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MiszaBot I (talk | contribs) m Archiving 2 thread(s) from Talk:Monty Hall problem/Arguments. |
MiszaBot I (talk | contribs) m Archiving 2 thread(s) from Talk:Monty Hall problem/Arguments. |
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::::::::::::This seems like a topic for [[talk:Monty Hall problem]], not here. I would encourage you to bring this up there. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 15:08, 5 May 2011 (UTC)
== From the main talk page ==
This discussion has been taken from the main talk page
:::I would be very interested to see the sources that support your statements above. Let us start with, 'in either a frequentist or subjectivist view the player's probability is a function of the host's preference between two "goat doors", and if the player doesn't know this preference the best we can say is the probability of winning by switching is an unknown value between 1/2 and 1' and consider the subjectivist (Bayesian) view. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 16:01, 5 May 2011 (UTC)
::::As you are well aware Morgan et al. (which is based on a Bayesian view) says this virtually verbatim, as does Gillman (specific references in the article). Puza et al. [http://onlinelibrary.wiley.com/doi/10.1111/j.1467-9639.2005.00190.x/abstract] and Eisenhauer [http://onlinelibrary.wiley.com/doi/10.1111/1467-9639.00005/abstract] as well. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 18:47, 5 May 2011 (UTC)
{{od}}
I found '''[[Probability]]''' to be incredibly helpful in understanding this. Is says,
"''The word ''probability'' does not have a consistent direct definition. In fact, there are two broad categories of '''[[Probability interpretations]]''', whose adherents possess different (and sometimes conflicting) views about the fundamental nature of probability:''
#'''''[[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.''
#'''''[[Bayesian probability|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.''"
An interesting aside: in my engineering work, we pretty much follow the '''[[Propensity probability]]''' interpretation. It's extremely useful for getting work done, but is it the way the real world works? Hard to tell. --[[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 19:26, 5 May 2011 (UTC)
:Thanks Guy for reminding us all what the Bayesian interpretation means. It means the probabilities are based on (or defined to be) a degree of belief given the evidence. If there is no evidence at all we must take all outcomes as equally likely. Thus, as Whitaker's statement does not tell us how the car was placed, how the player chooses, and how the host chooses, we ''must'' take all possibilities to be equally likely. That is pretty well the meaning of the Bayesian definition. Thus using a Bayesian definition, based strictly on Whitaker's statement alone, the answer is obviously exactly 2/3. It can be nothing else.
:Where exactly in Morgan does it say that they are using a Bayesian interpretation in the main body of their paper?
:In fact in their conclusion they say [my italics], 'In general, we cannot answer the question "What is the probability of winning if I switch, given that I have been shown a goat behind door 3?" unless we either know our host's strategy or ''are Bayesians with a specified prior'' '. Or to put this more clearly ''If we are Bayesians we can answer the question in general'' The appropriate prior, as they point out elsewhere, is a noninformative one. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 21:42, 5 May 2011 (UTC)
In moving some comments to here they have become separated from other comments which may have been intended as responses to me. I did not move them because I was not sure if they were intended as suggestion on how to improve the article or responses to me. Please feel free to move them her if they were intended to be responses to me. Otherwise, I await response from Rick or indeeed anyone else. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 13:00, 8 May 2011 (UTC)
::Martin - what is your source for your claim that "if there is no evidence we must take all outcomes as equally likely" or that this is the meaning of the Bayesian definition? Regarding where Morgan et al. say they are using a Bayesian interpretation they (of course) don't specifically say this - however they're not talking about measurements taken from repeated experiments so given the dichotomy you seem to think exists between frequentist and subjectivist they are rather obviously on the subjectivist side. They make very clear "the probability" (without distinguishing frequentist or subjectivist - which might reasonably lead us to believe they think this distinction is not important) of winning by switching, given the clarifications vos Savant made explicit in her later columns (initial car placement is uniformly random, host must open a goat door and must make the offer to switch) is 1/(1+''q''). Bayesians can in addition assume a specific prior to arrive at a single numeric answer - but assuming a prior is an additional assumption just as much as assuming the host chooses uniformly between two goats. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:06, 8 May 2011 (UTC)
:::As the question gives us no information as to the way the host may choose a legal door door we naturally want an uninformative prior. The simplest and most obvious of the possibilities is a uniform distribution. This is the same prior that Morgan use on their Bayesian calculation earlier in the paper to get a (erroneous) single numeric answer. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 17:57, 8 May 2011 (UTC)
::::Are we disagreeing about anything here? Given all of vos Savant's clarifications, the Bayesian answer is 1/(1+''q''). Additionally assuming an uninformative prior this works out to 2/3. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 18:20, 8 May 2011 (UTC)
:::::Strongly disagree. Quite the contrary. You wrote ''"Additionally assuming an uninformative prior of q=1/2".'' This is an incorrect and misleading formulation / assertion.
:::::"Additionally assuming an uninformative prior of q=1/2" in effect is ''no additional assumption at all.'' It's just a ''given fact (!)'' based on your "lack of knowledge". It never can be any "additional assumption". Such assertion is misleading and not correct.
:::::I repeat: ''No "additional assumption".'' – As ''q'' is totally ''unknown,'' there is NO USE "to use" any ''superfluous'' "assumption". ''You don't need to assume''. Just honestly cut down on what you really know.
:::::As you actually don't know anything about the value of "q", nor will ever know but in "using q", though, you are ''strictly forced'' to take q as 1/2. That's a fact. Once more: In case you superfluously should "like" to ''use'' "q", you are ''forced'' to attribute the value of 1/2 to q. Period.
:::::And if you should "like" to attribute any ''other'' value to "your q" for the actual game show in question, then you will just get what you "like" to get. Quite outside the MHP. Never addressing the famous "MHP-question" anymore, but just "testing" Bayes' capability to come up ''with any "assumptions and illusions" you like to assume.'' Unnecessarily for the MHP. Quite outside the famous MHP-question.
:::::''Because, as long as you do not know'' the exact "known" host's bias and its direction, it doesn't matter whether the host has some bias or not. So it's superfluous to speculate about that. – On the other hand, if you like to use Bayes and any "q", you are free to speculate and to do what you "like". But it's an illusion to think that you could get something reasonable to answer the famous question on the actual game show in question. On no account. You will just always get just the illusion of ''"what you like to get",'' and that will be and forever will remain ''within the given strict bonds.'' Not addressing the famous question about the actual game show in question anymore, but just addressing Bayes' theorem, and just showing that conditional probability can be applied wherever and whenever you want to, whatever unproven "assumptions" you should like to include. Outside the famous question about the actual game show in question. But in any case obviously remaining within the strict bonds ''shown and given by just simply asking "what if he always" and "what if he never": Probability to win by switching will remain within the range of at least 1/2 to 1.'' Read the actual sources. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 19:13, 8 May 2011 (UTC)
::::::We're talking here specifically about what Morgan et al. say, which I believe is exactly what I've said above. If you'd like to bring up what ''other'' sources say feel free, but please name them. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 19:20, 8 May 2011 (UTC)
:::::Rick, I think you are talking about a different bit from me. I am referring to the bit at the bottom of the left column on page 286 which starts, 'This provides an excellent opportunity to bring in the Bayesian perspective...', it then mentions a noninformative (uniform) prior, and ends with a single figure answer of 0.693. This was the figure which Nijdam and I showed should have been 2/3. This is the bit that Morgan refer to in their conclusion. You will not that it gives a single value answer to the question. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 21:57, 8 May 2011 (UTC)
::::::We're talking about the same thing. The 1/(1+''q'') answer is a perfectly Bayesian answer. What they're saying is that Bayesians can ''additionally'' obtain a single numeric answer by assuming a particular prior. With a noninformative prior their answer is .693 (which Puza et al., or your recent letter, correctly show is actually 2/3). Per the top of the next column, with different priors Bayesians can arrive at other answers - but never less than 1/2. Again, are we disagreeing about something here? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 22:29, 8 May 2011 (UTC)
:::::::What other prior would you propose on the basis only of the information given in Whitakers question? [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 22:57, 8 May 2011 (UTC)
::::::::We are still talking about Morgan et al., right? Are you asking me to make a conjecture about something that this paper doesn't say, or are you claiming this paper says anything other than "the probability" (even the Bayesian probability) is 1/(1+''q'') and that if one insists on a single numeric answer a Bayesian can get one by also assuming a specific prior (and the probability one arrives at depends on what prior is assumed)? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 01:57, 9 May 2011 (UTC)
:::::::::I am asking you what other prior there could possibly be other than a noninformativce one, considering we are given no information. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 22:01, 9 May 2011 (UTC)
::::::::::So you're not talking about sources here, but you want my personal opinion? No thanks. You asked me to substantiate, from sources, a claim I made. I've done that, and you picked Morgan et al. to discuss further (not me). You are now apparently trying to change the subject to THE TRUTH. Will you please respond to my question to you. Are you claiming Morgan et al. says anything other than "the probability" (even the Bayesian probability) is 1/(1+''q'') for the vos Savant/Whitaker version (with vos Savant's clarifications) and that if one insists on a single numeric answer a Bayesian can get one by also assuming a specific prior (and the probability one arrives at depends on what prior is assumed)? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 04:30, 10 May 2011 (UTC)
:::::::::::Well, of course, we are free to talk about The Truth here on the talk page and indeed some understanding of probability would be assumed by Morgan for their intended audience. There is nothing new or unexpected in assuming a uniform prior in discrete probabilty theory where there is no information to indicate otherwise. It has been a standard part of probabilty theory since Laplace (seeTruscott, F. W. & Emory, F. L. (trans.) (2007) [1902]. A Philosophical Essay on Probabilities. ISBN 1602063281) and has remained so ever since.
:::::::::::We can both read what Morgan write. They give only one numerical answer and this for the uninformative prior. They also discuss priors where the player may think that the host has a particular strategy (for example with a large weight near q=1). There is nothing in Whitaker's question to suggest that this may be the case, it gives us no information on the matter, hence the noninformative prior would be the standard choice and that is the one that Morgan actually use in calculating a single numerical solution. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 08:45, 10 May 2011 (UTC)
== http://en.wikipedia.org/wiki/Monty_Hall_problem ==
http://en.wikipedia.org/wiki/Monty_Hall_problem
Please note I am not a wordsmith and I am just a layman fascinated by the mathematics quoted in the above article which seems 100% correct in the context used.
However to me with a basic understanding of mathematics the original choice given you have a 1 out of 3 chance of "winning" is not correct.
The actual chance you have of "winning" are greater at the outset of 1 out of three because if 1 of the 3 choices of "winning" is already arranged to be removed prior to your choice of a "pick" of 1 out of 3 (if you are incorrect in picking a winner at the beginning) then the mathematic formula’s quoted surely is incorrect
Tony Gray
New Zealand <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/114.134.7.115|114.134.7.115]] ([[User talk:114.134.7.115|talk]]) 22:40, 6 May 2011 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->
:The statement means that right at the start, before any doors have been opened, you have a 1 in 3 chance of picking the car. It does not refer to your probability of going on to win the game. If you are smart, this is 2/3. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 22:54, 6 May 2011 (UTC)
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