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== Gerhard's question ==
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== A Fresh Start ==
I just archived this page so everybody can make a fresh start after the Arbitration. I encourage all involved editors to put away any hard feelings from the past and to make a fresh start. Please remember that this article is now on probation and that editors making disruptive edits may be blocked by an administrator. I would strongly encourage all editors to stop and review [[WP:NPOV]], [[WP:OR]], [[WP:V]], [[WP:OWN]], [[WP:CIVIL]], [[WP:NPA]], and [[WP:EW]]. Even if you have read them before, read them again, and treat them as your road map for editing this article, its talk page, any any discussions about this article on your user page. Thanks! [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 04:12, 25 March 2011 (UTC)
== One of two doors ==
Here's a quick way to argue against the current article's conclusion. It makes what appear to me to be some of the same subtle errors that I believe occur in the arguments for the currently-presented solution, but it has the virtue of being brief, and ( as it appears to me ) of producing the correct result. Let's ourselves be the contestant here. Assuming as much, then:
* At the outset of the problem, before we've made any choice at all, consider that the car is behind one of two doors.
::Yes, one of two. <u>We just don't know ''which'' two</u>, but it certainly is behind one of two. And ''of those two'', based on what we know at this point, the chance that it's behind one is as good as the chance that it's behind the other.
* Next, just to humor Monty, we "pick" one of the three available doors, and we try to keep his spirits up by pretending that our "choice" at this point matters. But we know it ''doesn't'' matter, because the outcome of the ensuing sequence will be the same regardless of which of the three doors we select at this point. After our apparent "choice", nothing relevant to the outcome has changed. The car is still behind one of two doors, and we still don't know which two those are.
* Then Monty opens one of the unchosen doors, and now something ''has'' changed: <u>We now know "which two"</u>.
::Monty says, "Do you want to switch?" which translates into, "You now have the opportunity to pick between two doors. The car is behind one of them. Which one do you pick?" This is the first real choice we have, the first choice we can make that's relevant to the final outcome.
::<small>( There's zero chance that the car is behind the now open door, and no reason to believe that the now-dead chance that it ''was'' should be magically and preferentially transferred to any remaining door. There's a very subtle flaw involved here, imo, that I won't try to elaborate on, but that I think is due in part to a semantic confusion and in part to not properly accommodating time and instantiation as factors in this reasoning. Or maybe it would be more fun to observe that there was ''never'' any chance that the car was behind the now open door, but that we just weren't aware of that until now. ;-)</small>
::Remember that at the outset we knew that the car was behind one of two doors, that we didn't know which two, and that either of the two was as good as the other. Now we know "which two", and choosing one is ''still'' as good as choosing the other.
That being the case, there's no advantage to "switching". As I said, this is an abbreviated argument that I know incorporates certain subtle errors, but can anyone refute its conclusion in this same idiom? Apologies if this has been proposed before, especially if I'm wrong in my conclusion. ;-) Many thanks for reading, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 07:52, 1 April 2011 (UTC)
:[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] your argument above shows exactly why the Monty Hall problem is so famous. Marylyn vos Savant got the answer exactly right, you should switch and you double your chances of winning if you do so. This is not in dispute. Please have a read through the article and see if anything there convinces you otherwise. Do you see anything wrong with any of the solutions given? If not, why are you not convinced by them? We would all be very interetsed to see how this article looks to a new reader. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 09:21, 1 April 2011 (UTC)
::Thanks for replying, [[User:Martin Hogbin|Martin]]. I didn't know if anyone would. I did read through the article, of course. But I thought it would be quicker to express this positive argument than to try to make a negative one, i.e. than to use the idiom of the current article's examples and show why I suppose they're in error. I'll give that a try, though, and would be pleased to hear what you think about that effort. I'll also post, of course, if I persuade myself that the solution examples presented there ''are'' correct.
::Likewise, if you feel inclined, I'd be pleased to know where you think I've gone wrong in what I presented above. No need to do so before I present my critique of existing solution examples, of course. But I wonder, for example, whether there would be general agreement with the proposition: "The first 'choice' ( i.e., before the contestant is offered the chance to 'switch' ) has no impact on the final outcome. Only the stay/switch choice is relevant." Best, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 13:37, 1 April 2011 (UTC)
:::I would be happy to explain where you have gone wrong but this should probably be on my or your talk page as these pages are intended for discussions on how to improve the article. If you do not find the solutions in the article convincing then in my opinion the article need improvement. You will see from the article that most people get the answer wrong but one thing you can be sure about is that you have.
:::Have a look at the section beginning, 'Another way to understand the solution is to consider the two original unchosen doors together' and tell me why you do not agree with it. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 20:47, 1 April 2011 (UTC)
::::'''What a delightful problem!''' I read quickly through the article a couple of days ago, and was unconvinced. Then, last night, without re-reading it, but after rapidly scanning through [http://www.marilynvossavant.com/articles/gameshow.html Marilyn Savant's presentation] and remaining unconvinced, I posted the above. I was really ''very'' sleepy, though, both when I initially read the article and again, last night, when I scanned Savant's explanation and posted the above. ( I find Savant's self-promotion in very poor taste, and I suppose I'm quite prejudiced against her ideas. ) But when I awoke this morning it was the first thing on my mind, and I "grokked" it within the next couple of minutes. But how to express that "grok" concisely? Are you interested in hearing about my mental process to get there? I assume you might be, since you said you were curious about how the article would look to a new reader.
::::When I awoke, I found I did so with an image of a "shell game" present to mind, but one modified to include a gazillion shells with still just one pea. Obviously, I had combined one of Savant's suggestions with the presentation I vaguely recall from the article ( I'll look at both again when I'm through writing this ) having something to do with a million doors. I imagined Monty sweeping all the other shells but two right off the table, one of those two being the one I'd previously chosen, of course. Then it "hit" me: I'd been ignoring the information value of Monty's removal of ( a gazillion minus 2 ) shells.
::::I'd been preoccupied with my own subjective knowledge as a contestant, and missed the fact that Monty, transferred from his studio to the setting of my shell game, wasn't making his removal of all but one of the remaining shells ''at random'', that his doing so was informed by intelligence I didn't have access to. '''''I realized that, in effect, by sweeping all the <u>other</u> shells off the table, <u>Monty, also, was making a choice as to which shell the pea was under</u>''''', being constrained only by the (almost) negligible restriction that he couldn't choose the one I'd already chosen. Like me, he'd manifested a choice as to which shell the pea was under; he just had better information to inform his choice than I'd had.
::::I'm not certain how to formalize that insight without so-encumbering it with explanation that the "intuitive flash of recognition" it presents gets too submerged in notation or words or tables to be easily grasped. Decision tables don't seem to do it for me; it's one thing to grasp an idea, but another thing entirely to prove or demonstrate it, of course, and it's not often the case that the proof communicates the grasp, communicates that initial "intuitive flash of recognition" very economically, I believe...
::::I suppose one could explain that in the "gazillion shells" scenario, if ''my'' probability of making the correct choice was one-over-a-gazillion, that ''Monty's'' probability of doing so would equal 1 minus one-over-a-gazillion, i.e. that his probability wouldn't equal 1 only because he couldn't select from among ''all'' the gazillion to make his choice. But I'm not sure this would leave the typical reader very much enlightened, even if he were to accept the truth that the same formula must apply as the number of initial selections is (gradually?) reduced in successive example trials from a gazillion to just three. But it seems a promising avenue to me; is that in the article already? I'll be interested to see if it is, in that form. It wouldn't surprise me at all if I were recollecting it from there and assuming I'd come up with it originally.
::::I'll be curious to see how re-reading the article, and reading Gerhard's post, below, will affect my view of the problem. I purposely haven't done either yet, because I wanted to be able to express how I came to this with my presentation of it uninfluenced by doing so. Just in closing, I'll mention that one of the things I love about the way I came to see this, to the extent I presently do, is that it depends on the "reformulating the problem according to an extreme case" approach. Using a "gazillion" shells or doors, in other words, demonstrates Polya's statement to the effect that if you can't solve a problem, that there's also a simpler one (or an extreme case, as I'd put it, in this instance) that you can't solve, and that you should find that simpler problem, solve it, and use that knowledge to move on to the more puzzling one.
::::I see I no longer have time just now to read Gerhard's post with the attention it deserves, or to re-read the article carefully, either, but I'll do so soon, and will reply further. I expect the small degree of vague unease I still feel over the "three door" or "three shell" reductionist case, as opposed to the "gazillion" case, will be dispelled by the understanding I expect I'll gain in doing so. Thanks so much; this has been loads of fun so far. It really is a delightful and fecund problem! Thanks, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 04:55, 2 April 2011 (UTC)
:::::Ohiostandard, I am interested in how you came to understand the correct solution as I think i will help improve the article. The way in which you understood the answer also explains another aspect of the problem that some people find troubling, which is that it matters whether Monty knows where the car is and does not just pick an unchosen door at random which happens to reveal a goat. I find your description of the transfer of information from Monty to you an interesting way of explaining why this is so.
:::::If you read the explanations in the article you will probably find them all convincing now you have grokked it, that is why I was so interested in your thoughts earlier as I want to find out how effective this article is for newcomers and how it might be made more so. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 08:47, 2 April 2011 (UTC)
::::::Thanks again, Martin, very much, for being patient with ''yet another'' newcomer who initially thought he knew better. :-) I can entirely understand why this article has resulted in such a shouting match; a person would need to have the patience of a saint to keep trying to explain, over and over again, for years on end.
::::::I have now re-read the article, and Gerhard's and Rick's very kind explanations, as well. I'm ''very'' grateful for the work everyone has put into the article, and especially for Gerhard's and Rick's generosity, and for yours as well. But without the least possible slight toward any preceding explanation, I have to admit that none of those really do it for me. They're fine as proofs or demonstrations, but for me ... well, they don't really nudge me along toward that "flash of insight" (aka "grok") that seems so necessary in this case.
::::::I guess my major criticism of the article in its current state is that it focuses too little on helping mathematically unsophisticated readers get the "intuitive flash" they need. As I'm sure you're all aware, many, many mathematicians have written about the disparity between the insight that motivates a proof and the proof itself, and have observed that the proof isn't always or even usually the most efficient means of communicating that "flash".
::::::I love well expressed proofs dearly, of course, and I understand why it's absolutely crucial that they be painstakingly exact in the ideas they present and in the language and symbols they use to present them. It's just that the insight that motivated them in the first place is often times submerged, especially for the non-mathematician, in the necessary rigor.
::::::With a view toward trying to help address that, I've posted to the article's main talk page with a suggested re-write for the "increasing the number of doors" approach, and would be very pleased to learn what you all think of that. Thanks again, to everyone here, for your generosity and patience, and refusal to bite! Cheers, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 21:29, 2 April 2011 (UTC)
'''@Ohiostandard:''' Will that [http://en.citizendium.org/wiki/Monty_Hall_problem#Explicit_computations table] in your opinion help to show that the contestant will win by ''staying'' in 1/3 of cases, but will win by ''switching'' in 2/3 of cases, <br />no matter at all which door she should have chosen (1, 2 or 3), and no matter which ''other'' door whatsoever the host has opened, showing a goat:
{| class="wikitable" style="margin:auto; text-align: center;"
|-
! Supposed door<big> A</big> was chosen by contestant: <br /> ||colspan="4"|<br />"Another" door was opened by the host, showing a goat:||colspan="2"|<br /><br /><br /> contestant's final decision:
|-
! Initial arrangement behind doors A, B, C<br/>(probability)|| Open Door A <br/>(probability)|| Open Door B <br/> (probability)|| Open Door C <br/>(probability)||Joint<br/> probability ||Win by<br/> staying ||Win by<br/>switching
|- align="center"
|rowspan="2" |<big><b>Car</b></big> Goat Goat (1/3)||No||<b>Yes<br/>(1/2</b>)||No||1/3 x 1/2||<b>Yes<br>(1/6</b>)||No
|- align="center"
|No||No||<b>Yes<br/>(1/2</b>)||1/3 x 1/2||<b>Yes<br>(1/6</b>)||No
|- align="center"
|Goat <big><b>Car</b></big> Goat (1/3)||No||No||<b>Yes<br/>(1</b>)||1/3 x 1||No||<br /><b>Yes<br />(1/3)</b><br />
|- align="center"
|Goat Goat<big><b> Car</b></big> (1/3)||No||<b>Yes<br/>(1</b>)||No||1/3 x 1||No||<br /><b>Yes<br />(1/3)</b><br />
|- align="center"
|colspan="5"|<b>Note that the host has no choice when the contestant in 2/3 of all cases fails and <br />chooses a goat, as in these cases the host is just bound to show the second goat.||<b>Win by<br/>staying<br />1/3||<b>Win by<br/>switching<br />2/3
|}
And note: If, in only 1/3 of cases, the contestant luckily should have chosen the car, only then the host has a choice indeed and can open either of his two doors, both containing goats.
Some sources now said that the host eventually could be ''"biased"'' to some extent to open just only ''one special "preferred" door if ever possible,'' avoiding to open the other door, avoiding his ''"unwanted"'' door.
But for the decision asked for, to stay or to switch, the "door number first chosen" by the contestant and "the door number opened by the host" is of no relevance at all, as you do not exactly know about such "bias" and about its "direction", and even if you exactly knew about such bias, its extent and its direction, it would never "help better" to make the right decision. Let us assume even an extremely biased host:
If, in 1/3 of cases, the host got both goats, then he can and will open his preferred door, and switching will loose. Probability to win by switching: Zero.<br />And if, in another 1/3 of cases, he got the car and one goat, and the ''goat'' being behind the door he usually prefers to open, then he can and will also open his preferred door, but in this case switching will win, probability to win by switching: 1.
So, whenever in these 2/3 of cases an extremely biased host opens his preferred door, probability to win by switching will be at least 1/2. And that means staying never can be any better than to switch, so you should switch.
If however, in the last 1/3 of cases, he got one goat and the car, but the ''car'' being behind his preferred door, then he exceptionally will be opening his "avoided" door and, by doing so, be showing that the car is very likely to be behind his preferred but now still closed door, and switching is very likely to win the car, and switching is imperative.
You never can nor will ''know'' any ''"assumed closer is better" -probabilities'' for any "actual game". Conditioning on "just assumptions". Unneeded diligence. Results of "conditional probability theorems" will forever remain within the strict bounds of at least 1/2, but never less, to 1. Irrevocably. You know that already from the outset. So conditional probability calculus is quite irrelevant for the question asked for, even if one should exactly know about ''"host's bias, its extent and its direction",'' as staying never can be any better than always to switch, and you know for sure that switching will win the car in exactly 2/3 of cases. Full stop. Irreversible. No further "proof by conditional probability calculus" needed.
Will that table help the readers to see that switching will double the chance to win from 1/3 to 2/3? Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 18:16, 1 April 2011 (UTC)
:Gerhard, I'm extremely grateful to you for taking the time to respond so carefully here: It's very generous in you. You may notice the response I top-posted above, after Martin's reply to me. I hope you won't be offended that I (still) haven't read your own reply; I certainly will do so, but it would be disrespectful not to do so with the care it merits, and I have to fly out the door just now. I'll read through it carefully soon, though, and will reply. Again, many thanks for taking the time to post this. Best regards, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 04:55, 2 April 2011 (UTC)
::Gerhard, I've read carefully through your post, now, and would like to again thank you for it. As I wrote above, though, I'm afraid I have to admit that I'm too dense for tables to be of much help to me. I certainly admit they can ''demonstrate'' a solution but, for me, they're not much assistance in understanding why that solution is as it is; they're not much help for me in trying to ''understand'' the problem, that is. There's just too much for me to look at, too much to take in ... I think it was Poincare' who wrote that you don't really understand a proof until you see it as a single idea. I don't know if I'd go quite so far as to say that for all problems, but I think it does apply to this one. Sorry to disappoint; I'm sure there are others who would read through your presentation above and at some point in their reading get that "Aha!" moment of understanding. I'm just not one of those, I'm afraid. Thanks very much for your presentation, however; as I said before, I'm very sensible of your generosity in making it. As I mentioned to Marin, up above, I've posted a section to the article's main talk page about a possible re-write based on the "increasing the number of doors" idea. I'd be pleased to know whether you think it has any value. Best regards, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 21:29, 2 April 2011 (UTC)
:::But I am interested to hear from you whether it is plausible that the above table clearly shows that the probability to ''win by staying "1/6 + 1/6 = 1/3"'' is only one half of the probability to win by switching of "1/3 + 1/3 = 2/3". Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 23:34, 2 April 2011 (UTC)
::::Maybe it does show that, Gerhard, but I'm afraid I can't say that it does so ''clearly'', for me. Perhaps it would do so for someone who had taken a probability class recently, though, and who was thus already familiar with the interpretation of such tables. For my part, I spent 30 minutes before I understood what the "(1/2)" in the top cell of the third column was intended to mean, and then only did so by reading the text of [http://en.citizendium.org/wiki/Monty_Hall_problem#Explicit_computations the Citizendum page] that includes basically the same table. And without previous acquaintance with Bayes' rule ( "posterior odds equals prior odds times likelihood ratio", as Citizendum expresses it ), the "Joint probability" column wouldn't make sense to the uninitiated, and the rest of the table would be equally opaque, as well, I believe. To speak very candidly, I think the table would be very likely to confuse most readers, even assuming they'd take the time to try to understand what it's intended to communicate. Sorry to disappoint; perhaps other readers would find it easier to interpret than I do. Best, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 02:20, 3 April 2011 (UTC)
@OhioStandard - Responding to your original request (in the same idiom as your original argument), if we follow your argument through it's fairly simple to point out where it goes wrong.
* At the outset of the problem, before we've made any choice at all, consider that the car is behind one of two doors. We don't know which, and the chances that it is behind one is the same as the chances it is behind the other.
::Perfectly true. It's even true that if you pick ANY two doors at this point the chance that the car is behind either one is the same. However, the only way to make this true is to consider the probability that the car is behind any of the ''three'' doors is the same, i.e. p(door 1) = p(door 2) = p(door 3), and they must add up to 1, so they must all be 1/3. If you pick any two, the chances are even at 1/3:1/3 which only adds up to 2/3. This reflects the uncertainty that you've picked the right two. The chances that the car is behind a randomly selected two doors at this point is only 2/3.
* Next, just to humor Monty, we "pick" one of the three available doors, and we try to keep his spirits up by pretending that our "choice" at this point matters. But we know it ''doesn't'' matter, because the outcome of the ensuing sequence will be the same regardless of which of the three doors we select at this point. After our apparent "choice", nothing relevant to the outcome has changed. The car is still behind one of two doors, and we still don't know which two those are.
::Also perfectly true. The key is we still don't know how to pick two doors guaranteeing the car is behind one of those two. So the probability of each of the three doors is still 1/3. The probability of any two (at this point) is 2/3.
* Then Monty opens one of the unchosen doors, and now something ''has'' changed: <u>We now know "which two"</u>.
::Yes, now something has changed. But it's more than you think. The probability that the car was behind the two remaining doors that used to be 2/3 is now 1, and the probability the car is behind the open door that used to be 1/3 is now 0. How this is possible is that we're now not talking about the "original" probabilities - but something different. Specifically, we're now talking about the [[conditional probabilities]] in the case that the host has opened (say) door 3. This doesn't mean the "original" chances that the car was behind the two remaining doors was 1 or that the original chances that the car was behind the door that has been opened was 0. The original chances were (and still are!) 1/3 for each door, 2/3 for any two doors, and 1 for all three doors. If you pick (say) door 1, when the host opens a door these original chances are split into two cases, one where the host opens door 2 and the other where the host opens door 3. The probabilities in each of these cases are conditional probabilities. The original probabilities of door 1 and door 2 were indeed equal, at 1/3 each. And now, if we're in the case where the host opened door 3, we know its (conditional) probability is 0. We also know the conditional probabilities of the two remaining doors will add up to 1 (just as they do in the case where the host opens door 2). Where you're going wrong is thinking that because the original probabilities of these doors was equal, then the conditional probabilities of these doors must be equal as well. If the car is behind door 2 the host MUST open door 3. However, if the car is behind door 1 the host chooses whether to open door 2 or door 3. Assuming this is a fair choice, this means the host opens door 3 only half the time the car is behind door 1. Half the time behind door 1 and all the time the car is behind door 2 means you have a 2 to 1 advantage if you switch.
::If you pick door 1 the original 1/3 + 1/3 + 1/3 = 1 splits (unevenly) into
::1/6 + 1/3 + 0 = 1/2 (if the host opens door 3)
::and
::1/6 + 0 + 1/3 = 1/2 (if the host opens door 2)
::Expressing these as conditional probabilities, you get 1/3 + 2/3 + 0 (considering only the case where the host opens door 3), or 1/3 + 0 + 2/3 (considering only the case where the host opens door 2).
::The essential confusion is the difference between the original ''unconditional'' probabilities and the ''conditional'' probabilities in effect after the host opens one of the doors. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 05:40, 2 April 2011 (UTC)
:::Thanks so much, Rick, for your generosity in replying like this, "in my idiom", so to speak. This is a really clear explanation of where I went wrong, and it was great fun to read and understand. I agreed with everything you'd written up to this point, but when I read, "Where you're going wrong is thinking that because the original probabilities of these doors was equal, then the conditional probabilities of these doors must be equal as well" I thought to myself, "that's a beautiful and concise explanation!" Before coming back to this I'd read James & James' definition/entry for "Probability", since it's been so long since I've thought about the subject ( and was never any good at computational or applied math, anyway ) and that refreshed my memory re the distinction between mathematical or (same) ''a priori'' probability, and conditional probability, an important distinction, of course, and one that the wikilink you provided points out, as well.
:::Again, I appreciate your kindness and effort in presenting this explanation, Rick, very much. You'll see that I've also responded at some length to Martin, above (sorry for the top post), although my reply there really is meant for you and for Gerhard, as well. As I said (as if) to Martin, I'd also be pleased if you'd care to look at the "Increasing the number of doors rewrite?" section I added to the article's main talk page, and see whether you think it has any value. Perhaps it just represents the way my own peculiar brain came to the "flash" or "grok" or "Aha! moment" of seeing the solution; I'd be pleased to hear objective opinions about whether it might be more universally helpful. Best regards, – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 21:29, 2 April 2011 (UTC)
Btw, since this is so long a thread, I'll have no objection if regulars here want to collapse it, or parts of it, or just move it manually to archives at some point when everyone has been able to reply. I'll leave that up to all y'all. Thanks again, everyone, for your very generous comments so far! – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 21:40, 2 April 2011 (UTC)
===A different table===
OhioStandard - can you comment on this table showing how many times we'd expect the car to be behind each door in a sample of 300 games, and the outcomes when switching (assuming the player picked door 1 initially)?
{| class="wikitable" style="margin:auto; text-align: center;"
|-
! colspan=4 | Situation BEFORE the host opens a door
!colspan=4 | Situation AFTER the host opens a door
|-
! Door 1 || Door 2 || Door 3 || total cases
!colspan=2 | host opens Door 2
!colspan=2 | host opens Door 3
|-
! || || || || cases || result if switching || cases || result if switching
|-
| Car || Goat || Goat || 100 || 50 || '''Goat''' || 50 || '''Goat'''
|-
| Goat || Car || Goat || 100 || 0 || N/A || 100 || '''Car'''
|-
| Goat || Goat || Car || 100 || 100 || '''Car''' || 0 || N/A
|}
The point is to show that by opening either door the host splits the original 300 cases into two subsets. The (conditional) probability of winning by switching (if the host has opened, say, door 3) is evaluated in the context of one of these subsets, not in the context of the entire original sample of 300 shows. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 16:17, 3 April 2011 (UTC)
:Thanks, Rick. I sincerely hope it won't offend anyone if I say so, but this is the first table I've seen that's simple-enough and clear-enough that I could understand it quickly, and see how it demonstrates the 2/3 versus 1/3 advantage re switching. I'm getting the impression that most regular contributors here have their favorite way of illustrating that, but I have to say that I think this is a really good one. It's much (!!!) easier for me to understand than either of the tables currently in the article, or than the decision tree either, for that matter.
:I know that someone must have put a great deal of work into the graphics for the illustrated table especially, and I certainly don't intend to disparage that effort: Quite the contrary; I honor it. If I'm to answer candidly, though, I'm afraid I have to say that while I imagine the decision tree has value to the uninitiated, I can't support the same statement for the two tables presently in use in the article, and I'd prefer to see the article simplified by using the one above, instead.
:I've top-posted this, above Guymacon's flush-left/outdented comment below, btw (sorry, Guy) because his outdenting made it impossible for me to otherwise indicate which post I was responding to. Thanks Rick, for this table. I think it's extremely helpful, and that it provides the strongest assist I've seen so far to help the average reader who has no training in probability or statistics understand the solution. – <font face="Cambria">[[User:Ohiostandard|<font color="teal">'''OhioStandard'''</font>]] ([[User talk:Ohiostandard|talk]])</font> 15:11, 4 April 2011 (UTC)
::Thanks for correcting the outdent mistake. It really is a quite good table. [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 22:59, 4 April 2011 (UTC)
<small>''( ← outdenting )''</small> Let me ask a question here (there is no ulterior motive or implied criticism: I really do not know). Are we as a group slipping back into the sort of [[Wikipedia:Truth]] discussion that led to stalemates and frustration in the past, or are we moving forward with the sort of [[Wikipedia:Verifiability, not truth]] discussion that leads to [[Wikipedia:Consensus]]? [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 14:10, 4 April 2011 (UTC)
:This is the "Arguments" page. IMO, here (but not on the main talk page) discussions are explicitly permitted to drift more toward "truth" - the topic is mathematical arguments concerning the MHP as opposed to content of the article. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 14:25, 4 April 2011 (UTC)
::Ah. Of course. (note to self: next time smoke crack ''after'' posting to Wikipedia...) [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]])
== Talk about notation ==
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)
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