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:: In Selvin's second note, this is long before Marilyn Vos Savant, Selvin quotes from a letter sent by Monty Hall himself, who did solve the problem in one sentence from the point of view of the host. For the host, the ___location of the car is known, and the way he will choose a goat-door if necessary is his own business. The only thing random for him, is which door the player chooses, which from his point of view has a odds of 2 to 1 against of being correct. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 11:09, 10 April 2011 (UTC)
== Will the real probability please stand up ==
One of the continuing problems here, in my opinion, is the assumption by some editors that there is some 'real' meaning of the the term 'probability' that everybody knows and understands. The truth is much closer to the reverse; there are two ''definitions'' of probability in common use and nobody understands what the word 'really' means.
Richard Gill and I have tried several times to discuss this important subject but to no avail. Might I suggest that it would avoid many pointless arguments if people would state which definition of probability they are using and stick to it throughout their argument. It is generally unhelpful to talk the results of 'repeating the experiment' when discussing a subject from a Bayesian perspective. Similarly, states of knowledge are of little relevance in calculating a frequentist probability.
Now, my personal opinion, at least, is that it is always possible to make the two interpretations of probability agree by carefully making sure that the frequency distributions in the frequentist interpretation match the state of knowledge in the Bayesian interpretation. However that is a subject for another time and place.
What would greatly help here would be for editors to clearly state which interpretation they are using and then stick to it. Unfortunately, not many sources do this either. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 08:57, 9 April 2011 (UTC)
:The following is a quote from [[Probability interpretations]]:
:''"The terminology of this topic is rather confusing, in part because probabilities are studied within so many different academic fields. The word 'frequentist' is especially tricky. To philosophers it refers to a particular theory of physical probability, one that has more or less been abandoned. To scientists, on the other hand, "frequentist probability" is just what philosophers call physical (or objective) probability. Those who promote Bayesian inference view 'frequentist statistics' as an approach to statistical inference that recognises only physical probabilities. Also the word 'objective', as applied to probability, sometimes means exactly what "physical" means here, but is also used of evidential probabilities that are fixed by rational constraints, such as logical and epistemic probabilities."''
:Also see [[Probability]], [[Probability interpretations]], [[Frequency probability]], [[Bayesian probability]], [[Pignistic probability]], [[Algorithmic probability]], [[Philosophy of probability]], [[Sunrise problem]], [[Two envelopes problem]], [[Necktie paradox]], [[Exchange paradox]], and [[Doomsday argument]]. [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 14:50, 9 April 2011 (UTC)
::Guy, I am not sure what you are trying to tell me. Your quote above confirms my point above that there is no general agreement on exactly what probability 'really' is.
::The [[probability]] article says:
::'' 1. Frequentists talk about probabilities only when dealing with experiments that are random and well-defined. The probability of a random event denotes the relative frequency of occurrence of an experiment's outcome, when repeating the experiment. Frequentists consider probability to be the relative frequency "in the long run" of outcomes.[1]
::'' 2. Bayesians, however, assign probabilities to any statement whatsoever, even when no random process is involved. Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, or an objective degree of rational belief, given the evidence.''
::My point is that in discussing probability issues regarding the MHP it would be very helpful and would avoid pointless disagreements if editors would adopt a particular perspective and stick to it for the duration of the argument.
::My personal opinion is that the Baysian approach is the most appropriate approach for the MHP but I do wish to try to force that opinion on anyone else and I am happy to discuss the problem from a frequentist perspective.
::I am not sure from your post whether you agree with me or not Guy. Are you asserting that there are other important interpretations of probability or do you agree that there are only two?
::Do you agree that in a discussion about probability it is best to stick to one interpretation for the duration of a logical argument rather than change the meaning of the word part way through a discussion? [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 15:39, 9 April 2011 (UTC)
:::Yes. I fully agree that we need to pick a definition and stick to it. The definition I use as a working engineer is not the definition a philosopher uses, and I suspect that a statistician will have a third definition. Any of the different definitions will work if everybody uses them , but having different people use different definitions is an invitation to misunderstanding and strife. [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 20:16, 9 April 2011 (UTC)
:::: Yes the words are confusing and are used to mean the opposite by different communities. Maybe we should just stick to using the adjectives "epistemological" and "ontological" ;-) [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 06:56, 10 April 2011 (UTC)
:Responding to Martin's original post in this thread - IMO, what we mean by probability is not even remotely the issue here. What IS the issue is what probability we mean (per my response to Martin in the thread above). Bayesians and frequentists are both completely capable of distinguishing P(win by switching) from P(win by switching|player picks door 1 and host opens door 3). If we want to reduce the confusion here, we should clarify which one of these we're talking about. Are we interested in the probability of winning of a strategy of (always) switching, or the probability of winning given the player has picked a particular door (say door 1) and has seen the host open some other door (say door 3)? Which of these we're talking about is and always has been the issue - not the meaning of probability. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 23:59, 9 April 2011 (UTC)
:: Rick, that was not a response to Martin's point at all, I think. Of course everyone ought to be able to agree on a distinction between P(A) and P(A|B). And both kinds of probability satisfy the same rules so the arithmetic is the same too. But the words around the arithmetic and the distinction of concepts *is* important. And the literature is confusing since different writers do clearly have different intended meanings of probability in mind. Moreover, the assumptions which some people want to add into the problem in order to be able to solve it their way have to be justified or explained, and the justification or explanation is different per different understanding of what the word probability is supposed to mean.
:: So an editor of the page his well advised to be aware of the distinction and he or she has to be master of using formulations which are probability interpretation free. It can be done but it requires careful and clear thinking. Never talk about "how many times". Aways talk about "so many times more likely". Basicly it forces you into the language of [[odds]], not of probabilities. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 06:51, 10 April 2011 (UTC)
::Rick I am not saying that the issue we are arguing about is related to the different interpretation of probability, just that not making clear what interpretation you are using makes it very hard to discuss the subject without continual misunderstandings. I am not even asking you to stick permanently to one approach, all I am asking is that we talk about them one at a time. I have not been able to respond to your reply to me above because you seem to continually flit between one approach and another whilst using concepts from one approach in the context of another.
::Let me say again, I do not want to have a philosophical discussion about the meaning of probability on this page all I ask is that during any given discussion we adopt one interpretation and stick to it. Can there be any valid reason not to do that? [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 21:14, 10 April 2011 (UTC)
:::And, what I'm saying, is that for the purposes of any discussion we've ever had, which interpretation of probability we're talking about makes no difference whatsoever. In my reply above (that you haven't responded to) I don't "flit between one approach and another" but connect what we ARE talking about to both. The OP above was asking why his professor insisted on using conditional probability. The explanation (whether you're a frequentist or a Bayesian) is the same - i.e. that at the point of making the decision whether to switch, the player knows which specific doors are involved (and the doors are not "otherwise indistinguishable"). Whether you're a Bayesian or a frequentist, this means you're interested in the conditional probabilities. Can there be any valid reason for continuing to harp on a point that has no relevance to what we are talking about, and refusing to address the actual issue?
:::Richard says above "everyone ought to be able to agree on a distinction between P(A) and P(A|B)". So, Martin, are we talking about P(A) or P(A|B)? Harder question - do the "simple" solutions address P(A), or P(A|B)? If you think it's necessary, feel free to specify what interpretation of probability you're talking about. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 23:20, 10 April 2011 (UTC)
::::Rick, I am not harping on, I am asking you to make your arguments clearer by stating which interpretation of probability you are using. This is simply to help other editors understand exactly what you mean. You may know what you mean but Richard and I need to as well. I am quite ready to address the actual issue provide we agree at the start to use and stick one interpretation of probability. We may well arrive at the same conclusion whichever path we take but it will be an easier journey, for me at least, if we stick one philosophy throughout. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 13:27, 11 April 2011 (UTC)
::::Regarding your question, let us agree to take a Bayesian perspective on Whitaker's question. You asked whether simple solutions calculate the value of P(win by switching)? I think we can all agree the answer is 'yes'.
::::Having said 'yes', I should point out that the simple solutions are in no way rigorous and include many assumptions that are not stated. As I think Boris said, there is a very rigorous for of mathematics in which even the most simple solutions become inordinately complex and completely unsuitable for anyone except mathematical specialists. For our expected readership and especially those who may read only the first part of the article, the simple solutions give, in my opinion, perfectly adequate calculation of P(win by switching).
::::If you ask whether the simple solutions calculate P(win by switching|player picks door 1 and host opens door 3) my answer would be the same except that they are marginally less good in that we do not actually state the obvious fact that as we have no information to allow us to distinguish between doors P(win by switching|player picks door 1 and host opens door 3) must be equal to P(win by switching). And as we have agreed that the simple solutions are satisfactory for calculating P(win by switching) we know that they also calculate P(win by switching|player picks door 1 and host opens door 3). It is obvious that, from a Bayesian perspective, P(win by switching|player picks door 1 and host opens door 3) = P(win by switching|player picks door 1 and host opens door 2) = P(win by switching). [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 14:00, 11 April 2011 (UTC)
== Question to Glopk and to Rick and any others regarding success rates ==
Article understandable to the reader, desirably to grandma and grandson likewise. Therefore my question:
MHP – "decision to switch or to stay" – gains and losses as results of effectively taken decisions of frequentists
:a) decision based exclusively on the so-called flawed simple solution, after the host had shown a goat and thereafter had offered to switch to his still closed door
:b) decision based exclusively on the so-called only correct conditional solution as per updated actual probability calculus
Please can anyone help me to find some reliable source that ever has determined the respective results of decisions to switch (resp. to stay), that ''effectively had been taken'' on solely "a", compared to solely "b" and thereafter has presented the specific results, for any combination of door chosen by the guest and door opened by the host likewise?
Is it conceivable that decisions that had exclusively been made based on the only correct conditional probability are of any higher success rate?
Is there evidence that specific decisions for any combination of doors, that had been made exclusively based on the only correct conditional probability "b", correctly updated for every show, has a higher success rate, compared to the specific results of decisions that have strictly been made on the so called flawed simple solution as per Marilyn vos Savant, i.e. to switch in the respective "actual" game show after the host has made his offer? And if so, to which degree? Because that must necessarily be added to the article. And otherwise, the contrary as well.
It is about the possible superiority of the so called conditional solution. Is its success rate really indeed so significantly higher for any specific combination of doors, as it is claimed by some?
Please help, because afaik there is no better possible decision than just "to switch" in the specific actual game, irrelevant of door numbers. Is there reliable evidence in reliable sources that "updated b" is of remarkably higher success rate than "a"? Otherwise "updated b" impossibly can be presented as a relevant topic here, but just the quite other aspect that the MHP is a suitable method to train conditional probability theorems. Thank you. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 17:32, 11 April 2011 (UTC)
: The simple solution shows that always switching is a much better strategy than always staying. The conditional solution shows that there is nothing better still. But of course nobody would imagine this, anyway. This is true if all doors are initially equally likely to hide a car. It doesn't matter whether or not the host has some bias. It's easy to see that the more biased the host is the more favourable it is to the player. And with a maximally biased host you can easily see (just work out the two possibilities in the Monty Crawl game) that switching is never unfavourable. Yet also in Monty crawl always switching only has success rate 2/3. This proves rigorously that there is no better strategy than always switching. (This argument was discovered by Lambiam, it is inspired by game theory, and it makes the explicit computations of Morgan et al. superfluous and allows you also to explain the biased host case to your Grandma or Grandpa. I have written it up in the longer version [http://www.math.leidenuniv.nl/~gill/mhp-statprob.pdf] of my article in the StatProb encyclopedia [http://statprob.com/encyclopedia/MontyHallProblem.html]. It's also written out on my University of Leiden home page. @Lambiam got a prize from me for finding this proof. [User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 16:48, 14 April 2011 (UTC)
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