Talk:Monty Hall problem/Archive 22

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Archive 20

This has to end

All is said over and over. The only serious opponent of the expert discussiants is Martin. I, and others, have tried to convince him of the error in the simple solution, but some of his answers show Martin does not have a consistent idea of the problem and the solutions. My suggestion is, we stop discussing. We present in the article the standard form of the problem, give the right solution, and in a subsequent section mention the simple solutions, with a note of their alleged shortcoming, but that they may be understood as a way of understanding. Anyone may then add other aspect, other forms, other solutions, as long as they are well sourced. I don't take any notice of remarks made by glkanter or GerardValentin; they are laymen with no understanding of probability.Nijdam (talk) 15:30, 14 July 2010 (UTC)

This is Wikipedia, you know, 'where anyone can edit'. Not Nijdampedia, where only one person makes decisions. Besides, you 'professionals' don't always get it right, do you? Oh, and then there's those 'professionals' named Morgan, et al. Give me a break. Glkanter (talk) 17:48, 14 July 2010 (UTC)

On the contrary Nijdam, it is you who has been inconsistent. I have shown you an urn problem where you agreed that it would be acceptable to present a simple solution based on an obvious symmetry rather then present the complete solution. That is all that is required for the symmetrical conditional MHP.

I have also shown in the section 'One for the pedants' that the solutions that you support are not complete in that they do not take full account of the door originally chosen by the player but also rely on symmetry.

You need to be consistent in your approach. Martin Hogbin (talk) 16:04, 14 July 2010 (UTC)

Martin, keep to the following:

in the MHP the player is offered to switch, after the host has opened the door with the goat
a probabilistic solution has to be based on the (posterior/conditional) probability given this situation
a simple solution is a solution that does not make mention of such a probability, but simply states in one or another form that the probability to find the car behind the chosen door is 1/3
symmetry has nothing to do with the difference between simple and correct solution

Nijdam (talk) 19:32, 17 July 2010 (UTC)

Yes, I know, and you know I know, and I know you know I know...
What is called the probabilistic solution here is incomplete as Gill and I have explained above. It does not take account of the doors that the player might have chosen. Considering just the doors actually chosen is not good enough, as you well know. If you really need me to explain that to you I will.

>>>It depends on the formulation of the problem

Unless the game rules state that the player is only allowed to choose door 1 (and I have never seen this suggested anywhere) the fact that the player chooses door 1 is a condition which must be applied by conditioning of the full sample set of all possibilities. Martin Hogbin (talk) 16:58, 19 July 2010 (UTC)

3 I agree that the simple solution does not mention the specific conditions stated in the probabilistic solution.

4 Symmetry shows us that the answer given by the simple solution (which we agree correctly answers what is generally called here the unconditional problem) must be equal to the answer given by what is called here the probabilistic solution (which answers a different and specific conditional problem). I think we all agree on what I have said above. The only real disagreement is over the best way to present the two solutions bearing in mind all the facts surrounding the problem, which we probably all agree about. I can state them all again if you like. Martin Hogbin (talk) 10:34, 18 July 2010 (UTC)

>>>That does not make the simple solution a correct solution to the MHP. So what is your point?Nijdam (talk) 14:45, 19 July 2010 (UTC)

That depends on your answer to my reply to 2 above. A solution must start with the full sample set as permitted by the game rules and condition it according to the player's initial choice and the host's door choice in order to be complete.

Of course, you could argue that certain conditions in certain formulations clearly do not affect the probability of interest due to symmetry and thereby simplify the problem greatly. Martin Hogbin (talk) 16:58, 19 July 2010 (UTC)

@Martin - You seem to be confusing solving a particular case that applies without loss of generality to any other particular case (by renumbering the doors if necessary) with solving the average of all cases (the unconditional problem) by always calling the door the player picks door #1 and the door the host opens door #3. Although they sound similar, these are completely different. Consider the size of the sample set. In the former case (where we've renumbered to force the door the player has picked to be #1 and the door the host has opened to be #3) there are still 6 separate combinations of player pick and door the host opens. If we start with 900 iterations of the show where the initial player pick was random and the host acted in the "standard" way (picks randomly between two goats), this renumbering means we're talking about only (roughly) 150 of the 900 iterations. In the latter case (the doors are not initially numbered and we always call the door the player picks #1 and the door the host opens #3), we're talking about all 900 iterations. If we're talking about all 900 iterations, we're treating the doors as if they're indistinguishable - i.e. we've made the problem unconditional. This (clearly) is what you're trying to do - but you're solving an entirely different problem. It's not the probability given the player has picked door #1 and the host has opened door #3, but the average probability across all initial player picks and doors the host opens. -- Rick Block (talk) 22:50, 19 July 2010 (UTC)

@Martin, @Rick - From this and all previous threads, I think Martin's confusion is actually compounded. Yes, I do believe that, as Rick notices, Martin routinely confuses a solution based on an arbitrary numbering of the doors (which only requires the host and the player to agree on which door is which), with a solution marginalized with respect to all possible door numberings - i.e. the so-called unconditional solution. But, in addition, he also seems to routinely confuse "symmetry of the problem" and "maximal state of ignorance of the player", as if the former implied the latter. It does not, read any reference on the maximum entropy principle for why. The unconditional solution should always include the caveat "jointly averaged over the population of all the players and all hosts, for a given door numbering." glopk (talk) 00:43, 20 July 2010 (UTC)

Glopk, Rick, I am not confused about anything, "jointly averaged over the population of all the players and all hosts, for a given door numbering." makes as many unjustified assumptions as the simple solution does.

I have never confused "symmetry of the problem" and "maximal state of ignorance of the player", all I ask for is consistency. If the host's goat door choice is to be taken as non-uniform then so must the initial car placement; no information is given about either in the Whitaker problem statement.

I suggest continuing this discussion on the /Arguments page. Martin Hogbin (talk) 11:04, 20 July 2010 (UTC)

The opening remarks of this thread seem to indicate that Martin is the only proponent of the simple solution. This is obviously not accurate. There have been many proponents of including the simple solution and most people who have taken part feel we need to include both. It is just that Martin has been the only one with the wherewithal to keep debating, while most of us have gotten tired of our points being ignored and moved on to other things. I can see that in the mathematics world issues like this can be a big deal to those involved, but that should not prevent you from opening your mind and at least attempting to see the problem from another angle.

That being said I really like the shape of the article right now. It flows nicely from simple to more complex math and is easy to follow for those of us who are not already familiar with the nuances of probability theory. I must leave the details of the math to others as I am no mathematics expert. If I had anything else to add it would be a expanded section on the psychology of the MHP. Considering the large number of papers written on the subject there is not much coverage here. Colincbn (talk) 01:44, 20 July 2010 (UTC)

Mediation

OK Rick. You have convinced me. I will wait until mediation begins before I expect you and Nijdam to play by Wikipedia's rules. As interpreted by someone other than you guys. Glkanter (talk) 21:01, 9 August 2010 (UTC) (actually about an hour earlier.)

Per AGK's further update we now have two mediators. -- Rick Block (talk) 00:09, 10 August 2010 (UTC)

As the mediation is now in progress, I've archived the discussions of late July and early August. A clean slate, so to speak. If there are issues unrelated to the mediation that need to be dealt with, please contact me. Sunray (talk) 15:37, 12 August 2010 (UTC)

Some statistics, just for fun: User:Tsirel#170,000 words on a single talk page. Boris Tsirelson (talk) 13:53, 8 September 2010 (UTC)

Hi Boris! Why don't you come and join the mediation? Gill110951 (talk) 08:43, 9 September 2010 (UTC)

No! I would be horrified by "only" 17,000 words... See also Proof by verbosity (not deleted yet). Boris Tsirelson (talk) 09:02, 9 September 2010 (UTC)

The rate of increase is decreasing at the moment. The big question seems to be: is the simple unconditional explanation for 2/3 simultaneously a complete answer to MHP? Or more precisely: is this a reliably sourced opinion? Of course, the answer depends on what you mean by MHP. Rather, it depends on what reliable sources mean by MHP. And that depends on what you mean by reliable sources. Is a non-statistican also a reliable source on MHP? Gill110951 (talk) 19:03, 15 September 2010 (UTC)

In think the answer there is from the WP perspective. Yes (the unconditional solution can be extended in a complete solution) and not yet (that approach is not yet described in the current reputable sources. And of course is reputable source not required to be a statistician, in fact a large number of the current (reputable) sources are not statisticians anyway.

I don't think there was even much of an disagreement, that the unconditional solution can given as a complete solution, the (opposing) argument merely states that most of the unconditional given in sources are not complete solutions as stated, meaning the additional arguments to view it as a complete solution provided in the various discussions here, are not a part of the unconditional solution stated in the sources.

It would be great, if Boris were to find time to participate in the mediation. A distinguished third party expert might help to overcome the entrenched positions of both parties and provide some outside sanity check. He could also serve as a neutral technical advisor to the mediators in case they require some.--Kmhkmh (talk) 01:00, 16 September 2010 (UTC)

I agree, Kmhkmh. But I think the way you state things is hiding the biggest of all the issues. You write about solutions and whether they are complete or not, or can be extended in some way to be complete. This presupposes that we (or rather, the reliable sources) agree on what is the problem. But is that the case? What do we have? We have the words attributed to Whitaker in vos Savant's column in Parade. They do not state a mathematical problem, a problem *inside* pure mathematics. They do clearly present an invitation to be mathematized. People from all kinds of different backgrounds have explicitly or implicitly done that, by presenting mathematical arguments which are supposed to convince us that the player should switch doors. To mention two extremes: mathematical economists pull game theory out of their tool chest, professional statisticians and probabilists pull out Bayes theorem. The different mathematizations use different mathematical assumptions and derive, given those assumptions, a mathematical concusion which is supposed to be seen as an argument for switching. The assumptions are different and the conclusions are different. Then, aside from the specialists, all kinds of highly reliable people present short intuitive arguments (usually of the unconditional kind) to which one could easily add a mathematical formalism and a mathematical problem, for which they become correct and complete, as mathematics. People mostly didn't bother to do this since it's a waste of time or because they wanted to push alternative approaches. Since the formalized problem is simpler the mathematics is simpler too, the assumptions are weaker, and the conclusion is weaker. That's a fact of applied mathematical life: the more model assumptions you put in, the more model conclusions you can get out. But the more you put in, the harder it is for the real world to match your model. Then there is the hidden dimension of what you mean by probability in the real world. Depending on the real world meaning you give to the word "probability", your assumptions can be more or less convincing, and your conclusions can have different meanings too. The book by Rosenhouse has a whole chapter on this facet of the MHP. In my opinion, realizing the existence of this dimension helps one to understand why different reliable sources have such different ideas about MHP. Gill110951 (talk) 05:57, 16 September 2010 (UTC)

Actually my comment above referred only to the conditional interpretation of the problem. If you go for unconditional interpretation or simply a heuristic argument, then the simple solution (by vos Savant or others) is complete in that context, But i think we have a general agreement there as well. What the opposing side here merely demands, is that the these fine distinctions need to be transparent to readers. The thing we need to avoid, is that readers may (easily) get the impression, that the unconditional solution given in the sources is as stated a complete solution to the conditional problem. That's at least my only personal concern here and how I've read the summary of other opposing voices like glopk, coffetotheorems or nijdam.--Kmhkmh (talk) 09:53, 16 September 2010 (UTC)

Dear Kmhkmh, I agree entirely with you! Gill110951 (talk) 21:04, 16 September 2010 (UTC)

But isn't "The thing we need to avoid, is that readers may (easily) get the impression, that the unconditional solution given in the sources is as stated a complete solution to the conditional problem." representative of a single POV? Don't the reliable sources that provide simple solutions offer the simple solutions as *a complete solution to the MHP?* That's certainly a reliable sourced POV, why should it be considered an 'inferior' reliable sourced POV? Isn't calling the 'conditional solution' alone 'complete' a POV? Glkanter (talk) 10:44, 16 September 2010 (UTC)

Garry, Kmhkmh is right again, helping the obscure article to improve. For the reader it should be accessible what the MHP is all about, it should be transparent to the readers. The "poor unconditional solution", based on Whitaker's question only, leaving completely unanswered whether the host just may have offered to switch because the guest's first choice per chance was the car, and leaving unanswered whether there could be any "famous host's whistle-blowing tattletale bias" in opening his unfavored door, showing the car actually was behind his preferred door, etc., is one thing. That's the crux. To find sources helping to "show correct simple solutions" is needed. --Gerhardvalentin (talk) 14:14, 16 September 2010 (UTC)

Selvin, vos Savant, Devlin, Adams, Carlton, eventually Morgan, et al, etc., etc., etc... I don't see your point at all. Glkanter (talk) 14:42, 16 September 2010 (UTC)

Yes, whether you look at the problem from a mathematical/logical standpoint, like Gill, or based solely on what the sources say, there is not just one correct solution, the simple solutions are valid either way. Martin Hogbin (talk) 21:57, 16 September 2010 (UTC)

There are lots of sources which present a simple solution as the whole solution. Many of them are moreover logically self-consistent and entirely convincing (in their own terms). Some are obviously flawed (as of course are "obviously" the source, all those maths PhDs, who wrote in to say that vos Savant was stupid and got it completely wrong). We don't have a duty to report "every" solution. Let's use our sound editorial sense to focus on sources which we think are useful for understanding, don't waste time and bandwidth by reporting sillyness. Since different editors find different sources useful, likely different readers will also be attracted to different solutions. Wikipedia editors do do own research at the meta-level, sifting and compiling the evidence, and trying to present a coherent informative overview. All the controversies around MHP are part of the MHP story.Gill110951 (talk) 05:23, 17 September 2010 (UTC)

Question re: the 'Why the probability is not 1/2' sub section

The following is the opening sentence. I find it erroneous, meaningless and confusing:

"The critical fact is that the host does not always have a choice(whether random or not) between the two remaining doors."

This needs attention. Glkanter (talk) 20:22, 19 September 2010 (UTC)

Garry, it could better read as follows:
The critical fact is that the host (whether random or not) only has a choice between his two remaining doors in 1/3, if both of his doors are hiding goats, but he does not always have a choice, because in 2/3 one of his two doors will be hiding the car. He always chooses a door that he knows hides a goat after the contestant has made their choice. He always can do this, since he knows the location of the car in advance. If the host chooses completely at random when he has a choice, and if the car is initially likely to be behind any of the three doors, it turns out that the host's choice does not affect the probability that the car is behind the contestant's door. But even if the host has a bias to one door or another when he has a choice, the probability that the car is behind the contestant's door, so it turns out, can never exceed 1/2, again as long as initially all doors are equally likely.
Do you think this first sentence would be better? Gerhardvalentin (talk) 12:09, 21 September 2010 (UTC)

I agree it is confusing. How about "The critical fact is that the host always can and will open a door different from that chosen by the player and revealing a goat. The probability that the original door hides a car is not changed by receipt of this "certain" information, hence it remains equal to 1/3. The probability that the now open door has a car behind it has manifestly become zero. The fact of opening that door and revealing a goat has given us information about its contents and changed the probability that that particular door conceals a car from 1/3 to 0. The probability that the "other door" hides the car must therefore also have changed from 1/3 to 2/3, since probabilities have to add to 1. This change in probability certainly "feels" counterintuitive (the same door, still closed!). However it is not in contradiction with our first remark that the host's action *doesn't* change the chance that the *initially chosen* door hides a car. That is because the identity of the "other door" has been determined by the host's action, it was not fixed in advance. To see this in another way, initially our chosen door hid the car with probability 1/3, and it was behind one of the two other doors with probability 2/3. The host can and will always open one of the other two doors revealing a goat. The probability remains 2/3 that there is a car behind one of them, and what we have learnt is that it is not behind the now opened door." (Now add references to reliable sources).

Next notice that in elementary argument we didn't specify the identity of the opened door nor of the intially chosen door. We only used the fact/assumption that the probability that the initial choice was right, is equal to 1/3. If we had mentioned the identities of both of the two chosen doors, things would have changed. If I don't give you more information/facts/assumptions, you cannot say anything about the probability that door 2 hides the car given that you chose door 1 and that door 3 was opened by the host. Some people may find that fact unintuitive, but it is easy to give examples where the conditional probability is different from 2/3 and hence also the chance that your initial choice was correct, has changed. We are no longer conditioning on a probability 1 event. This gives a useful bridge, I think, to the conditional approach. Gill110951 (talk) 14:36, 20 September 2010 (UTC)

How about we start with a source that discusses this, and work on how to paraphrase what the source says? -- Rick Block (talk) 15:01, 20 September 2010 (UTC)

I'm sorry Rick, I'm bloody lazy, that's why I became a mathematician. How about people who feel strongly about this, work on the sources? My day job is to read all these sources and replace them with a single new source which I write myself, which makes all the old stuff superfluous. I agree that I'm not allowed to work in that way on wikipedia. Gill110951 (talk) 15:48, 20 September 2010 (UTC)

I have a day job, too, and it doesn't include finding sources for what Richard Gill thinks should be added to the Wikipedia article on the MHP. As an academic, I believe your job is more or less to produce what Wikipedia calls OR. However, unless you're actually familiar with what others have published you can't know whether your thoughts are original or not. If you're reading the sources anyway, it seems the differential effort on your part to simply cite your references is nearly 0. Of course if you're saying you're thinking about things from first principles without attempting to find out if your original thoughts have already been published - then, yes, it would be a chore to find references. -- Rick Block (talk) 18:18, 20 September 2010 (UTC)

Of course not Rick! Sorry, I told you I am lazy. Indeed, nobody is obliged to take over my job, whether my day-job or my wikipedia hobby. I just tell you it will take days, weeks, for me to get around to doing the literature searching. On the other hand lots of people here are more familiar with some of the literature than me, so maybe they recognise some of "my" arguments in existing sources. I am not claiming to be producing anything new here at all, indeed, I learnt a lot which was new to me from the discussions. I do try to create my own synthesis of what is "out there". I think I'm making progress. The three levels of assumptions which I identify in [1], and the corresponding three levels of conclusions, show to me that all the approaches are different and valuable and meaningful. We need active (I choose a door randomly) unconditional, passive (I know all doors are equally likely) unconditional, and passive conditional, to tell the whole story that is out there in all the sources. (I used to call "active unconditional" game-theoretic but the term might be off-putting - indeed it is just elementary probability). Knowing there is a clean mathematical synthesis might aid the editorial job of overviewing and structuring the whole enormous mess in a neutral way. Don't call it OR. Call it the work of the scribe who has to collate the material to be copied. Gill110951 (talk) 06:18, 21 September 2010 (UTC)

Something to include

I'd like to suggest some minor improvement for this article. Please, if you you think it has some value, include the following table:

|me|car|strategy 1|strategy 2|
| 1|  1|      lose|       win|
| 1|  2|       win|      lose|
| 1|  3|       win|      lose|
| 2|  1|       win|      lose|
| 2|  2|      lose|       win|
| 2|  3|       win|      lose|
| 3|  1|       win|      lose|
| 3|  2|       win|      lose|
| 3|  3|      lose|       win|

Explanation: Imagine the selection is done by me and by the car. These are all possible outcomes. And that's the end of the theory of probability. Now check what outcome strategy 1 (I change my selection) and strategy 2 (I do not change my selection) give. We don't need to bother with Monty Hall's selection, because either ha doesn't have a choice (all outcomes except 11, 22, 33), or it doesn't matter what he choose. So the truth is we can safely ignore Monty's selection, as the result doesn't depend on it. Donno if it makes sense, but if it does please include it.123unoduetre (talk) 20:40, 17 September 2010 (UTC)

Isn't this effectively the same as the table already in the "Simple solutions" section? The existing table shows only the first three lines, but assuming the player has picked door 1 these are the only lines that are relevant. -- Rick Block (talk) 23:56, 17 September 2010 (UTC)

Yes, but in this version you can see the whole sample space. And it also stresses all the independent choices which give the sample space. (first and second column show directly the sample space) In the table in "Simple solutions" the choices and the sample space are not obvious from the table. It's only about a presentation of the data. If you think the table from "Simple solutions" is easier to understand for most people, leave it as it is.123unoduetre (talk) 08:49, 18 September 2010 (UTC)

Is this not the table given by Selvin in one of his letters? Here is how we might have it as a table. It is well-sourced and intuitve to me. Martin Hogbin (talk) 09:11, 18 September 2010 (UTC)

Door chosen by player	Door hiding car	Player switches	Player sticks
1	1	Lose	Win
1	2	Win	Lose
1	3	Win	Lose
2	1	Win	Lose
2	2	Lose	Win
2	3	Win	Lose
3	1	Win	Lose
3	2	Win	Lose
3	3	Lose	Win

What a stupid table. Does it clarify anything? It says no more than: if you originally pick the door with the car, then switching means loosing. Nijdam (talk) 10:04, 18 September 2010 (UTC)

I can't comment on 'stupid', but your explanation of the table seems complete, valid, and well sourced. What is it lacking? Glkanter (talk) 15:32, 18 September 2010 (UTC)

I agree that the table shows that, 'if you originally pick the door with the car, then switching means loosing'. The point is that, for some reason, the vast majority of people who see the problem for the first time cannot see and and understand this simple fact and thus they cannot see that the probability of winning by switching is 2/3. That is why the problem is so well known. It is why we have an article on it.

The effect of any information that might be revealed (in slightly different circumstances) by the number of the door opened by the host provides an interesting secondary problem for students of conditional probability. Martin Hogbin (talk) 17:15, 18 September 2010 (UTC)

Let's formalise it slightly: C='you originally picked the door with the car' S='you switch' L='you lose'.

The table shows exactly:

(C and S and L) or (C and (not S) and (not L)) or ((not C) and S and (not L)) or ((not C) and (not S) and L) which is equivalent to C<=>(S<=>L), which means 'if and only if you originally picked the door with the car, switching is equivalent to losing'. And yes, it is EXACTLY why i make such proposal for inclusion of the table. 'if you originally pick the door with the car, then switching means loosing' is ambigious, because it might mean implication or equivalence. This table shows it in such a way, you don't need to know anything about logic.123unoduetre (talk) 15:27, 19 September 2010 (UTC)

Martin, I don't think the conditional approach is only academic. You know the law of total probability: probability of winning is equal to sum over distinct cases, of probability of winning in each particular case, times the probability of that case. We know that the "switcher" gets the left hand side in this equation equal to 2/3. The only way to get a higher probability overall than 2/3, would be by getting a better conditional probability of winning, by staying rather than switching, in one or more of the six particular cases: (initial door chosen by player, door opened by host). Moreover, the player has to *know* that the probability is better for staying rather than switching. In the minimal set-up where the player chooses a door completely at random but otherwise knows nothing, he doesn't know enough about the conditional probabilities in order to know for any particular case that it would be definitely better to stay rather than switch. In the case when we know more, when we know that the three doors are initially equally likely to hide the car (and the player might as well choose door 1 and not bother to randomize), one can use Bayes to show that each of the conditional probabilities is at least 1/2, so in each particular situation it does not pay off to stay instead of to switch. I would like to find a simple argument why no strategy can beat the overal probability of 2/3, in the scenario that the car is equally likely behind each door. If we could find a short simple and mathematically rigorous argument for that, the difference between the conditional and unconditional approach would vanish. We understand that the "switcher" wins overall 2/3 of the time, under the minimal assumption that his initial choice is correct 1/3 of the time. The conditional approach adds to this, that under the stonger condition that initially all doors are equally likely, you cannot do better overall than 2/3. If only we could find an easy way to prove this, we wouldn't need the conditional approach at all. I know that it can be done with von Neumann's minimax theorem, you can prove theorems in probability from theorems in game theory (!), but the proof of the minimax theorem is not elementary, so that does not bring us closer to the goal of a *complete and completely elementary* solution. Gill110951 (talk) 16:39, 19 September 2010 (UTC)

The table in the article is simpler, conveys the same information, and is what vos Savant used to explain her solution. Since there are 3 rows, rather than 9, it is more obvious that the probability of the player's initial choice being the car is 1/3. I don't see any reason to use the expanded version. -- Rick Block (talk) 17:28, 18 September 2010 (UTC)

As 123unoduetre points out, the above table shows the whole sample space. We all know that what might have happened is just as important in probability problems as what actually did happen. This table shows all the possibilities and is supported by a very reliable source. Martin Hogbin (talk) 09:16, 19 September 2010 (UTC)

For quite a few people, writing out a table in full makes clear what only a few people grasp more rapidly: those who always switch, win when those who always stay lose, and vice versa. The probability a switcher wins is therefore equal to the probability that a stayer loses, 2/3. This is an unconditional probability, taking no notice of the door numbers. I would like to point out that the table actually makes more assumptions than the result which we get from it, uses. You don't need to know that all doors are equally likely to hide the car, to get the "good" result. You just need to know that your initial choice hits the car with probability 1/3. Given that fact, a stayer wins with probability 1/3, a switcher wins with probability 2/3. So actually the table says more (assumes more), as Nijdam points out. I wouldn't say the table is stupid, lots of people seem to find it useful and I don't want to say that they are all stupid. By the way, lots of people thinkor automatically believe that the initial probability of 1/3 is engineered by the player himself or herself choosing a door at random. Then we don't have to make any assumptions about prior beliefs or knowledge.

Let me say part of this a different way, and add a different dimension: for most people, writing out a table in full is a useful first step towards comprehending the problem and finally discovering themselves that there are shorter and equally convincing ways to get to the good answer. Now, quite a few people actually think that the question we should be answering is: what is the probability the car is behind door 2 given we chose door 1 and the quizmaster opened door 3? That requires a more eleborate table to answer, and to fill in that table you'll either have to make even more assumptions, or you'll be obliged to do some algebra (using symbols for "unknowns" which fortunately, it turns out, you don't have to know if you only want to know whether you should switch or not). Gill110951 (talk) 13:59, 19 September 2010 (UTC)

I like the overview you provide here, Richard. It strikes right to the heart of how any person can readily come to understand the paradox. And it's supported by Selvin's original letter to the mighty and heralded American Statistician professional journal. Glkanter (talk) 14:55, 19 September 2010 (UTC)

There is no mentioning of any probability in the table, yet several editors deduce a probability from it! Nijdam (talk) 14:36, 19 September 2010 (UTC)

That's a good point, Nijdam. Writing down a table is specifying a sample space. It enables one to "see" the identity of certain events (eg of the event "stayer loses" with "switcher wins"). After that you can fill in probabilities as you think is warranted, and even leave probabilities unknown which you don't know. Just as long as "stayer loses" has probability 2/3, you have proved (under the assumptions you have made) that "switcher wins" has probability 2/3. Elementary statistics texts sometimes give the impression that you have to specifiy *all* the probabilities before you can do anything. But that is not true. Once you have fixed enough probabilities to fix the probability of the event which interests you, you are done. Gill110951 (talk) 16:20, 19 September 2010 (UTC)

Quite interesting, Richard. Again you come up with numbers, apparently on the basis of this very, very interesting table, whereas the table itself shows no number at all. Nijdam (talk) 09:52, 20 September 2010 (UTC)

No Nijdam. I just say that if you write down the sample space and then, *if* you fix *some* of the probabilities, some other probabilities might well be fixed too. A clever mathematician investigates what are necessary and sufficient conditions to get desired conclusion. In ordinary words, what are the minimal conditions needed to draw a desired conclusion? I said that *if* a stayer has probability 2/3 to lose the car, *then* a switcher has probability 2/3 to gain the car. Moreover the identity of the events (independent of any probability assigments) establishes that the condition is both necessary and sufficient. Gill110951 (talk)

Nijdam,I would have expected there to be a word of explanation in the text regarding the table. The car placement and the player's initial choice are uniform (and independent) so putting probabilities to the table entries is not too hard. Martin Hogbin (talk) 16:45, 19 September 2010 (UTC)

Not really! You cannot put probabilities, and that's why I didn't. Consider the following scenario: the car is always in gate one. Still switching strategy is better than sticking strategy. But (!) you cannot assign any probabilities, if you don't know what is the probability of the car being in each gate. That's why I switched columns. This solution doesn't depend neither on what monty does, nor on what probability there is for the car beeing in each gate. The only thing we know is, that in each situation, wherever the car is, you better do by switching than by sticking. This is best strategy in each case, and therefore it's best overall strategy. Please change explanation 2 if it doesn't make it clear.123unoduetre (talk) 10:02, 20 September 2010 (UTC)

Proposal by 123 uno due tre

Yes you are absolutely right. You don't need to know the probability of where is the car. The explanation I gave makes such assumption. But still the table doesn't list any probabilities. Thanks to your suggestions, now I think the following table would be better:

Clearly represented table of whole sample space

Door hiding car	Door chosen by player	Player switches	Player sticks
1	1	Lose	Win
1	2	Win	Lose
1	3	Win	Lose
2	1	Win	Lose
2	2	Lose	Win
2	3	Win	Lose
3	1	Win	Lose
3	2	Win	Lose
3	3	Lose	Win

Explanation 2:

Consider three situations where the car is (first column of the table). In each case if you have equal probability to select correctly you win two times if using switching strategy, and only once if using sticking strategy. Notice that your winning or losing doesn't depend on what Monty does. When you've chosen wrongly, Monty has to select a gate without the car. If you've chosen correctly, he can select either one, but it doesn't influence outcomes of any of the strategies. There are also possible other strategies like: 'if there is second gate to switch to, always switch to it, but if there isn't stick to one you've chosen' and they DO depend on what Monty does.123unoduetre (talk) 15:27, 19 September 2010 (UTC)

Please reword explanation in final version ;-)123unoduetre (talk) 15:32, 19 September 2010 (UTC)

+1, table very intelligible and imho a great benefit to all (new) readers. Gerhardvalentin (talk) 17:52, 19 September 2010 (UTC)

What's in the article now is

The solution presented by vos Savant in Parade (vos Savant 1990b) shows the three possible arrangements of one car and two goats behind three doors and the result of switching or staying after initially picking Door 1 in each case:

Door 1	Door 2	Door 3	result if switching	result if staying
Car	Goat	Goat	Goat	Car
Goat	Car	Goat	Car	Goat
Goat	Goat	Car	Car	Goat

A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.

Just to be clear, you're suggesting replacing this table and text with the expanded table and text you're suggesting above (presumably referenced to Selvin). Is this correct? Given that this is in the section of the article showing simple solutions, I think the simpler table and much shorter explanation is clearly better. If instead you're suggesting adding another table and more text, where (exactly) would this go? Wherever it might be added, isn't it effectively redundant with vos Savant's table? It is mathematically saying exactly the same thing, just using way more words and a 3x larger table. What is it adding that isn't in vos Savant's explanation? -- Rick Block (talk) 16:41, 19 September 2010 (UTC)

As I said before it's a matter of presentation, but as Gill110951 showed, also not making unnecessary assumptions. Yes I suggest replacing this table with my version. Information below the table from the article makes assumption about equiprobability of these situations. As Gill110951 noticed it doesn't need to be true. The probability of beeing in one of the gates doesn't need to be 1/3. Also the table from article do not show complete sample space of solution. Additionally in my explanation there is an information concerning what influence makes a selection by Monty. —Preceding unsigned comment added by 123unoduetre (talk • contribs) 16:52, 19 September 2010 (UTC)

I think that the suggested table does have some advantages in that it shows all the possible doors that the player might have chosen. This might make the solution seem more complete to some people. I do not much like the current table and think it would be better along the lines of the longer table thus:

Door hiding car	Door chosen by player	Player switches	Player sticks
1	1	Lose	Win
1	2	Win	Lose
1	3	Win	Lose

Despite the long arguments about conditional probability, I have only one agenda, and that is to make the article explain to readers as clearly as possible why the answer is 2/3 and not 1/2. This may take more than one type of explanation.

Once that explanation is complete, I am happy to cover all the other issues fully. Martin Hogbin (talk) 16:56, 19 September 2010 (UTC)

Of course you may use shorter table. I thought longer is better because of only two issues:

1. It shows WHOLE sample space

2. It allows you to think about other strategies (like in my explanation 2)

Of course none of this points is critical, and if you prefer shorter one, It's equivalent. (If you give description like 'and other six rows are similar')123unoduetre (talk) 17:04, 19 September 2010 (UTC)

How about the following:

The solution presented by vos Savant in Parade (vos Savant 1990b) shows the three possible arrangements of one car and two goats behind three doors.

Door 1	Door 2	Door 3
Car	Goat	Goat
Goat	Car	Goat
Goat	Goat	Car

A player who stays with the initial choice, say Door 1, wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.

This shows the whole sample space (all possible locations of the car). I distinctly dislike Martin's table since I think the car's location is the primary variable here, not the user's choice. From a referencing standpoint as well - who approaches it by analyzing all choices with a given car location? As far as I know, for sources using this sort of approach it's absolutely standard to show the 3 equally probable locations of the car and then to work through the effect of a given user's choice (i.e. holding the choice constant and varying the location of the car, not the other way around). -- Rick Block (talk) 17:19, 19 September 2010 (UTC)

It is a matter of opinion whether the car's location or the player's choice is the primary variable. Both opinions are supported by different reliable sources. This matter is further complicated by the obvious fact that *if* the car is hidden uniformly at random, *then* the player might as well choose door 1 (or *if* the player knows nothing and uses probability in a subjective Bayesian sense, then she might as well pick door 1 since "1" is her lucky number). Emphasizing the possible randomness of the player's choice is the big contribution of the mathematical economics and game theory literature to MHP. It shows that there is a rational solution to the puzzle which does not rely on making unwarranted assumptions about the host's behaviour. The player has two choice opportunities, not one. Most people think the first choice is not important. But not all people. Gill110951 (talk) 12:17, 20 September 2010 (UTC)

Ok, now I understand where is the disagreement. We disagree what sample space is. In my version of sample space there are 9 elements. The sample space is {<1,1>,<1,2>,<1,3>,<2,1>,<2,2>,<2,3>,<3,1>,<3,2>,<3,3>}. In your version (if I understand it correctly) it is {1,2,3}. The truth about second version is, you need three different sample spaces: one for the sentence 'Let's assume you've selected gate 1', second one for the sentence: 'Let's assume you've selected gate 2', and third one for the sentence: 'Let's assume you've selected gate 3'. They are identical, but only because we assume they do not cheat (moving car). Of course under this simple assumption the solution is equivalent, so it's possible to use second version. But still you have to use something similar to explanation 2, not the one used.123unoduetre (talk) 17:32, 19 September 2010 (UTC)

The quiestion is: do we really need to make such an assumption. In my opinion we don't. So first solution is more general.123unoduetre (talk) 17:39, 19 September 2010 (UTC)

Sorry, little correction. Sample spaces would be identical in each case. But probability spaces would differ. If you acknowledge that, both solutions are identical. It's only matter of using proper explanation.123unoduetre (talk) 17:47, 19 September 2010 (UTC)

Let me get this straight. All of you (including Martin, Glkanter, Gerhard, and Richard) think the 9 row table and "Explanation 2" above is more understandable than the existing 3-row table and text as above and support replacing the existing table and text with this? I'm finding this a little bit hard to believe. -- Rick Block (talk) 22:06, 19 September 2010 (UTC)

Don't panic Rick. I have only said the 9-row table has some merits. The others do not seem to have said much about it at all, or have I missed something? Martin Hogbin (talk) 22:40, 19 September 2010 (UTC)

Why would that be hard to believe, Rick? A table with 9 equally likely outcomes was Selvin's preferred solution when he originated the MHP in the American Statistician peer reviewed professional journal in 1975, 15 years before vos Savant/Whitaker and 16 years before Morgan, el al's paper, and 35 years before Morgan, et al's corrective letter. Glkanter (talk) 23:15, 19 September 2010 (UTC)

They don't have to be equally likely at all!!! We only assume your choice is equally likely, but the choice where is the car could not bee equally likely. For example the car could always be in gate 1. But still the switch strategy is better! In other words, the probability has to be the same for all YOUR three choices in each group. That's why I've switched the first and second column. Please read carefully explanation 2, and correct it if it's not obvious that the probability of car being in each gate don't need to be the same.123unoduetre (talk) 09:36, 20 September 2010 (UTC)

I find it hard to believe because you and Martin have been arguing for two years that the solution should be simple. Replacing a perfectly clear simple solution with one that is less clear and less simple strikes me as something you would be dead-set against. -- Rick Block (talk) 02:35, 20 September 2010 (UTC)

Well, the 9 entry table *is* a simple solution, as we've been using that term for some time. And it's the only solution Selvin presented when he originated the puzzle. What I *am* against is the repeated and intrusive inclusion of criticisms prejudicing the reader towards a POV in violation of Wikipedia's NPOV and UNDUE policies. Glkanter (talk) 04:55, 20 September 2010 (UTC)

Rick, please consider also the point Gill110951 made (concerning the fact we don't need to know probabilities of the car beeing in gates), and also my answer to you, and explanation 2 (concerning other strategies). But in my humble opinion, this table is also pragmatically better. (Of course it could be shortened (3 rows, like Martin's one), if presented with proper description.) I think all of you know my point of view, so I'm waiting for your suggestions now. Explanation 2 might be reworded for better presentation. Thanks for discussion.123unoduetre (talk) 23:44, 19 September 2010 (UTC)

Maybe we should back up a bit. Is this solution from a source, perhaps Selvin's 1975 letter as Glkanter is suggesting, or is it your own analysis? -- Rick Block (talk) 02:35, 20 September 2010 (UTC)

Table from the answer (by von Savant) to Selvin 1975 (a) is:

Keys are in box	Contestant chooses box	Monte Hall opens box	Contestant switches	Result
A	A	B or C	A for B or C	loses
A	B	C	B for A	wins

etc. My version is almost the same, but without Monte Hall opens box column, and with additional Result column for other strategy.123unoduetre (talk) 08:18, 20 September 2010 (UTC)

Rick are you trying to use the OR rule (no "own research") to ban a useful insight? But don't worry. This solution is the game theorist's (eonomist's) solution as I've been saying all along. Plenty of reliable sources which support it. Gill110951 (talk) 12:12, 20 September 2010 (UTC)

No. I'm trying to establish what source (or sources) this solution is attempting to represent - the text in particular. It is sounding like a game theoretic solution but the suggestion is to replace vos Savant's simple solution. If we're clear about what source(s) we're representing I think it will help. For example, Selvin doesn't present anything remotely like a game theoretic solution. He asserts each of the 9 rows of his table are equally likely, so the player has a 6/9 chance of winning by switching (and clarified, later, that he was assuming the host picks randomly between two losing alternatives and showed a solution using Bayes theorem as an alternative solution - implying he was thinking about the conditional probability). -- Rick Block (talk) 14:17, 20 September 2010 (UTC)

Explanation 2 is not a game theoretic solution. It's plain probability theory. In it you reason about unknown probability space with some known properties. If you need math, here it is:
Omega={<1,1>,<1,2>,<1,3>,<2,1>,<2,2>,<2,3>,<3,1>,<3,2>,<3,3>} (sample space)
P is unknown (probability measure), but (!) we do know something about it/
Consider three subsets A={<1,1>,<2,1>,<3,1>}, B={<1,2>,<2,2>,<3,2>}, C={<1,3>,<2,3>,<3,3>}.
A means "we have chosen gate 1", B means "we have chosen gate 2", C means "we have chosen gate 3".
We assume we select each gate with probability 1/3.
Therefore P(A)=P(B)=P(C)=1/3
Let's check the strategy "we stick".
D={<1,1>,<2,2>,<3,3>}
D means "we win", because as we discussed earlier, what Monty does is unimportant.
We need to know P(D).
Consider the sets E={<1,1>,<1,2>,<1,3>}, F={<2,1>,<2,2>,<3,2>}, G={<3,1>,<3,2>,<3,3>}.
E means "the car is in gate 1", F means "the car is in gate 2", G means "the car is in gate 3".
Now the tricky thing. We assume they do not cheat, which means they do not change the gate the car is in. (independence)
It means P(A product E)=P(A)*P(E) and P(B product F)=P(B)*P(F) and P(C product G)=P(C)*P(G).
But D=(A product E) sum (B product F) sum (C produc G), and because they are mutually exclusive, we have
P(D)=P(A pruduct E)+P(B product F)+P(C product G).
therefore
P(D)=1/3*P(E)+1/3*P(F)+1/3*P(G)=1/3*(P(E)+P(F)+P(G))=1/3
When we use strategy "we switch" we get
P(Omega \ D)=1-P(D)=2/3.
Ok, now let's discuss assumptions we have made:
1. We assumed P(A)=P(B)=P(C)=1/3.
It was because we have chosen gates with equal probabilities. It's obvious.
2. We assumed P(A product E)=P(A)*P(E) and P(B product F)=P(B)*P(F) and P(C product G)=P(C)*P(G).
It is because we assume they do not cheat. If we allow them to cheat, they could for example use following strategy:
"Always move the car to the gate the player had chosen". In this situation with sticking strategy you always win, and with switching strategy you always lose. Therefore we have to assume they do not cheat, which I think is standard assumption about this task, and it means independence.
Now let's think about alternatives. The question from this task was: "which strategy is better, switching or sticking". The reason why Monty is unimportant and we could ignore his decisions is exactly because we think about these two strategies. BUT if point 1 isn't correct, or point 2 isn't correct or we think about other strategies, the answer is different. But we need only to signalize it to a reader, not explain it fully, because it's not the original problem, and it shouldn't go to the Simple Solution section.
Now you'll say: OK, I agree, but it's ORIGINAL RESEARCH. And of course I have to agree with you. That's why if anybody would like to contribute something she could check the sources, and find the one which has similar explanation to explanation 2 and copy-paste. It seems not copy-pasting won't satisfy some people. The table is almost copy-pasted (with omission of selection of Monty, and addition of the other strategy). I'm tired now, and I think I've said enough. It seems to me, that to satisfy some people, if we need for example give some proof Newton first did, we should use calculus of infinitisimals, because he used it, and copy-paste from Philosophiæ Naturalis Principia Mathematica, and use his language of disappearing non-zero infinitisimal quantities. If I could publish it I would, but this problem is damn simple problem from first class of probability. EOT 123unoduetre (talk) 16:35, 20 September 2010 (UTC)

As of the last featured article review, this version of the article had a reference for essentially every independent thought in the entire article, which is the standard for FA. There are 100s of reliable sources about the MHP. The chances that anyone can come up with something original is very small. The point is not that we should only copy-paste, but that each independent thought in the article should be referenced to a reliable source - and that what the article says reflects what the sources (in the aggregate) say. If the point you're adding is obscure enough that it's hard to find a source, it probably doesn't belong in the article. The point you seem to be interested in is whether the player's chances of winning are influenced by Monty's actions. Many sources say they are - if you delay the point of when you're deciding to switch until AFTER the host opens a door. If you decide beforehand, many sources say your chances of winning are 2/3. I believe the article already addresses this topic (using sources already cited in the article). Is there some point you think is missing? If so, what is it and what sources make this point? -- Rick Block (talk) 18:31, 20 September 2010 (UTC)

Ok, I agree that there are two different problems:

(problem 1) First is: Who wins more often: someone who always sticks, or someone who always switches.

(problem 2) Second on is: what is the best strategy (either to switch or to stay), after Monty opened one of the gates?

A solution of the first problem uses only probability (as I presented it). A solution of the second problem uses game theory.

I remind you we discuss only the part named: Simple Solutions.

The solution which starts with: "The solution presented by vos Savant..." and ends with "...winning by switching is 2/3." is an answer to the first question above. It doesn't answer second question. My proposal is to change it to my version. Additionally please if possible stress the difference between these two questions. The difference is tremendous. Answer to the first question is of course "someone who switches". But anwer to the second one needs the knowledge of Monty's strategy, and probabilities of car placement. Therefore it's not simple solution. We should therefore emphasize different questions we like to answer. Maybe in the header of simple solutions section, we could add: "If the question is understand to mean ...." or something similar. To summarize: problem 1 and problem 2 ARE COMPLETELY DIFFERENT PROBLEMS WITH DIFFERENT ANSWERS. First paragraph in "Simple Solutions" solves problem 1 only. Similarly further paragraphs. Only paragraph starting with"As Keith Devlin says..." says it solves problem 2 but actually it solves problem 1. My queston for you is: you said about deciding to switch AFTER the host opens the door. But afair there is no complete explanation of problem 2 in the article. It doesn't discuss questions like: what if you know hosts strategy, what if you don't? Therefore I think answering problem 2 is either besides the topic of this article, or should be included. In my opinion we should only discuss first problem, and my version concerns only this one. Of course I haven't given any sources, but let's first agree what to search in these sources...123unoduetre (talk) 19:43, 20 September 2010 (UTC)

A note on terminology: your "problem 1" and "problem 2" above are called on these pages the "unconditional problem" and the "conditional problem" respectively. The distinction between the two is quite clear in the sources (e.g Morgan, Falk), and the two problems have numerically the same solution (switching wins with prob = 2/3) under the standard unequivocal interpretation of the MHP (the Krauss & Wang one). Their solutions are, likewise, the "unconditional" and "conditional" solution respectively. Both problems are solved using elementary probability theory (though seeing them both as a game theory problems may provide some insight). For example, the article section entitled "Mathematical formulation" gives a common probability-theoretical solution to the conditional problem, which can be immediately generalized to the unconditional one by marginalizing (i.e. averaging) with respect to the initial location of the car C and the player's initial selection S. I some other editors agree with you that the article in its current form does not stress enough the difference between the two problems, and appears to favor the unconditional version. This is, in fact, part the issuer currently under WP mediation. glopk (talk) 03:39, 21 September 2010 (UTC)

Except that the NPOV and UNDUE violations in the article favor the sources that criticize the simple (unconditional) solutions. Glkanter (talk) 04:15, 21 September 2010 (UTC)

@123unoduetre - The confusion between these two problems and what sources address what problem is one of the main reasons we're in formal mediation about this article. In the current version of the article, the "Conditional probability solution" section presents a solution to your problem 2 (under the standard assumptions that the car is uniformly located to begin with and the host picks evenly between two goats). As of the last featured article review, the "Solution" section included both a "simple" solution and a conditional solution with the following two paragraphs in between:

The reasoning above applies to all players at the start of the game without regard to which door the host opens, specifically before the host opens a particular door and gives the player the option to switch doors (Morgan et al. 1991). This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).

A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below).

The intent was to make the distinction you're asking for. There's a faction of editors who apparently believe that saying anything about this distinction is incredibly biased, and claim the simple solutions solve the problem where the user is deciding AFTER the host opens a door. I think everyone agrees the usual wording of the problem puts the decision point AFTER the host opens a door, so agreeing these are two different problems effectively means the simple solutions are addressing a slight variant (Morgan et al. calls them "false solutions"). The editors favoring simple solutions are unwilling for the article to a) state as a matter of fact (even sourced) that these two problems are different, b) state that the "simple solutions" address your problem 1 and not your problem 2 (even if sourced to peer reviewed math articles and textbooks). -- Rick Block (talk) 14:28, 21 September 2010 (UTC)

I think the whole situation is clarified by taking a look at The Truth. Here it is:

Vos Savant/Whitaker don't ask for a probability but for a decision. Many of their readers interpret: "say Door 1" and "say Door 3" as remarks in parentheses intended only to aid your imagination. Rosenhouse even says in this book that that would semantically be the right interpretation. Marilyn's own "unconditional" solution and her proposal for simulation experiments supports that interpretation as her intention.

I agree that *if* you can compute a conditional probability, it could be wise to do so. (In the situation of the *standard problem* - everything specified, everything uniform, totally symmetric) you can *know* it without computation and know it is irrelevant). However

If *all* you assume is that your initial choice has chance 1/3 to hit the car (e.g. because you yourself chose your own door at random, which is a legitimate interpretation made by many people) then *all* you can conclude is that "always switching" beats "always staying" (2/3 versus 1/3).

If you are able to assume that initially *each door* has probability 1/3 to hide the car, then you can easily prove with Bayes that "always switching" not only beats "always staying" but it also beats any mixed strategy. Equivalently, all conditional probabilities are at least 1/2.

If you *also* assume neutral behaviour of the host, *then* all conditional probabilities are 2/3.

I am thinking here of probability in the frequentist or ontological sense. Something which objectively exists "out there" even if I don't happen to know it.

If however you use probability in a subjectivist sense then not knowing anything else, all doors are equally likely for you to hide the car, and the host is equally likely to open either door at his disposal if he has a choice. "Equally likely" simply means that *you* have no information to break the symmetry. It means that you are indifferent. It is about you much more than about the quiz show. The specific door numbers are irrelevant to you since your beliefs are invariant under permutation of the door numbers. Hence your beliefs are independent of the specific door numbers. If you are consistent in your subjective beliefs it follows from your earlier summarized beliefs that the "other door" is two times more likely to hide the car than "your door". This just means that you ought to be indifferent to take either sides of a bet at 2:1 odds that the other door hides the car. It also means that you *ought* to switch *if* you want your decisions to be consistent with your other beliefs, and if there are no costs involved. One of the reasons people don't switch is because it brings a cost to their ego.You don't compute a conditional probability because of your total indifference to the door numbers. You know in advance that the conditional probabilities will equal the unconditional. You also don't condition on the fact that today is Wednesday.

I think the hidden reason for the many disputes about MHP lie in different interpretations of probability.

OK I have to find reliable sources for all this, and if I fail, I'll have to write one myself. Gill110951 (talk) 09:54, 22 September 2010 (UTC)

Richard, would you agree that Morgan, J. P., Chaganty, N. R., Dahiya, R. C., and Doviak, M. J. said it best (and pithier)?:

"We take this opportunity to address another issue related to our article, one that arose in vos Savant’s (1991) reply and in Bell’s (1992) letter, and has come up many times since. To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period." The American Statistician, May 2010, Vol. 64, No. 2 193

Glkanter (talk) 11:18, 22 September 2010 (UTC)

Well it is certainly a pithier statement but I entirely disagree with it. Firstly, the question asks for an action, not a probability. Secondly, I disagree that one particular set of conditions is implicit in the problem. I think therefore that the best answer to the question is: under various supplementary conditions, the answer is "switch". If you can make assumption A, then the reason for this is X; if you can make assumptions A+B, then you have even better reasons Y; if you can make assumptions A+B+C, then the situation is particularly nice and the reason for switching is Z.Gill110951 (talk) 14:34, 22 September 2010 (UTC)

Would you say that the statement I quoted, with which you disagree, comes from a reliably published source? Glkanter (talk) 18:47, 22 September 2010 (UTC)

Framework for discussion proposal

I've been thinking about this problem for a while, and would like to point out, there could be many different problems, and original question is ambigious. Following are only examples how we could understand original question. First I'd like to summarize them in english, and then in math:
PROBLEM A ("unconditional"): Which people win more often, when given selection of some gate, choose one of them, and after that Monty opens some other gate? Those who always switch, or those who always stick?
PROBLEM B ("conditional"): Let's think about the following situation: when playing the game like the one above, player selected gate 1. Monty opened gate 3. Now let's ask: Which people win more often, when given such a situation, those who switch, or those who stick?
PROBLEM C: What is the best strategy in some game theoretic sense, for winning most often.
PROBLEM D: Given the situation in PROBLEM B, what is the best strategy in some game theoretic sense for winning most often.
Now please let me formalise problems A and B. Firstly I'd like to present the whole sample space for PROBLEM A. Sample space for PROBLEM B is part of this sample space:

No	Car	Me	Monty	I stick	I switch
1	1	1	2	I win	I lose
2	1	1	3	I win	I lose
3	1	2	3	I lose	I win
4	1	3	2	I lose	I win
5	2	1	3	I lose	I win
6	2	2	1	I win	I lose
7	2	2	3	I win	I lose
8	2	3	1	I lose	I win
9	3	1	2	I lose	I win
10	3	2	1	I lose	I win
11	3	3	1	I win	I lose
12	3	3	2	I win	I lose

After introducing the table, please let me introduce the following notation:
[1,3,5] means the set {<1,1,2>,<1,2,3>,<2,1,3>}.
Let me define the following sets:
Om=[1,2,3,4,5,6,7,8,9,10,11,12] --- the sample space
S1=[1,2,3,4] --- it means "the car is in gate 1"
S2=[5,6,7,8] --- it means "the car is in gate 2"
S3=[9,10,11,12] --- it means "the car is in gate 3"
J1=[1,2,5,9] --- it means "I've selected gate 1"
J2=[3,6,7,10] --- it means "I've selected gate 2"
J3=[4,8,11,12] --- it means "I've selected gate 3"
W1=[1,2,6,7,11,12] --- it means "I win using strategy 'I stick'"
W2=[3,4,5,8,9,10] --- it means "I win using strategy 'I switch'"
B=[2,5] --- situation when I've selected gate 1 and Monty've selected gate 3
a=P([2])/P([1,2]) --- probability that Monty selects gate 3 when in situation [1,2]
Now let me specify what do we know:
1. For all i, for all j P(Si product Jj)=P(Si)*P(Jj) --- because our selection of gate is independent of where the car is. ("axiom")
2. P(W1)+P(W2)=1 ("theorem")
3. P([2])=a*P([1,2])=a*P(S1 product J1)=a*P(S1)*P(J1) ("theorem")
4. P([5])=P(J1 product S2)=P(J1)*P(S2) ("theorem")
Now we are ready to ask questions:
PROBLEM A: P(W1)=?
P(W1)=P((S1 product J1) sum (S2 product J2) sum (S3 product J3)). Because of beeing mutually exclusive... Because of independence...
P(W1)=P(S1)*P(J1)+P(S2)*P(J2)+P(S3)*P(J3)
PROBLEM B:
P(W1 product B)/P(B)=?
P(W1 product B)/P(B)=P([2])/P([2,5])=P([2])/(P([2])+P([5]))=a*P(S1)/(a*P(S1)+P(S2))=1/(1+ P(S2) / (a*P(S1)) )

Now let's discuss some additional assumptions we might make (not making them now):
Assumption 1: P(J1)=P(J2)=P(J3)=1/3
Assumption 2: P(S1)=P(S2)
Assumption 3: a=1/2
Assumption 4: a=1

If we make assumption 1 only: PROBLEM A:
P(W1)=1/3*P(S1)+1/3*P(S2)+1/3*P(S3)=1/3

If we make assumption 2 only: PROBLEM B:
P(W1 product B)/P(B)=1/(1+ 1/a)=a/(a+1)

If we make assumptions 2 and 3 only: PROBLEM B:
P(W1 product B)/P(B)=1/(1+2)=1/3

If we make assumption 2 and 4 only: PROBLEM B:
P(W1 product B)/P(B)=1/(1+1)=1/2

Now let me discuss about PROBLEMS C and D. These problems are DIFFERENT problems than PROBLEM A and PROBLEM B. To solve them we need to think about knowledge of strategy of an opponent. Does Monty know your strategy? Could he manipulate car placement etc. Do you know Monty's strategy? Is it important if you know his strategy, or he knows yours? Some trivial examples: if you know probabilities P(S1), P(S2), P(S3) you select the most probable gate, and stick with it. Trivial example: P(S2)=P(S3)=0, and you know it, you can use strategy: "always select gate 1". Sorry, I know nothing about game theory, therefore I can't tell what is correct solution as far as game theory is concerned (although I'm still learning ;-). Also I hope I haven't made some mistake (please review if you have enough time ;-)

Now after presenting whole problem (not really, I haven't touched PROBLEM C and PROBLEM D), please let me explain what do I think. I wonder if it is possible to use this presentation somehow when creating wiki page. I think the most important is to present how people might understand MHP. Full solution might be given, but I wonder if it is possible to use english instead of math, but still be strict enough. I think this presentation is more useful for people who would like to understand MHP in Discuss pages, not in wiki page. So my opinion is:
1. tell people that original problem is ambigious and tell them what are possible ways to understand it, but not necesarilly use math to explain it, we might explain it in english if possible, but it MUST be strict enough, for (smart) people not to be able to misunderstand each problem.
2. tell them solutions to at least PROBLEM A and PROBLEM B, maybe with some sort of explanation, but giving necessary assumptions.
123unoduetre (talk) 20:49, 22 September 2010 (UTC)

Suspending discussion here during mediation

Article talk pages are intended for editorial decisions concerning the article. During a mediation, the editing process is usually suspended while issues are sorted out. I know it is hard, as we have a limit on length of discussion on the mediation talk page, however, would editors please refrain from discussion here? If participants would like to hold particular discussions during the mediation, I would suggest creating a subpage (or pages) of the mediation talk page. That way any consensus reached can be related back to the mediation. Sunray (talk) 14:18, 23 September 2010 (UTC)

Some suggestions for a compromise

As you know I do not want to encumber the simple solutions with talk of conditional probability or anything else that might complicate the explanation.

Although it is my personal opinion that the simple solutions are completely valid, it is universally agreed that the simple solutions are valid for the unconditional problem. My only objection to stating this is that to do so brings in complications that might confuse the reader. Is it possible that we describe the unconditional problem in our simple solution section but in a subtle way that will not draw attention to the distinction between the conditional and unconditional problems at that stage?

For example, rather that say 'players whose strategy is to swap' we might just say 'players who swap'. This can still be argued to be technically correct but avoids bringing up the complication issue of strategy vs decision-made-at-the-time. Maybe the 'conditionalists' should write the simple section in language that they are happy with then the 'unconditionalist' could attempt to tone down the language whilst still keeping it technically correct. Later in the article, of course, the distinction could be discussed fully. That would include stating that many sources do give simple solutions to the problem 'as stated'. Martin Hogbin (talk) 17:54, 19 September 2010 (UTC)

Excuse me, but I do not uderstand who was the target of this comment. If it was about "Something to include" section, I don't understand what you are saying. We certainly do NOT discuss conditional probability.123unoduetre (talk) 18:00, 19 September 2010 (UTC)

We have a long standing dispute about sources that say the problem is asking about the conditional probability P(win by switching|player picks door 1 and host opens door 3) as opposed to the unconditional probability P(win by switching) or the "semi-conditional" P(win by switching|player picks door 1). I can only surmise Martin is thinking what I'm suggesting here is somehow related to this discussion - perhaps because the analysis I'm suggesting is technically P(win by switching|player picks door 1). I'm suggesting this analysis NOT because of this dispute, but because it is what I think most sources presenting this sort of analysis say. It is clearly what vos Savant says. -- Rick Block (talk) 19:38, 19 September 2010 (UTC)

123unoduetre, I started a new section because this was, exactly as Rick has said, a suggestion that concerns a very long running dispute about conditional/unconditional probability that is currently the subject of formal mediation. The problem is that it is very hard for the current editors to discuss what might be best in the article without seeing it from their particular POV. My suggestion was for a compromise that might help end the dispute so that the existing editors could all work together and consider new suggestions, such as yours, from the same perspective, that of improving the article for the general reader. Martin Hogbin (talk) 21:11, 19 September 2010 (UTC)

It would indeed help us a lot further if a "conditionalist" would draft with as much sympathy as he or she can muster the simple solution (or the first part of it), and if an "unconditionalist" would draft with as much sympathy as he or she can muster the conditional solution (or the first part of it). And both parties would allow the others to propose rewordings which remove hidden assumptions or hidden bias. (I am disqualified since I am neither. And sadly no one else yet has felt inclined to support the "economist" point of view.) Gill110951 (talk) 12:40, 20 September 2010 (UTC)

I think it is also useful to realise that the conditionalist's solution makes more assumptions in order to gain stronger conclusions. @Nijdam recently challenged me to explain what I thought was wrong with the conditionalist's solution. I answered him here [2], on my talk page. Actually I don't think there is anything wrong with it, though I don't like the way it is sometimes sold (whether on wikipedia or in "reliable sources"). My approach as a professional writer of reliable sources in probability and statistics is always to try to figure out and understand "The Truth" first, in the hopes of finding a way of making it more easily to apprehend, despite wikipedia policy that "The Truth" is irrelevant. In the case of MHP I come to the conclusion that The MHP Truth is multi-facetted (and this is exactly what makes it such a great topic!). Now people who don't follow my arguments may complain that I am pushing Own Research. That's fine, still, it's my opinion that in the field of mathematics it does pay off to search for The Truth, and if it turns out that The Truth is easily and immediately recognisable, then it would be sad that wikipedia would have to wait 10 years for it to appear on wikipedia pages. Fortunately, if The Truth is indeed so easily and immediately recognisable then people will come up with other fundamental wikipedia policies which justify its inclusion. It's like the situation when different fundamental rights of man are in conflict with one another (eg free speech versus a ban on racial or religious or gender discrimination). I do think we're making real progress on MHP thanks to all the dedicated editors here! Gill110951 (talk) 12:32, 20 September 2010 (UTC)

I agree that it is essential to understand all the issues relating to the MHP to write a good article on it. A cut-and-paste approach to sources does not work. Neither is it the way that WP is intended to work. Sources are meant to support the information given by editors to show that it is correct. They do not tell us how to write the article.

I am not trying to push either the simple or conditional solutions, I am just trying to make the simple solutions as unconfusing as possible for the average reader. Talk of 'players whose strategy is...', 'solutions that do not address the stated problem' etc do nothing to help the reader understand why the answer is 2/3 rather than 1/2. on the other hand we may be able to make the simple solutions refer to the unconditional problem more subtly, with talk of 'players who switch' or the like. The lead section, which nobody has criticised, makes quite a good job of this. Martin Hogbin (talk) 22:33, 20 September 2010 (UTC)

Rick: Influenced by Monty's action? Indeed there is some point that you are missing. As I read Ruma Falk, she clearly is saying that the opening of one door by the host cannot change the odds of the door selected by the guest, unless the host is biased to open his favored door, "and you know about this bias", and she says that – without such a given and known host's bias, the odds of the door selected by the guest remain unchanged, because we’ve learned nothing to allow us to revise the odds.
Are there reliable contradicting sources that disagree with Falk?
You say there is an influence. We should pay attention to the fact that – without expressively mentioning of such a given and known host's bias – the opening of a door by the host never can be any relevant condition to change the odds. It has no influence.
"Using" of the door opened by the host as a so called "new condition" without mentioning the host's given and known bias, evidently emerges just as an utilitarian and practical method to provide an alleged "new condition" for the purpose of teaching conditional probability maths, but all of that without really addressing the MHP itself. We should pay regard to the fact that any alleged "influence" can only result from the given and known bias of the host to open his preferred door. Otherwise not. We should mention this fact. Gerhardvalentin (talk) 22:44, 20 September 2010 (UTC)

You're reading something into what Falk is saying that she doesn't actually say (and I'd bet she wouldn't even agree with). What she says is that if the host is biased and you know about it, then it is not true that (in vos Savant's words) "we've learned nothing to allow us to revise the odds". As logical propositions, the structure here is IF A AND B, THEN NOT C, where A is "host is biased", B is "you know about it" and C is "we learn nothing". What you're claiming she's saying is IF NOT B, THEN C, which is neither what she actually says or logically implied by what she says. -- Rick Block (talk) 14:01, 21 September 2010 (UTC)

There is no other way to gather what Falk clearly is saying: she clearly indicates just this unique reason, and no other reason, because there is no other reason. Without a given and known bias of the host, by opening of one door, you have learned nothing to revise the odds of the door selected by the guest. We should be careful and not turn around the meaning and and the significance of the statement. Gerhardvalentin (talk) 14:16, 21 September 2010 (UTC)

Falk understands probability in the subjectivist way. Rick understands it in the frequentist way. Different notions of probability make different assumptions reasonable or unreasonable, motivate different approaches to MHP. For Falk, probability is in her mind, her information. For Rick it is in the real world. Both ways of using probability are reasonable. Since they are different kinds of probability, assumptions concerning them have different meaning, and conclusions drawn with them have different meaning. The answer may be "switch", "2/3" but the meaning of "2/3" is completely different. Gill110951 (talk) 18:16, 21 September 2010 (UTC)

Frequentist

I disagree with you here Gill. The frequentist approach gives the same answer as the Bayesian approach, provided that the correct experiment is repeated. Martin Hogbin (talk) 21:45, 21 September 2010 (UTC)

Nonsense. Experiment and choice of priors are independent for the subjectivist Bayesian (as well as for the objectivists who believe in MAXENT, as MAXENT is not a necessary constraint on the choice of priors). There is no such thing as a "repetition of correct experiments" for the subjectivist. There are only explicit statements of what the priors and the likelihoods are. glopk (talk) 14:12, 22 September 2010 (UTC)

I have not mention subjectivists. I said that a frequentist approach would give the same result as a Bayesian approach if the correct experiment is repeated. If you disagree, perhaps you could give me an example where the Bayesian and frequentist approaches give different answers to the MHP. Martin Hogbin (talk) 17:32, 22 September 2010 (UTC)

(1) My answer covered both subjectivist and objectivist Bayesian (problems with your reading glasses?). (2) Example: use the prior P(C) = 0 for all c.glopk (talk) 22:00, 22 September 2010 (UTC)

I thought you were going to give me an example of a case where the Bayesian and frequentist answers (probability of winning by switching) were different. Martin Hogbin (talk) 22:39, 22 September 2010 (UTC)

I thought you knew that frequentists have no notion of "priors" (hence the frequentist response is unaffected), whereas the Bayesian answer with prior P(C) = 0 is undefined (0/0 naively, see Borel–Kolmogorov paradox): different enough? glopk (talk) 23:03, 22 September 2010 (UTC)

Sorry, I should have said realistic version of the MHP. I am not interested in talking about a version with no prize. Martin Hogbin (talk) 23:33, 22 September 2010 (UTC)

Ooooh, now you want "realistic", and your criterion for being realistic is that there is a prize (in the player's opinion, since we are talking priors here). OK, I'll make it easy and really really close to the standard version. Please conjure a physically feasible repeated experiment whose relative frequencies of winning by switching converge to a value equal to the Bayesian posterior for the same event, given the following prior: P(C=c) = { 1/3 - 2x iff c=1; 1/3 + x iff c = 0,2 }, where x = 1/exp(N!) and N is the Avogadro number, and the "realistic" convergence threshold is less than x. glopk (talk) 05:09, 23 September 2010 (UTC)

BY realistic I mean some version of the MHP. There has never been a version suggested with no prize. The experiment does not need to be physically feasible. In principle, any experiment can be repeated as many times as we wish. Martin Hogbin (talk) 08:56, 23 September 2010 (UTC)

As usual, you didn't even try to read my response, so I have to spell it out for you. I gave you above an example with prizes which is vanishingly close to the standard version of the MHP. Yet, a frequentist analysis for it has to conjure an experiment that is infeasible to such a degree as to make the whole notion of "experiment" redundant: if you tried random sampling for the prior I gave you above, you'd have to collect data for much longer than the age of the universe before you'd get a statistically significant set. So this is the mirror image of the "no prize" example I gave you above: here it's the Bayesian guy that gets the answer in a straightforward manner, whereas the frequentist guy is left in a hopeless situation. I ask again: different enough? glopk (talk) 17:57, 23 September 2010 (UTC)

I don't know what you mean here, Martin. Of course all probabilities, whether ontological / objective or epistemological / subjective (whether supposed to be properties of objects in the real world like time, mass, length, or supposed to be a reflection of particular person's knowledge), can be thought about by imagining appropriate repetitions. Both kinds satisfy the same axioms. However, in the real world, motivating assumptions about probabilities does involve settling also this question. If you want to motivate an assumption that some probability is 1/3, you have to say 1/3 of what kinds of repetitions. It means 1/3 of the times, but what "times" are we are talking about? The "times" could consist of viewing a movie of a possible parallel world, the parallel worlds of our imagination corresponding to our beliefs about the real world based on our information about it. Conversely, "objective" probabilities can be thought of as the subjective probabilities of an imaginary subject who knows a great deal more than you or I. The philosophers have been going round in circles for 300 years and will no doubt continue to do so for another 300 years. I think ultimately it is a matter of taste (politics, religion, culture) whether you take one or the the point of view as primary, since you can always embed the other view inside your own. But it is very important to get out onto the table, what repetitions people have in mind when they talk about a probability. What stays the same, what varies; how and why does it vary. Gill110951 (talk) 14:19, 22 September 2010 (UTC)

Gill, your last two sentences were exactly the point that I was making. With the appropriate repetitions in the frequentist result is the same as the Bayesian result. Martin Hogbin (talk) 17:32, 22 September 2010 (UTC)

Frequentist - Mediation Link Placeholder

The tree derived from Carlton's simple solution leads me to conclude that the 1/3 v 2/3 results are identical for the simple and the conditional solutions. Any number (the 1/3 v 2/3 before a door has been revealed, aka Carlton's simple solution) multiplied by 1 (the 100% likelihood that the host will reveal a goat) remains unchanged (the 1/3 v 2/3 after a door has been opened to reveal a goat, aka the non-traditional conditional solution). Glkanter (talk) 14:28, 22 September 2010 (UTC)

More nonsense. Of course it is possible for the conditional solution to yield a different value for the probability of winning by switching - Whitaker's statement of the problem is compatible with any assignment of the prior distribution P(C) for the location of the car, and with any assignment for the host's strategy P(H | C, S) for selecting the door to open. With P(C) uniform (= 1/3) only the optimal decision need be the same between the simple and conditional solutions, since the posterior P(C | H, S) >= 1/2 regardless of the host's strategy, but the probability of winning by adhering to that decision can still be any number in [1/2, 1]. It's just that in the conditional formulation more "stuff" is made explicit that the simple solutions conveniently sweep under the carpet, and therefore the conditional formulations allow more flexibility for reasoning about the assumptions while remaining consistent with them. glopk (talk) 16:02, 22 September 2010 (UTC)

It's not 'nonsense'. It just neatly disproves your arguments. Can you demonstrate how that decision tree is flawed in some way? Or that it doesn't address the MHP as put forth by Selvin, vos Savant and Whitaker? But not Morgan's 'slightly different problems'. The article isn't about those. Besides, are you taking issue with the simple solutions, or with the *presentation* of the simple solutions by the reliable sources? Glkanter (talk) 18:53, 22 September 2010 (UTC)

I am taking issue with the nonsensical statement above: "the 1/3 v 2/3 results are identical for the simple and the conditional solutions". The conditional solution can yield any value in [1/2, 1] for the probability of winning by switching with a uniform prior for the initial placement of the car, depending on the host's strategy P(H|C,S). It's only in the K&W formulation of the MHP (specifically, assuming no host bias), that the simple and conditional solution yield numerically equal values for the probability of switching. I don't have to demonstrate any flaws in any OR "decision tree" pictures to see that: it's math. glopk (talk) 21:55, 22 September 2010 (UTC)

Yes, "it's math". And "logic". I'd like to see you prove the decision tree wrong for the MHP about a game show. Based on the premises as derived from Selvin, vos Savant and Whitaker.Glkanter (talk) 23:40, 22 September 2010 (UTC)

But not Krauss & Wang? glopk (talk) 00:00, 23 September 2010 (UTC)

Sure, incorporate K&W. They're not an 'original source', so I left them out. But use the whole thing from K&W. No requirement for doors 1 & 3 ("Imagine that you chose Door 1 and the host opens Door 3, which has a goat." is what they say for clarity *after* listing the premises), everything is 'random' or 'uniformly at random', no need for a solution to repeat the premises given. So, tell me where the decision tree derived from Carlton's simple solution fails. Glkanter (talk) 00:09, 23 September 2010 (UTC)

Plus, just because they're provided, not every piece of information *must* be used. For example, Selvin says in his 2nd letter that the host chooses between two goats uniformly at random. Or something to that affect. K&W pick this up as a premise. Carlton's simple solution shows that that information is not required to solve the puzzle. (In fact, Rick Block's old 'broken clock' analogy applies here. That conditional solution you're so fond of ONLY works when the host chooses equally. Carlton's, and other simple solutions show that constraint isn't necessary.) That's not a flaw in the solution. That's elegance. Glkanter (talk) 01:55, 23 September 2010 (UTC)

I thought I could resist commenting here, but this is just too much. Like many things, you have this exactly backwards. Assuming we're interested in the probability of winning by switching AFTER the host has opened a door, i.e. the counterintuitive 2/3 probability a player has when deciding whether to switch looking at two closed doors and one open door showing a goat, the conditional solution always works since this is precisely what the conditional solution addresses. The simple solution only "works" (works meaning results in the correct numeric value for the probability) if the host chooses equally. The simple solution is the broken clock here. It's always (correctly) telling you the probability of winning if you decide to switch BEFORE the host opens a door. But if you want to know the probability AFTER the host has opened a door (when you're looking at two closed doors and one open door), using the simple solution doesn't tell you the right answer unless the answer happens to be 2/3 (which it can be but doesn't have to be - exactly like a broken clock correctly tells you the time, but only if the time is what the clock says). -- Rick Block (talk) 02:59, 23 September 2010 (UTC)

Well Rick, I've said for a long time that you and I see (nearly) all things 180 degrees differently. Maybe you could explain how that decision tree derived from Carlton's simple solution is flawed as a solution to the MHP? But no 'it fails for slightly different problems', please. But I'm especially interested in how it fails to address the 'after' problem. It shows a door has been opened to reveal a goat. 100% of the time. Exactly as K&W state the premise. No more. No less. Glkanter (talk) 03:26, 23 September 2010 (UTC)

I'm not going to play this game. See you in mediation. I suggest you bring your sources. -- Rick Block (talk) 03:49, 23 September 2010 (UTC)

You should have followed your original instinct Rick, and not got involved in this thread. You called me out in a big, tough paragraph based on your personal opinions. Which is OK, most everybody has been offering personal opinions in this thread. I asked you to validate your 'broken clock' claims regarding the decision tree, and you accuse me of 'playing games', you decline, and say something about 'sources'. OK. Carlton, K&W, Selvin, vos Savant, Whitaker and basic math are the sources for that tree. Which sources was your comment supported by? Glkanter (talk) 04:10, 23 September 2010 (UTC)

Let's go for all the marbles. Maybe Nijdam would address the decision tree in regards to his longstanding claim that the 'before' 1/3 & 2/3 are not the same as the 'after' 1/3 & 2/3. This shows pretty clearly that they are the same. Glkanter (talk) 04:29, 23 September 2010 (UTC)

Rick has already correctly answered your comments to my notes above. I have nothing to add, see ya in mediation. glopk (talk) 04:43, 23 September 2010 (UTC)

Gamesmanship. Hard to hold onto that presumption of 'Good Faith' from all editors. Glkanter (talk) 04:56, 23 September 2010 (UTC)

Much easier than my having to hold onto a presumption of good faith AND competence for a certain pair of editors. glopk (talk) 06:15, 23 September 2010 (UTC)

Consistency

Why do you choose to take P(C) as uniform but P(H|C,S) to be non-uniform? Martin Hogbin (talk) 17:38, 22 September 2010 (UTC)

Because I can: I only need one counter-example to falsify Glkanter's general statement that "the 1/3 v 2/3 results are identical for the simple and the conditional solutions". And for the umpteenth time I have to repeat to you the simple mathematical fact that the two solutions don't have to yield equal numerical values for the probability of winning by switching. One must assume uniform car placement and lack of host bias to get equal numerical values. Most simple solutions in the sources are incomplete because they do not explicitly state these necessary assumptions, whether they purport to solve the decision problem of switching after the host's question, or whether they compute the marginal probability of winning for a population of switchers. glopk (talk) 21:55, 22 September 2010 (UTC)

Yes, I know one must assume the unstated distributions to get and answer to the MHP but one should be consistent in one's assumptions. No information is given in Whitaker's question about either the initial car placement or the host's goat-door choice. There are two self-consistent ways to treat the problem: take both distributions to be non-uniform (and unknown), in which case the problem is insoluble, or take both distributions to be uniform, in which case the answer is exactly 2/3. There is no justification for doing anything else. Morgan et al have now agreed with this. Martin Hogbin (talk) 22:46, 22 September 2010 (UTC)

Wrong again. (1) Exactly because no information is given about those distribution's in Whitaker's statement, there are infinite self-consistent ways to treat the problem: one may assume any mathematical form for them, and still remain consistent with the statement. (2) "Consistent" is not the same as "justified": a baroque choice for the priors may be hard to justify (did Monty feel baroque that night?), but still yield a solution consistent with the problem statement. (3) Please don't mix an argument about math with one about sources, unless you'd rather hide behind an authority. (4) Perhaps the Argument pages is more appropriate for this discussion, or we may accept Rick's plea to continue on the mediation page. glopk (talk) 23:50, 22 September 2010 (UTC)

If course, if you want to contrive an interesting problems for statistics students, you are free to treat the unknown distributions differently but there is no mathematical, philosophical, or real-world justification for doing so.

Regarding outside discussions during mediation, I am happy to accept the mediators' advice on the matter. Martin Hogbin (talk) 08:50, 23 September 2010 (UTC)

Why don't we use The Truth here to guide our discussions? Different sources think that Whitaker asks different questions. Different sources find it more or less reasonable to make different supplementary assumptions. (Sometimes because, IMHO, they have different understandings of the word "probability", but that's a side issue; and sometimes because they think of the guest as an active player, not as a passive player). We all agree that we know that the host always opens a door revealing a goat and can do this because he knows where the car is hidden.

Do we all agree that the following (uno due tre) are Mathematically Indisputable True Facts? Do we all agree that each makes in succession a stronger assumption than the previous one?

Uno: IF you only assume player's initial choice is correct with probability 1/3 THEN you know that switching gives him the car with probability 2/3

Due: IF you actually assume that *all* doors are initially equally likely to hide the car THEN not only is the preceding true, but we also are guaranteed that all conditional probabilities are at least 1/2 (and on average 2/3 because of "uno"). Not only is "always switching" a good idea, but also there is no strategy (sometimes switching, sometimes not, depending on the door numbers in question) which is better.

Tre: IF you not only assume that all doors are initially equally likely to hide the car but ALSO assume that either choice of Monty is equally likely when he has a choice, THEN the door numbers are irrelevant by symmetry. The car is behind the other door with probability 2/3. Both unconditionally, and conditionally given the specific doors. You know in advance that you can ignore the door numbers. You can also ignore the day of the week - it doesn't matter if it's Tuesday or Wednesday, either.

That's basically all there is to say. All that remains is to add the references to the sources, say that they tend to quarrel about the right assumptions to make, and to organise the whole thing into a nice story for wikipedia readers. Gill110951 (talk) 12:19, 23 September 2010 (UTC)

Exactly. Uno: True (because there is only one car, but the player had no knowledge of where it is. That is known only by the host and his staff, and nobody else knows.)
Due: True (probability from at least 1/2 (in 2/3) to max. 1 (in 1/3), so on average 2/3, as long as nothing else is evident.)
Tre: True (for as long as you have no knowledge whatsoever about the host's behaviour, you impossibly can declare this behaviour to be "evident". So in opening of one door, no matter which "number", you are absolutely out of position to "assume better information".) – But you can speculate about it (if the host should be extremely biased either in 2/3 the odds of both doors are min. 1/2 each, or in 1/3 the chance for the second closed door will be max "1". But, for this one single game, in the situation given, you have no knowledge whatsoever on that, and so all you can say is: Pws will double your chance to 2/3. Anything more would be an arrogant presumption. Gerhardvalentin (talk) 14:42, 23 September 2010 (UTC)

I have not been a party to this mayhem, but as I work with stats I wanted to interject a quick opinion, I swear it will be my last.

The unconditional solution can be better expressed with cards. Say you have three scrambled cards face down in front of you: a jack, a queen and a king. The king is the winning card. Now someone takes a card and turns it face up and it is the jack. You are now to choose between the remaining two cards, one of them that wins, and one of them that loses. Your chances of winning are 1/2.

I understand there are conditions that make your chances of winning 2/3 but they need to be explained. Otherwise, the chance of getting the car is 1/2.108.65.0.169 (talk) 09:28, 17 October 2010 (UTC)

In both the conditional and unconditional formulations of the MHP. The player first chooses a door then the host opens one of the other doors that he knows hides a goat. In your card version, the player needs to identify a card that they have chosen and the host must then turn over one of the other cards that they know not to be the king. If the player swaps to the remaining card, they have a 2/3 chance 0f getting the king. Martin Hogbin (talk) 09:36, 17 October 2010 (UTC)

Need for page protection

The article was recently placed under temporary protection due to edit waring. That protection has since expired. Further edit waring will result in the article being protected indefinitely. Hopefully the mediation will produce a climate of collaborative editing. In the mean time, would editors be willing to refrain from editing the article? Thank you. Sunray (talk) 21:17, 6 October 2010 (UTC)

Would you categorize the deletions I made last week that were reverted and led to the page protection as 'major edits'? Glkanter (talk) 21:40, 6 October 2010 (UTC)

Or replacing the paraphrased solution with the exact quote? Glkanter (talk) 17:33, 7 October 2010 (UTC)

I think it best not to edit the article at all, so I've changed the wording to reflect that. Sunray (talk) 21:12, 7 October 2010 (UTC)

I regret to inform you that I have a 'policy' of not agreeing to self-censorship on Wikipedia. I suggest you have the page protection restored. I'll wait a reasonable amount of time. Glkanter (talk) 21:18, 7 October 2010 (UTC)

Monty Hall code

To help show the logic I submit a Perl program that simulates the Monty Hall problem. Sorry if there is a better way to submit this.

Code

#!/usr/bin/perl

# Monty Hall Problem
# Jason Holm - jason@jasonholm.com
# 29 October 2010

$number_of_plays = 1000; # How many times the game is played.
$verbose = 1;            # Display plays

print "Playing $number_of_plays time(s)\n\n";

for ($plays=0;$plays<$number_of_plays;$plays++) {
 $winning_door = int((rand() * 3) + 1);
 $player_pick = int((rand() * 3) + 1);
 $player_switch = int((rand() * 2) + 1);
if ($verbose == 1)
{
 print "Player picks door number $player_pick\n";
}
 do
 {
 $goat_door = int((rand() * 3) + 1);
 } while ($winning_door == $goat_door || $player_pick == $goat_door);
if ($verbose == 1)
{
 print "Player is shown door number $goat_door as having a goat...\n";
 print "User is asked if they want to pick the other door...";
}
 if ($player_switch == 1)
 {
if ($verbose == 1)
{
 print "player wants to switch doors!!!\n";
}
 do
 {
  $switch_to_door = int((rand() * 3) + 1);
 } while ($player_pick == $switch_to_door || $goat_door == $switch_to_door);
 $player_pick = $switch_to_door;
 }
 else
 {
if ($verbose == 1)
{
 print "player does not want to switch doors...\n";
}
 }
if ($verbose == 1)
{
 print "The winning door is number $winning_door!!!\n";
}
 if ($player_pick == $winning_door)
 {
 if ($player_switch == 1)
 {
if ($verbose == 1)
{
  print "Players wins after switching doors!!!\n";
}
  $SAW++;
 }
 if ($player_switch == 2)
 {
if ($verbose == 1)
{
  print "Players wins after keeping their choice the same...\n";
}
  $KAW++;
 }
 }
 else
 {
  if ($player_switch == 1)
 {
if ($verbose == 1)
{
  print "Players loses after switching doors!!!\n";
}
  $SAL++;
 }
 if ($player_switch == 2)
 {
if ($verbose == 1)
{
  print "Players loses after keeping their choice the same...\n";
}
  $KAL++;
 }
 }
if ($verbose == 1)
{
print "---------------------------------------------------------------------------------\n";
}
}
print "* * * RESULTS * * *\n";
print "\n";
print "Switch and Win: $SAW\n";
print "Kept and Wins: $KAW\n";
print "Switch and lose: $SAL\n";
print "Kept and loses: $KAL\n";

— Preceding unsigned comment added by 131.151.49.156 (talk • contribs) 18:15, 29 October 2010

There are implementations in a variety of languages at wikibooks:Algorithm_Implementation/Simulation/Monty_Hall_problem. -- Rick Block (talk) 01:43, 30 October 2010 (UTC)

Further simplification of problem to bare skeleton

This problem can be simplified as follows. Instead of arranging doors, cars and goats, host just randomly chooses a number within the set (1, 2, 3). Participant simply needs to guess that number in two chances. Participant tells his guess to host. Host then reveals a number within the set which is neither he himself had chosen nor participant had guessed and adds that number is not the correct answer. At this point he asks participant if he wants to switch this choice. At this point if participant decides to switch then has 66% chance of being correct otherwise 33%.

( I tried to add above section, but got deleted. Can somebody explain, how is this simplification wrong ?) —Preceding unsigned comment added by 120.61.32.199 (talk) 11:43, 14 November 2010 (UTC)

I reverted it for two reasons. It had very poor grammar with many mistakes, so was difficult to read and unclear. And the probabilities are 2⁄3 and 1⁄3, not 66% and 33% (which don't even add up to 100% so are clearly wrong). Further it is redundant as the information is already in the article.

But see also the note at the top of this talk page: the article; it is under mediation, so editors should not be making significant changes to the article at the moment. That means even if your change was well written and error free it is wrong to add it now.--JohnBlackburne^words_deeds 12:07, 14 November 2010 (UTC)

120.61.32.199, your contribution was also not supported by a reliable source. Martin Hogbin (talk) 12:16, 14 November 2010 (UTC)

rudeness to von Savant

In the explanation of "Why the probability is not 1/2", it mentions that if the host doesn't know which one has a goat, von Savant "correctly replied ...". It's rude to question von Savant's judgement here, especially when he was right. This mathematician revealed a paradox which couldn't be explained by many great minds and now some keyboard warrior is passing approval for him. Why don't we give him a star as well? What a good boy von Savant is.. Owen214 (talk) 11:25, 27 October 2010 (UTC)

This comment is making no sense at all (vos Savant is neither a mathematician nor male and nobody is really questioning her judgement (at best her explanation from a very particular point of view). Also WP is not in the business of handing out stars (to whoever) ...--Kmhkmh (talk) 15:05, 27 October 2010 (UTC)

I'm reading this as a suggestion to delete "correctly" as it sounds like we're passing judgment on vos Savant's reply. -- Rick Block (talk) 16:50, 27 October 2010 (UTC)

If that's the case he was switching subject and object around constantly. leaving us to pure guessing regarding his intended message. Btw. what's going with the mediation?--Kmhkmh (talk) 16:58, 27 October 2010 (UTC)

Yes, I got mixed up and thought it was a man called von Savant. Either way, it's still rude. If people are sending her mathematical problems to answer, one would think that she could be called a mathematician. Kmh, you seem not to understand how subtelties can make a sentence rude. Mentioning that "vos Savant correctly replied.." implies there was reason to think that her response may not have been correct. It's also rude to talk about me as "he" when I'm actually part of this conversation. Owen214 (talk) 08:14, 5 November 2010 (UTC)

Considering your inability to write a clear sentence, your contention that subtleties can make a sentence rude is probably not going to be listened to. Especially since the sentence was not rude. It stated the simple fact that vos Savant's reply was correct, a point that is necessary for the discussion to make sense. Finally you were being spoken of, not spoken to so the use of a pronoun is not rude but only an example of saving time and effort. If there had been a fourth poster involved, using he to refer to you could have been ambiguous but it still would not be rude. --Khajidha (talk) 16:14, 24 November 2010 (UTC)

many people say switching is better, but are still wrong

many people will say switching is better because your odds are 1/2. This is wrong of course, because your odds are 2/3. I think this particular confusion needs to be made explicit in the article —Preceding unsigned comment added by 76.126.238.69 (talk) 23:41, 9 February 2011 (UTC)

How about the information angle ?

How about first establishing what actual information is present after the host opens the door instead of playing with probebilities that become irrelevant once the door gets opened? Call it a methodological problem if you like. Probabilities of guesses can be mapped to presence and absence of information. The difference is that information based reasoning points clearly what's relevant and what's irellevant. Relevant is that player had new info which gives him 1/2 chance if he tosses a coin and chooses again at random. His first choice is irellevant since he didn't get any info about that choice, he only got infor that new choice with better ods is possible. If we believe that a random choice under uncertainty gives the best odds, then he will improve his chances only if he makes a new random choice, not a forced one.

Information-wise the 2 cases from initial state are merged when new information is provided - they don't exist anymore in the new state, they are indistinguishable. Same applies to the case with 1 mil doors. If player makes forced choice he remains in the prior state with much lower odds - he's not using new information. If he wants to use new information then he has to toss the coin again in order to realize the new state.

Another way to look at it is to realize that in an assembly only a random choice can select new configuration and all forced, non-random transitions are equivalent and confined to the same configuration.

His actual odds still improve from 1/3 to 1/2. ZeeXy (talk) 13:21, 3 November 2010 (UTC)

(Attempt to) Pick wrong and increase the chances of switching right, or (Attempt to) pick right and not know until it's too late. Metaphysically speaking, I'd rather wing it. 70.15.11.44 (talk) 05:28, 4 November 2010 (UTC)

When wondering where the car is, you shouldn't just use the hard information which you have in front of you, but also the likelihood that that information came to you under the different scenarios which concern you. You chose Door 1. The host is twice as likely to open Door 3 if the car is behind Door 2 than if the car is behind Door. 1. When the game is repeated many times, the car will be behind Door 2 twice as often as it is behind Door 1, within those occasions that you chose Door 1 and the host opened Door 3.

Forget about probability, forget about information. This is about very simple arithmetic. Richard Gill (talk) 07:20, 5 November 2010 (UTC)

Wrong. If a host opens #3 you don't know if that's because you missed or guessed #1 and no imagination can help you. If you automatically switch to #2 that's equivalent to picking #2 in the first place -- he'd still open #3. Now what? :-) He could even let you keep switching till you turn blue. That's why he can be "generous" - coz you are at 50% ignorance and there's nothing that can help you. You could have started with 100 doors and he could have let you switch every time he closed a door and you'd still be in exactly the same state. That's why I said that information-wise prior states are merged as a result of the new info - they are indistinguishable, there is no observable difference. Your first 98 choices are irellevant for the new state -- you are still in a state of 50% ignorance. The only thing you can do to improve your chances in any 50% ignorance state is to toss a fair coin.

Wrong, wrong. We are told as part of the problem statement that the quizmaster knows where the car is hidden, that he will always open a door revealing a goat, and that he will always offer us the opportunity to switch to the other still closed door. Richard Gill (talk) 13:22, 6 November 2010 (UTC)

Here's example with a real statistical ansamble

Say you were choosing among 3 presidential candidates and one got killed. Does that automatically make the one you haven't picked a 2/3 winner? :-) Then it has to apply to all other people who picked one of these 2. All these people are now real statistical ansamble - genuine massive sample of random choices with equal probabilities and you can clearly see that your particular initial choice is irellevant. Assume totally split election, every candidate having 1/3 before one got killed and assume no one really cares - everyone just wants to vote for the winner since then they get a coin if they voted for a winner. Suppose they all follow your automatic switch tactics and the 3rd voter set gets split equally. No one wins. What's the best chance of voting for the winner you have? 50% What the sole way to insure that someone does win and thereby 50% of you achieve the goal -- that no one does anything authometically but everyone tosses a coin. Why? Because fair coin toss is never 100% fair with finite number of tosses. Only the act of everyone tossing insures the winner. This is probably the best illustration how irellevant and blocking automatic switching is and how you do need to toss in order to actually realize the chance presented by new information.

This is all very different from a situation in which there would be some underlying cause and your sampling is really just measuring it since then you would expect sampling to converge with a very high probability. Without underlying cause it takes infinite number of samples to realize your statistics and that's strictly historical and completely irellevant for a particular singular trial. That's the part people easily forget when they start deluding thelselves with abstract statistics -- it's irellevant for a particualr sample unless there's underlying cause which will ensure rapid convergence. That's the sole thing that makes sampling worth anything -- if there is underlying cause there will be fast convergence. Have to remind you that theoretical limit for a big number rule to apply is infinite number of samples.

Go tossing coins and see how long runs of equal values you are going to get and how huge deviations from imagined 1/2 you are going to get. ZeeXy (talk) 12:44, 6 November 2010 (UTC)

Maybe it would be best if the participants in this discussion only made reference to reliably published sources, as a method of discussing changes to the articles. I'm sure there are various more appropriate forums elsewhere on the internet that welcome spirited debate on the mathematics and logics of the MHP. Glkanter (talk) 12:52, 6 November 2010 (UTC)

This another typical response to the MHP article. Everyone wants to know why/how/if the answer is 2/3 rather than 1/2. Martin Hogbin (talk) 10:36, 6 November 2010 (UTC)

@ZeeXy - if you'd like to discuss the mathematics behind the problem, I'd suggest we move this thread to the /Arguments subpage. If you're suggesting a change to the article, please say what change and on what source (or sources) you'd base that change. -- Rick Block (talk) 15:16, 6 November 2010 (UTC)

Rick, there shouldn't *be* an 'arguments' page. Arguing the math/logic of the MHP is no more appropriate for a Wikipedia article talk page than discussing the greatness of your favorite musical performer with like-minded fans. The 'arguments' page should be deleted, rather than encouraged. Talk pages are for discussing editing Wikipedia articles. Glkanter (talk) 15:31, 6 November 2010 (UTC)

The point of the /Arguments page is to have a place for these sorts of discussions, which are not directly related to editing the article, to be held. It is more or less like the Wikipedia:Reference desk - but with a specific focus on the MHP. There's one for various other articles on controversial topics, like 0.999.... You are absolutely correct that THIS page is for discussing editing the article. If you strongly feel the need to see a community consensus about whether the Arguments page should be deleted, please open a discussion at Wikipedia:Miscellany for deletion. -- Rick Block (talk) 17:23, 6 November 2010 (UTC)

Thanks for the suggestion. I may do that. If its a Reference desk item, well, then it belongs at the Reference desk. Or, I could just take the page off my Watchlist... Glkanter (talk) 21:55, 6 November 2010 (UTC)

Recent overhaul and state of the mediation

First of all thanks to all who put effort in the recent overhaul, which from my perspective works well overall.

I minor nitpicking I'd have though is the (incomplete) quotation of Behrends in the sources of confusion section. It should be mentioned while Behrends considers both answers as correct he does consider them as 2 slightly different problems or questions at least.

Another I'd like to know is whether the conflicting parties in the mediation are happy with the current version (or at least can live with it) or whether we still have (major) disagreements and an potenial editing conflict down the line. --Kmhkmh (talk) 15:46, 14 November 2010 (UTC)

I think I am reasonably happy with the article as it is now. I did not realise that the article was being actively edited during the mediation so I am assuming any edits made during mediation to be non-contentious ones.

As you will see on the mediation page, I have suggested that we start discussion based on the article as it is now, rather than rewriting large chunks of it from scratch. I guess you support this proposal. Martin Hogbin (talk) 10:41, 9 December 2010 (UTC)

Just out of curiosity, what was the mediation over? The MHP is a stats problem, which does not strike me as something prone to violent arguments. --Ludwigs2 17:39, 8 February 2011 (UTC)

Best look through the last couple of years' talk pages. I do not think anyone wants a re-run. Martin Hogbin (talk) 22:02, 8 February 2011 (UTC)

Without looking through the archives, why is there a long list of references, but no in-line citations? Cla68 (talk) 00:07, 10 February 2011 (UTC)

The article uses Harvard style referencing. There are plenty of inline references. -- Rick Block (talk) 00:37, 10 February 2011 (UTC)

I'm pretty much happy with the present article. But I did some more OR in the direction of creating a synthesis between simple and conditional solutions, see [3]. Richard Gill (talk) 22:50, 10 February 2011 (UTC)

False dates, missing references

The new recently added alternatives solutions seem to cite publications from The American Statistician with false dates (1980, 1981 - probably meant to be 1990,1991?). Also both papers are not listed under references either.--Kmhkmh (talk) 04:33, 15 February 2011 (UTC)

Yep, exactly what I meant by saying that they are "unsourced" above. glopk (talk) 07:27, 15 February 2011 (UTC)

Sorry, those were typos. I will fix the dates and add the references, if we don't have them yet in the big list. They are both part of the reactions to Morgan et al. and well worth reading. Richard Gill (talk) 08:01, 15 February 2011 (UTC)

@Kmhkmh and @glopk, are you telling me you don't actually know the literature on MHP? And these are two very readable papers at the very centre of the Morgan et al. controversy. Fie on you! Do your homework! I have a collection of pdf's of all this literature which I can email to you privately if you like. Alternatively, I could set up a dropbox.com shared folder for the use of us wikipedia MHP editors - anyone interested?. Richard Gill (talk) 08:29, 15 February 2011 (UTC)

I'm telling you that the correct/complete references were missing. Whether I've read them or not, whether they are particularly readable or not or whether you have personal pdf copies is rather irrelevant. What's relevant here, is that if you use them please reference them properly. And yes in case I haven't read all the possible interesting publication on MHP nor do I intend (= I got better things to do). However if you don't mind I'd indeed appreciate a copy of the pdf files (I#ll send you an email).--Kmhkmh (talk) 15:51, 15 February 2011 (UTC)

Great! Sorry for annoying you, I should at least have added ;-) to my reaction above. Richard Gill (talk) 16:13, 15 February 2011 (UTC)

Too many assumptions

The implicit assumptions list in the lead was too long--- the assumption that the car is placed randomly is unnecessary, if your first guess is uniformly random, it doesn't matter how the car is placed. There is no need to assume that the host will chose the goat door to open at random either. The only assumption that is needed is that the host will open one goat-door no matter what your initial guess. That's it.69.86.66.128 (talk) 07:00, 16 February 2011 (UTC)

Splendid, Mr (or Mrs) 69.86.66.128 ! I have been pointing out this solution for several years and it is mentioned several times in the literature but no one here is interested in this point of view. Finally I wrote a couple of reliable sources giving this solution and linking it to game theory. Richard Gill (talk) 07:09, 16 February 2011 (UTC)

Again there is a connection with the interpretation of probability. If you are a frequentist you don't know enough to solve the problem so you take action yourself - you randomize and switch and win the car with probability 2/3. The 2/3 is a property of the apparatus you use to choose your door (dice, tossing coins...). If you are a subjectivist you know nothing and hence your subjective probability that switching will give you the car is 2/3, independent of which door numbers are seen chosen and opened. The 2/3 is a property of your (non)knowledge. Richard Gill (talk) 07:17, 16 February 2011 (UTC)

Proposal to add some alternative conditional solutions

I would like to see some more mathematical solutions to the conditional problem, just as there are various informal solutions to the unconditional problem. I think they all give additional insight into MHP. There exist at least three solutions which follow a simple chain of logical reasoning, and which can be converted step by step into equivalent mathematical formalism (this is a useful exercise for the beginning student of the formal probability calculus who has to learn how to connect the formalism with ordinary logical reasoning and insight into the structure of the problem), but which avoid calculations or formula manipulation. These are: simple plus symmetry, Bayes' rule, and symmetry plus simple.

Simple plus symmetry: by symmetry the probability that the car is behind door 1 cannot depend on whether the host opened door 2 or door 3. The unconditional probability was 1/3. Therefore the two conditional probabilities are equal to 1/3 too. Reference: Bell (1982).

Bayes rule: the odds that the car is behind door 1 (the door chosen by the player) are initially 2 to 1 against. Whether or not the car is behind door 1, the chance that the host opens door 3 is the same, 50%. (In the one case because if the car is behind door 1, the host is equally likely to open either other door, in the other case, because if the car is not behind door 1, it is equally likely behind either other door, and the host's choice is forced.)

Symmetry plus simple. Pretend for a moment that the player's choice is also completely random. After the host's action, we can refer to the doors as: door chosen by player X, door opened by host H, door remaining closed (to which the player may switch) Y. From the simple solution we know that the door hiding the car, C, is either door X or door Y, with probabilities 1/3 and 2/3 respectively. By symmetry, the triple of door numbers (X,H,Y) is a completely random permuation of the numbers (1,2,3), and C either equals X or Y, with probabilities 1/3 and 2/3, independently of which of the six permutations is the permutation (X,H,Y). This tells us that the specific door numbers are irrelevant to the player who wants to maximize his chance of getting the car. The actual numbers are completely independent of the relationship of C to X,H,Y.

Nothing is changed by fixing the value of X, X=1 say. Now there are just two permutations possible, (1,2,3) and (1,3,2). By symmetry they are equally likely, and this is independent of whether or not C=X.

Each of these alternative proofs has pedagogical value for students of probability and statistics since they use extremely valuable tools. I think they each give further insight into "why you should switch". Each of the proofs is intuitive, you don't need a formal mathematical training to appreciate the ideas used in them. All the proofs explain why the ordinary lay person is perplexed by the argument that the simple solutions are "wrong" and that you have to learn Bayes's theorem and formal probability calculus to solve MHP properly - because each of the proofs make clear in a different way why the ordinary lay person is completely right not to be too bothered about the specific door numbers. Each of the proofs ties in with Vos Savant's wording "say, Door 1", and "say, Door 3", since we see that the specific door numbers are indeed irrelevant to the chance that switching will win and to the decision process of the player.

Reliable sources: the various uses of symmetry go back to discussants of the Morgan et al. (1981) paper, especially Bell (1982). The use of Bayes' rule is promoted by Jeff Rosenthal (2006?) (who by the way is a prominent mathematician and probabilist, as well as a prominent popularizer of probability and statistics whose "popular" writings are appreciated both by lay persons and by experts). See also a prepublication by me, [4], which is based entirely on what I learnt from fellow editors on the MHP page. Richard Gill (talk) 09:16, 11 February 2011 (UTC)

I think it is generally useful to understanding have several ways of looking at a problem covering all levels of understanding and interest. The only thing I would want to keep in mind is that the MHP is essentially a simple problem that most people get wrong so we should start with simple and convincing solutions> After that, something for the experts. Martin Hogbin (talk) 11:06, 11 February 2011 (UTC)

Those are all great ideas, Richard. Too bad the Conditional solution section is so grossly polluted with variants, hypotheticals, OR, NPOV and UNDUE WEIGHT violations, and other assorted crap intended to diminish the simple solutions, but which serves only to confuse the reader.

By the way, is that great authority, Jeff Rosenthal, the same one who says of a simple solution?:

"This solution is actually correct, but I consider it "shaky" because it fails for slight variants of the problem. For example, consider the following:"

"Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door?"

Maybe you could clarify for me what the English term, 'actually correct' means and what bearing 'it fails for slight variants of the problem' has as to the 'actually correct'-ness of that simple solution as a technique for solving the Wikipedia article's subject, which is the symmetrical MHP? Glkanter (talk) 12:00, 11 February 2011 (UTC)

@Martin: yes indeed. My proposal would be to place these solutions in place of the present formal mathematical proof via Bayes' theorem. That proof can be replaced by a reference to the article Bayes theorem where it already figures as an example.

@Glkanter. Yes, Jef's words are incomprehensible. What is wrong with something which is right? What is the relevance that "it" fails for a different problem? I think Jef was using the word "solution" in two different sences without realizing it. The "answer" (2/3) is correct but the "argument" is not. Because the same argument gives the same answer, 2/3, for a different problem Monty Crawl, where 2/3 is the wrong answer. And the argument is not even applicable to Monty Fall. But why don't you ask him yourself? So I guess this was a momentary lapse. But I don't have e.s.p., so I can't tell what was going on in his mind. I can only guess. Richard Gill (talk) 19:05, 11 February 2011 (UTC)

*I* don't have any problem comprehending his words. Nor do I find any 'lapse'. The only reason I bring it up, again, is because you have repeatedly agreed with Rick Block and others that those words constitute a 'criticism' of all simple solutions. That's an unsupportable conclusion by you people. His only point is to show why his 'discovery', or whatever, has utility in some other applications, which the simple solutions, of course, would fail in. This (these?) other method(s) is the whole point of his paper. It is in no way a 'criticism' of simple solutions as a solution to the symmetrical MHP at all. Hopefully nonsense like this will be exposed in arbitration, don't you think? And wouldn't it be nice if the Conditional solution section, which also talks about variants for 90% of the time was cleaned up to make room for your suggestion? Glkanter (talk) 19:18, 11 February 2011 (UTC)

If you are expecting arbitration to settle the longstanding content disputes on this article, you will probably be disappointed. Content disputes are outside ArbCom's remit, and while sometimes they try to help resolve a content dispute by giving disruptive editors a "time out," as a rule they won't address the content dispute directly. Woonpton (talk) 05:40, 12 February 2011 (UTC)

So the whole arbitration is about Rick Block's complaint about my conduct? Glkanter (talk) 06:58, 12 February 2011 (UTC)

@Glkanter: I think @Woonpton is right, and also @David Tombe knows about these things through personal experience and probably has good advice on how to survive. @Rick Block has complained about your behaviour, you can see his complaint somewhere. Richard Gill (talk) 12:29, 12 February 2011 (UTC)

@ Richard. I note that implicitly you now take the "conditional formulation" (decide after the door has been opened) as the MHP to be presented. Your "simple plus symmetry" solution, better is called "conditional using symmetry" solution, as the characteristic thing of a simple solution just is not considering any conditional probabilities. This solution and the one using Bayes' have always been considered correct solutions. Let us not discuss your third option, as IMO it is not suited for Wikikpedia. As for the term "simple", let's keep it for the simple solution, the one not being a correct solution to the conditional formulation of the MHP. The article should mention it with the criticism. Nijdam (talk) 11:15, 12 February 2011 (UTC)

Again incorrect? Not paying regard to "which door" has been opened? Please avoid that misleading, fallacious and nebulizing strategem. Based on the question asked, the identity of the door opened cannot give any "secret additional info" on the actual location of the car, unless you "assume" something totally unknown. So "before" and "after" are identical. Anyone is free to "assume" what he likes, but in that case he preferably should clearly disclose his assumption. Gerhardvalentin (talk) 12:50, 12 February 2011 (UTC)

And, just to acknowledge but also to solve that world famous conflict on Wikipedia: Have a look there:

"From the mathematical point of view, one can completely solve MHP by first using symmetry to show that the relationship between the doors when identified only by their role - door chosen by player, door hiding car, door opened by host - is statistically independent of the numbering of the doors (at least, this is true when the player's choice is also random). This implies that the player's decision whether to stay or switch might just as well be made in advance,
ignoring the door numbers in question. They give him no further information."

". . . They give him NO further information." But every now and then here unchangibly is repeated, and repeated, and repeated, and repeated, and repeated: "Most people will be confused by the unconditional formulation of the simple solution",
or: "the final answer about the required probability depends on the way the host acts",
or: "the simple solution is not a solution to the MHP, and that's the crux of the problem",
or:"There is a difference between the probability before the host opens a door and the probability thereafter",
and: "Anyone with basic understanding of probability theory should know the difference",
or: "It's some reasoning, which fails in the simple solution.",
or: "By watching the shows, and keeping track of how many times ... - an outsider sees the result of the host's bias. It is this history..."
or: "these tallies may show different odds of winning by switching (depending on which door the host opens).",
or: "the individual chance (by door) may be different (depending on the host bias).",
or: "the simple solution, the one not being a correct solution . . . The article should mention it with the criticism",
and so on, and so on, and so on, and so on, and so on, and so on, and so on, for more that two years now. – Most gratifying for all observers. :::::Gerhardvalentin (talk) 21:22, 12 February 2011 (UTC)

@Gerhard, I am trying to solve Vos Savant's question and I am using Laplace's definition of probability (subjectivist-Bayesian) based on the primitive concept of "equally likely". I do this because this is how the man in the street approaches probability and because MHP is a popular brain-teaser. "Equally likely" means equally likely from your (subjective) personal point of view. All you know is what Vos Savant's tells you. You have not watched the show a hundred times in the past! "Equally likely" is not something which is verified by statistics from past observations. It refers solely to your personal expectations of one single play of the game. The definition of "the probability of X is 2/3" is no more and no less than "there are three equally likely outcomes and for two of them X is true, for the other one it isn't". No more and no less. The car is behind the initially chosen door with probability 1/3 given your initial choice and the choice of the door opened by the host means no more and no less than the situation that your initial choice was door 1, and the door opened by the host was door 3, can be decomposed into three equally likely situations, in two of which the car is behind door 1 and in one of which it isn't. Because we know nothing, to begin with the car is equally likely behind each door. For us it is equally likely, when we have chosen door 1 and the car is behind door 1 too, that the host opens door 2 or door 3. It is nothing about historically observed repetitions of knowledge about how the host's brain works. It is about our lack of any information to believe any more strongly that door 2 or 3 would be opened.

With this approach, we might as well imagine the player choosing his initial door at random, and afterwards consider the special case that it was Door 1. Since from our initial information it makes no difference at all how we initially choose our door. Now we notice that because of our total lack of any information, all six possible values of (door chosen, door opened, door left closed) are equally likely: (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2), or (3,2,1). Whether or not door chosen=door hiding car cannot, by symmetry, depend on what is the numbering of our triple f (door chosen, door opened, door left closed) . Obviously the probability is 1/3 that the initially chosen door hits the car. Therefore the probability is also 1/3 that the initially chosen door hits the car, given that (door chosen, door opened, door left closed)=(1,3,2). The total symmetry tells us that the decision to switch or not can be taken independently of the actual numbers of the doors chosen, opened, left closed in the case at hand.

I think that the concept of independence is more fundamental than the concept of conditional probability. Conditional probability is a derived concept. So a solution based on independence and symmetry is better for a broad public, than a proof based on conditional probability. Richard Gill (talk) 15:48, 13 February 2011 (UTC)

Are you aware of this arbitration? Glkanter (talk) 21:30, 12 February 2011 (UTC)

@Glkanter. I'm sorry, I disagree. Jeff is a mathematician. He is interested in correct arguments just as much as correct answers. I find it annoying that he doesn't explain *why*, from his point of view, you *have* to compute a conditional probability, but I do believe that that is his point of view. I also know good reasons for having that point of you. Of course, mathematics can not ever tell you what you have to do. It has no legal or moral authority. But it can tell you what it would be wise to do. Richard Gill (talk) 12:29, 12 February 2011 (UTC)

Yes, but you ignore his English language answer in order to disagree. I am not required to, nor should I, make the same error. Glkanter (talk) 21:30, 12 February 2011 (UTC)

Since his written English language taken at face value is self-contradictory, it could be wise to try to figure out what he might have meant. It could be wise to try to appreciate the context in which he is writing, to appreciate the likely understanding of his intended audience. My claim to have some understanding in this direction (I'm a member of the same academic and professional community as Jeff Rosenthal) lead you to accuse me of claims of extra sensory perception! Richard Gill (talk) 16:08, 13 February 2011 (UTC)

@Nijdam. I am tired about bickering about what is THE Monty Hall Problem. You and I disagree. I think there are many mathematizations and any decent mathematization of course allows many different decent solutions. You may call the approaches which I listed by any name you like. I think the names I gave will be understandable to everyone interested in MHP and active on these pages, not just to the people who share your point of view, which I find dogmatic and inflexible. PS my third solution is the most beautiful of all especially since it is a mathematization of the (correct) intuition of all intelligent laypersons that specific door numbers are irrelevant, the only thing that counts is the probabilistic relation between the roles - both manifest and (for the player) hidden - of the doors. Indeed, the relation between the roles is independent of the numbering. Vos Savant's "say, Door 1" and "say, Door 3" was spot on. These words can be placed in parentheses, they can even be deleted altogether - at least for a rational Bayesian/subjectivist, like Laplace and like all ordinary people. Laplace built his theory of probability on symmetry. He would like solution number 3. Richard Gill (talk) 12:34, 12 February 2011 (UTC)

Now I'm even more confused about what the bone of contention here is. It seems to me that the problem in the MH problem is not statistical (the statistics are actually quite clear on the matter), it's that people (even experts) make the error of treating a non-probablistic act (the opening of a door with a goat) as though it were in fact a probabilistic act. I mean, this would be obvious for two doors (you know there's a goat behind one door and a car behind the other, you open the door to reveal the goat, what's the probability that the car is behind the other door?); This is more like a word game than a problem in statistics. are you all just arguing over your personal favorite ways to talk about the stats? --Ludwigs2 22:39, 12 February 2011 (UTC)

The fights about solutions are "much ado about nothing". Fights about whether the solution which you might want to give to students in a mathematics class on introductory probability, should be seen as superior to the solutions which ordinary folk can understand. So it's also a fight about demarcation, about ownership, about arrogance of power. What's the use of a wikipedia page which gives a list of solutions which everyone can understand, tells them these solutions are wrong, and then presents solutions using concepts which ordinary people don't know about? And moreover, does this from a position of authority and dogmatism. Never explaining *why* they think that one shoud solve the problem in a particular way. Mathematics can never tell you what you *must* do. It can only tell you what it is *wise* to do. And the good mathematician should be able to explain why it is wise.

In a number of publications I took the trouble to explain *why* it could be wise to solve the problem in a certain way. I also pointed out that the simple solutions make less assumptions hence are of wider applicability. I also pointed out that how you want to solve the problem, and what assumptions you are able to make, could well depend on what you understand by "probability". Which is an ongoing unresolved debate lasting for at least three hundred years, and no sign that it is about to stop. My own opinion is that it is a matter of taste. Richard Gill (talk) 16:00, 13 February 2011 (UTC)

Richard, You sound a little annoyed, and I in your place also would be annoyed with myself. But don't blame the messenger. You d... well know one big issue here is about which solution suits which version. So, behave scientific, and do not mention just solutions without referring to the version of the problem. It's not in the interest of the article, nor in the interest of the readers, and in the end not in your interest if you leave this point unclear. Nijdam (talk) 09:12, 13 February 2011 (UTC)

That's only one of many issues. MHP is not owned by any particular person or community. It is certainly not owned by the community of teachers of Bayes' theorem in introductory discrete probability. I am in favour of diversity. Against dogmatism. Richard Gill (talk) 16:00, 13 February 2011 (UTC)

Richard, strange that you can be sharp as a knife and precisely to the point, but often also seem to have trouble to stick to the subject of discussion. And that is: speaking of solution, without reference to what is is supposed to solve, Got it? For the rest: which are the other other issues? List them in short as a help in the arbitration. Did I say MHP is owned by some particular person?? I also like diversity, as long as it is not confusing. Nijdam (talk) 22:07, 13 February 2011 (UTC)

There are countless reliably sourced solutions that are 'conditioned' on the 100% likelihood that the host will reveal a goat behind another door & offer the switch. And unless the contestant knows these facts before he selects a door, the odds aren't necessarily 2/3 & 1/3. So he can start solving the puzzle without waiting for the host to open a door.

2/3 of the time the contestant will select a goat.

Therefore he should switch.

...Is reliable sourced, mathematically correct, and meets the narrative of the puzzle, "Suppose you're on a game show...". Conditional door-based solutions also solve the same problem statement, but are not required. No matter what these guys tell you. We're in arbitration about this right now. Glkanter (talk) 23:56, 12 February 2011 (UTC)

You are right, Glkanter, because the symmetry of the problem, when we are using probability in the man-in-the-street subjectivist sense (Laplace, 1814), tells us that the specific door numbers of door chosen and door opened are irrelevant. The pedantic maths teacher will say that your solution is not complete if you don't mention this fact. He will say that your answer is correct but your reasoning is not complete. Possibly you were indeed smart enough to see that you don't need the door numbers, but if you don't write this down explicitly, he can't decide whether you are smart but fast, or careless and not aware that you are missing a possible issue. The correctness and completeness of the argument to get to the answer is as important as the answer itself, for the pedantic teacher, in the maths or logic classroom. Richard Gill (talk) 16:14, 13 February 2011 (UTC)

I don't have to do anything. The reliable sources are what matter. Your opinion about them, not so much. I don't see any ambiguity in what Rosenthal writes. Glkanter (talk) 17:03, 13 February 2011 (UTC)

Further, my original point was that the Conditional solution section is horrible, and that your proposed stuff would be a lot better than the horseshit in that section today. Glkanter (talk) 17:07, 13 February 2011 (UTC)

Based on your responses, Richard, it would seem Rosenthal could be handled in one of 3 ways:

Based on the English words he wrote - he is *not* a critic of simple conditional solutions - (Glkanter)
Declare his writings incoherent - [at best, then, his paper should be disregarded] - (Richard says the paper is contradictory)
Decide for him what he really meant to say - despite what he wrote, "This solution is actually correct, but I consider it "shaky" because it fails for slight variants of the problem.", he is a critic of the simple conditional solutions - (Richard, The Most Noblest Of The High Priests)

The first one seems right to me. You have argued for the 2nd option, (but not the conclusion in the brackets), in order to support #3. I don't see how. And you have also told me the third option is the best, because you have 'lived the life', and that you alone, can know what he 'really' means. Preposterous for Wikipedia purposes. Or for any discussion, anywhere else on the planet, really. Glkanter (talk) 08:48, 14 February 2011 (UTC)

I think you are jumping to conclusions, @Glkanter. If one sentence appears incoherent but the rest of the text appears very professional and correct, and if there is a simple rewrite of that one sentence which makes it coherent with the rest of the text, then I would go for the rewrite. But, if *you* are unable to understand the rest of the text then *you* are stuck between Scylla and Charybdis: either you trash the whole text, or you trust an expert. (Fortunately Bayes' rule is intuitive and easily "internalizable" so you should not find it hard to understand the rest of Rosenthal's paper). Richard Gill (talk) 11:21, 14 February 2011 (UTC)

Jeff Rosenthal wrote me the following in an email today and I quote it here with his permission Richard Gill (talk) 17:47, 14 February 2011 (UTC)

"I apologise that my article's one sentence about the "Shaky Solution" wasn't sufficiently clear. What I meant was that this solution does give the correct answer, but only because it so happens that in this particular case, conditioning on the fact that the host has opened a non-car door does not change the probability that the original guess was correct. This last is a very subtle point which requires justification (e.g. using Bayes' Rule). I believe that many people who quote the Shaky Solution do not realise the importance of this point nor how subtle it is. As illustration, I believe that many people who quote the Shaky Solution would believe it also applies to the Monty Fall Problem even though it does not (and gives the incorrect answer in that case). In summary, I would say that the Shaky Solution can be "made" to be correct, but only by providing a clear justification for why this conditioning does not change the probabilities; without such justification, the solution is incomplete and insufficient."

Thank you for pursuing that, and sharing it, Richard.

My opinion is unchanged. His paper is *not* a criticism of simple conditional solutions as interpreted and prostelicized by Rick Block and Nijdam.

What you have shared is simply his opinion, in a private correspondence, that the version of the simple solution and/or the problem statement, he, himself restated in his own words, was 'incomplete and insufficient'.

Which was his intention all along:

4 Monty Hall Revisited

The Proportionality Principle makes the various Monty Hall variants easy. However, first a clarification is required. The original Monty Hall problem implicitly makes an additional assumption: if the host has a choice of which door to open (i.e., if your original selection was correct), then he is equally likely to open either non-selected door. This assumption, callously ignored by the Shaky Solution, is in fact crucial to the conclusion (as the Monty Crawl problem illustrates). Monty Hall, Monty Fall, Monty Crawl Jeffrey S. Rosenthal (June, 2005; appeared in Math Horizons, September 2008, pages 5{7.)

The generally accepted (Selvin, K & W) MHP makes the 50/50 host bias premise very clear, along with revealing the goat behind another door and the certainty that the switch will he offered. Glkanter (talk) 18:24, 14 February 2011 (UTC)

For some puzzle solvers, "Suppose you're on a game show..." in the problem statement says all they need to know about symmetry. Other will only feel 'complete' when the solution says '...and therefore, due to symmetry..." The entire world will fall somewhere in-between these points, inclusively. I reject any assertion as to the 'only right' way to phrase a solution. Who says 'certain ' premises (Suppose..., uniformly at random car & goats, 50/50 host bias) must be repeated in the solution? As per whose doctrine?

Rosenthal's telling of the problem leaves the phrase 'Suppose you're on a game show...' out. It's not really the MHP as per Selvin, vos Savant, or K & W, is it? Nor was that really his intent, obviously: "...callously ignored by the Shaky Solution..." Otherwise, there's no raison d'être for his paper, is there? Glkanter (talk) 12:55, 15 February 2011 (UTC)

I recommend you read Rosenthal's very nice popular book "Struck by Lightening". It has a good chapter on MHP. (Also I recommend you read Jason Rosenhouse's popular book on MHP, especially the chapter on probability interpretations). Jeff Rosenthal, a professional / academic mathematician / probabilist, and a prominent popularizer of probability and statistics, knows that our ordinary probability intuition often lets us down. As Persi Diaconis said (interviewed in the NY Times article on Vos Savant's article), "our brains just aren't wired for probability". If Rosenthal is preaching anything, he's preaching Bayes' rule. Which I would like to preach, too. If you don't have the right insight to solve the problem in a flash all on your own, you had better learn a reliable route to get to the right answer.

Once you've found the right answer, you can start wondering if there was a short cuts to find it.

Bayes' rule (as opposed to the horrible formulas usually known as Bayes' theorem) is simple to state in words, insightful, easy to internalise.

How do mathematicians find beautiful theorems.? First by stumbling across them by accident after a lot of sweat and hard work, seeing a pattern and recognising that it isn't just chance, there is something wonderful going on there, thereby discovering the deep connections which make the result obvious, once you do the necessary "parallel thinking". Bayes' rule should be called Bayes' theorem. What is usually called Bayes' theorem is just the result of substituting, twice, the definition of conditional probability. The formula gets longer on each substitution... big deal! Bayes' rule follows by dividing two instances of "Bayes theorem" by one another (conditioning two different "target" events both on the same "given" event). A load of crap cancels out; what is left is beautiful and memorable and powerful and useful. Richard Gill (talk) 14:47, 19 February 2011 (UTC)

"In its usual form the problem statement does not specify this detail of the host's behavior"

So what?

The last paragraph of the MHP page summarizes the conditional/unconditional controversy. It seems to me that this whole section is written with a strongly frequentist slant. Randomness is seen in the host's behaviour, and probability is seen as measuring randomness. Yet according to a subjectivist view of probability, probability is a measure of our knowledge or lack thereof. Probability does not measure physical randomness, but it measures our personal uncertainty. For a subjectivist, the host's behaviour is irrelevant. The subjectivist who hears Vos Savant's words has those words, and those words only, to go on, plus some general/common knowledge about quiz shows etc. The problem description gives us no reason to assign any special meaning to any particular door numbers. "Say, Door 1", and "say, Door 3" could just as well have been any other pair of doors numbers. For a subjectivist, all three doors are equally likely to hide the car, not because of any particular randomization procedure used by the quiz-team, but just because of lack of information to the contrary. Similarly, if coincidentally the player initially would pick the door hiding the car, the host would be equally likely to open either door, not because Monty uses a fair coin toss to determine his choice, but because the player would bet at equal odds for or against either choice, because for him the two doors are exchangeable.

Thus for a subjectivist, the fact that the problem statement does not specify the host's behaviour is totally irrelevant.

I think this needs careful thinking about. Rosenhouse has a whole chapter on this topic in his book. Either the text in the article should be neutral as to probability interpretation, or there needs to be a small discussion about the issue. Laplace (1814) builds his whole theory of probability on top of the *subjective* ("primitive") notion of "equally likely" and shows how symmetry, whether of knowledge or of physical laws, determines subjective probabilities in a whole range of problems. Richard Gill (talk) 13:46, 15 February 2011 (UTC)

But as I have said many times before, even from a frequentist POV Morgan's original conclusion is not justified. From a strict frequentist POV the problem is simply insoluble as we are not told how the car is initially placed.

Host bias is only important if we choose to make it so by selecting it for special attention from all other factors that might affect the outcome. Martin Hogbin (talk) 11:24, 19 February 2011 (UTC)

Yes. And with a subjectivist's (ordinary man-in-the street's) interpretation of probability, and with only Vos Savant's info to go on, what we think about host bias doesn't matter a bit - even if we believe that he might well have a bias, we are still totally ignorant of its direction, whatever its size, because our opinions are neutral to relabelling of the doors. This should have been the big point of Morgan et al's computation, the one they got wrong and which you, Martin, corrected: with a symmetric prior on host bias, your personal expectation that switching gives the car remains 2/3. As must also be the case, by symmetry. Richard Gill (talk) 13:58, 19 February 2011 (UTC)

Conditioning on the day of the week and other acuteness

The famous PARADOX on the one hand, and quite other dissentient issues on the other hand, that do not relate to the famous paradox question, actually are "mixed together" in the article. In order not to befog, such strange sidelines and stratagems should not appear every now and then in the lemma, they imho should be presented and treated in a different section, because that issues really belong to a different section, e.g.:

Some just feel free to predetermine high-handed and licentious haphazard, shaky and unproven assumptions. Quite outside the famous PARADOX. They may be free to do so.

In the absence of permanence lists they decide for example (just for their own pleasure) that the host on Sundays never does pay respect to the actual placement of the car, and this way on Sundays therefore in 1/3 of cases just will show the car instead of a goat. And thus on Sundays he limits Pws to 1/2.

And on every Wednesday, they let him only open his nearby adjacent door, whenever possible. And they decide that they know exactly about that fact. On Wednesdays he only will exceptionally open the distant door if his nearby door actually is hiding the car, and then Pws clearly is "1". And, to get the appropriate cognition, are applying conditional probability terms just for practice. So they like to need the appropriate day of the week as a condition, and they draw conclusions of "coincidence and evident inference" depending on the day of the week. And they are conditioning on the appropriate day of the week and other acuteness to handle such coincidental.

Such forgery and adulteration, even if it has been said so, never is element of the "famous paradox" itself. Not to befog, all of that should not be interspersed in the article, but be shown in a separate section. Gerhardvalentin (talk) 18:51, 15 February 2011 (UTC)

Nice paradox: a frequentist cannot solve MHP because he doesn't know anything (unless he stops being passive, and randomizes himself). A subjectivist can solve it, for the very same reason, namely, that he knows nothing! Ignorance is bliss. It seems to me that Vos Savant's question has to be solved with subjectivist (aka Bayesian) probability because she does not give us any information. If anyone wants to quote this, it is the concluding sentence of my recent paper in Statistica Neerlandica. But I can't push "own research" on wikipedia. Hopefully this wisdom will permeate into the standard literature on MHP in a few decennia. ;-) Richard Gill (talk) 20:32, 15 February 2011 (UTC)

If the above paragraph is High Priest talk for "when you're on a game show, of course it's reasonable to assume uniformly at random distributions unless told otherwise', I've heard it (and said it) 1,000 times before.

Does that change if the contestant makes his selection by 'lucky number' rather than by using a random number generator? I didn't think so. Glkanter (talk) 20:43, 15 February 2011 (UTC)

Subjective probability in, subjective probability out. Objective probability in, objective probability out. But yes @Glkanter you may take this as High Priestly authorization of the usual assumptions for the purposes of solving a well known brain teaser in pubs, at parties, etc. The biased host is irrelevant when we solve the problem using ordinary man in the street probability. He doesn't know anything about it, either way, so for him, for this specific game, it's 50-50. And that's all we are talking about. For the rest, you are welcome to your opinion about the difference between choosing by lucky number or by random number generator. (But I thought you didn't care for the opinion of high priests). Richard Gill (talk) 21:08, 15 February 2011 (UTC)

Richard, I would not say so. No, the biased host is irrelevant not only for the use of the ordinary man in the street probability. Any pretense of bias is useless and completely in vain to "solve" the paradox, because you have no permanence list for that "one and only game" the question is about, that incidentally never was on stage in reality, in exactly this manner. It's not about the "door numbers #1 and #3", it's not about the "necessity of conditional probability theorems", it's just about worthless and cheeky assumptions that never can be given. It's just about cheeky, unproven assumptions. Because, in effect, you really do know nothing at all about all of that. Marilyn explicitly excluded such "additional hints". Such senseless assumptions therefore never affect the famous question in any way. You have no permanence list! Such assumptions are an alien issue, completely outside the famous paradox. Such assumptions are really a waste and never can "help to solve the paradox".

To say "If you knew that the host uses to give closer information on the actual location of the car in each and every game, then you would know better" is a brain teaser quite outside the famous paradoxon. And for that you do not need indispensable conditional probability theorems, all what you need is your assumptions. Full stop. That never are to be given. Only those "assumptions" are the "condition" you base on, then. Not "before and after", not door numbers, not "on Sunday or on Wednesday". Your only condition is your actual "shake off the cuff never to be given assumption". That's the only thing you *need*. But it will not help you for your decision, as an answer that is based on unproven assumptions will never be "a better answer" than the plain simple solution is giving: You should switch.

And "selling unproven assumptions" belongs to a separate section, not to intersperse them over the whole article. Conditional probability is fine, as long as we aren't using it to sell unproven assumptions, slyly without naming them as what they are. Gerhardvalentin (talk) 23:28, 15 February 2011 (UTC)

The article is already structured so that the first part (up to "Variants") discusses solutions using the usual (symmetric) assumptions. What change are you suggesting? -- Rick Block (talk) 01:01, 16 February 2011 (UTC)

Except for the 90% of the Conditional solution section that talks about host bias variants, and all the stuff in Sources of Confusion that calls simple and/or unconditional solutions 'the sources of confusion', Rick. Glkanter (talk) 01:11, 16 February 2011 (UTC)

@Rick, those famous lousy "variants" that only can be based on senseless "unproven assumptions" (because everyone is in a complete lack of knowledge thereon), and that not ever can be able to contribute to any serious "solution", and that all belong to a separate "variants-section", begin with the "Sunday's version" (host does not pay regard to what's behind his doors) in line 61:

This is different from a scenario where the host simply always chooses between the two other doors completely at random and hence there is a possibility (with a 1 in 3 chance) that he will reveal the car.

And again in line 67:

This example can also be used to illustrate the opposite situation in which the host does not know where the prize is and opens doors randomly.

And in line 84 it reads:

Some sources, however, state that although the simple solutions give a correct numerical answer, they are incomplete or solve the wrong problem. - and:

but without additional reasoning this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the host opens.

And more weasel words, without expressly naming the underlying irresponsible "assumptions" in line 86:

the conditional probability may differ from the overall probability and the latter is not determined without a complete specification of the problem

Btw it's not just "the conditional probability" that differs, it's just the underlying assumptions! - And in line 92:

This analysis depends on the constraint in the explicit problem statement that the host chooses uniformly at random which door to open after the player has initially selected the car (1/6 = 1/2 * 1/3). If the host's choice to open Door 3 was made with probability q instead of probability 1/2, then the conditional probability of winning by switching becomes (1/3)/(1/3 + q * 1/3)). The extreme cases q=0, q=1 give conditional probabilities of 1 and 1/2 respectively; q=1/2 gives 2/3. If q is unknown then the conditional probability is unknown too, but still it is always at least 1/2 and on average, over the possible conditions, equal to the unconditional probability 2/3.

Line 195:

. . . but if the host can choose non-randomly between such doors then the specific door that the host opens reveals additional information. The host can always open a door revealing a goat and (in the standard interpretation of the problem) the probability that the car is behind the initially-chosen door does not change, but it is not because of the former that the latter is true. Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially-chosen door are very persuasive, but lead to the correct answer only if the problem is completely symmetrical with respect to both the initial car placement and how the host chooses between two goats

Line 196:

depending on how the host chooses which door to open when the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3. In its usual form the problem statement does not specify this detail of the host's behavior, nor make clear whether a conditional or an unconditional answer is required, making the answer that switching wins the car with probability 2/3 equally vague. Many commonly presented solutions address the unconditional probability, ignoring which door was chosen by the player and which door opened by the host; Morgan et al. call these "false solutions"

And only then, one line later, follows the relevant section:

Variants – slightly modified problems

All of that should be presented in one separate section, as a dazzling display of fireworks in literature, not to be tangent to the famous question. All of that is equal senseless as words such as: "By watching the shows, and keeping track of how many times ... - an outsider sees the result of the host's bias. It is this history..."

We have no knowledge on the correctness of any of those "possible assumptions", and - as they never may be given - we don't need them for getting guaranteed "better results". It's just as similarly helpful as a sure-fire roulette system. – Because it's about ONE game, and everything else is completely unknown.

Once more: It's not on "conditioning", it's about irresponsible assumptions. It is very important to show all of that as "interesting meanders", in a later and separated section, those allegations, pretending to "know" what no one can ever nor will ever know. To show that clearly. The lemma should not obfuscate. It clearly should underline them as being unproven assumptions, and show what they really are. Never being able to contribute to any better decision than the only one "correct" solution that is honoring what really matters: Our complete unawareness of any further "conditions". Regards, Gerhardvalentin (talk) 03:06, 16 February 2011 (UTC)

Nicely done, Gerhardvalentin! Glkanter (talk) 03:21, 16 February 2011 (UTC)

Unfortunately this polemic piece regarding conditioning and assumption misses an important point entirely. It is not up to us to decide whether assumption are meaningful/useful/helpful or whatever, but it is up to reputable sources. What you or I personally think of various assumptions or solutions doesn't really matter much as far as WP (and this article) is concerned. As long as people are pushing for "their way" to treat the problem rather than the "sources" way, there is no way to resolve the edit conflicts of this articles. An evangelist approach doesn't work, if different faiths have to cooperate. All we do this way is recreating and extending the conflicts and bickering of the academic community in here, while our real job should be to neutrally report on their (endless) bickering rather than creating one of our own.--Kmhkmh (talk) 07:34, 16 February 2011 (UTC)

Thank you, Kmhkmh, for commenting on my really polemic piece. And yes, I fully agree with you. But that's the problem here: I ask you to just take into consideration that, as all of the important sources and aspects have to be shown, those aspects should be presented successional and in a coherent manner. Clearly arranged for the reader. To make the lemma meaningful and readable. And we should stop to present a motley conglomerate of secret and hidden shaky assumptions, clad in mathematical theorems, neither naming them as what they are, not showing what is meant and what is addressed. Just boldly annunciating the importance of "conditional probability" to solve the problem, without naming, but hiding the real underlaying absurd shaky "assumptions", naming them conditions.

The MHP is perceived with little, but *necessary* assumptions. First, that it is a question about just ONE game show. Not on a dozen, and not on millions. Secondly, that the host not shows one goat and offers to switch just because the contestant has luckily chosen correctly. That's the first two steps, just to provide any sense. But as you *never* will dispose of "permanence log lists", and the behavior of the host is completely unknown - as a third step - that, besides of showing a goat, he will not be giving away any additional hint whatsoever on the actual location of the car, behind the two still closed doors. Not by words, not by gestures and not by some bizarre hint concerning the door he opens, if in that game he should have disposed of two goats: He will "equally likely" have opened one door. These assumptions make sense, as you indeed have no knowledge whatsoever about anything else.

All of that has to be supposed to make the question meaningful, otherwise you could grasp it just as a bad joke. If he's supposed to give additional hints, or just offers to switch because the guest has first chosen the car.

But then, being without further info, some reliable sources liked to turn things back again, from a meaningful question about an obvious paradox to a bad-joke-question. Adding additional shaky "assumptions", contradicting and going far beyond and grossly exceeding your just defined primary knowledge about ONE hypothetical game show.

The circumstances should be presented in a clear way. saying what the sources mentioned really do address. Not mixing contradicting aspects motley in a discretionary way. No, we should show that in a very clear way, naming and frankly saying what the sources expound. We should present that in a succession of all that additional, contradicting "assumptions" they present.

Please help to finally stop presenting motley mixed "contradicting assumptions" without naming them as what they are. That's a matter of courtesy to the reader.

And once more: All important sources should be shown, in the "variants of contradicting assumptions"-section. And YES, mathematics is helpful and should be shown as necessary, preferably in odds-form. But, when dealing with additional contradicting assumptions you should distinguish mathematics and assumptions and underline that it is not about mathematical "truth" then, but about "additional, boldly contradicting assumptions". Kind regards, Gerhardvalentin (talk) 13:56, 16 February 2011 (UTC)

Well personally I don't care much for the particular ordering and what goes into which chapter as long as everything is covered. Your argument however has the same problem as before, it is really not up to you (or me) to decide what the "real" MHP and what merely a "variant" is. The same holds again for your notion (or mine) of reasonable/unreasonable or required/unnecessary assumptions, they are irrelevant, It only matters what the reputable sources consider as reasonable or not (yes the reputable sources don't really agree either, but that in doubt the article needs to reflect that). There is no objective/universally agreed method to settle which approaches, aspects and assumptions are "bad", "good" or "best", since the decision ultimately depends heavily on personal taste, interest, as well as personal schools of thought one adheres to (for instance frequentist version bayesian) and last but least ego. In short we can argue the various aspects until the end of time, which is each side believing (with some justification) to be right. It's almost like arguing faith/religion and that's why I said before the evangelist approach won't work here as long as involved parties adhere to different faiths.--Kmhkmh (talk) 03:45, 17 February 2011 (UTC)

Thank you again, Kmhkmh, and you are right again. Wikipedia is a forum where you can decide whether edits are good or bad, and it's not up to the editor to decide whether the sources are good or bad. But we should not hide what the sources are actually talking about, their underlying initial assumptions concerning the rules of the game. Once more: their differing underlying assumptions. And my strong "belief" is that the lemma, last but not least, is there for the reader, from student to Grandma, this should not be left out. The aim is to help the reader to grasp what the sources actually are talking about, their differing underlying initial "rules of the game".
That it is possible and conceivable to view and to evaluate the paradox from different sides, based on differing initial conditions, from differing rules. And to clearly distinguish those two groups of sources and to clearly emphasize this difference by distinctly separating them in the lemma.

Are they starting from (1: "the standard problem"), or do they allow for (2: "an additional host's hint") on the actual location of the car behind the two still closed doors that the host can and will be giving, by his considered possible special "behavior" in opening one of his two doors. Will (1) the conceivable probability to win by switching of 2/3 on average be the only reasonable and admissible answer for the actual game show, or can it be recommended and useful to include the (2) host's hint in your determination of the actual probability to win by switching. To clearly keeping apart that two kinds of "sources" will be of real benefit for the reader. For the reader, the difference of those two quite differing "solutions" should distinctly be made obvious and clear. A Gordian knot to be offered, as in the past, should be avoided at all costs. Whether the sources are good or bad is up to the reader to decide. But it's up to the editors to decide whether it is "good" to hide everything behind the alleged "truth of mathematics". It would be fine if you could help to clearly show the difference of the two groups of sources. Your help is needed. Regards, Gerhardvalentin (talk) 09:02, 17 February 2011 (UTC)

The MHP paradox is why it's 2/3 & 1/3 rather than 1/2 & 1/2. Any premises that change that are variants. Glkanter (talk) 05:56, 17 February 2011 (UTC)

I agree completely, that the different approaches and assumptions should be treated separately and clearly for what they are and where they differ. However imho one of the main reasons, why this has not worked out that well in the article so far, is that various factions here insist on their approach or assumptions being the "real ones" and that they need to be treated as "the" essential MHP and that's exactly the evangelists again. They care less for an overall lucid explanation of all aspects, but rather that their favoured version is most prominently featured. The current partially less lucid state, is result of all those evangelists trying putting their version on the top creating a "obfuscating" mix of everything.--Kmhkmh (talk) 14:05, 17 February 2011 (UTC)

Kmhkmh, thank you so much for your neutral evaluation of the current status. And yes, your words are *the* long-awaited clearance and liberation. You see the field and are raising your voice from a higher vantage point, and I would like to hope this will be of appreciable effect. And I am hopefully that we are closer to an end of the long lasting solidification now. Thank you once more! Regards, Gerhardvalentin (talk) 15:01, 17 February 2011 (UTC)

Yes!

Kmhkmh, your approach is the basis of my current suggestion in arbitration, that we have the article in two parts, for the benefit of our readers. The first part has only simple solutions, without disclaimers, not because they are 'The Truth' but because they are what most people will want to understand. After we have dealt with that, we we can have a scholarly expostion of all the other issues relating to the MHP.

This is not some dastardly plot to promote my truth above all others, it is the way that most good text books and encyclopedia article work. Give a simple explanation first, maybe glossing over some technical details, then discuss the more complicated stuff. This effectively brings this long argiument o a close and enables us to work cooperatively again.

In the past you expressed your approval of this approach, which is pretty much how the article is now. Perhaps you could confirm that you still support this idea. I believe that it is the only way to move on. Martin Hogbin (talk) 11:18, 19 February 2011 (UTC)

I don't think my "approach" extactly matches your suggestion. I have no issue with the "simple solution" being presented first or without caveats/pointers to morgan style crticism. I do however have an issue with it being represented in an obfuscating manner (there I agree somehwat wih nijdam). Meaning we should clearly state what the original simple solution computes the overall probability for winning. No caveats but clearly stating what it does. You may also argue that the overall probability is identical with the conditional one under certain assumptions, but then again that needs to described in lucid manner (rather than what you personally consider as "obvious"). However the detailed argument here might better be pushed to a later article, to keep the "simple solution" indeed short and simple.

Having said since I dont intend to play the role of a "lucid description everywhere" evangelist. I'd live with your (imho mathematically obfuscating) suggestion just for the compromise's sake. I.e. if the rest of the editors is fine with your approach I won't object as long as a more lucid description regarding the details is further down in the article in a separate chapter. But alas so far neither the other "conditionalists" are willing to budge nor are you.--Kmhkmh (talk) 16:13, 19 February 2011 (UTC)

Kmhkmh, could you please indicate which reliable sources that actually offer a simple solution say the things you describe above about 'overall probability of winning'? If it comes from critical reliable sources, please reference them. Thank you. Glkanter (talk) 16:29, 19 February 2011 (UTC)

In the mean time I hear that I am being crucified on the arbitration page for being an expert and for trying to give expert advice where relevant and for patiently trying to explain the expert's point of view on the talk pages. It seems that a lot of people don't have much of a sense of humour, and in particular, are not able to appreciate it when people do not take themselves seriously, but in fact make jokes at their own expense. I guess it is rather British. Richard Gill (talk) 16:06, 17 February 2011 (UTC)

No, that's not accurate. You are not being crucified on the arbitration page for being an expert, far from it, although a false accusation has been made to that effect... As a statistician and a forensic expert, I would think you would be able to tell the difference between real evidence and a false accusation. My answer to that false accusation here As for the sense of humor issue, I had ignored that as it doesn't adequately explain away the evidence IMO, but I will respond to that accusation on my talk page, where you first raised the issue. Woonpton (talk) 16:31, 17 February 2011 (UTC)

Yes, you are right @Woonpton! One must distinguish between evidence and accusations. And certainly, at face value, the evidence presented is strong. I need to make more use of a sandbox when composing some text for wikipedia. By the way, you are falling into my own humour-trap here, because my calling what is going on "being crucified" is poetic license, hyperbole, exageration for dramatic effect. (Though, those were the very words a worried friend of mine used, to me.). I'm a typical somewhat autistic mathematician, poor communication skills. As long as I can remember, people have been alternately angry with me for appearing to be arrogant, or angry with me for being over modest, self-effacing. Fortunately, now I'm nearly 60 I don't have to care so much what people think. In Dutch there is a phrase "whether or not the host trusts his guests tells us whether he is an honest man himself". I honestly think that I act on wikipedia in good faith and I assume good faith on the part of anyone else. And to be sure, I'm human with all human failings, like all of us. Richard Gill (talk) 14:17, 19 February 2011 (UTC)

It shows that Rick Block's claim about variants not polluting the article is his usual hogwash. The variants monopolize the Conditional solution section at the expense of clarity. That's fact, not dogma. The positioning of extensive discussions of variants in the article as a means of criticizing the simple solutions is also UNDUE WEIGHT and a POV that is not supported by a significant minority of reliable sources. That makes it OR. Glkanter (talk) 07:41, 16 February 2011 (UTC)

FAQ

When you fire up wikipedia MHP Talk page from a smart-phone you get to see the FAQ in all its glory, the rest is initially hidden. So I finally noticed we had an FAQ and I read it and added some stuff on symmetry. There is also a lot of symmetry at my latest peer-reviewed earth-shattering brilliant masterpiece contribution to the Annals of MHP Studies ;-) [5]. Extended version at [6]. Those texts are based on a lot of interaction with Boris Tsirelson at [7]. Richard Gill (talk) 10:27, 20 February 2011 (UTC)