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The present article contains in the conditional probability section a fully written out proof of Bayes' theorem in the context of MHP. (IMHO, totally superfluous, since also present in the wikipedia article [[Bayes theorem]]). But anyway there were discussions about the proper notation to be used in this proof. I have written out an alternative [http://en.wikipedia.org/wiki/User_talk:Gill110951#MHP_notation_for_Bayes.27_theorem_proof here] (on my talk page) and written a general essay about notation [http://en.wikipedia.org/wiki/User:Gill110951/Probability_notation here] also in my user area. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 09:51, 5 April 2011 (UTC)
== Increasing the number of doors rewrite? ==
<br />
I think the "Increasing the number of doors" approach that's currently [http://en.wikipedia.org/w/index.php?title=Monty_Hall_problem&oldid=421992876#Increasing_the_number_of_doors (permalink)] in the article is probably the one that's most likely to be accessible to our widest swath of readers. But I also think it needs to be rewritten to accomplish that. Here's how I'd use the idea to explain the solution to someone who knew only simple algebra:
<div class="toccolours">
:<big>'''Look at it from Monty's perspective'''</big>
:To understand this, you need to set aside your natural preoccupation with your own perspective as a potential contestant, and look at the situation from Monty's perspective instead. Specifically, you need to recognize that you're not the only one who expresses a choice about which door the car is behind. It's easier to understand that if you imagine there's some impossibly large number of doors, at first; let's say a "gazillion".
:So you recognize that when you first choose one door, your chance of doing so correctly is really small, one-in-a-gazillion, right? Then switch perspectives, and put yourself in Monty's shoes: Ask yourself "What is Monty's chance of choosing the correct door, the one with the car behind it?" "But wait!", you say, "Monty doesn't choose anything!!"
:''Monty <u>does</u> make a choice, though, and this is the key to understanding the problem'': When Monty opens all the other doors but one, he eliminates all those doors because he knows the car isn't behind them. ''Assuming you didn't guess correctly the first time (a reasonable assumption, given your one-in-a-gazillion chance) then '''Monty is effectively choosing the door the car is behind, by keeping that one door closed''''' and opening all the others. The rules he has to operate under make it impossible for him to do anything else. Further, his expressing that choice, even though he does so indirectly, discloses information to you that you'd be smart to take advantage of when he eventually asks if you want to switch from your original choice to his.
:<big>'''The mathematics of this insight, and what they show for the three-door case'''</big>
:The mathematics of this become pretty simple to understand if you reduce the number of doors at the outset from a "gazillion" to just 100. In that case, when you first choose a door, it's pretty clear that you have a .01 or (same thing) 1% chance of choosing the right one. But then ask yourself, "What chance does ''Monty'' have of choosing the right door?" ( Recall that he has to express his choice not by picking the door directly, of course, but by opening all the ''other'' ones, except yours. )
:Since Monty knows where the car is, the answer would be that he'd choose correctly 100% of the time <u>if</u> he were not constrained by your having eliminated one of his possible choices when you picked first. So that constraint means that he has only a .99 or (same thing) 99% chance of choosing the door with the car behind it. His chances of picking the correct door are, in this example, 99 times better than your chance of having done so.
:To express that insight in a formula, if we let
:* N = the number of doors at the outset,
:* P<sub><small>Contestant</sub></small> = the contestant's probability of choosing right the first time
:* P<sub><small>Monty</sub></small> = Monty's probability of choosing the right door, i.e. of leaving closed the door with the car behind it
:Then it's clear that
:* N = 100
:* P<sub><small>Contestant</sub></small> = 1 / N = 1/100 = .01 = 1%
:* P<sub><small>Monty</sub></small> = 1 - ( 1/N ) = .99 = 99%
:* P<sub><small>Monty</sub></small> = 1 - P<sub><small>Contestant</sub></small>
:This last statement or formula isn't strictly necessary to solve the problem, of course, but it might help you understand it better: It says that Monty's probability of choosing correctly is 1 minus the contestant's probability of choosing correctly, because the contestant might have got lucky and picked the correct door before he could, thus eliminating Monty's ability to do so.
:Now these formulae continue to hold if, instead of starting with a gazillion doors, or even 100 doors, we reduce the number of doors at the outset to 50, or 25, or even to three. So <u>with three doors</u> at the outset, we have:
:* N = 3
:* P<sub><small>Contestant</sub></small> = 1/3
:* P<sub><small>Monty</sub></small> = 2/3
:It may seem contrary to common sense in this case, when we've reduced the number of doors to just three, but the same reasoning and the same formulae do continue to hold, and if you switch to the door Monty "chose" (by leaving it closed), you do in fact double your chances of winning.
</div>
I don't intend to introduce this proposed explanation into the article myself: That would seem the height of presumption to me, given that some of you have been herding cats here for years! ( Thanks for that! ) But I do think it presents the clearest chance that a person who's relatively unsophisticated in math has of getting the "flash of insight" that's necessary to understanding the problem. I don't claim it's the best way to ''prove'' the solution, or even that it's a proof at all, though, just that it's likely to be pretty accessible to a typical reader. What does everyone else think about that? – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 19:50, 2 April 2011 (UTC)
:<small>( On re-reading this, I have some misgivings over saying that Monty has only a probability of .99 in choosing the "right" door in the case where we start with 100 doors at the outset, or .66 in the case where we start with three. Obviously his probability rises to unity, to 1, if the contestant's first pick was incorrect. But I won't (yet?) complicate the presentation I made above by correcting it to accommodate cases based on whether or not the contestant's first choice is correct. I know that might be helpful, of course, because Monty's "chances" of picking the correct door drop to zero if the contestant gets lucky on his first pick. But if people think the approach has value, maybe we can work together to clarify the two cases, and dot the "I's" and cross the "T's". Best regards, all, and thanks for your ongoing work on this article, very much. Btw, special thanks to Martin, Gerhard, and Rick, for their patient and generous help on the "Arguments" subpage. Cheers, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 22:12, 2 April 2011 UTC )</small>
:''"Monty is effectively choosing the door the car is behind, by keeping that one door closed [...] impossible for him to do anything else."'' — Incorrect, because if you should have chosen the winning door, the host will not dispose of the car. So what is written here is just incorrect. And if the host "should be biased to just ''never'' open the door he left closed, if any possible", then the chance that the still closed door hides the car could be only 1/2 at least (as per Ruma Falk), even in your gazillion example. And btw: "Then it's clear that ..." is never enough, the content of the article must be sourced. Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 23:10, 2 April 2011 (UTC)
::Thanks, Gerhard, but I think you missed the condition that the boldfaced part of the sentence is preceded by, viz. ''"Assuming you didn't guess correctly the first time (a reasonable assumption, given your one-in-a-gazillion chance)..."'' Also, I'm afraid I don't understand your objection about the possibility of the host's bias. As I understand the problem, Monty has no opportunity for bias; as I understand it he has no choice but to open all but one door, excluding your own. Could you explain further?
::I've moved your comment to just above, btw, from being interleaved with the presentation I made in the "callout box". If everyone did that, then the original intent of that post, and what I'd written versus what everyone else contributed, would become obscure. Also, if we can all agree on an aid to understanding (not a proof) then I don't see any reason why that would have to be be "sourced". Or was that a part of the arbitration requirements? – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 00:00, 3 April 2011 (UTC)
::: Good point, @Ohiostandard. Host bias (or not) is irrelevant to the simple solutions. Those who would always switch will win with unconditional probability 2/3. For frequentists, this holds whether or not the host is biased, and for subjectivists this holds whether or not their opinions about possible host bias are symmetric regarding the direction of any bias. Gerhard Valentin refers to Ruma Falk for the host-biased conditional result but it was already in Morgan et al, but more easily derived with Bayes' rule (odds form of Bayes theorem), for instance, cf. Rosenthal's paper and book. But you are talking about an argument for the simple (unconditional) result. It's indeed a valuable aid for understanding why the "naive" argument "no need to switch because Monty has not given you any information about your initial choice" is faulty. It's given by many sources. In the unbiased case, Monty has given you no information about your intitial choice, so the probability that that was right is still 1/3. There is still 2/3 chance left and that stays where it was, with the two other doors, one of which Monty has kindly shown to you does not hide a goat. As Gerhard mentions, in the case of a biased host, the identity of the door which Monty opens can contain information. In the most extreme cases, the chance your initial choice was right can rise to certainty, or fall to 1/2. But it is never unfavourable to switch, since there is no way to improve the 2/3 overall success chance of "always switching". Editor @Lambiam recently found an alternative proof that 2/3 cannot be beaten, and hence that all conditional chances of wining by switching are at least 1/2, by showing that a known host bias can only be advantageous to the player. Yet even with a maximally biased host, it's easy to see that 2/3 (overall) can't be improved. I wrote it out in [http://www.math.leidenuniv.nl/~gill/mhp-statprob.pdf]. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 10:47, 3 April 2011 (UTC)
:::: Thanks, Richard. Your kind response reminds of one of the many things I love about Wikipedia: There are experts here in just about any subject that could interest a curious person, and they're usually willing to share their expertise, even with untrained persons, quite generously and patiently. However it's communicated, I do wish it could be made explicit in the article that Monty is actually making a choice, too, but one that's based on better information, and that his doing so has information value for the contestant.
:::: Also, I don't know whether it can make it into the article or not ( I'll leave that to you who've contributed here for so long ), and without the ''least'' wish to disparage any other presentation, all of which I honor, I feel that candor requires me to disclose that the first immediately convincing presentation that has worked for me for the simple version is the one ([http://en.wikipedia.org/w/index.php?title=Talk:Monty_Hall_problem/Arguments&oldid=422161032#A_different_table permalink]) that Rick Block posted to [[Talk:Monty_Hall_problem/Arguments | the ''Arguments'' page]] under the heading, "A different table". Best regards, all. – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 17:25, 4 April 2011 (UTC)
:::::OhioStandard, I am a little concerned that you are deferring to other editors based upon them being here longer. Please read [[WP:NVC]] [[WP:OWNERSHIP]] and [[WP:ODNT]]. (I do agree that it is best to discuss changes and and seek consensus rather than just jumping in and changing things, but we are all equal here.) [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 07:52, 10 April 2011 (UTC)
::::::That's a fair point, Guy, based on the language I used above. If it'll make you feel better, I don't mind disclosing that my apparent humility is much more formal than real. "Jungle manners" on entering new or disputed territory, as Marie Louise von Franz used to say. ;-) The links are helpful, nevertheless, but please let me assure you that if something comes up that seems important to me, I'm perfectly willing to make any amount of noise or to (figuratively) bloody my knuckles to be heard about it. I genuinely appreciate your remarks, though; thanks. – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 10:16, 10 April 2011 (UTC)
== Probability concepts ==
IMHO, one can see from any written source about MHP whether the writer is a subjectivist or a frequentist in his or her use of probability. Often of course without even knowing about the distinction. I think that the present page contains quite a few subtle biases to one or the other concept of probability. For instance the just mentioned proof of Bayes' theorem explicitly put randomness in the physical procedure used by game-show team and the host to hide the car and open a door respectively. Probability is in the real world. Ontological probability (about things). Whereas the "usual assumptions" of equal probabilities really (IMHO) only make sense for the subjectivist (aka Bayesian) viewpoint, whereby probability describes our information or lack thereof. It's in our mind. It describes our personal ignorance or knowledge. Epistemological probability (about our knowledge). I wrote a little essay about this on my talk page, [http://en.wikipedia.org/wiki/User_talk:Gill110951#MHP:_it_all_comes_down_to_what_you_probably_mean_by_probability here], it was originally a response to a correspondent. Read [[Probability interpretations]] on wikipedia.
So far editors of the page have shied away from this issue. Maybe they are scared of opening a can of worms? Or they have a POV and don't know any other? I think the issue can't be avoided in any decent discussion of MHP. You can only draw conclusions by making assumptions and where do those assumptions come from? The man in the street is a subjectivist, the academic literature is biased to the frequentist position. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 10:00, 5 April 2011 (UTC)
:I am very happy to discuss the subject with you. It has indeed caused a lot of confusion in recent arguments.
:I do not think the distinction is as great as you suggest. The two approaches can always be made to agree by making sure that the set-up of the frequentist approach matches the state of knowledge in the Bayesian approach. If you disagree, please give an example of where the two approaches give different answers. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 22:03, 7 April 2011 (UTC)
== Better shoes argument? ==
If it won't annoy people, may I ask whether any sources have tried to give the reader the necessary "flash of insight" by asking him to put himself not in the position of the contestant, as is usual, but in the position of the host, instead? As in the following, for example:
:''Imagine the producers of the show let the host keep the car for himself personally if the contestant picks the wrong door at the outset and the host then (indirectly) "picks" the correct one, i.e. by choosing to leave that one of the remaining two doors closed. Now suppose the producers, at the start of the event, offer to let you play the role of the host instead of that of the contestant. The contestant gets to pick a door first, but you'd be told in advance which door the car is behind if you played the host. Which role would you prefer, and why?''
This seems pleasantly concise, and (to me) very likely to at least give the reader pause, to give him a salutary doubt that his immediate "50/50" assumption re the stay/switch decision is correct. It also has the pleasant consequence, I think, that if people think about it, it will bring them to the conclusion that it's more advantageous to know where the car is than it is to pick a door first. One could then go on to explain that by choosing "switch" one effectively puts himself into the host's shoes, and thus benefits by the host's foreknowledge of where the car is. I'm mostly just curious to know whether this has been argued before. Thanks, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 00:34, 8 April 2011 (UTC)
:Krauss and Wang (reference in the article) set out to try to get people to arrive at the correct answer more frequently. Asking people to take the perspective of the host is one of the things they tried (experimentally). It indeed does help. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 06:24, 8 April 2011 (UTC)
:: 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)
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