would recommend you read several more sources before participating.SPACKlick (talk) 00:06, 29 August 2014 (UTC)
 * First and foremost it is absolutely necessary to plainly distinguish the paradox from its wide spread parody that is commonly used in textbooks. Otherwise the article will continue and carry on to unnecessarily confuse the reader. Gerhardvalentin (talk) 08:06, 29 August 2014 (UTC)


 * And that is acheived with the standard assumptions. As long as the following obtain you have the original paradox
 * There are three doors
 * Concealed behind the three doors are a car and two goats
 * The locations of the above are known to the host and the contestant has no prior information about them
 * The contestant is given free choice of one of the three doors
 * The host always then opens a door that the contestant didn't choose
 * That door is always one that conceals a goat
 * The method by which the host chooses the door is unknown to the contestant
 * The contestant is then always offered the choice between their original door and the remaining door.
 * Even if the car and goats were distributed non-randomly and even if the host had strong door bias it doesn't stop it being the original paradox as long as the contestant doesn't have that information.
 * The standard "parodies", or variants, all involve a change in at least one of these rules. Often however there is conflation between the host having a preference for one door over another and the contestant knowing that. SPACKlick (talk) 08:32, 29 August 2014 (UTC)

Gerhards point is central to the problem
The point that Gerhard makes is, in my opinion, central to the way the MHP should be dealt with. Here is the question, 'Do we allow the host do do anything at all (that he is not required to do under the rules of the game) that might give the player a clue as to where the car is?'. I am sure that if you asked people this qustion, the answer would be, 'No, of course not'. The problem becomes pointless if we assume the host can help the player in this way and it is unlike any game show that I have ever seen. Of course most people would have in their minds the host saying something like, 'I think you might stand a better chance with this door' or the host might wink and point to one of the doors but, in fact, the host could give equally valuable information by opening a door and saying, 'You know, I never open this door unless I absolutely have to'. Absurd though these suggestions are, they are essentially what is implied by the Morgan solution, in which the host's door preference must be somehow known to the player so that, when the probabilities are calculated from the player's point of view, the host's door preference can be parameterised. There seems little point in introducing a parameter only to point out that we have no idea what it is.

It seems to me that any breach of this requirement turns the problem into a different one, quite unlike any game show that I have ever known and, I am sure, not intended by the problem authors. A more realistic scenario in which this secrecy requirement might be breached is that the host has a secret but consistent door preference. Over a series of games, some people would spot this preference and it could therefore be used to the the advantage of the player. In reality, it is not good enough that host player does not reveal his choice preferences; it is essential that he always chooses randomly, with the express intention of not, by any of his actions, giving a clue as to where the prize might be. I am sure that, like the standard game rules, this secrecy principle is assumed by most people when they read the problem, even if they do not forsee all the ways the host might improperly help the player. All modern day TV game shows, in the UK at least, go to great lengths to avoid modification of the player's chances of winning by the actions of the host and I am sure the same principle was applied in the days of 'Lets Make a Deal'. Martin Hogbin (talk) 08:56, 29 August 2014 (UTC)


 * To take your question Do we allow the host do do anything at all (that he is not required to do under the rules of the game) that might give the player a clue as to where the car is? I would agree that the answer is no. However I disagree that the host being non uniform in his door/goat preference or even non-uniform in his car goat distribution DOESN'T give the player a clue as to where the car is. It is only transmitting that information to the contestant that gives the contestant a clue.
 * The Morgan solution very much DOESN'T imply that the host is giving extra information to the contestant. The Morgan solution is showing that if the contestant considers that the host may be non-uniform he can still detect that whatever preference the host has, it is still almost always better and never worse to switch. Morgan doesn't imply that the contestant actually has information about the specifics of the hosts behaviour, just that he knows that the probability of him opening the specific door given the choice is between 0 and 1. Doing so, and fully admitting we don't know where p falls between 0 and 1 allows us to show the range of the relevant probability is 0.5 to 1 and therefore almost always better and never worse than even odds.
 * As I already said above as long as the information is not available to the contestant we still have the MHP. SPACKlick (talk) 09:18, 29 August 2014 (UTC)


 * You have just argued (and I agree with you) that we are to answer the question from the player's state of knowledge. If the player does not know the host's door preference the probability of winmnig by switching (from the player's POV) is exactly 2/3 (even Morgan later agreed this).  If the player does not know the host's preference the parameter p is meaningless.  This your own argument.  Martin Hogbin (talk) 09:52, 29 August 2014 (UTC)


 * No, p is not meaningless, p is merely unknown. Using p we can see that 0.5<= 1/(1+p) <=1 for 0<=p<=1. This is information we don't have without considering p. By considering all the hosts preferences about which we can make decisions we see that no matter the host strategy "switching is always the maximal strategy for the contestant". This is additional information over "The average odds for the switch strategy across all host strategies is 2/3". It's the difference between expected return and return limits. For a single play game this is relevant information. SPACKlick (talk) 10:09, 29 August 2014 (UTC)
 * The player has no idea what value p has. You have very strongly used the argument that what the player does not know cannot be accounted for in probability calculation.  Now I would argue, like you, that it is obviously mathematically possible to imagine that the player has more knowledge than he does and introduce a parameter p that represents the (unknown) host door preference.  The player can then apply his lack of knowledge of the host's preference and therefore take p to be 1/2 giving the answer 2/3.   THis gives the same result as treating p as meaningless.  It is just the same as with the goat revealed.  Even if the player is able to gain no information at all from the (properties of the) goat revealed he is still able to mathematically include the two notional goats in a probability calculation.  He can then use his lack of knowledge of the host's goat preference, and if you like his inability to distinguish between the goats, to show that the revealing of a goat does not change his probability by winning by switching from 2/3.  This was a point that I mad a while ago.  Mathematically you can do what you like, including distinguishing between goats, so long as you apply your actual knowledge to the calculation.   Either p is meaningless or we can (mathematically) condition on the (properties of) the goat revealed.  You cannot have it both ways, Martin Hogbin (talk) 08:52, 30 August 2014 (UTC)
 * I'm not having it both ways. It would be meaningless, from contestant POV within game, to calculate the probability "given a value for p". because the contestant can never know that he is in a situation where p has that value. Introducing p, as a label for "the hosts preference" is like introfucing "Goatn" as a label for the specific goat revealed. Where you run into meaninglessness is calculating, from contestant POV within a game, for a given value of either n or p.
 * My point is that pd and pg can both, in principle, be defined but neither is of any value unless we know soemthing about the host's preferences.
 * And again, "porperties of the goat" =/= "the goat"&#32;SPACKlick (talk) 22:13, 30 August 2014 (UTC)
 * For the time being let us agree to disagree on this. When I say 'goat' you can take it to mean the discernable and relevant properties of one goat. Martin Hogbin (talk) 09:16, 3 September 2014 (UTC)
 * I'm happy to do that for now but with the reservation that if at any point it makes a difference I'll bring it back up. The reason it could make a difference is that "this goat" refers to 1 goat and "a goat with the discernable properties of this goat" refers to infinite goats in potentia&#32;SPACKlick (talk) 09:34, 3 September 2014 (UTC)

Summary
I believe that the spirit of the puzzle requires the host not to do anything (unless it is required under the game rules) that might give a clue as to where the car is. This is Gerhard's point. I cannot say whether Henze says this because I do not speak German but if Henze does not say it then someone should.

If we relax this rule and allow the host's choice to give us a clue, and we also stick only to possibilities mentioned in the problem statement, then it becomes possible that the host's choice of door matters. It also becomes possible that his choice of goat matters. Do you agree? Martin Hogbin (talk) 09:19, 28 August 2014 (UTC)


 * To your first paragraph, I'm not sure I agree because I don't think the MHP is affected by host preference (it's unavailable to the contestant so makes no difference to the maths). If the host always opened the highest numbered door he could, that would only give the contestant a clue if the contestant somehow knew that fact.
 * The argumnent I wish to make is against the host doing anything that might give a clue as to where the car is. I have always said that it does not matter how the host chooses if the player does not know his policy.  If the player does not know how the host chooses he must take the choice to be uniform.
 * The original Morgan argument implied that the player knows the host's door choice preference, otherwise it must be taken as uniform. They later agreed that in the conditions implied by the problem the probability of winning by switching was 2/3. Martin Hogbin (talk) 10:58, 28 August 2014 (UTC)
 * To the second paragraph, as with my response to the first, in my interpretation of the word matters (as in effects the calculation) neither the door preference nor the goat preference can affect the contestant's maths without significant additional information not available in the puzzle.
 * Agreed.
 * From the host POV both his preferences affect how likely a contestant is to win by switching given the specific door was chosen and specific door/goat combo was opened. But none of that can make a difference calculating as the contestant. SPACKlick (talk) 10:39, 28 August 2014 (UTC)
 * Yes, I agree also. You seem to be missing my point.  The point I have been making all along is that, if we accept the original Morgan solution (which I do not) which gives the probability of winning by switching as 1/(1+q), where q is a parameter based on the host's door preference, then, by the same argument, we must take into account the (properties of) the goat that has been revealed. Martin Hogbin (talk) 10:58, 28 August 2014 (UTC)
 * No, you have "all along" been conflating "the goat" with "the properties of the goat" which are very different things. As I have said repeatedly if we are to condition on things in the problem we can only condition on The door the contestant chooses, The door the host opens, the remaining door [any two of those specify the other one] and what properties the revealed goat has.
 * So where Doorz is the chosen door, Doorx is the opened door, Doory is the remaining door, Doora is the door concealing the car, Goatn is the revealed goat and Properties(I) are the properties of I which can be observed by the contestant. The full question is P(a=y|z=1,x=3, Properties(Gn) and the Morgan solution ignores a term, the term about the PROPERTIES of the goat. The question you have been asking all along whether it is rendered P(a=y|z=1,x=3,n=n) or P(a=y|z=1,x=3,n=1) is the one I have been objecting to. SPACKlick (talk) 11:11, 28 August 2014 (UTC)
 * I disagree and will respond to this below. Martin Hogbin (talk) 08:54, 30 August 2014 (UTC)

A geometry problem
We have a triangle ABC. A line connects point C to the mid point (D) of AB. What is the length of CD?

The lengths of the sides of the triangle are AB=8 and AC=BC=5.

What is your solution? Martin Hogbin (talk) 10:06, 24 August 2014 (UTC)


 * If CD is a straight line CD=3. But what's the purpose of stating this problem? Nijdam (talk) 23:47, 24 August 2014 (UTC)


 * Dear me! You have got the right answer but your method is incorrect.  You appear to have assumed that the angles CDB and CDA are right angles!  Where is you calculation showing this?


 * The correct method is to apply the cosine rule in the two triangles CDB and CDA and then use the fact that cos(x) = -cos(180-x) to solve the resulting simultaneous equations for CD. You have just been lucky in that the two sides CA and CB were the same length in this particular example of the problem.  You got the right answer, but only by fluke (or maybe you noticed an obvious symmetry which made the two relevant triangles right-angled). Martin Hogbin (talk) 09:17, 25 August 2014 (UTC)
 * Are you clairvoyant? Nijdam (talk) 16:51, 25 August 2014 (UTC)
 * No, I do not really know how you solved the problem but I would bet that it was not by applying the cosine rule in the two triangles CDB and CDA and solving the resulting simultaneous equations for CD.


 * I think that you would agree though that, if this was a question in a geometry exam, a perfectly correct solution would be to point out that CDB and CDA were right angles by symmetry and then solve the problem using Pythagoras' theorem. Using Pythagoras without mentioning the symmetry might result in the loss of a mark or two.  Of course, students who solved the problem the long way round would get full marks but they might not have time to finish the paper.  Martin Hogbin (talk) 19:15, 25 August 2014 (UTC)
 * I solved it using symmetry. I agree with your analysis. What is the point you want to make?Nijdam (talk) 09:30, 31 August 2014 (UTC)
 * The point is that simplifying a problem by using symmetry and then solving it is not bad mathematics, it is a perfectly proper way of solving problems. The 'proper' way of solving my problem, which has two unknown angles, is not necessarily to apply the cosine rule in the two triangles CDB and CDA and then solve the resulting simultaneous equations for CD. There is no rule in mathematics that you must always do things the hard way.


 * In the case of the MHP it is obvious to most people that the problem is symmetrical with respect to door number, in other words, if the question had stated that the player had chosen door 1 and the host opened door 2, the player had chosen door 2 and the host opened door 1 the answer (chances of winning by switching) would be the same. The door numbers can therefore be completely ignored when solving the problem. Martin Hogbin (talk) 10:53, 3 September 2014 (UTC)
 * There you go again. No one ever said that using symmetry is bad mathematics. On the contrary, if you'll go though years of discussion, you may find several references to the solution based on the symmetry. But this does not mean the door numbers can be ignored. No, it means that the appropriate conditional probability can be calculated without using Bayes' formula, but by using the symmetry in the problem. There is no way of circumventing the mentioning of the door numbers, how much you try. The simple solution keeps being incorrect! Nijdam (talk) 09:28, 4 September 2014 (UTC)


 * To avoid this conversation going at cross purposes, Nijdam can you outline precisely what you think the simple solution is? I suspect this is a matter of what's being called the simple solution rather than any disagreement over the maths. &#32;SPACKlick (talk) 09:41, 4 September 2014 (UTC)


 * Nijdam, symmetry with respect to door number means that the door numbers can be changed without affecting the solution. In other words, the door numbers can be ignored.
 * Martin, formulate in proper formulas what you intend to say. Nijdam (talk) 19:08, 4 September 2014 (UTC)


 * Nijdam and Spacklick, by simple solutions I mean things like Morgan's 'Solution F2'. We konow that the doors were intended to form no part of the problem except to prevent the player from knowing which prize they had originally picked, from one car and two goats.  Martin Hogbin (talk) 18:26, 4 September 2014 (UTC)

A simple solution ignoring the doors
My sample space consists of two outcomes, each including two events (after which the player may switch or not).

The first outcome (G) contains the event that the player picks a goat (behind any door, it makes no difference because there is symmetery with respect to door number). It also contains the event that the host reveals a goat without revealing the player's original choice and removes this goat from the player's final choice. This event could be omitted because it occurs with certainty. This outcome has probability 2/3, because there are three 'prizes', two of which are goats.

The second outcome (C) contains the event that the player picks the car (behind any door, it makes no difference because there is symmetery with respect to door number). It also contains the event that the host reveals a goat without revealing the player's original choice and removes this goat from the player's final choice. This event could be omitted because it occurs with certainty. This outcome has probability 1/3, because there are three 'prizes', one of which is a car.

There is no more probability involved. If the player now chooses to swap they win the opposite of their original choice therefore they win the car with probability 2/3 and a goat with probability 1/2.

The above sample space includes all possibilities and uses all information available to the player. Martin Hogbin (talk) 08:20, 5 September 2014 (UTC)
 * Sorry, Martin, I'll not comment on this. Nijdam (talk) 09:34, 5 September 2014 (UTC)
 * As Richard Gill says, the MHP is an exercise in mathematical modelling, not in probability. Martin Hogbin (talk) 11:05, 5 September 2014 (UTC)
 * You are, of course, always correct if you chose to ignore the cases in which you are wrong. Martin Hogbin (talk) 08:10, 11 September 2014 (UTC)
 * Martin, Mine would be very similar but slightly differently described.


 * Whereas from the information available all doors are symmetric; and
 * Whereas from the information available all goats are symmetric;
 * The first event node is the distribution of the prizes behind the doors, irrespective of distribution, the doors being symmetric, all distributions collapse and have properties 1/3 doors winner 2/3 doors loser
 * The second event node is the contestant's door selection, all doors being symmetric, the specific door collapses but the prize behind it does not, sample space divides into 1/3 player has chosen the car and 2/3 player has chosen a goat, the two goats having collapsed into a single branch.


 * [1/3] Whereas the player has chosen the car
 * The third event is the hotsts opening of a door. The host may open either door revealing either goat, the doors and goats being symmetric the choices collapse under symmetry.
 * All events being ended the player will lose if they switch doors.


 * [2/3] Whereas the player has chosen the goat;
 * The third event is the hosts opening of a door. He may only open the door with the other goat. The goats being symmetric, this provides no new information.
 * All events being ended the player will win if they switch doors.


 * So; I agree in that the only event which does not collapse under symmetry is the players Door selection and this is therefore the determining event. I add the specification that the reason the reveal of the final goat collapses is symmetry. I think it is clear from these why the simplest way for some people to understand the problem is in the grouping the doors into two sets version. I also understand the impulse to reach for conditional formatting because it looks like the reveal of a goat provides information, however it doesn't provide any relevant information. &#32;SPACKlick (talk) 11:14, 5 September 2014 (UTC)


 * I understand what you are saying and I agree with your conclusion but you are using somewhat unconventional language (event node?), which I doubt will convince Nidjam.


 * I am trying to mathematically represent the problem in a conventional way, with a sample space of all the posssible outcomes. I agree with Nijdam that, if you formulate the problem in terms of door numbers, you are inevitably lead to a conditional solution, even if the condition makes no difference to the answer. The answer, in my opinion, is to ignore the door numbers from the start. Martin Hogbin (talk) 14:10, 5 September 2014 (UTC)


 * That's how we always referred to the nodes on a tree diagram of a sample space. They were event nodes because they were nodes of the tree which branched at events in the universe. I don't know what the current term for them would be. I mean to put it in plain english

It doesn't matter which prize is behind which door because there's no difference and it doesn't matter which door you picked, just whether you got the car or the goat If you got the car then the host had a choice of doors but it doesn't matter which he  opened because you don't know anything about either of them or the goats behind them so whatever happened you lose if you switch If you got a goat then he can only open the other goat and you win if you switch Since you have a 2/3 of getting a goat you have a 2/3 of winning by switching


 * I think we basically agree on what the simple solution is. I'm not sure I agree Nijdam's solution below is the simple solution. &#32;SPACKlick (talk) 14:23, 5 September 2014 (UTC)


 * I agree with Nijdam that, if you chose to model the problem mathematically using numbered doors, at some stage in a calculation using a sample space you must have outcomes which include events such as 'the host opens door 3'. You need to assign a probability to this event and, using Bayesian inference or symmetry, you can give this probability a value of 1/2 but you still need to consider these events in your calculation.  If I were teaching probability, I would certainly mention a solution based on numbered doors but I would mention its weaknesses as well as its strengths. I do not agree that such as solution is the only 'correct' one.


 * In less formal arguments, such a yours above, you can use symmetry in the argument to show why the door number opened by the host can be ignored. Martin Hogbin (talk) 08:28, 11 September 2014 (UTC)


 * And I think that's what makes the simple solution a solution. It uses the fact that the door opened by the host and the goat revealed don't provide information due to symmetry to avoid a lot of unnecessary calculation. Not all presentations of the simple solution show why they collapse under symmetry or why the event is information neutral but they all use the fact that it is. SPACKlick (talk) 09:14, 11 September 2014 (UTC)


 * The door actually opened gives no irregular update of knowledge by any irregular evidence as to the actual location of the car. You may call it "the host's secrecy", or "as if he tossed a coin" or, as Morgan et al. (1991) say: "used a randomization device". Hence the ratio "lucky guess scenario : wrong guess scenario" remains 1:2, as a consequence of intended symmetry. Obviously no difference between "Always switching wins 2/3 times" or "Whatever the host does you have a 2/3 chance of switching". This is the constitutive basis of the paradox, framed by its intuitive implicit restrictions: No clairvoyance, no betrayal of secrets. No matter whether this is derived from the host's secrecy or from a randomization device, or from the intended obvious symmetry. Anything else is a different problem, inconsistent with the intended paradox. Gerhardvalentin (talk) 10:42, 11 September 2014 (UTC)