![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 5 | ← | Archive 7 | Archive 8 | Archive 9 | Archive 10 | Archive 11 | → | Archive 15 |
When Monty opens a door, he doesn't tell us anything we didn't already need to know. He ALWAYS shows a goat. It makes no difference to this puzzle WHICH remaining door he shows. So it starts out as 1/3 for your door + 2/3 for the remaining doors = 100%. Then he shows a door, but we knew in advance that he was going to show a goat. The odds simply haven't changed following his action. They remain 1/3 for your door + 2/3 for the remaining doors (of which there is now just 1).
That's why on the show, Monty didn't HAVE to show a door (unlike this puzzle), and he could bribe you with cash. Otherwise, there's no show. You always switch.
I'm new to this. Is there a way to get this 'solution' onto the article page?
75.185.188.104 (
talk)
17:09, 25 October 2008 (UTC)
I agree that the presentation of this article leaves something to be desired. I would think that most users come to this page to settle an arguement, or to enlighten themselves on how they don't 'get' the solution. I know that's why I came here.
So, an elegant, simple solution would serve the readers. I'm a stats and probability geek, but I don't like having to 'prove' something by running simulations (since I don't know who created the simulation, I can't trust it). And, if there is an easier to understand explanation that doesn't require umpteen diagrams, the typical reader would prefer that, as s/he will probably try to explain what s/he read here to other people. 75.185.188.104 ( talk) 12:27, 26 October 2008 (UTC)
Re: The 'Combining Doors' comment above. Yes, it's similar. imho, these should be a prominant part of the 'Solution', not relegated to 'Aids to Understanding'. Glkanter ( talk) 15:12, 26 October 2008 (UTC)
I find the explanation in the combining doors section somewhat abstract. Here's the explanation I've been using with people who have a hard time understanding why switching is the best strategy -- I know this probably shouldn't go on the main page (I'm the only reference I know of) but I wanted to share it.
I'm going to write it in the form of a dialog because that is the shortest form of this explanation and also the form in which it has been used:
Am I correct that this would be inappropriate for the main page? Stepheneb ( talk) 17:08, 2 January 2009 (UTC)
I follow the reasoning all the way, yet the following twist to the scenario puzzles me. Can someone explain what I'm missing?
As we know from the article, the first contestant has a 2/3 probability of finding the car. But for the second contestant it's a straight choice between 2 doors, so he has a 1/2 chance of finding the car. So it appears that at the moment the two contestants make their final choices, p(finding the car | choosing door 2) has two different values simultaneously depending on who does the choosing, whereas intuition (mine, anyway) says that the probability should have a unique value, and that it should depend only on the possible permutations of the original distribution of car and goats, and the shared knowledge that door 3 revealed a goat. Help! -- Timberframe ( talk) 09:25, 26 October 2008 (UTC)
Mr. Hogbin is correct. Note that the 2nd player has less likelyhood of picking correctly. 50% vs 67%. This is because he has less knowledge of how the 2 doors came to be. He can only assume randomness, but the first player knows Monty did not act randomly when he opened the door revealing a goat. 75.185.188.104 ( talk) 12:17, 26 October 2008 (UTC)
Thanks guys, got it now. The falacy in my proposition is in the assertion that because there are two closed doors and the car is behind one of them, the second contestant has a 50:50 chance of finding it. This is only true from the second contestant's limited perspective. The conditions of the scenario is posited have in reality already shifted the odds to 2:1. Thanks again -- Timberframe ( talk) 19:29, 26 October 2008 (UTC)
The following is prominant in the article:
"Solution The overall probability of winning by switching is determined by the location of the car...
This result has been verified experimentally using computer and other simulation techniques (see Simulation below)."
This first sentance doesn't mean anything, and in fact is incorrect. The probability is NOT affected by the location of the car. One instance of playing the game is affected, but not the probability.
Further, a valid proof, which I believe my entry is, does not require being "verified experimentally using computer and other simulation techniques". A valid proof stands on it's own.
And lastly, this at the top of the Discussion Page is at a minimum, off-putting:
"Please note: The conclusions of this article have been confirmed by experiment
There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment."
As I stated above, valid proofs do not rely on 'experiments'. Yet, there are two references to 'experiments'. Me thinks the lady doth protest too much.
75.185.188.104 (
talk)
13:34, 26 October 2008 (UTC)
Can I mention again that I have set up this page: user:Martin Hogbin/Monty Hall problem (draft) for those, like myself, who do not find the current solution clear or convincing to develop a better basic solution. The concept is to develop a very clear solution that is verifiable and which can be eventually added to this article. There is also a talk page for discussion on how we might achieve this. Martin Hogbin ( talk) 16:58, 26 October 2008 (UTC)
"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)."
I have no idea what this means, and fail to see how it part of a solution to the problem. People come to this page for an explanation. The above paragraph, plus whatever follows it, are not part of the solution to the Monty Hall Problem, and, imho should not be present there. Glkanter ( talk) 14:59, 26 October 2008 (UTC)
(outindent) The question is if it's advantageous to switch after we've picked a door and the host has opened one of the remaining doors. Your solution is the following:
I've highlighted the problematic parts. These are assertions, not deductions based on what is given in the problem statement. They're true statements, but stating them without justification is equivalent to assuming the solution. We're not given that it makes no difference which door the host opens and we're not given that when the host opens one of the two remaining doors the initial 1/3 odds don't change - this is essentially what we're asked to show. It's like in a geometry proof using that an angle in the diagram is a 90 degree angle because it looks like one without showing why it must be one. If indeed it must be one because of what else is given it's certainly true, but using it without justification is at best incomplete. The simple solution currently in the article (3 equally likely scenarios, switching wins in 2 and loses in 1) doesn't have this problem. Do you find this solution hard to understand? -- Rick Block ( talk) 19:21, 26 October 2008 (UTC)
Several editors believe that, although the current article may meet the FA criteria, it still remains unconvincing and also concentrates far to much on variations of the basic problem that are irrelevant to the core problem.
It is therefore proposed that the current solution be replaced with a new version giving simple, intuitive, and verifiable solutions to the basic problem.
Preparation for this new section is taking place on a [ [1]] . All interested editors are asked to make their comments on the suitability of this section for inclusion in the current article and to make improvements to the new section. Martin Hogbin ( talk) 20:54, 6 November 2008 (UTC)
Perhaps you might be able to improve the development article. Martin Hogbin ( talk) 09:26, 7 November 2008 (UTC)
I got it! Start a new Wikipedia article called 'The Equal Goat Door Constraint Paradox'! Marilyn vos Savant can write a column about it in her general interest Sunday magazine column. People all over the world will marvel at how counter-intuitive the solution is, when all along, after a trip to the nearest university library, Morgan, et al and Bayesian statistics were there to explain it. Glkanter ( talk) 16:17, 8 November 2008 (UTC) Glkanter ( talk) 16:26, 8 November 2008 (UTC)
The following introduction is still confused suffering from poor English, a lack of clarity and some misconceptions.
"Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3. Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3. With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty. The total probability of winning when switching is thus 2/3."
The object of the problem is not whether the person wins the automobile but how the probability shifts when a door is exposed (with a goat). If the person selects the correct door or not on the first guess is irrelevant to the problem - especially as an introduction.
A breakdown:
(1) Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter.
Is fine
(2) In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.
Does not need "in the usual interpretation of the problem" in the introdcution.
(3) Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3.
Should read, "If the player initially chooses the winning door then changing is clearly disadvantageous" I don't even think it is necessary as this is confusing in the introduction.
(4) With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty.
Is badly constructed. It should read, "If the player has initially chosen one of the two losing doors: when one of the losing doors is revealed then switching wins the player the automobile."
However, I would argue that this isn't even needed anyway.
(5) The total probability of winning when switching is thus 2/3.
Is a non-sequitur as the paragraph is referring to two different scenarios (ref to 2 and 3). Both scenarios, incorrect first choice and correct first choice cannot lead to the same probability of winning when switching. -- Candy ( talk) 22:52, 24 November 2008 (UTC)
Reversion of good faith edits should not be marked as minor. Martin Hogbin ( talk) 20:27, 11 December 2008 (UTC)
I've moved the following link which was added to the article here. There's nothing on this page that can't be incorporated directly into the Wikipedia page, per WP:EL that means it is not an appropriate external link.
-- Rick Block ( talk) 14:39, 19 December 2008 (UTC)
--- I believe the above link should be on the main page for the Monty Hall Problem. It cites important academic journals and articles on the issue. We cannot ignore literature that discuss this problem as it was a hot topic of discussion amongst prominent economists and mathematicians. Some of the articles mentioned in the above link do form a basis to the solution that is widely accepted today. - Aadn
What if I want the goat and not the car? Should this probability also be addressed, even if it's not part of the original Monty Hall problem? 76.95.124.146 ( talk) 15:09, 8 January 2009 (UTC)
I've undone this edit which adds an introductory paragraph to the Solution section with numerous WP:MOS issues discussing an alternate problem (boxes rather than doors). The addition is essentially the same as the existing "combining doors" section. The text is in the history if anyone wants to talk about it. -- Rick Block ( talk) 06:01, 24 January 2009 (UTC)
I've deleted the addition of the following new section.
Although interesting, it's unreferenced and sounds like WP:OR (and requires an understanding of Minesweeper). If anyone can find a reliable source including this analogy we can talk about re-adding it. -- Rick Block ( talk) 01:43, 25 January 2009 (UTC)
The, what is often called, "simple solution", saying: the player originally has a chance 1/3 of winning, so switching increases this chance to 2/3, is theoretically wrong! This explanation is also offered directly under "Solution" , and although it offers numerically the right answer,it is not correct, as may be seen by comparing it to the statement of the problem right above it. In the problem statement the question is: Should the player switch to Door 2? And in the solution the player is allowed to switch to Door 3 as well. The point is, and I think it is mentioned somewhere in the above discussions, that the posed question can only ask for a conditional probability, given the situation, i.e. Door 1 chosen and Door 3 opened revealing a goat. This is also clear from the equivalence with the Three Prisoners problem and Bertrand's box paradox. Nijdam ( talk) 11:26, 3 February 2009 (UTC)
This is an other way of putting it. But as the article is so precise in defining the problem, with all guarantees of no possible misinterpretation, the given "simple solution" is no solution to this precise stated problem. Nijdam ( talk) 16:04, 3 February 2009 (UTC) —Preceding unsigned comment added by 82.75.67.221 ( talk)
Any of you recent contributors may not have read this page in its entirety. It's pretty long. So, just to let you know, many, many people are very unsatisfied with the entire article. It's long winded, confusing, and outright wrong in places. But we respect the process, so here we are. Search for my signature, as one example. Glkanter ( talk) 03:30, 4 February 2009 (UTC)
I tried to follow the discussion above between the two of you, and I actually got confused, not about the problem or the solution, but about you. I'm not very happy with Rick's revision of my extension to the "simple solution". My extension shows in essence the flaw in this way of reasoning. I showed that the "simple solution" is not a solution to the stated problem. It's like the man, searching in the evening for his lost wallet under a streetlamp where things are well visible, allthough he knows he lost the wallet in a dark side alley. So I invite either one of you to read my extension an comment on it. Nijdam ( talk) 10:49, 4 February 2009 (UTC)
I did a little more reading of the above discussion about this topic and I think I find Rick on my side, but Glkanter opposed because of the reasoning: 1/3 + 2/3 = 1, meaning: the initial probablility is 1/3 of picking the car, so whatever happens, as (!?????) this has not changed, the final probability of the alternative should be 2/3. Well as you may guess, I already indicate where it goes wrong. The "initial probability changes" under the assumed condition. Among all possible cases, the ones with the car behind Door 1 and Door 3 openend, only form 1/6th. Nijdam ( talk) 11:28, 4 February 2009 (UTC)
Nijdam, please show me the error in the 1/3 + 2/3 = 1 (100%) solution. I'd also like to know what is unacceptable about Martin's solution, 2/3 of the time I choose a goat, and 100% of those times Monty show me the other goat? Everything else, from Morgan, et al, to Bayesian, to "We have run simulations that prove this" is wasted binary digits.
Glkanter (
talk)
23:12, 4 February 2009 (UTC)
In a moderating kind of way, I have two suggestions here. 1) Please comment on content, not on other editors' mental facilities. WP:COOL has some good suggestions. 2) As far as possible, please support arguments with references rather than arguing what you simply know to be the Truth. There are no shortage of references about the Monty Hall problem - any significant point has almost certainly been published (at least once). -- Rick Block ( talk) 02:00, 5 February 2009 (UTC)
Rick, I'm still waiting for your clarification of the above statement. As it is central to your claim of the non-validity of my proof. What does this mean:
Are they talking about the Monty Hall Problem or some varient? Or is this the 'before switching' beast you speak of? MOST IMPORTANTLY, what possible relavence can it have to my proof?
Glkanter ( talk) 07:30, 8 February 2009 (UTC)
Well, don't be offended; take it as a complement instead, because from the discussions above I concluded both of you are well skilled at least in mathematics. Yet you keep asking for arguments, which I gave in the extension of the article and extensively above. Moreover, Rick quotes the same arguments I gave. So what more needs to be said? If the "simple solution" as a solution to the well stated problem, is supported by any source, the conclusion can only be, this source is not reliable. The only necessity is changing he article, as I suggested. 82.75.67.221 ( talk) 10:01, 5 February 2009 (UTC)
Well, I find it unacceptable to start with a "solution" that isn't a solution. My suggestion (please improve my English, as I'm foreign):
According to the problem statement above, a car and two goats are arranged behind three doors and then the player initially picks a door. Assuming the player's initial pick is Door 1 (the same analysis applies for any other door the player picks):
This leads to the following situations:
Players who choose to switch win if the car is behind Door 2. This is the case in 2 out of 3 cases, so with 2/3 probability.
This means if a large number of players, after choosing Door 1 and being showed a goat behind Door 3, 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).
Nijdam ( talk) 17:56, 5 February 2009 (UTC)
According to the problem statement above, a car and two goats are arranged behind three doors and then the player initially picks a door. Assuming the player's initial pick is Door 1 (the same analysis applies for any other door the player picks):
This leads to the following situations:
Looking at the figure above, in cases where the host opens Door 3 (which is the case according to the problem statement) switching loses in a 1/6 case where the player initially picked the car and wins in a 1/3 case where the player initially picked a goat. Players who choose to switch win twice as often as players who stick with the original choice, which is a 2/3 chance of winning by switching.
This means if a large number of players randomly choose whether to stay or switch after initially choosing Door 1 and being shown a goat behind Door 3, then approximately 1/3 of those choosing to stay and 2/3 of those choosing to switch would win the car.
Sorry, among all the almost simultaneous contributions to this page, I didn't notice this part. I'm sorry too to say I prefer my suggestion. Especially because the exact formulation of the problem, as given in the article, states Door 1 is chosen and Door 3 opened. Nijdam ( talk) 12:23, 8 February 2009 (UTC)
Note: There's a summary below, see #Request for brief clarifications. -- Rick Block ( talk) 20:55, 8 February 2009 (UTC)
I'll be honest. I don't understand most of what you guys are talking about. It's a simple puzzle. Do you increase your chances of winning the car by switching? It's been years since I took a Logic course, or a Probability and Statistics course, so I can't cite chapter and verse. But the goal is to present a valid proof with as little complication as possible. So you take shortcuts, which don't affect the proof. Two goats? I only care about the car. Ignore the goats in the proof. 3 doors? There'll only be one 'other one' left when I'm asked to switch. Ignore the door #s in the proof. Conditional probability? Don't know what it means. Don't see how anything is conditional here. Two closed doors when I'm asked if I want to switch, exactly one has a car. No conditions exist.
I do remember something about a requirement for proofs being valid and true. 2/3 of the time the remaining door I didn't choose has a car. Both valid and true.
That's why the real game show didn't work like this. Glkanter ( talk) 03:16, 6 February 2009 (UTC)
Okay Rick, at least I see you completely knows the ins and outs. So what to do? I cannot accept a reasoning that's logically wrong. What about the suggestion I made above, by changing the given "simple solution" in such a way that some possibilities are ruled out as not happened? Nijdam ( talk) 11:54, 6 February 2009 (UTC)
There are really only two situations to consider. Not 3, not 6.
2/3 of the time I will choose a goat.
100% of those times Monty reveals the other goat, leaving the car.
=>2/3 of the time I should switch.
1/3 of the time, I will choose the car.
100% of those times Monty reveals one of the goats. It doesn't matter which.
=>1/3 of the time I should not switch.
2/3 is greater than 1/3.
Therefore, by switching, I increase my chances of getting the car when offered the switch.
I earlier said a good proof uses 'shortcuts'. A better phrasing would have been 'eliminates the unessential' from the proof. That is, a goat is a goat, and a door is a door.
Martin's proof, and the one above, address the problem as defined. To deny that is incorrect. [User:Glkanter|Glkanter]] ( talk) 13:34, 6 February 2009 (UTC) 63.97.108.114 ( talk) 14:19, 6 February 2009 (UTC) Glkanter ( talk) 14:22, 6 February 2009 (UTC)
Thank you for the specific response to my question. I'll try to look at those papers. I suppose Morgan, et al is one of those mathemeticians, who's the other?
Do I understand your response to Nijdam correctly? Does "many published sources (possibly all popular expositions of this problem) include only an unconditional solution" mean that the solution I've been advocating is what most other references provide as the only solution?
Would you indulge me? In your own words, tell me which of the eight statements, beginning with '2/3 of the time I will choose a goat' and ending with 'Therefore, by switching, I increase my chances of getting the car when offered the switch.' is either untrue, invalid, or not germain to the question 'Now that Monty has revealed a goat, do you increase your chances of winning a car by switching'?
Thank you. Glkanter ( talk) 07:06, 7 February 2009 (UTC)
I'll just address 3 & 4, my answer renders 1 & 2 moot.
Twice, my proof includes the words "100% of those times Monty reveals...(a) goat." I don't see how my proof can be of anything but the after-Monty-opens-a-door scenario.
Enough about before or after! A premise of the puzzle is that Monty will reveal a goat. There is no 'reveals car' scenario to consider! That would not be the Monty Hall Problem! So of my 8 line proof, 2 of those lines, for clarity purposes, repeat a premise. I'm surprised that this has come up again, we've already been around this block.
So, if that's all you got, let's declare this a valid proof and move forward. Especially eliminating that 'flawed solution' talk from the 'Solutions' section.
Glkanter ( talk) 00:39, 8 February 2009 (UTC)
Glkanter ( talk) 01:06, 8 February 2009 (UTC)
I deny the existance of different 'before' and 'after' questions. There is only one. It goes, "After Monty has revealed a goat, do you increase your chances of winning by switching?"
That is the only problem I have addressed since joining the discussion.
That is the only problem I have provided a proof for.
That is the only problem which I have asked you to find a flaw in my proofs.
And it is the only problem I will discuss on the discussion page of the Monty Hall Problem.
You're last gasp at finding a flaw was that I'm solving the 'before' puzzle. I pointed out the error in that statement, that twice the proof says "Monty reveals a goat". Which is a premise, for crying out loud!
So, whether or not there are two puzzles, I am solving the 'right' one. What's your critisism now?
Glkanter ( talk) 06:21, 8 February 2009 (UTC)
Your question is immaterial to the Monty Hall Problem we are solving. I have no idea or interest in it's clarity. It is obsfucation. You are avoiding the Monty Hall Problem question. I have presented a proof, then sought meaningful critique. I have explained how a proof is validated. Yet you repeatedly refuse to remain engaged in a constructive discussion of my proofs.
Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.
Glkanter ( talk) 06:43, 8 February 2009 (UTC)
I read every word you post. Believe me. Look, here's how it works. Someone puts forth a proof. Then everyone else looks for where there may be errors. Then when everyone is satisfied that it's correct, it is accepted. Once it's accepted, there is no saying about it: "Although the reasoning above is correct it doesn't answer the precise question posed by the problem, which is whether a player should switch after being shown a particular open door (Morgan et al. 1991)."
Step 1. Somebody puts forth a proof. I've done that.
Step 2. Everybody else looks for where there may be errors. In the proof. Nowhere else.
All you have ever done is say that I am not solving the appropriate question. To that, I again say, Hogwash! Each of those 8 lines passes. And the conclusion: 'since 2/3 is greater than 1/3, switching wins more often', answers the question being asked.
Enough, already!
Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.
Why don't we admit that neither of us is going to be persuaded by the other? I've recently requested that Martin take this to the next level of Conflict Resolution. I'm tired of the circular nature of the argument, but I think I'll stick around.
Glkanter (
talk)
18:35, 8 February 2009 (UTC)
MCDD is Morgan and three other guys. <Oh! The three other guys are the 'et al'!> Glkanter ( talk) 22:00, 8 February 2009 (UTC)
"In fairness, MCDD do moderate their tone later on, writing, "None of this diminishes the fact that vos Savant has shown excellent probabilistic judgment in arriving at the answer 2/3, where, to judge from the letters in her column, even member of our own profession failed.""
http://www.math.jmu.edu/~rosenhjd/ChapOne.pdf page 48
Who is John Gault?
Glkanter ( talk) 21:00, 8 February 2009 (UTC)
I'll do one effort to explain the problem. The situation is already conditional on the choice of Door 1 by the player. And we know Door 3 has been opened revealing a goat. Let us assume the game has been played 3000 times. What might have happened?
Player has chosen Door 1 | |||
---|---|---|---|
Car hidden behind Door 1 | Car hidden behind Door 2 | Car hidden behind Door 3 | |
1000 times | 1000 times | 1000 times | |
If nothing else is revealed | |||
Player wins the car if he doesn't switch | Player wins the car if he switches | ||
But player gets "information" | |||
Host opens Door 2 | Host opens Door 3 | Host must open Door 3 | Host must open Door 2 |
500 times | 500 times | 1000 times | 1000 times |
This case didn't happen | One of these cases happened | This case didn't happen | |
Player wins the car if he doesn't switch | Player wins the car if he switches |
Hence only in half the cases did the situation arise in which the player finds himself. So we see it cannot be unconditional, because only a part of all possibilities could have happened. Clear? Nijdam ( talk) 18:05, 6 February 2009 (UTC)
Rick, I thought that it had been agreed that, for the fully defined problem (host alway offers the switch and always opens a goat door), there is no distinction between the conditional and unconditional cases. Martin Hogbin ( talk) 22:56, 6 February 2009 (UTC)
Well, to be honest, I'm not. I strongly oppose starting with a reasoning that is not a solution to the stated problem. The right solution, as I suggested above, is quite accessible to the layman. If anything has to be said about the "simple solution", it should be as a solution to a (slgihtly, but essentially) different problem. It doesn't seem right to me, to have thoughts like "keep the laymen in the dark, it is no use to explain things properly to them".[Sorry, forgot to login] Nijdam ( talk) 12:59, 7 February 2009 (UTC)
Nick, in what way are the conditional and unconditional cases distinct in the case of the fully defined problem? Martin Hogbin ( talk) 14:44, 7 February 2009 (UTC)
Here's something to think about:
The presentator has a simple strategie: when possible he opens Door 3, otherwise he tries Door 2 and is that also impossible he opens Door 1. Quite easy. See what happens:
Yet the initial prob. of winning the car is in all cases 1/3 and according to the "simple reasoning" switching would increase it to 2/3. Notice BTW that the overall prob. of winning the car when switching is (of course) 2/3. And also conditional given the chosen door. But (!) not conditional given the chosen and the opened door. Don't let it keep you out of your sleep. Nijdam ( talk) 22:56, 7 February 2009 (UTC)
Interesting that you put is this way. Because what is the meaning of "no information is revealed"? This actually means that the possible outcomes are not restricted. And here they are: allthough the conditional probability is the same as the unconditional, the information revealed is for instance that the car is not behind Door 3. Nijdam ( talk) 11:12, 8 February 2009 (UTC)
Alas(?!), no. You really miss the point. The information revealed - with the proper strategie of the host - does indeed not affect the probability of winning by switching. But the meaning of such a statement is that the conditional probability given this information, is the same (numerically) as the unconditional one. And as I said before: the revealed information restricts the possible outcomes. I get the feeling that all efforts are made, against better judging maybe, to avoid rejecting the "simple solution". I sincerely hope this is not the case. Nijdam ( talk) 22:58, 8 February 2009 (UTC)
Well, some things happen and do not limit the possible outcomes. Like the started rain outside or the joke the host tells in between. But, as I over and over argued, the open door does. I gave a "simulated" example further down, in reply to Gkanter. Have a look there and be convinced. I really can't think of anything more that should be said. Maybe the real issue is some discussiants are not familiar with probabilities, let alone conditional probabilities. Nijdam ( talk) 00:00, 9 February 2009 (UTC)
I agree with Glkanter: "Everybody else looks for where there may be errors. In the proof. Nowhere else." He may not be responding to the conditional problem, as presented by Rick Block, but should he? People try to explain the conditional situation, but why do they fail to explain to him why the particular problem cannot be unconditional? Where are the errors in the simple solution, as an answer to the given question? I found some probable 'answers' to that:
The question is: should the requested chance be regarded as a conditional probability, and if so, why? Morgan et al. say it should, but their only argument or hint is that the information in the number of the door is otherwise not used. What information is in the number? Does it matter if the (blind) player has knowledge of that? What other information might be relevant? Should it be limited to events which restrict the possible outcomes? Even if it doesn't affect the outcome?
The article Conditional probability states that P(B|A) = P(B) if A and B are statistically independent. In other words: conditional and unconditional chances are the same if A does not change the probability of B. And it doesn't. Heptalogos ( talk) 15:34, 10 February 2009 (UTC)
I've been re-reading some past postings. According to Rick, this article has been reviewed on 2 occasions as a 'Featured Article', and that much of what I find inessential actually was a (by)-product of those reviews. Rick is proud of 'shepherding' this article through at least one of those reviews.
So, in some ways, I seem to be arguing against Conventional Wisdom. But I don't feel that way. I have a few college courses on this topic, over 30 years ago, and a lifetime of being a data analyst. My viewpoint is, 'There is no possible way I am wrong about this'. To me, this whole discussion has as much to contribute as a discussion of whether the sun will rise in the east tomorrow morning.
How does a single voice effectively confront the Conventional Wisdom? This is a question not just for Wikipedia, but any societal system. In the US, a swindler set up a Ponzi scheme on Wall Street. Individual investors went to the regulatory agency numerous times, but to no avail. The guy didn't actually get caught. He turned himself in! How does a minority, but important, voice get heard?
Yes, I look at this entire article, excepting maybe 5% of it, as an elaborate hoax. I think everyone went along because they did not want to admit to limited knowledge of the subject matter. Everybody drank the kool-aid. And the emperor is wearing no clothes.
2/3 of the time I will select a goat. Therefore I should switch. Glkanter ( talk) 15:32, 7 February 2009 (UTC)
And really, who wants to argue when at the top of the page it says:
Please note: The conclusions of this article have been confirmed by experiment |
There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment. |
Two mentions of proof via experimentation. I'll bet not a single reference source uses 'experimentation' as their solution. Because 'experimentation' is not a statistical or logical proof. An error like this confirms my above stated concerns. I'm curious, is it common for Wikipedia discussions to have such a banner? I've never seen anything like it in any 'academic' setting before. Anywhere. Glkanter ( talk) 15:54, 7 February 2009 (UTC)
And this, it's the 2nd paragraph after the 1st chart in the Solutions section. It's one of the more prominent items in the Solutions section.
What sort of a 'Solution' section leads off with a proof it is going to immediately (and fraudulently) discredit? And then goes on endlessly doing so? For who's benefit? The once-interested reader is by now, long gone.
Glkanter (
talk)
17:06, 7 February 2009 (UTC)
Here's my only edit to the actual Article. On October 23, 2008 I deleted the following erroneous line:
The above statement was the very first words of the Solutions section.
Imagine. Being THAT wrong on the very first line.
Glkanter ( talk) 17:57, 7 February 2009 (UTC)
So, I posted the following two days ago, at 13:34, 6 February 2009:
Eight lines. Including a premise repeated twice for clarity. Since then, there have been countless postings by Rick. Running off on tangents to and fro. But we NEVER focus on the 8 statements. It's always some BS about 'before or after', or whatever the mcguffin of the day is.
It's a valid proof, and could be shorter. Mr. Block! Tear down your wall!
Glkanter ( talk) 07:46, 8 February 2009 (UTC)
This is why I say it's time to end this charade, now.
In October, 2008, Martin posted this to Rick:
So, it's not just me. It's been enough people, for a long enough time, that the offical Wikipedia RFC process was put in motion. Good. It's time for the next step in the process. BTW, who do suppose could be the 'overly protective current editors' referred to?
Glkanter ( talk) 15:34, 8 February 2009 (UTC)
Do I sound frustrated? Well, I am. Here's another one from October. Sound familiar?
What was that completely erroneous statement? Just this:
So, yes, I'm personally invested. I'm interested. I just read that there are 'archive' pages for this discussion. So what we see here, is just the most recent discussions of this lunacy. The tip of the iceberg, so to speak.
So, enough of the charade!
Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense. —Preceding unsigned comment added by Glkanter ( talk • contribs) 15:54, 8 February 2009 (UTC)
You know what? It's been long enough. Martin, I understand that at some time last year you were specifically requested to get involved, in order to facilitate the resolution of some conflict(s) relating to the Monty Hall Problem.
I ask Rick to invalidate my proof, he tells me I'm not addressing the Monty Hall Problem.
It is obvious these discussions have gotten nowhere, and are going nowhere. So I request that you declare this situation as 'at an impasse', and escalate this to the next level of Conflict Resolution.
Thank you.
Glkanter ( talk) 17:05, 8 February 2009 (UTC)
Since October 25, 2008 I have posted on this discussion page 70 times. 66 as Glkanter, and 4 previous times as an ip address. In addition, there was a separate discussions page that was set up, and I posted there many times as well.
70+ postings over a 2 line proof! When I pick a goat, Monty leaves a car. I pick a goat 2/3 of the time.
I feel that this has gone on long enough! I honestly wonder if there isn't a bot out there generating responses from a limited selection of inaccurate/meaningless/non-relavent statements.
This has been going on with countless other editors since at least 2005. Check out the archives sometime.
To the powers-that-be at Wikipedia, I implore you, please bring this to a conclusion!
Glkanter ( talk) 15:43, 10 February 2009 (UTC)
I just went to the page where Rick requested assistance. One of the people there, who described himself as 'an expert in probability' had this to say:
Does this mean what I hope it means?
Glkanter ( talk) 22:37, 10 February 2009 (UTC) moved comments 76.243.187.238 ( talk) 01:29, 11 February 2009 (UTC) doh! Glkanter ( talk) 01:50, 11 February 2009 (UTC)
After just 4 months, I've run out of new things to say. So rather than clog up other people's exchanges, I'll just add commentary here. I don't see the point in further arguing, but I'm not giving up the quest.
I see the 'random goat constraint' has reared it's (their) ugly head again. In my proof, which I maintain is sufficient for this puzzle, the goats are immaterial. The puzzle asks about increasing the liklihood of selecting the car, not the location of a goat.
Does the probability change if instead of goats there is nothing? So would we have a 'random nothing consraint'?
My understanding of Probability Theory says if an item isn't specifically addressed, it's considered random. And anything that's random is not a constraint. And if it's not a constraint, it doesn't need to be addressed.
For those keeping score, I was indeed reported to the authorities. I, and others with my criticisms were described as a 'cadre', 'wanting to "dumb down" ' the solution by Rick.
Glkanter ( talk) 14:36, 11 February 2009 (UTC)
And, the doors and their numbers are meaningless. The doors only exist to shield the contestant's view of the car and the two goats.
Let's focus on why this is such a famous puzzle. Because it's easily understood (2/3), and easily misunderstood (50/50). So, no reader really understands the solution until he has mastered the 'random goat constraint'? Pure folly.
Nobody cares about the specific case of doors 1 and 3, any more than they would care about suitcases 11 and 25 in Deal or No Deal. That's not what Probability Theory is about. Here, I disagree, until my last breath, with Rick. Probability, and this puzzle are always solving the 'over time' question. If repeated endlessly over time, on aggregate, what would the results be?
Glkanter ( talk) 14:54, 11 February 2009 (UTC)
I was struggling for the right word earlier, but it came to me. Despite what Rick is posting elsewhere about 'physical doors' and such, all Probability can do is create a mathematical model that approximates real life. That's what a proof sets out to do. And if valid, it serves as a proxy for 'real life'. From which we can analyze and draw conclusions.
I, of course belief I have presented various proofs that meet this criteria. Rendering all that other stuff, at best, redundant and confusing, and at worst, erroneous.
Glkanter ( talk) 16:19, 11 February 2009 (UTC)
The Monty Hall Problem gained much (most) of its current noteriety when it was written about by Marilyn Vos Savant in Parade Magazine. I read about it there. We all know the story about how a thousand PHDs challanged her, etc., etc.
But ultimately, Vos Savant was acknowledged as having the right answer at 2/3.
Let's hear from the great lady, herself, courtesy of Wikipedia'a Marilyn Vos Savant page:
Or this, from the Newspaper of Record:
So, even as we argue semantics on this page, it's unambiguous what problem she was solving. With all that hub bub, what proof did Marilyn and all these people use to come to the conclusion that 2/3 is right? Well, it wasn't Morgan et al, via the 'random goat constraint'.
So, if the unconditional probability proof was good enough for Marilyn, Monty, 1,000 PHDs, and 10s of millions of Parade magazine's general interest readership, why is it not good enough for the Wikipedia readers of today?
Glkanter ( talk) 08:01, 12 February 2009 (UTC)
I've just learned that Rick's comment that being 'Right' is trumped by being 'Verifiable' for Wikipedia purposes has some merit. I further understand that being in the mainstream (my term) trumps being in the minority (my term).
We're still arguing over who's 'right'. But apparently, that's not even the salient point.
Would anyone describe the 'random goat constraint' as being in the mainstream of the published literature? How about after reading this?
Glkanter (
talk)
13:35, 12 February 2009 (UTC)
I've posted here nearly 80 times trying to prove my proof is valid, and that I'm 'right'. Turns out, that was never the argument I needed to win.
So why, since 2005 at least, hasn't the response been that the more meaningful (for Wikipedia purposes, anyway) arguement is, to paraphrase, 'popularity of the published material'? Not until about two days ago, anyways.
So, thanks for wasting 4 months of my time. And 4 years of various other people's time.
Now, can we start fixing the Article?
Glkanter ( talk) 14:48, 12 February 2009 (UTC)
Pretty interesting exchange relating to my 'being reported' by Rick at the bottom of Rick's talk page. First, he gets spanked for crying wolf, then he writes this:
Maybe next I'll be reported for trying "to gain consensus for their desired change". Does trying to build consensus make me a bad Wikipedian?
Glkanter ( talk) 17:35, 12 February 2009 (UTC)
Would you consider the following statement as an indication that an editor has claimed 'ownership' of an article?
Could this be an 'Aid To Understanding' for why there have been 4 years, 7 archives, and thousands of postings calling for significant changes to the Monty Hall Problem article, all to no avail?
Let me ask the 'cadre', is there 'consensus' that this editor should be reported for violating the Wikipedia Ownership policy as it relates to the Monty Hall Problem article? If forced, I guess I could be the one to make the report.
Glkanter ( talk) 18:06, 12 February 2009 (UTC)
Does anybody know what form any sanctions would take? I hope it would include being blocked from editting both the MHP Article and this talk page.
Glkanter (
talk)
18:45, 12 February 2009 (UTC)
Here's where Rick first asked for assistance to aid in Resolving our Conflict.
All these other Wikipedia Math gurus already knew about Rick's MHP article Ownership issues!
I'm a first-timer here. It's been way too long, but is has been instructive as to how horribly mishapen things get when an editor claims ownership of an article.
Glkanter (
talk)
19:25, 12 February 2009 (UTC)
If you're still reading these edits, you night appreciate my very first posting. I created a new section that day, but I still don't know how to link to it. "Monty's Action Does Not Cause The Original Odds To Change."
We were all so civil then! I was all ready to make my first edits to the Article! Glkanter ( talk) 19:47, 12 February 2009 (UTC)
You guys should read the Featured Article Reviews. There's a link to each of them at the top of this page. In addition to some of the inane discussions we take part in here, the FA reviewers also argue about footnotes.
And there is one recognizbale name constantly making edits. That would be Rick. More documentation as to his 'Ownership' of the article.
Now, after his encouragement to suggest edits, he says this:
Looks like an 'Ownership' issue to me.
Basically, Rick is saying that the Article is already perfect. Therefore, any changes will make it un-perfect. And he knows, because he's written most of it, nominated it for FA, 'sheperded' it through FA, and personally responded to every FA critism himself. Through his 4 years of Ownership-fueled protectionism, he also has granted the existing Article 'tenure'. I wonder if there is such a policy as WPTENURE? Glkanter ( talk) 04:02, 13 February 2009 (UTC)
I'll probably have to document my claim. Here's another good one from Rick:
Thanks, dude. You're too generous. No, really. Glkanter ( talk) 08:01, 13 February 2009 (UTC)
"This page is 495 kilobytes long. It may be helpful to move older discussion into an archive subpage." Plus 7 archive pages dating as far back as 2005.
With my edits to this talk page, I have been accused of violating WP:CIVILITY, and worse. I'm comfortable with everything I've written. As I wrote at the start of this section, it's difficult for a minority voice to be recognized as credible, when it defies the Conventional Wisdom. So, I am trying to accomplish what seems, so far, to be a difficult task. The only tool I have is my ability to post on various Wikipedia pages. Talk about a probability problem! How do I know who to turn to?
So, I post here. I can't post in that pseudo-collegial, 'oh, beg to differ, chum', style that seems so popular. And I can't out-reference or out-WP:Policy Rick. He's got like 10,000 article edits to his name. No, all a punter like me can do is write with common sense and clarity, and try to draw some attention. So I have my own 'style'. Now you know why. Glkanter ( talk) 14:49, 14 February 2009 (UTC)
Let me ask it this way, in the 1/3 of the time when I select the car, how is Monty's motivation or actual actions going to change the fact that the car was my choice? And that by switching, I would give up the car? How is the 'equal goat door constraint' the solution to that question? It isn't. For probability proof purposes, like the most simple 2 or 3 line proof, we never even need to go there. I reject anew Morgan et al.
Glkanter (
talk)
16:13, 14 February 2009 (UTC)
Here's my new proof.
Look at that. No Monty. No Monty's behaviour. Thank you, thank you very much.
Glkanter (
talk) 16:21, 14 February 2009 (UTC)
Glkanter (
talk)
16:37, 14 February 2009 (UTC)
I think this is even better.
No Monty, and now no goats. Glkanter ( talk) 16:53, 14 February 2009 (UTC)
Those of you who reject the existance of Original Research are excused, and free to resume cataloging references to Bigfoot, Unicorns, etc. Glkanter ( talk) 17:17, 14 February 2009 (UTC)
I just realized that any answer other than 'never switch' inproves your liklihood of winning. The worst you can do is not switch, and go 1/3. Do anything else, I'm just thinking 'go random', and yours odds increase, in this case to 50%.
Glkanter (
talk)
23:13, 14 February 2009 (UTC)
I came here from a request on WT:WPM. But the conversation is too long to read. Could each person post a 1-paragraph summary of what she or he thinks are the main issues under discussion? thanks, — Carl ( CBM · talk) 20:30, 8 February 2009 (UTC)
In September or October I used the Monty Hall Problem Article to further my understanding of the puzzle. I found about 1000% more editorial matter than I would have expected. It's roughly 1% useful, 99% BS. Since then, I've been trying to establish why I believe it's 99% BS in order that this be deleted. I have chosen to use Probability Theory as the (only) appropriate method to convice the other editors. I have proferred what I am confident is a valid proof, and have requested that other editors find any errors. This has gone on for 4 or 5 months now. I am not unique in this. There is about 4 years of history attempting to do what I have set out to do. Rick says I'm not addressing the right problem. I insist that I am. Here's the proof. It's longer than it needs to be for clarity:
Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.
There's another dozen points of disagreement. There's no point getting to those until we resolve the question of whether or not the 1st Solution (or something like it) is correct, and that all that other stuff is BS.
But, we're basically at a standstill. Neither Rick or I is going to pursuade the other. I want to see the page improved, finally, and requested that our Conflict get some Escalated Resolution help. Glkanter ( talk) 21:06, 8 February 2009 (UTC)
Glkanter I totalt agree with you the current article is Bull crap! 99% useless and 1% useful.
It does not do a good job communicating a simple and straightforward unconditional solution!
I totaly agree with you and I want you to change it!! I have however started to lose interest because RickBlock is married to the current article and he interpret any attempt of changes as an attack on him. I just wonder why is his name RickBlock by the way?
I thought the whole point was to collaboration?! I am actually a bit insulted by his choice of name!!--
92.41.214.24 (
talk)
22:14, 8 February 2009 (UTC)
I'm still not up to speed on this, but I have a few general comments.
I also have a few questions if none of you objects:
— Carl ( CBM · talk) 23:03, 8 February 2009 (UTC)
@Glkanter: look at what happens in 18 realisations of the game;
I sorted the outcomes for a better overview:
Choice 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 Car 1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3 Open 2 2 3 3 3 3 3 3 2 2 2 2 3 3 3 3 1 1 3 1 1 1 1 1 2 2 2 2 1 1 1 1 1 1 2 2
6 times there was the same situation as in the player's case; here they are:
Choice 1 1 1 1 1 1 Car 1 1 2 2 2 2 Open 3 3 3 3 3 3
The conditional (!) probability to win the car by switching is (surprised?): 4/6 (BTW: the unconditional probability is 24/36, different, but numerically the same!) After all that is allready said, what needs to be said more? If you're not convinced, I guess you don't want to be convinced, or worse. Nijdam ( talk) 23:13, 8 February 2009 (UTC)
It is not the general relativity theory either, yet the above figures explain it all. Please scream, if it make you feel better, I'll go to sleep. Nijdam ( talk) 00:03, 9 February 2009 (UTC)
What would make this article better would be a rearrangement of content that puts like things together, progresses from simple to complex, and uses headings that clearly define the issues. Here is a suggestion:
1 History of the problem (This would not be about the Three Prisoners problem or other “pre-history” problems, but about what happened when the problem appeared in Parade and in follow-up columns. This satisfies the needs of the casual reader.)
2 Explanations clearly consistent with the original solution (This would include 2.1 Name for the first explanation, 2.2 Name for the second explanation, etc.)
3 The common objection (This would be about the difference between asking the more general question whether one should switch, versus the more specific question whether one should switch when, say, door #3 is opened.)
4 Explanations that meet the common objection (This would include 4.1 Name for the first explanation, 4.2 Name for the second explanation, etc. Bayes would be here.)
5 Another objection (This would be about the objection that one does not know if the host is as likely to open one wrong door as another when the player chooses the right door, and the Morgan, et al. solution.)
6 Similar problems (The Three Prisoners problem would be moved into this section.)
7 Theories about the problem
Omit objections that clearly contradict the problem as stated, or put them into a final section 8: Other objections and why they are irrelevant
This proposal, or something like it, meets the needs of both the casual reader who has heard of the problem and wants to know what the deal was, and the reader who is interested in a greater understanding of the issues. Simple314 ( talk) 01:10, 9 February 2009 (UTC)
The Monty Hall Paradox is only a paradox for people who like to compare apples to oranges. If a player were to pick one door out of three, two out of three times that player would be wrong. So a player should always switch if they get a second chance. With respect to the POSSIBLE OUTCOMES of this game, there are more scenarios of a player who did switch won a car. But that is not to say that the act of 'switching' improves probability. The probability of picking one door out of two is ALWAYS 1 to 2. This is true by definition. Trying to say othrwise is simply meaningless, and you can not prove it with a computer simulation. Just as you can not prove that the probability of flipping heads of a fair coin is 1 to 2. Try it, it is only true by definition. —Preceding unsigned comment added by 68.198.159.238 ( talk • contribs)
Total tries | Total times you had the ace |
---|---|
Number of times dealer showed the heart two | and you had the ace |
Number of times dealer showed the diamond two | and you had the ace |
Total tries | Total times you had the ace |
---|---|
60 | 20 (1/3 of total tries) |
Number of times dealer showed the heart two | and you had the ace |
40 (2/3 of total tries) | 20 (half of these tries) |
Number of times dealer showed the diamond two | and you had the ace |
20 (1/3 of total tries) | 0 (all of these) |
Your last paragraph confuses me. Isn't it a premise of the puzzle that the dealer discards a two (reveals a goat) from among his two cards? There's never anything random about this action. It may be random how he chooses which of the two two's, but the fact that he's going to discard a two is a premise. Let's switch back to Monty Hall. Unless the contestant knows something about how Monty chooses 'which goat', he gains no new knowledge when Monty reveals the goat. If he knew something about how Monty chose goats, then that's a new premise or constraint. Which makes it a completely different puzzle.
We have a saying in my business. "If it looks like a duck, quacks like a duck, and smells like a duck, it's a duck."
At the risk of showing my ignorance, I don't think there is a difference between the conditional and unconditional problems. It's already agreed that they have the same statistical probabilities. Now, it looks to me like you're using the exact same simulation that I would use to 'aid in the understanding' of my proposed proof. Since the probabilities are the same, I would suggest that there is no difference in the total knowledge (useful information) on the contestant's part. (That is, the premises and constraints are identical, unlike the 'Monty always chooses the left-most goat-door' variant). Lastly, In order for the two puzzles to be different, yet have the same probabilities, it seems that those probabilities would diverge at some time, yet co-incidentally return to the same values. I don't see that happening. They just both start out at 2/3 and never go anywhere.
Glkanter ( talk) 17:04, 9 February 2009 (UTC)
To summarize: in response to my question, you found it necessary to add a constraint, in order to make it a conditional probability problem. That makes it a different puzzle than the Monty Hall Problem. Further, one would not say that the probabilty for this new puzzle is 2/3. Going back to the Monty Hall Puzzle, since this new constaint (left-most goat-door) is known by the contestant (or else, what's the point?), there are scenarios when he knows exactly where the car is. So the answer would be stated "100% when Monty reveals door #3, 100% when I've chosen door #3 and Monty reveals door #2, etc." So this contestant makes more informed choices than the contestant in the real puzzle. So the contestant will win MORE than 2/3 of the time.
Glkanter ( talk) 17:50, 9 February 2009 (UTC)
Two things. In the specific door 1, door 3 question, why isn't it 2/3? Doesn't the contestant have a 2/3 chance of picking a goat? Then Monty shows the other goat, leaving a car. That's all we know at this point, right? That fits in perfectly with my new and improved, customized Super Proof! (Tip of the hat to a certain MH, who I shall keep out of this.)
The second thing is, I reject the door 1, door 3 literal interpretation. (Not that it makes any difference, as I just proved.) That's not how Probability Theory works. And that's not what makes this an internationally known paradox. Heck, the real Monty didn't even use doors. He used curtains.
Glkanter ( talk) 05:16, 10 February 2009 (UTC) Glkanter ( talk) 06:14, 10 February 2009 (UTC)
The third point is, the 'left-most door variant' is a different puzzle. It adds a new constraint on Monty. Which, in some cases, causes Monty to reveal the exact location of the car. My proof was never intended to solve that problem. I wish you would stop using this as your 'proof' that my proofs are flawed. I think it's this action on your part that I react the most strenuously against.
Glkanter ( talk) 05:24, 10 February 2009 (UTC) Glkanter ( talk) 06:14, 10 February 2009 (UTC)
I've wondered how you would respond when the inevitable happened, and your arguments were proven to be meritless. Now we know. You attack the messenger, and start posting that for Wikipedia purposes, the Truth (being right) is less important than having been published.
I've asked for Wikipedia intervention for days now. I welcome any actions you take to bring closure to this festering sore.
Glkanter ( talk) 18:35, 10 February 2009 (UTC)
There may be a kind of solution that comes close to the "simple intuitive solution", but is really a solution to the problem. Suppose the player is in the situation of having chosen door 1 and being opened door 3. We indicate this situation as [13]. Of course other players may find themselves in [12], [21], [23], [31] or [32]. Together all these 6 situation represent the unconditional one. With the "normal" strategie of the host, there is no fundamental difference between these 6 situation, hence the (conditional) probability of winning the car by switching is the same for each. But the overall probability, the unconditional, is 2/3, as we know from the intuitive (simple) solution. But then all the conditional probabilities must aldo be 2/3. Will this be satisfactory? Nijdam ( talk) 15:38, 10 February 2009 (UTC)
Hey Rick, which "peer reviewed academic sources" did you get this gem from?
That was sentence #1 of the Solutions section until I deleted it last year. It is 180 degrees counter to every bedrock principle upon which Probability Theory rests.
Glkanter ( talk) 21:46, 10 February 2009 (UTC)
I am trying to get agreement on a hypothetical question. Suppose the question posed was this:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door and the host opens another door which has a goat and he does this in a way the does not affect the probability that your original choice was a car. He then says to you, "Do you want to pick the other unopened door?". What is your advice?
I know that this is not the question actually posed but in the above case do you accept that the conditional case is identical to the unconditional one? Martin Hogbin ( talk) 22:09, 10 February 2009 (UTC)
I think that the statement of
Boris Tsirelson confirms the point that I am making.
In that case, why do we not consider all the other actions that occur between the original choice and the decision to swap as conditions? 09:39, 11 February 2009 (UTC)
What I would like to have is an addition to this article of a very simple and convincing explanation of the essential Monty Hall problem.
I believe that, for this purpose, it is acceptable to consider the unconditional problem for the following reasons:
1 The inability of most people to get the right answer, however the problem is formulated, is what makes it notable.
2 The academics who have studied the problem have to some degree (understandably) obfuscated the central problem with unwarranted and unjustified complications, including making it necessarily conditional.
3 Even if the conditional version is considered to be the best one, it is not uncommon in mathematical expositions to start by answering a related, and often simpler, problem and then to explain how this relates to the full blow problem.
The more academic discussion can remain as it is for those who are interested. Martin Hogbin ( talk) 23:09, 10 February 2009 (UTC)
We can start by observing that, if the player switches, he always gets the opposite of his original choice. The are no conditions involved, no probabilities, no dependence on host action - this is a certainty. Obviously of he does not switch he keeps his original choice.
All we have to show now is that the action of the host does not affect the probability the he had originally chosen a car, or to put it another way, that no information is revealed that might indicate the players original choice was.
Let me ask this question, looking at the Parade statement, what is the information that is revealed? Martin Hogbin ( talk) 23:23, 10 February 2009 (UTC)
For clarity in the discussion: I will remove any comment of someone else besides RickBlock and Martin Hogbin given between this header and the corresponding trailer and move it above this header. Also I put this section allways at the end of this page.
@RickBlock. It's sad and frustrating all these people, hardly knowing anything about probability, but yet thinking they just know all the ins and outs of this problem. From now on, I'll only discuss matters with you (as one expert), if necessary, and with MartinHogbin, who seems to me at least responding in an adequate manner. Others may profit by reading the comments, I will no longer respond to them. Also I think it is about time this discussion ends. My suggestion above under "work around" (maybe not so well chosen title) is a sound reasoning, in the line of the "simple solution". It lookes if you crititizised it. It is (seemes?) easy to proof and may be helpfull in understanding the conditional nature as well. Nijdam ( talk) 22:55, 10 February 2009 (UTC)
Like you. I also strongly object the 'unconditional' non-solution, that's the whole point in this discussion. Maybe I didn't express myself clear enough. The "work around" idea is to incorporate the basic idea of this unconditional reasoning as to explain the right conditional explanation. Of course in the 'random goat' strategy, only then there symmetry in the 6 basic cases. And this symmetrie is not a conclusion of the calculation of the condtional probs. The suymmetry leads to equalness of the conditional probs. Hence it can be used ([123] means chosen 1, opend 2 car 3) as follows: for all a,b,c all different:
Hence all conditional probs equal 1/3. But it remain (are) cond. probs.
The 'random goat' strategy is part of the premisses and I think the 'unconditional' non-solution doesn't find support in other cases. or is this the point you are making, that people reason in the same way even when the host has another strategy? Nijdam ( talk) 12:26, 12 February 2009 (UTC)
Okay! Indeed is the problem symmetric only with the 'random goat' strategy. But this strategy is part of the formulation of the problem. The symmetry doesn't stem from the equality of the probs, but from the fact that the problem is then invariant under permutations of the doors. And hence all the conditonal probs must be the same. And this implies they are equal to the unconditional 1/3. But they are of course conditional. So I don't say the 'simple reasoning' is correct, but a kind of 'simple reasoning' may be used. And yes, I know, without the 'random goat' there is no symmetry. And also then it are the conditional probs that had to be considered. Nijdam ( talk) 12:26, 12 February 2009 (UTC)
What is there to be proven? In the 'random goat' strategy there is no reference to any specific door, hence no door plays a different role than another one. Hence all the situations on stage chosen a and opened b are equivalent. Is this original research? I think, this is what the advocates of the 'non-solution' really have in mind wthout being able to formulate it. Nijdam ( talk) 17:07, 12 February 2009 (UTC)
@Martin Hogbin. I repeat the following points:
chosen 1 1 1 1 2 2 2 2 3 3 3 3
car 1 1 2 3 1 2 2 3 1 2 3 3
opened 2 3 3 2 3 1 3 1 2 1 1 2
prob 1 1 2 2 1 2 2 1 2 2 1 1 /18
The columns form the outcomes with their probability (/18 as indicated)
In which situation is the player? That is which event has occurred? Well in the stated problem [13], which is:
chosen 1 1
car 1 2
opened 3 3
prob 1 2 /18
consisting of 2 outcomes, and really a proper part of the sample space. In the second column the car is behind the not chosen door, and the well known conditional (!) probability of winning the car by switching is: (2/18)/(3/18) = 2/3.
Nijdam (
talk)
23:41, 10 February 2009 (UTC)
A am suggesting that the formulation of the problem by Morgan et al introduced unnecessary uncertainties. Here is the Parade statement for reference:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
What does it say about the host offering the choice, it says: 'He then says to you, "Do you want to pick door No. 2?" '. Note that it does not say 'sometimes says to you' or 'may say to you' but 'says to you'. There is no mention in the statement of the problem that the host may not offer you the choice. That possibility is an invention of Morgan.
It says the host, 'opens another door'. It does not give the way that the host decides which door to open. We must therefore take this as random. This is perfectly normal in probability problems, in cases where the method of selection is not given we take it as random, this is not an assumption it is an obligation.
Does the host always reveal a goat. The statement says, 'which is a goat'. It is therefore pure invention to assume that it might not be.
There are an infinite number of perverse actions that could occur in the problem - what if sometimes there are three goats? We do not address these possibilities because they are not part of the stated problem.
Martin Hogbin ( talk) 11:34, 11 February 2009 (UTC)
To support a point that I make above I ask people to answer this question:
A ball is chosen from a bag containing only red and white balls. What is the probability that it will be red?
Please answer the question as posed. Martin Hogbin ( talk) 22:11, 11 February 2009 (UTC)
You're aiming at the answer 1/2. Probability is here to be understood as relative frequncy in all possible realisations of the experiment. The symmetry in the colours brings the answer. Nijdam ( talk) 12:38, 12 February 2009 (UTC)
To the question, as posed, I believe that 0.5 is the correct answer. To say that I am aiming at that answer suggests that ther may be more than one answer. I think there is only one correct answer to my question and that it is 0.5. Are we all agreed on that or not? Martin Hogbin ( talk) 20:26, 12 February 2009 (UTC)
'What is meant by probability'? Surely we are not meant to be discussing that here.
This talk page is actually for talking about changes to the article, not for talking about the ins and outs of the Monty Hall problem or probability theory or the meaning of what a "bag" is. There are really only three legitimate topics for discussion here.
1) Things not in the article that should be added.
2) Things in the article that should be deleted.
3) Things in the article that should be changed.
The 0.999... article has a subpage of its talk page called "Arguments", see Talk:0.999.../Arguments, and a FAQ. I've set up such a structure here, in the hopes that it might be helpful. If anyone wants to move a section they created here to that page, please go right ahead. I'd appreciate everyone's help in moving sections from here to there when it seems appropriate. Thanks. -- Rick Block ( talk) 01:35, 13 February 2009 (UTC)
Yes, I would like to see the paragraphs in the Solutions section that says the first solution offered does not solve the Monty Hall Problem be removed. I believe is it not consistant with what the majority of the published references says.
I would also like to see the references to the minority point of view as espoused by Morgan et al moved out of the Solutions section.
I believe both of the changes are consistant with Wikipedia policy regarding regarding authoritative sources.
Rick, how will we know when it's proper to edit the Article? Would that be when a consensus is reached? Would any editors reserve a veto power?
Glkanter ( talk) 02:19, 13 February 2009 (UTC)
I fell for it. I let him use the old 'rope-a-dope' on me. Next thing you know, instead of argueing about my proof, we'll be argueing about 'who is more mainstream', or 'what is prominant'. We are not arguing about your proof. Your proof is WP:OR and of no relevance, no matter how correct it may be. Please realise that. Leaving out the hyperbole would be helpful. To comment on the principles: no, this is not a democracy, yes all have the same potential to contribute, no Rick doesn't have a veto, no Rick did not say that he had a veto, no the argument that the content has been in place for ages is not a compelling reason to keep it William M. Connolley ( talk) 08:31, 13 February 2009 (UTC)
I guess I'll have to understand and follow Wikipedia protocol if I'm going to be successful in ending Rick's Ownership of this Article. This is from Wikipedia:Ownership of articles:
Rick, as you are aware, I believe you have improperly taken ownership of the Monty Hall Problem article to the detriment of the article. I estimate this has been the case for nearly 4 years. I would like to find a solution to this, in order that I, and others could edit the article, without fear of either your veto, or entering into a reversion war.
I lay out my reasoning beginning with my post of 17:35, 12 February 2009 (UTC), which comes in the section title Conventional Wisdom.
I look forward to your response to this serious issue.
Thank you Glkanter ( talk) 07:36, 13 February 2009 (UTC)
"I have done my level best over many days now to help you understand why the change you seek won't happen" (The Dictator aka Rick Block, 2009) '
This to me implies that "Rick block" actually belives that he can decide by himself what changes should be made to the Monty hall article !!!! Rick Block should decide everything and if someone does anything that "Rick Block" does not like then "Rick Block" is going to mobilize enough support to force such a person out of the game by telling people to revert any changes to the article. Abuse and Bullying on a higher level !! Definitely ownership with capitalized 0 --
92.41.74.138 (
talk)
21:36, 13 February 2009 (UTC)
William M. Connolley if you remove my comment another time (just cause you cant stand the truth) I will remove ALL your comments :-) --
92.41.74.138 (
talk)
21:36, 13 February 2009 (UTC)
There is another source of unnecessary complication introduced by the academic papers on this subject and it is this:
The Parade statement is clearly written from the point of view of the player, 'Suppose you're on a game show...What is your advice?'. The questioner is obviously asking for advice as to what they should do if they were a contestant on the show.
Some academics, however, have reformulated the question from the perspective of some imaginary third party. Someone who may know, for example, that the host has a preference for a particular door. This is something that the player does not know.
As a result of this reformulation, there have been countless arguments on this page over whether it matters if the player knows about the host's preferences. This reformulation is a major obstacle to understanding the solution. Martin Hogbin ( talk) 09:30, 13 February 2009 (UTC)
I have an alternate suggestion regarding the conditional vs. unconditional question that keeps coming up. How about if we add a more comprehensive discussion about this, perhaps like the following (a slightly edited version of an explanation I originally posted above that might have been buried before many people noticed it). If we don't add it to the article, I suggest that it be included in the FAQ linked at the top of this page. The wording may not be quite Featured Article quality, but I think it is close. -- Rick Block ( talk) 05:28, 14 February 2009 (UTC)
Morgan et al. and Gillman claim the Monty Hall problem must be solved as a conditional probability problem. As usually stated, the problem asks about a player who has picked a door and can see which of the remaining two doors the host opens in response, and is then given the opportunity to switch doors. The question pertains not to all players in general, but only to the subset of players who have picked a door (for example door 1) and then have seen the host open a different door (for example door 3). The point concerns the difference between
and
is the unconditional probability of winning by switching. It is roughly what fraction of all players who play the game will win by switching. Gillman suggests this would be the probability of interest in a revised version of the problem where "you need to announce before a door has been opened whether you plan to switch." [italics in the original!]
is the conditional probability of winning by switching given that the player initially picked door 1 and the host has opened door 3. It is roughly what fraction of all players who pick door 1 and see the host open door 3 will win by switching.
These are different questions which may or may not have the same numeric answer. As usually stated, Morgan et al. and Gillman both say the Monty Hall problem asks the conditional question, not the unconditional one.
An unconditional solution such as "a player has a 2/3 chance of initially picking a goat, all of these players who switch win the car, so 2/3 of all players who switch will win the car" correctly answers the unconditional question, but doesn't necessarily say anything about the conditional problem. For example, this solution says the answer for both the Parade version and the Krauss and Wang version of the Monty Hall problem is 2/3. The Morgan and Gillman papers both introduce a different variant, identical to the Krauss and Wang version except instead of
they specify
In this variant, an unconditional solution also correctly answers the unconditional question. The player still has a 2/3 chance of initially picking a goat, and in this case the host must reveal the other goat, so the solution above still applies and 2/3 of all players who switch will win the car. However, in this variant, if the player initially picks door 1 and the host opens door 3 the player knows the car is behind door 2 and has a 100% probability of winning by switching. If the player initially picks door 1 and the host opens door 2, either the player's initial pick was the car (a 1/3 chance) or the car is behind door 3 (also a 1/3 chance). This means, given that the host has opened door 2, the player's chance of winning by switching is 1/2. The point of introducing this variant is to show the difference between the unconditional and conditional questions. In this variant, these questions have different answers exposing the difference between unconditional and conditional solutions.
Using an unconditional solution produces the correct answer for the more exact Monty Hall problem as stated by Krauss and Wang, but this is in effect a coincidence. The problem with the approach is hidden because in this version the conditional and unconditional answers are the same. Applying the same solution to a variant where the unconditional and conditional solutions are different, like the "leftmost door variant", reveals the problem.
This is exactly the same as responding to the question "what is 22?" by saying "2 plus 2 is 4, the answer is 4". This is the right answer, and a true statement, and related to the right reasoning, but not actually right. This wrong reasoning can be exposed by asking, "what is 33?" Similarly, the problem with using an unconditional solution to answer the Monty Hall problem can be exposed by asking about a slightly different variant, where the wrong reasoning produces the wrong answer.
So, how does this work in practice? There's no vote taking, and consensus doesn't necessarily rule the day. But Rick can still change his article, right?
Glkanter (
talk)
10:14, 14 February 2009 (UTC)
Actually I think we now should change the conditional solution to an unconditional one. Concensus has been reached (4 against 1)
Glkanter, Heptalogos, Martin Hogbin and 92.41.236.242 against "Rick Block" which means that the unconditional solution is the winner
What do you want to include in the unconditional solution guys? --
92.41.236.242 (
talk)
12:52, 14 February 2009 (UTC)
To help structure this discussion, I'm adding subsections. Please stay on topic within the subsections. -- Rick Block ( talk) 18:18, 14 February 2009 (UTC)
The reason I am suggesting adding this content is because the basic point made in these two papers is clearly not widely understood. If the article includes a comprehensive discussion of what they say, the wording can be improved by anyone who may be able to express the point better. We should not add the content if the point is insignificant, or mathematically discredited. It is my belief that these two papers, nearly simultaneously published (Morgan et al. was first by 2 months), are the first mathematically rigorous treatments of the Monty Hall problem which makes them crucial to include in this article. I also believe these papers were published specifically in response to the nationwide furor over this problem that followed the Parade articles published in 1990-1991. The Morgan et al. paper is cited 49 times according to Google scholar. The Gillman paper is cited 30 times. As far as I know, the central point they make is not reflected in non-academic literature, but this doesn't mean we shouldn't include what they say. -- Rick Block ( talk) 18:18, 14 February 2009 (UTC)
You started by saying the fully defined problem is directly equivalent to the 3 prisoners problem. This is somehow contradictory to your opposition to accept that it is about conditional probability. The 3 prisoners problem is! Nijdam ( talk) 21:10, 14 February 2009 (UTC)
If you think the suggested wording doesn't fairly summarize what the papers are saying, please say how or where (the analogy at the very end is original, but everything else is directly from the papers). The papers are short, but I could add references to specific paragraphs if that would help. A point they both make, not in the proposed wording above, is that viewing the problem as a conditional probability problem means the Parade version of the problem lacks a condition necessary to justify the 2/3 answer. Specifically, they BOTH mention this version does not say the host picks between two goat doors randomly. BOTH of these papers make this point, and provide a solution where the host's preference for one door or another is left as a variable q. They BOTH show the probability of winning by switching is then in the range of .5 to 1, exactly equal to 1/(1+q). I think this additional point should likely be added as well, but first things first. -- Rick Block ( talk) 18:18, 14 February 2009 (UTC)
Arguing that what these papers say is wrong in a mathematical sense has no bearing on whether the content should be included or not. If you would like to discuss the points they make, please post at /Arguments or Wikipedia:Reference desk/Mathematics. -- Rick Block ( talk) 18:18, 14 February 2009 (UTC)
I think making the two versions of the problem clear in the article might prevent a lot of argument. Martin Hogbin ( talk) 20:34, 14 February 2009 (UTC)
Rick thanks for making the players plural in my recent addition. I agree that is is a better mathematical representation and it is more PC.
I have restored the emphasis on 'always' to indicate that there is no doubt, uncertainty, or probability other than 1 for a swap to change a car to a goat and a goat to a car. I think this needs to be spelled out to the reader so that there are no lingering doubts. I was going to add either a footnote or a picture to make this absolutely clear. What do you think.
I have also put back the probabilities of swappers and non-swappers getting a goat. You may consider this obvious and superfluous but to a newcomer to the problem this may not be so. Martin Hogbin ( talk) 09:05, 14 February 2009 (UTC)
Thinking about this some more, I am not sure that this solution should even be in the lead section, which is meant to be a summary of the article as a whole. Would it be better to make a very brief reference in the lead and move this explanation into the start of the 'solution' section, maybe with pictures? What we have now (and had before) is one simple solution in the lead and a different one in the body. Martin Hogbin ( talk) 17:14, 14 February 2009 (UTC)
Quote: "But if that event cannot in any way possibly affect the outcome then it is a null condition". Martin, which rule says so? Heptalogos ( talk) 22:02, 14 February 2009 (UTC)
I do agree with you, and actually I was hoping for you to say that 'reducing sample space' is not a rule for conditional probability, but 'affecting probabilities' is. Somehow nobody reacts to that statement. Can you please, on the Arguments page? Heptalogos ( talk) 10:50, 16 February 2009 (UTC)
Have you guys had a chance to see my newest solution (it's not a 'proof', after all)? If I'm right, it shows the unconditional solution solves the conditional problem. And, I won't quote anybody here, but if you follow my 'contribs', you will find an editor-of-apparent-authority saying that the solution is a valid unconditional proof of the MHP. All that would remain is to find a published source, which, since it's right, must be out there. But, see for yourself. Make your own judgements... Glkanter ( talk) 12:37, 15 February 2009 (UTC)
This is the suggested unconditional solution. Feal free to make any improvments to the layout etc. This is the best I could do.--
92.41.74.110 (
talk)
21:03, 14 February 2009 (UTC)
There are three doors. Two doors have a goat behind them and one door has a car behind.
(Remember goat is bad and car is good)
1) Choose one door
2) Game host opens one door containing a goat
3) Do you change door?
For example:
Door 1 | Door 2 | Door 3 |
---|---|---|
Goat | Goat | Car |
Now you have two opions: Not switch or Switch
Not switch
pick | show | outcome |
---|---|---|
1 | 2 | lose |
2 | 1 | lose |
3 | 2 or 1 | win |
If you do not switch the probability of winning is 1/3
Switch
pick | show | outcome |
---|---|---|
1 | 2 | win |
2 | 1 | win |
3 | 2 or 1 | lose |
If you switch the probability of winning is 2/3
You should therefore choose to switch doors
92.41.74.110 (
talk)
21:03, 14 February 2009 (UTC)
pick (No. 1) | show (No. 3) | outcome |
---|---|---|
car (1/3) | goat (1) | lose (1/3) |
goat (2/3) | goat (1) | win (2/3) |
It's one thing to mention the "chances" as you called them, another is what there meaning is and where they come from. Hopefully this leads you to finally understand that what you called "chances" here, are the conditional probabilities. How else could door No. 3 have "chance" 1 to reveal a goat, whence the (unconditional) probability of having a goat behind it, is 2/3? Nijdam ( talk) 13:29, 15 February 2009 (UTC)
I don't know what the meaning is of this 'Bert' stuff. May be you're watching way too much Sesam Street, I at least don't see any connection with the discussion. And you may persist in thinking the doors are not labelled, why not, you may even think there are no doors at all. Or 3 doors and 7 little goats and a wolf, or prince William and his girlfriend behind one door and his grand grand uncle behind the curtains. If it makes you feel better, please do. Nijdam ( talk) 17:56, 16 February 2009 (UTC)
Rick went to http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics today to drum up support for his proposed changes.
I followed him by informing them of my allegations of Ownership by Rick. I suggested that today's behaviour alone shows this problem. Glkanter ( talk) 22:21, 14 February 2009 (UTC)
Rick proposed a major addition to the article today. Just hours after about 4 of us were discussing the merits of a 'formal' proposal for deleting stuff. Then he goes to your page to drum up support. What's up with that? Who's stirring the pot? Who's not showing good sense? I know of one highly respected math editor who today described 'the discussion is so turbulent'. And Rick proposes a major addition that he flat-out knows his primary critic is dead-set against? No. That's not normal.
Here's a probability: if you don't look into it, you will not find what I'm describing. However, If you look, you might find I'm right.
Glkanter (
talk)
04:54, 15 February 2009 (UTC)
You might even ask me if I have any documentation to support my claim.
Glkanter (
talk)
05:04, 15 February 2009 (UTC)
As the former leader of the free world once said, 'Fool me once, <pause>, uh, shame on you. Fool me again...won't get fooled again!'
I fell for the 'rope-a-dope' a second time! Turns out in C S's case, my apology was unnecessary. He, at least, had already made up his mind. Hostile mother.
And don't be coy. Tell me why I am 'so wrong you don't even know why you aren't going to garner any consensus.' I'll change my posting techniques if it's futile. Just tell me the rules. Glkanter ( talk) 13:07, 15 February 2009 (UTC)
The article says:
The host opens a known door with probability p, unless the car is behind it ( Morgan et al. 1991). | If the host opens his "usual" door, switching wins with probability 1/(1+p). If the host opens the other remaining door, switching wins with probability p/(1+p). |
The latter formula cannot be correct. In the extreme case, set p=1 and assume that the host does not open his usual door. This can only have been because the car is behind the usual door, so switching in this case wins with certainty. However p/(1+p) is only 50%.
According to my own analysis, the 1/(1+p) is correct, but the other formula should be 1/(2–p). (It is okay for the two probabilities not to add up to 1; conditional probabilities under mutually exclusive conditions have no necessary relation).
Is this an error in the source, in the reporting of it here, or in my understanding of what the behavior is supposed to be? – Henning Makholm – Henning Makholm 04:46, 15 February 2009 (UTC)
Is there a procedure for investigating WP:Ownership allegations? I'll be happy to share everything I can.
Otherwise, read the Conventional Wisdom section. That's where I lay everything out. —Preceding unsigned comment added by Glkanter ( talk • contribs) 06:03, 15 February 2009 (UTC)
Rather than argue about unconditional/conditional issues we could agree, for the moment at least, to continue what is already done in the article. This is what I have done in my changes to the lead.
I think it is agreed by all that if we consider the overall probability of getting a car of players who swap compared with players who stick there is no conditional issue. All that is needed is to talk in terms of players who swap and players who stick (in other words we consider players to be either stickers or swappers). I think it is generally accepted that, for any reasonable interpretation of the problem, stickers have a 1/3 chance of getting a car and swappers a 2/3 chance and that no conditional probability is required to show this.
The two problems with this approach that I see are firstly that I do not believe that it is actually necessary, and secondly that it detracts from the true nature of the problem as a game show in which the player is presented with a choice after a particular action of the host. This makes an overall probability solution justifiably less convincing to some, nevertheless it may suit others.
However, by being careful with our wording it might be possible to present a very simple and clear solution that is acceptable to all. As far as I am concerned the overall solution already given is fine in principle, it just needs improving in terms of words and pictures. Martin Hogbin ( talk) 10:21, 15 February 2009 (UTC)
(edit conflict) I'm late to this party, and the preceding discussion is very long and sprawling. Let me just dump my opinion, more or less from first principles, on what seems to be the matter:
The classical problem, as stated, is one where the plain and orthodox way to turn it into a mathematical question is to ask about a conditional probability. Calculating this conditional probability in the most by-the-book, pedestrian, not-trying-to-be-smart way will yield the correct answer. On the other hand, a mathematically mature reader will be able to see that because of symmetry, the same correct answer can be arrived at with considerably less calculation by imagining that the player makes his choice to swap or not swap in advance; then all one has to do is calculate an old-fashioned non-conditional probability. All of the foregoing is fact, and hopefully not in dispute; otherwise we're worse off than it seems.
Now, different readers will have different trouble with those facts, and we must try to satisfy all of them, even if that means that some readers need to read explanations that are not directly relevant to the understanding which that particular reader finds easiest. The choice between the two approaches is a trade-off between knowing more theory (conditional probabilities), or using less theory but applying more problem-specific cleverness (invoking symmetry). What one prefers is subjective, and not something that is objectively inherent in the problem.
The sticker-swapper argument is both popular and extremely attractive to readers who are capable of seeing that it is a valid approach, so it would be a bad encyclopedia that did not present it at least somewhere. On the other hand, seeing that it is valid is not entirely trivial, and there will be other readers who are more easily convinced by a straightforward conditional probability. So this must also be presented.
However, finding the right answer is not the end-all. The more curious reader will also want to investigate not only what is right, but also how he managed to be wrong when he initially thought it would be 50/50. And in order to discuss that -- an interesting and now famous perceptional problem that this article must address -- I cannot see how that can be discussed without the machinery of conditional probabilities.
What we then have to decide, from an editorial viewpoint, it not whether to present one or the other raw solution to the original problem, but merely which one to present first. Personally I would gravitate towards starting with the conditional probabilities, the one that requires less ad-hoc cleverness, but reasonable arguments for the opposite order can also be made. In the end I think it is more important the both explanations are good than which order they appear in.
Now have I misunderstood everything? – Henning Makholm 17:29, 15 February 2009 (UTC)
The policies say to resolve a dispute, 3rd party input from subject matter experts may help.
So, Rick and I have solicited this input.
Now, I've been criticized for mis-representing that feedback.
So, absent an edit to this talk page from the subject matter expert, how does one incorporate their contributions into the larger discussion?
I think one expert has unambiguously defended my proof. I think another has said he doesn't know what Morgan's problem with the unconditional solution is, so he'll research it further.
But I don't want to put words in people's mouths. I don't need to. It's out there in the edits.
As far as I can tell, the only tool available to me is to cut and paste quotes and links into new edits.
How do I incorporate those views expressed into this discussion, so as to build a consensus? Glkanter ( talk) 15:08, 15 February 2009 (UTC)
The feeling I got from discussing this page with subject matter experts is that they see it as a black hole, which they wish to fervently avoid. Please go to this page to determine for yourself the validity of this statement. http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics
I believe, then, that any efforts to encourage them to post on this page would be counter-productive.
Nobody has answered my question. 3rd party subject matter experts HAVE made comments, suggestions and opinions. How do we incorporate those into this discussion?
We've referred to Boris. I think he has a doctorate in Mathematics. Here's his page. Read the last section. I think it's reasonable to conclude that he's agreeing with me. But you decide! http://en.wikipedia.org/wiki/User_talk:Tsirel Glkanter ( talk) 19:24, 15 February 2009 (UTC)
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
CS, thanks for the explanation. It's all very obvious what you say, and actually that's not what I was looking for. That's why I referred to the Arguments page; here's where I ask the common rule for a problem to be conditional. Most of us are discussing the Krauss and Wang (2003:10) statement of the problem anyway, so there need not to be much discussion about what was meant by Marilyn. Even if so, it's another discussion outside maths. My question is what exactly makes the Krauss and Wang statement a conditional problem, following which rule. Heptalogos ( talk) 22:25, 15 February 2009 (UTC)
I'm trying to eat my humble pie quietly, but I just gotta ask, what do you mean when you say: "By the way, I hope I haven't given the impression that I think the problem must be understood as conditional."? I'm afraid that is what I think you're saying. I've admitted defeat here precisely because I thought you said that it must be. And I'm not arrogant enough to argue math with a math PHD.
Glkanter (
talk)
00:16, 16 February 2009 (UTC)
This may help me understand. You wrote: "If you assume the unconditional problem, this information is not necessary, since as you pointed out, you only had a 1/3 chance to begin with of getting the car, so if you can switch, you should." What is it about the conditional problem that makes that statement not applicable to it's solution as well?. Monty's motivation, behaviour and actions don't affect that 1/3 chance, do they? —Preceding
unsigned comment added by
Glkanter (
talk •
contribs)
01:26, 16 February 2009 (UTC)
I think an emphais on the unconditional solution for the Monty Hall Problem artcile is appropriate.
This would mean de-emphasising dramatically much of the remainder of the existing article. Glkanter ( talk) 17:08, 16 February 2009 (UTC)
File:MontyXXX.jpg-- Pello-500 ( talk) 14:56, 16 February 2009 (UTC)
What is the difference in content between this and the former "Suggested Unconditional Solution"? Heptalogos ( talk) 16:45, 16 February 2009 (UTC)
You choose a door | |||||
---|---|---|---|---|---|
You choose a door with a goat | You choose a door with a goat | You choose a door with a car | |||
You Stick | You Change | You Stick | You Change | You Stick | You change |
You get a goat | You get a car | You get a goat | You get car | You get a car | You get a goat |
If we were to put a simple unconditional solution this might be a better choice, taken from the 'Curious Incident...'
We can assume that the two goats and the car is located as follows: |
---|
Door 1 | Door 2 | Door 3 |
---|---|---|
Goat | Goat | Car |
We now have two opions: Not Switch or Switch |
---|
Not Switch | Switch | ||||
---|---|---|---|---|---|
Pick | Show | Outcome | Pick | Show | Outcome |
1 | 2 | lose | 1 | 2 | win |
2 | 1 | lose | 2 | 1 | win |
3 | 2 or 1 | win | 3 | 2 or 1 | lose |
If we do not switch the probability of winning is 1/3 | If we switch the probability of winning is 2/3 |
---|
We should therefore choose to switch doors if we want to increase our probability of winning the car |
---|
I know that I favour a simple solution but we do need to recognize that this is a featured article and we should tread carefully. The last few changes make the page much worse in my opinion and should be reverted. Let is try to reach a consensus on how to proceed. before we make such dramatic changes. Martin Hogbin ( talk) 23:13, 16 February 2009 (UTC)
Leaving out the door numbers in the wording of the problem, essentially changes the problem and takes away not only its charm, but also changes it to a form it never was intended to. I.e. it no longer would be equivalent to the three prisoners problem, from which it seemes to be derived, but especially it no longer reflects what happens in the Monty Hall show. Nijdam ( talk) 00:31, 17 February 2009 (UTC)
It's been 24 hours since I posted this:
applicable to it's solution as well?. Monty's motivation, behaviour and actions don't affect that 1/3 chance, do they?
Haven't heard a peep back yet.
So, it's over. You know it, and I know it. And I was right. From my very first post.
Rick Block, I commend you, sir. You were led, badly, by someone who you trusted, with good reasons. You were always, always considerate and professional. I hope someday you and I are on the same team.
C S, you are garbage.
I posted some fun stuff on my talk page. Check it out!
http://en.wikipedia.org/wiki/User_talk:Glkanter —Preceding
unsigned comment added by
Glkanter (
talk •
contribs)
03:08, 17 February 2009 (UTC)
So, now that it has been proven that the alleged conditional statements of the MHP can be solved using the unconditional solution, there really is no 'conditional problem'. Just so much noise. Because Monty can't change the past, so it's still a 1/3 chance I picked the car.
Let's focus on improving the Wikipedia reader's experience with this article. —Preceding unsigned comment added by Glkanter ( talk • contribs) 01:39, 17 February 2009 (UTC) Glkanter ( talk) 03:10, 17 February 2009 (UTC)
I can't follow y'all's heated arguments too good, but I tried to add a bit to make it clear that the assertion that the probability of getting a car is still 1/3 after Monty opens a door depends on some unstated assumptions about whether he always does this or not. The logic did not follow without this, and Martin took it out, so I just put a bit to say it depends on some assumptions. Feel free to further clarify, but don't go backwards to the condition of stating a "deduction" that doesn't follow. Dicklyon ( talk) 02:27, 17 February 2009 (UTC)
My statement of overall probability always holds, see Morgan et al. The host always opens a door, it says so in the question. This is not disputed by anyone except you. Martin Hogbin ( talk) 22:40, 17 February 2009 (UTC)
I just realized that a whole new crew of interested editors are joining the fray.
It's been happening for 4 years now, and will continue, with no reason for abating.
So, I'll eventually be gone, but someone new will take my place.
Why not try to reduce the 'heat' before it starts?
What would help is if the Math guys would, in writing, present the consensus 'house' line on the issue. Put it on the talk page. Not like the old banner that was torn down. Something that would pro-actively indicate the other subject matter experts' opinions are in line with the emphasis of the article. —Preceding unsigned comment added by Glkanter ( talk • contribs)
I've revised the new version of the initial solution a bit. I think rearranging the figure so the "host opens Door 3" and "host opens Door 2" cases are adjacent shows what's going on somewhat better. Actually, flipping these so the "host opens Door 2" cases are on the left might be even better (since Door 2 is shown on the left in the figure). -- Rick Block ( talk) 05:14, 17 February 2009 (UTC)
Any mileage in this? It does not rely on host action but does trick the reader.
Initially you have a 1/3 chance of picking a car and a 2/3 chance of picking a goat. After Monty has opened a door revealing a goat you will always get the opposite of your original choice if you switch. Martin Hogbin ( talk) 19:18, 17 February 2009 (UTC)
I only briefly comment on what I over and over read. Namely, that sticking to your (initial) choice, should logically lead to sticking to your chances, i.e. dealing with unaltered probabilities. I guess many of the people defending the "unconditional" solution think along this line, and never even thought it can be otherwise. But they should! Sticking to your choice is one thing, chances do not change, but situations do, and hence new probabilities have to be calculated: conditional ones of course, given the new situaton. Nijdam ( talk) 22:37, 18 February 2009 (UTC)
Section moved to Talk:Monty Hall problem/Arguments#Decision Tree. -- Rick Block ( talk) 14:20, 20 February 2009 (UTC)
Section moved to /Arguments#"Choice" vs. "Guess". -- Rick Block ( talk) 14:41, 24 February 2009 (UTC)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 5 | ← | Archive 7 | Archive 8 | Archive 9 | Archive 10 | Archive 11 | → | Archive 15 |
When Monty opens a door, he doesn't tell us anything we didn't already need to know. He ALWAYS shows a goat. It makes no difference to this puzzle WHICH remaining door he shows. So it starts out as 1/3 for your door + 2/3 for the remaining doors = 100%. Then he shows a door, but we knew in advance that he was going to show a goat. The odds simply haven't changed following his action. They remain 1/3 for your door + 2/3 for the remaining doors (of which there is now just 1).
That's why on the show, Monty didn't HAVE to show a door (unlike this puzzle), and he could bribe you with cash. Otherwise, there's no show. You always switch.
I'm new to this. Is there a way to get this 'solution' onto the article page?
75.185.188.104 (
talk)
17:09, 25 October 2008 (UTC)
I agree that the presentation of this article leaves something to be desired. I would think that most users come to this page to settle an arguement, or to enlighten themselves on how they don't 'get' the solution. I know that's why I came here.
So, an elegant, simple solution would serve the readers. I'm a stats and probability geek, but I don't like having to 'prove' something by running simulations (since I don't know who created the simulation, I can't trust it). And, if there is an easier to understand explanation that doesn't require umpteen diagrams, the typical reader would prefer that, as s/he will probably try to explain what s/he read here to other people. 75.185.188.104 ( talk) 12:27, 26 October 2008 (UTC)
Re: The 'Combining Doors' comment above. Yes, it's similar. imho, these should be a prominant part of the 'Solution', not relegated to 'Aids to Understanding'. Glkanter ( talk) 15:12, 26 October 2008 (UTC)
I find the explanation in the combining doors section somewhat abstract. Here's the explanation I've been using with people who have a hard time understanding why switching is the best strategy -- I know this probably shouldn't go on the main page (I'm the only reference I know of) but I wanted to share it.
I'm going to write it in the form of a dialog because that is the shortest form of this explanation and also the form in which it has been used:
Am I correct that this would be inappropriate for the main page? Stepheneb ( talk) 17:08, 2 January 2009 (UTC)
I follow the reasoning all the way, yet the following twist to the scenario puzzles me. Can someone explain what I'm missing?
As we know from the article, the first contestant has a 2/3 probability of finding the car. But for the second contestant it's a straight choice between 2 doors, so he has a 1/2 chance of finding the car. So it appears that at the moment the two contestants make their final choices, p(finding the car | choosing door 2) has two different values simultaneously depending on who does the choosing, whereas intuition (mine, anyway) says that the probability should have a unique value, and that it should depend only on the possible permutations of the original distribution of car and goats, and the shared knowledge that door 3 revealed a goat. Help! -- Timberframe ( talk) 09:25, 26 October 2008 (UTC)
Mr. Hogbin is correct. Note that the 2nd player has less likelyhood of picking correctly. 50% vs 67%. This is because he has less knowledge of how the 2 doors came to be. He can only assume randomness, but the first player knows Monty did not act randomly when he opened the door revealing a goat. 75.185.188.104 ( talk) 12:17, 26 October 2008 (UTC)
Thanks guys, got it now. The falacy in my proposition is in the assertion that because there are two closed doors and the car is behind one of them, the second contestant has a 50:50 chance of finding it. This is only true from the second contestant's limited perspective. The conditions of the scenario is posited have in reality already shifted the odds to 2:1. Thanks again -- Timberframe ( talk) 19:29, 26 October 2008 (UTC)
The following is prominant in the article:
"Solution The overall probability of winning by switching is determined by the location of the car...
This result has been verified experimentally using computer and other simulation techniques (see Simulation below)."
This first sentance doesn't mean anything, and in fact is incorrect. The probability is NOT affected by the location of the car. One instance of playing the game is affected, but not the probability.
Further, a valid proof, which I believe my entry is, does not require being "verified experimentally using computer and other simulation techniques". A valid proof stands on it's own.
And lastly, this at the top of the Discussion Page is at a minimum, off-putting:
"Please note: The conclusions of this article have been confirmed by experiment
There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment."
As I stated above, valid proofs do not rely on 'experiments'. Yet, there are two references to 'experiments'. Me thinks the lady doth protest too much.
75.185.188.104 (
talk)
13:34, 26 October 2008 (UTC)
Can I mention again that I have set up this page: user:Martin Hogbin/Monty Hall problem (draft) for those, like myself, who do not find the current solution clear or convincing to develop a better basic solution. The concept is to develop a very clear solution that is verifiable and which can be eventually added to this article. There is also a talk page for discussion on how we might achieve this. Martin Hogbin ( talk) 16:58, 26 October 2008 (UTC)
"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)."
I have no idea what this means, and fail to see how it part of a solution to the problem. People come to this page for an explanation. The above paragraph, plus whatever follows it, are not part of the solution to the Monty Hall Problem, and, imho should not be present there. Glkanter ( talk) 14:59, 26 October 2008 (UTC)
(outindent) The question is if it's advantageous to switch after we've picked a door and the host has opened one of the remaining doors. Your solution is the following:
I've highlighted the problematic parts. These are assertions, not deductions based on what is given in the problem statement. They're true statements, but stating them without justification is equivalent to assuming the solution. We're not given that it makes no difference which door the host opens and we're not given that when the host opens one of the two remaining doors the initial 1/3 odds don't change - this is essentially what we're asked to show. It's like in a geometry proof using that an angle in the diagram is a 90 degree angle because it looks like one without showing why it must be one. If indeed it must be one because of what else is given it's certainly true, but using it without justification is at best incomplete. The simple solution currently in the article (3 equally likely scenarios, switching wins in 2 and loses in 1) doesn't have this problem. Do you find this solution hard to understand? -- Rick Block ( talk) 19:21, 26 October 2008 (UTC)
Several editors believe that, although the current article may meet the FA criteria, it still remains unconvincing and also concentrates far to much on variations of the basic problem that are irrelevant to the core problem.
It is therefore proposed that the current solution be replaced with a new version giving simple, intuitive, and verifiable solutions to the basic problem.
Preparation for this new section is taking place on a [ [1]] . All interested editors are asked to make their comments on the suitability of this section for inclusion in the current article and to make improvements to the new section. Martin Hogbin ( talk) 20:54, 6 November 2008 (UTC)
Perhaps you might be able to improve the development article. Martin Hogbin ( talk) 09:26, 7 November 2008 (UTC)
I got it! Start a new Wikipedia article called 'The Equal Goat Door Constraint Paradox'! Marilyn vos Savant can write a column about it in her general interest Sunday magazine column. People all over the world will marvel at how counter-intuitive the solution is, when all along, after a trip to the nearest university library, Morgan, et al and Bayesian statistics were there to explain it. Glkanter ( talk) 16:17, 8 November 2008 (UTC) Glkanter ( talk) 16:26, 8 November 2008 (UTC)
The following introduction is still confused suffering from poor English, a lack of clarity and some misconceptions.
"Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3. Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3. With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty. The total probability of winning when switching is thus 2/3."
The object of the problem is not whether the person wins the automobile but how the probability shifts when a door is exposed (with a goat). If the person selects the correct door or not on the first guess is irrelevant to the problem - especially as an introduction.
A breakdown:
(1) Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter.
Is fine
(2) In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.
Does not need "in the usual interpretation of the problem" in the introdcution.
(3) Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3.
Should read, "If the player initially chooses the winning door then changing is clearly disadvantageous" I don't even think it is necessary as this is confusing in the introduction.
(4) With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty.
Is badly constructed. It should read, "If the player has initially chosen one of the two losing doors: when one of the losing doors is revealed then switching wins the player the automobile."
However, I would argue that this isn't even needed anyway.
(5) The total probability of winning when switching is thus 2/3.
Is a non-sequitur as the paragraph is referring to two different scenarios (ref to 2 and 3). Both scenarios, incorrect first choice and correct first choice cannot lead to the same probability of winning when switching. -- Candy ( talk) 22:52, 24 November 2008 (UTC)
Reversion of good faith edits should not be marked as minor. Martin Hogbin ( talk) 20:27, 11 December 2008 (UTC)
I've moved the following link which was added to the article here. There's nothing on this page that can't be incorporated directly into the Wikipedia page, per WP:EL that means it is not an appropriate external link.
-- Rick Block ( talk) 14:39, 19 December 2008 (UTC)
--- I believe the above link should be on the main page for the Monty Hall Problem. It cites important academic journals and articles on the issue. We cannot ignore literature that discuss this problem as it was a hot topic of discussion amongst prominent economists and mathematicians. Some of the articles mentioned in the above link do form a basis to the solution that is widely accepted today. - Aadn
What if I want the goat and not the car? Should this probability also be addressed, even if it's not part of the original Monty Hall problem? 76.95.124.146 ( talk) 15:09, 8 January 2009 (UTC)
I've undone this edit which adds an introductory paragraph to the Solution section with numerous WP:MOS issues discussing an alternate problem (boxes rather than doors). The addition is essentially the same as the existing "combining doors" section. The text is in the history if anyone wants to talk about it. -- Rick Block ( talk) 06:01, 24 January 2009 (UTC)
I've deleted the addition of the following new section.
Although interesting, it's unreferenced and sounds like WP:OR (and requires an understanding of Minesweeper). If anyone can find a reliable source including this analogy we can talk about re-adding it. -- Rick Block ( talk) 01:43, 25 January 2009 (UTC)
The, what is often called, "simple solution", saying: the player originally has a chance 1/3 of winning, so switching increases this chance to 2/3, is theoretically wrong! This explanation is also offered directly under "Solution" , and although it offers numerically the right answer,it is not correct, as may be seen by comparing it to the statement of the problem right above it. In the problem statement the question is: Should the player switch to Door 2? And in the solution the player is allowed to switch to Door 3 as well. The point is, and I think it is mentioned somewhere in the above discussions, that the posed question can only ask for a conditional probability, given the situation, i.e. Door 1 chosen and Door 3 opened revealing a goat. This is also clear from the equivalence with the Three Prisoners problem and Bertrand's box paradox. Nijdam ( talk) 11:26, 3 February 2009 (UTC)
This is an other way of putting it. But as the article is so precise in defining the problem, with all guarantees of no possible misinterpretation, the given "simple solution" is no solution to this precise stated problem. Nijdam ( talk) 16:04, 3 February 2009 (UTC) —Preceding unsigned comment added by 82.75.67.221 ( talk)
Any of you recent contributors may not have read this page in its entirety. It's pretty long. So, just to let you know, many, many people are very unsatisfied with the entire article. It's long winded, confusing, and outright wrong in places. But we respect the process, so here we are. Search for my signature, as one example. Glkanter ( talk) 03:30, 4 February 2009 (UTC)
I tried to follow the discussion above between the two of you, and I actually got confused, not about the problem or the solution, but about you. I'm not very happy with Rick's revision of my extension to the "simple solution". My extension shows in essence the flaw in this way of reasoning. I showed that the "simple solution" is not a solution to the stated problem. It's like the man, searching in the evening for his lost wallet under a streetlamp where things are well visible, allthough he knows he lost the wallet in a dark side alley. So I invite either one of you to read my extension an comment on it. Nijdam ( talk) 10:49, 4 February 2009 (UTC)
I did a little more reading of the above discussion about this topic and I think I find Rick on my side, but Glkanter opposed because of the reasoning: 1/3 + 2/3 = 1, meaning: the initial probablility is 1/3 of picking the car, so whatever happens, as (!?????) this has not changed, the final probability of the alternative should be 2/3. Well as you may guess, I already indicate where it goes wrong. The "initial probability changes" under the assumed condition. Among all possible cases, the ones with the car behind Door 1 and Door 3 openend, only form 1/6th. Nijdam ( talk) 11:28, 4 February 2009 (UTC)
Nijdam, please show me the error in the 1/3 + 2/3 = 1 (100%) solution. I'd also like to know what is unacceptable about Martin's solution, 2/3 of the time I choose a goat, and 100% of those times Monty show me the other goat? Everything else, from Morgan, et al, to Bayesian, to "We have run simulations that prove this" is wasted binary digits.
Glkanter (
talk)
23:12, 4 February 2009 (UTC)
In a moderating kind of way, I have two suggestions here. 1) Please comment on content, not on other editors' mental facilities. WP:COOL has some good suggestions. 2) As far as possible, please support arguments with references rather than arguing what you simply know to be the Truth. There are no shortage of references about the Monty Hall problem - any significant point has almost certainly been published (at least once). -- Rick Block ( talk) 02:00, 5 February 2009 (UTC)
Rick, I'm still waiting for your clarification of the above statement. As it is central to your claim of the non-validity of my proof. What does this mean:
Are they talking about the Monty Hall Problem or some varient? Or is this the 'before switching' beast you speak of? MOST IMPORTANTLY, what possible relavence can it have to my proof?
Glkanter ( talk) 07:30, 8 February 2009 (UTC)
Well, don't be offended; take it as a complement instead, because from the discussions above I concluded both of you are well skilled at least in mathematics. Yet you keep asking for arguments, which I gave in the extension of the article and extensively above. Moreover, Rick quotes the same arguments I gave. So what more needs to be said? If the "simple solution" as a solution to the well stated problem, is supported by any source, the conclusion can only be, this source is not reliable. The only necessity is changing he article, as I suggested. 82.75.67.221 ( talk) 10:01, 5 February 2009 (UTC)
Well, I find it unacceptable to start with a "solution" that isn't a solution. My suggestion (please improve my English, as I'm foreign):
According to the problem statement above, a car and two goats are arranged behind three doors and then the player initially picks a door. Assuming the player's initial pick is Door 1 (the same analysis applies for any other door the player picks):
This leads to the following situations:
Players who choose to switch win if the car is behind Door 2. This is the case in 2 out of 3 cases, so with 2/3 probability.
This means if a large number of players, after choosing Door 1 and being showed a goat behind Door 3, 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).
Nijdam ( talk) 17:56, 5 February 2009 (UTC)
According to the problem statement above, a car and two goats are arranged behind three doors and then the player initially picks a door. Assuming the player's initial pick is Door 1 (the same analysis applies for any other door the player picks):
This leads to the following situations:
Looking at the figure above, in cases where the host opens Door 3 (which is the case according to the problem statement) switching loses in a 1/6 case where the player initially picked the car and wins in a 1/3 case where the player initially picked a goat. Players who choose to switch win twice as often as players who stick with the original choice, which is a 2/3 chance of winning by switching.
This means if a large number of players randomly choose whether to stay or switch after initially choosing Door 1 and being shown a goat behind Door 3, then approximately 1/3 of those choosing to stay and 2/3 of those choosing to switch would win the car.
Sorry, among all the almost simultaneous contributions to this page, I didn't notice this part. I'm sorry too to say I prefer my suggestion. Especially because the exact formulation of the problem, as given in the article, states Door 1 is chosen and Door 3 opened. Nijdam ( talk) 12:23, 8 February 2009 (UTC)
Note: There's a summary below, see #Request for brief clarifications. -- Rick Block ( talk) 20:55, 8 February 2009 (UTC)
I'll be honest. I don't understand most of what you guys are talking about. It's a simple puzzle. Do you increase your chances of winning the car by switching? It's been years since I took a Logic course, or a Probability and Statistics course, so I can't cite chapter and verse. But the goal is to present a valid proof with as little complication as possible. So you take shortcuts, which don't affect the proof. Two goats? I only care about the car. Ignore the goats in the proof. 3 doors? There'll only be one 'other one' left when I'm asked to switch. Ignore the door #s in the proof. Conditional probability? Don't know what it means. Don't see how anything is conditional here. Two closed doors when I'm asked if I want to switch, exactly one has a car. No conditions exist.
I do remember something about a requirement for proofs being valid and true. 2/3 of the time the remaining door I didn't choose has a car. Both valid and true.
That's why the real game show didn't work like this. Glkanter ( talk) 03:16, 6 February 2009 (UTC)
Okay Rick, at least I see you completely knows the ins and outs. So what to do? I cannot accept a reasoning that's logically wrong. What about the suggestion I made above, by changing the given "simple solution" in such a way that some possibilities are ruled out as not happened? Nijdam ( talk) 11:54, 6 February 2009 (UTC)
There are really only two situations to consider. Not 3, not 6.
2/3 of the time I will choose a goat.
100% of those times Monty reveals the other goat, leaving the car.
=>2/3 of the time I should switch.
1/3 of the time, I will choose the car.
100% of those times Monty reveals one of the goats. It doesn't matter which.
=>1/3 of the time I should not switch.
2/3 is greater than 1/3.
Therefore, by switching, I increase my chances of getting the car when offered the switch.
I earlier said a good proof uses 'shortcuts'. A better phrasing would have been 'eliminates the unessential' from the proof. That is, a goat is a goat, and a door is a door.
Martin's proof, and the one above, address the problem as defined. To deny that is incorrect. [User:Glkanter|Glkanter]] ( talk) 13:34, 6 February 2009 (UTC) 63.97.108.114 ( talk) 14:19, 6 February 2009 (UTC) Glkanter ( talk) 14:22, 6 February 2009 (UTC)
Thank you for the specific response to my question. I'll try to look at those papers. I suppose Morgan, et al is one of those mathemeticians, who's the other?
Do I understand your response to Nijdam correctly? Does "many published sources (possibly all popular expositions of this problem) include only an unconditional solution" mean that the solution I've been advocating is what most other references provide as the only solution?
Would you indulge me? In your own words, tell me which of the eight statements, beginning with '2/3 of the time I will choose a goat' and ending with 'Therefore, by switching, I increase my chances of getting the car when offered the switch.' is either untrue, invalid, or not germain to the question 'Now that Monty has revealed a goat, do you increase your chances of winning a car by switching'?
Thank you. Glkanter ( talk) 07:06, 7 February 2009 (UTC)
I'll just address 3 & 4, my answer renders 1 & 2 moot.
Twice, my proof includes the words "100% of those times Monty reveals...(a) goat." I don't see how my proof can be of anything but the after-Monty-opens-a-door scenario.
Enough about before or after! A premise of the puzzle is that Monty will reveal a goat. There is no 'reveals car' scenario to consider! That would not be the Monty Hall Problem! So of my 8 line proof, 2 of those lines, for clarity purposes, repeat a premise. I'm surprised that this has come up again, we've already been around this block.
So, if that's all you got, let's declare this a valid proof and move forward. Especially eliminating that 'flawed solution' talk from the 'Solutions' section.
Glkanter ( talk) 00:39, 8 February 2009 (UTC)
Glkanter ( talk) 01:06, 8 February 2009 (UTC)
I deny the existance of different 'before' and 'after' questions. There is only one. It goes, "After Monty has revealed a goat, do you increase your chances of winning by switching?"
That is the only problem I have addressed since joining the discussion.
That is the only problem I have provided a proof for.
That is the only problem which I have asked you to find a flaw in my proofs.
And it is the only problem I will discuss on the discussion page of the Monty Hall Problem.
You're last gasp at finding a flaw was that I'm solving the 'before' puzzle. I pointed out the error in that statement, that twice the proof says "Monty reveals a goat". Which is a premise, for crying out loud!
So, whether or not there are two puzzles, I am solving the 'right' one. What's your critisism now?
Glkanter ( talk) 06:21, 8 February 2009 (UTC)
Your question is immaterial to the Monty Hall Problem we are solving. I have no idea or interest in it's clarity. It is obsfucation. You are avoiding the Monty Hall Problem question. I have presented a proof, then sought meaningful critique. I have explained how a proof is validated. Yet you repeatedly refuse to remain engaged in a constructive discussion of my proofs.
Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.
Glkanter ( talk) 06:43, 8 February 2009 (UTC)
I read every word you post. Believe me. Look, here's how it works. Someone puts forth a proof. Then everyone else looks for where there may be errors. Then when everyone is satisfied that it's correct, it is accepted. Once it's accepted, there is no saying about it: "Although the reasoning above is correct it doesn't answer the precise question posed by the problem, which is whether a player should switch after being shown a particular open door (Morgan et al. 1991)."
Step 1. Somebody puts forth a proof. I've done that.
Step 2. Everybody else looks for where there may be errors. In the proof. Nowhere else.
All you have ever done is say that I am not solving the appropriate question. To that, I again say, Hogwash! Each of those 8 lines passes. And the conclusion: 'since 2/3 is greater than 1/3, switching wins more often', answers the question being asked.
Enough, already!
Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.
Why don't we admit that neither of us is going to be persuaded by the other? I've recently requested that Martin take this to the next level of Conflict Resolution. I'm tired of the circular nature of the argument, but I think I'll stick around.
Glkanter (
talk)
18:35, 8 February 2009 (UTC)
MCDD is Morgan and three other guys. <Oh! The three other guys are the 'et al'!> Glkanter ( talk) 22:00, 8 February 2009 (UTC)
"In fairness, MCDD do moderate their tone later on, writing, "None of this diminishes the fact that vos Savant has shown excellent probabilistic judgment in arriving at the answer 2/3, where, to judge from the letters in her column, even member of our own profession failed.""
http://www.math.jmu.edu/~rosenhjd/ChapOne.pdf page 48
Who is John Gault?
Glkanter ( talk) 21:00, 8 February 2009 (UTC)
I'll do one effort to explain the problem. The situation is already conditional on the choice of Door 1 by the player. And we know Door 3 has been opened revealing a goat. Let us assume the game has been played 3000 times. What might have happened?
Player has chosen Door 1 | |||
---|---|---|---|
Car hidden behind Door 1 | Car hidden behind Door 2 | Car hidden behind Door 3 | |
1000 times | 1000 times | 1000 times | |
If nothing else is revealed | |||
Player wins the car if he doesn't switch | Player wins the car if he switches | ||
But player gets "information" | |||
Host opens Door 2 | Host opens Door 3 | Host must open Door 3 | Host must open Door 2 |
500 times | 500 times | 1000 times | 1000 times |
This case didn't happen | One of these cases happened | This case didn't happen | |
Player wins the car if he doesn't switch | Player wins the car if he switches |
Hence only in half the cases did the situation arise in which the player finds himself. So we see it cannot be unconditional, because only a part of all possibilities could have happened. Clear? Nijdam ( talk) 18:05, 6 February 2009 (UTC)
Rick, I thought that it had been agreed that, for the fully defined problem (host alway offers the switch and always opens a goat door), there is no distinction between the conditional and unconditional cases. Martin Hogbin ( talk) 22:56, 6 February 2009 (UTC)
Well, to be honest, I'm not. I strongly oppose starting with a reasoning that is not a solution to the stated problem. The right solution, as I suggested above, is quite accessible to the layman. If anything has to be said about the "simple solution", it should be as a solution to a (slgihtly, but essentially) different problem. It doesn't seem right to me, to have thoughts like "keep the laymen in the dark, it is no use to explain things properly to them".[Sorry, forgot to login] Nijdam ( talk) 12:59, 7 February 2009 (UTC)
Nick, in what way are the conditional and unconditional cases distinct in the case of the fully defined problem? Martin Hogbin ( talk) 14:44, 7 February 2009 (UTC)
Here's something to think about:
The presentator has a simple strategie: when possible he opens Door 3, otherwise he tries Door 2 and is that also impossible he opens Door 1. Quite easy. See what happens:
Yet the initial prob. of winning the car is in all cases 1/3 and according to the "simple reasoning" switching would increase it to 2/3. Notice BTW that the overall prob. of winning the car when switching is (of course) 2/3. And also conditional given the chosen door. But (!) not conditional given the chosen and the opened door. Don't let it keep you out of your sleep. Nijdam ( talk) 22:56, 7 February 2009 (UTC)
Interesting that you put is this way. Because what is the meaning of "no information is revealed"? This actually means that the possible outcomes are not restricted. And here they are: allthough the conditional probability is the same as the unconditional, the information revealed is for instance that the car is not behind Door 3. Nijdam ( talk) 11:12, 8 February 2009 (UTC)
Alas(?!), no. You really miss the point. The information revealed - with the proper strategie of the host - does indeed not affect the probability of winning by switching. But the meaning of such a statement is that the conditional probability given this information, is the same (numerically) as the unconditional one. And as I said before: the revealed information restricts the possible outcomes. I get the feeling that all efforts are made, against better judging maybe, to avoid rejecting the "simple solution". I sincerely hope this is not the case. Nijdam ( talk) 22:58, 8 February 2009 (UTC)
Well, some things happen and do not limit the possible outcomes. Like the started rain outside or the joke the host tells in between. But, as I over and over argued, the open door does. I gave a "simulated" example further down, in reply to Gkanter. Have a look there and be convinced. I really can't think of anything more that should be said. Maybe the real issue is some discussiants are not familiar with probabilities, let alone conditional probabilities. Nijdam ( talk) 00:00, 9 February 2009 (UTC)
I agree with Glkanter: "Everybody else looks for where there may be errors. In the proof. Nowhere else." He may not be responding to the conditional problem, as presented by Rick Block, but should he? People try to explain the conditional situation, but why do they fail to explain to him why the particular problem cannot be unconditional? Where are the errors in the simple solution, as an answer to the given question? I found some probable 'answers' to that:
The question is: should the requested chance be regarded as a conditional probability, and if so, why? Morgan et al. say it should, but their only argument or hint is that the information in the number of the door is otherwise not used. What information is in the number? Does it matter if the (blind) player has knowledge of that? What other information might be relevant? Should it be limited to events which restrict the possible outcomes? Even if it doesn't affect the outcome?
The article Conditional probability states that P(B|A) = P(B) if A and B are statistically independent. In other words: conditional and unconditional chances are the same if A does not change the probability of B. And it doesn't. Heptalogos ( talk) 15:34, 10 February 2009 (UTC)
I've been re-reading some past postings. According to Rick, this article has been reviewed on 2 occasions as a 'Featured Article', and that much of what I find inessential actually was a (by)-product of those reviews. Rick is proud of 'shepherding' this article through at least one of those reviews.
So, in some ways, I seem to be arguing against Conventional Wisdom. But I don't feel that way. I have a few college courses on this topic, over 30 years ago, and a lifetime of being a data analyst. My viewpoint is, 'There is no possible way I am wrong about this'. To me, this whole discussion has as much to contribute as a discussion of whether the sun will rise in the east tomorrow morning.
How does a single voice effectively confront the Conventional Wisdom? This is a question not just for Wikipedia, but any societal system. In the US, a swindler set up a Ponzi scheme on Wall Street. Individual investors went to the regulatory agency numerous times, but to no avail. The guy didn't actually get caught. He turned himself in! How does a minority, but important, voice get heard?
Yes, I look at this entire article, excepting maybe 5% of it, as an elaborate hoax. I think everyone went along because they did not want to admit to limited knowledge of the subject matter. Everybody drank the kool-aid. And the emperor is wearing no clothes.
2/3 of the time I will select a goat. Therefore I should switch. Glkanter ( talk) 15:32, 7 February 2009 (UTC)
And really, who wants to argue when at the top of the page it says:
Please note: The conclusions of this article have been confirmed by experiment |
There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment. |
Two mentions of proof via experimentation. I'll bet not a single reference source uses 'experimentation' as their solution. Because 'experimentation' is not a statistical or logical proof. An error like this confirms my above stated concerns. I'm curious, is it common for Wikipedia discussions to have such a banner? I've never seen anything like it in any 'academic' setting before. Anywhere. Glkanter ( talk) 15:54, 7 February 2009 (UTC)
And this, it's the 2nd paragraph after the 1st chart in the Solutions section. It's one of the more prominent items in the Solutions section.
What sort of a 'Solution' section leads off with a proof it is going to immediately (and fraudulently) discredit? And then goes on endlessly doing so? For who's benefit? The once-interested reader is by now, long gone.
Glkanter (
talk)
17:06, 7 February 2009 (UTC)
Here's my only edit to the actual Article. On October 23, 2008 I deleted the following erroneous line:
The above statement was the very first words of the Solutions section.
Imagine. Being THAT wrong on the very first line.
Glkanter ( talk) 17:57, 7 February 2009 (UTC)
So, I posted the following two days ago, at 13:34, 6 February 2009:
Eight lines. Including a premise repeated twice for clarity. Since then, there have been countless postings by Rick. Running off on tangents to and fro. But we NEVER focus on the 8 statements. It's always some BS about 'before or after', or whatever the mcguffin of the day is.
It's a valid proof, and could be shorter. Mr. Block! Tear down your wall!
Glkanter ( talk) 07:46, 8 February 2009 (UTC)
This is why I say it's time to end this charade, now.
In October, 2008, Martin posted this to Rick:
So, it's not just me. It's been enough people, for a long enough time, that the offical Wikipedia RFC process was put in motion. Good. It's time for the next step in the process. BTW, who do suppose could be the 'overly protective current editors' referred to?
Glkanter ( talk) 15:34, 8 February 2009 (UTC)
Do I sound frustrated? Well, I am. Here's another one from October. Sound familiar?
What was that completely erroneous statement? Just this:
So, yes, I'm personally invested. I'm interested. I just read that there are 'archive' pages for this discussion. So what we see here, is just the most recent discussions of this lunacy. The tip of the iceberg, so to speak.
So, enough of the charade!
Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense. —Preceding unsigned comment added by Glkanter ( talk • contribs) 15:54, 8 February 2009 (UTC)
You know what? It's been long enough. Martin, I understand that at some time last year you were specifically requested to get involved, in order to facilitate the resolution of some conflict(s) relating to the Monty Hall Problem.
I ask Rick to invalidate my proof, he tells me I'm not addressing the Monty Hall Problem.
It is obvious these discussions have gotten nowhere, and are going nowhere. So I request that you declare this situation as 'at an impasse', and escalate this to the next level of Conflict Resolution.
Thank you.
Glkanter ( talk) 17:05, 8 February 2009 (UTC)
Since October 25, 2008 I have posted on this discussion page 70 times. 66 as Glkanter, and 4 previous times as an ip address. In addition, there was a separate discussions page that was set up, and I posted there many times as well.
70+ postings over a 2 line proof! When I pick a goat, Monty leaves a car. I pick a goat 2/3 of the time.
I feel that this has gone on long enough! I honestly wonder if there isn't a bot out there generating responses from a limited selection of inaccurate/meaningless/non-relavent statements.
This has been going on with countless other editors since at least 2005. Check out the archives sometime.
To the powers-that-be at Wikipedia, I implore you, please bring this to a conclusion!
Glkanter ( talk) 15:43, 10 February 2009 (UTC)
I just went to the page where Rick requested assistance. One of the people there, who described himself as 'an expert in probability' had this to say:
Does this mean what I hope it means?
Glkanter ( talk) 22:37, 10 February 2009 (UTC) moved comments 76.243.187.238 ( talk) 01:29, 11 February 2009 (UTC) doh! Glkanter ( talk) 01:50, 11 February 2009 (UTC)
After just 4 months, I've run out of new things to say. So rather than clog up other people's exchanges, I'll just add commentary here. I don't see the point in further arguing, but I'm not giving up the quest.
I see the 'random goat constraint' has reared it's (their) ugly head again. In my proof, which I maintain is sufficient for this puzzle, the goats are immaterial. The puzzle asks about increasing the liklihood of selecting the car, not the location of a goat.
Does the probability change if instead of goats there is nothing? So would we have a 'random nothing consraint'?
My understanding of Probability Theory says if an item isn't specifically addressed, it's considered random. And anything that's random is not a constraint. And if it's not a constraint, it doesn't need to be addressed.
For those keeping score, I was indeed reported to the authorities. I, and others with my criticisms were described as a 'cadre', 'wanting to "dumb down" ' the solution by Rick.
Glkanter ( talk) 14:36, 11 February 2009 (UTC)
And, the doors and their numbers are meaningless. The doors only exist to shield the contestant's view of the car and the two goats.
Let's focus on why this is such a famous puzzle. Because it's easily understood (2/3), and easily misunderstood (50/50). So, no reader really understands the solution until he has mastered the 'random goat constraint'? Pure folly.
Nobody cares about the specific case of doors 1 and 3, any more than they would care about suitcases 11 and 25 in Deal or No Deal. That's not what Probability Theory is about. Here, I disagree, until my last breath, with Rick. Probability, and this puzzle are always solving the 'over time' question. If repeated endlessly over time, on aggregate, what would the results be?
Glkanter ( talk) 14:54, 11 February 2009 (UTC)
I was struggling for the right word earlier, but it came to me. Despite what Rick is posting elsewhere about 'physical doors' and such, all Probability can do is create a mathematical model that approximates real life. That's what a proof sets out to do. And if valid, it serves as a proxy for 'real life'. From which we can analyze and draw conclusions.
I, of course belief I have presented various proofs that meet this criteria. Rendering all that other stuff, at best, redundant and confusing, and at worst, erroneous.
Glkanter ( talk) 16:19, 11 February 2009 (UTC)
The Monty Hall Problem gained much (most) of its current noteriety when it was written about by Marilyn Vos Savant in Parade Magazine. I read about it there. We all know the story about how a thousand PHDs challanged her, etc., etc.
But ultimately, Vos Savant was acknowledged as having the right answer at 2/3.
Let's hear from the great lady, herself, courtesy of Wikipedia'a Marilyn Vos Savant page:
Or this, from the Newspaper of Record:
So, even as we argue semantics on this page, it's unambiguous what problem she was solving. With all that hub bub, what proof did Marilyn and all these people use to come to the conclusion that 2/3 is right? Well, it wasn't Morgan et al, via the 'random goat constraint'.
So, if the unconditional probability proof was good enough for Marilyn, Monty, 1,000 PHDs, and 10s of millions of Parade magazine's general interest readership, why is it not good enough for the Wikipedia readers of today?
Glkanter ( talk) 08:01, 12 February 2009 (UTC)
I've just learned that Rick's comment that being 'Right' is trumped by being 'Verifiable' for Wikipedia purposes has some merit. I further understand that being in the mainstream (my term) trumps being in the minority (my term).
We're still arguing over who's 'right'. But apparently, that's not even the salient point.
Would anyone describe the 'random goat constraint' as being in the mainstream of the published literature? How about after reading this?
Glkanter (
talk)
13:35, 12 February 2009 (UTC)
I've posted here nearly 80 times trying to prove my proof is valid, and that I'm 'right'. Turns out, that was never the argument I needed to win.
So why, since 2005 at least, hasn't the response been that the more meaningful (for Wikipedia purposes, anyway) arguement is, to paraphrase, 'popularity of the published material'? Not until about two days ago, anyways.
So, thanks for wasting 4 months of my time. And 4 years of various other people's time.
Now, can we start fixing the Article?
Glkanter ( talk) 14:48, 12 February 2009 (UTC)
Pretty interesting exchange relating to my 'being reported' by Rick at the bottom of Rick's talk page. First, he gets spanked for crying wolf, then he writes this:
Maybe next I'll be reported for trying "to gain consensus for their desired change". Does trying to build consensus make me a bad Wikipedian?
Glkanter ( talk) 17:35, 12 February 2009 (UTC)
Would you consider the following statement as an indication that an editor has claimed 'ownership' of an article?
Could this be an 'Aid To Understanding' for why there have been 4 years, 7 archives, and thousands of postings calling for significant changes to the Monty Hall Problem article, all to no avail?
Let me ask the 'cadre', is there 'consensus' that this editor should be reported for violating the Wikipedia Ownership policy as it relates to the Monty Hall Problem article? If forced, I guess I could be the one to make the report.
Glkanter ( talk) 18:06, 12 February 2009 (UTC)
Does anybody know what form any sanctions would take? I hope it would include being blocked from editting both the MHP Article and this talk page.
Glkanter (
talk)
18:45, 12 February 2009 (UTC)
Here's where Rick first asked for assistance to aid in Resolving our Conflict.
All these other Wikipedia Math gurus already knew about Rick's MHP article Ownership issues!
I'm a first-timer here. It's been way too long, but is has been instructive as to how horribly mishapen things get when an editor claims ownership of an article.
Glkanter (
talk)
19:25, 12 February 2009 (UTC)
If you're still reading these edits, you night appreciate my very first posting. I created a new section that day, but I still don't know how to link to it. "Monty's Action Does Not Cause The Original Odds To Change."
We were all so civil then! I was all ready to make my first edits to the Article! Glkanter ( talk) 19:47, 12 February 2009 (UTC)
You guys should read the Featured Article Reviews. There's a link to each of them at the top of this page. In addition to some of the inane discussions we take part in here, the FA reviewers also argue about footnotes.
And there is one recognizbale name constantly making edits. That would be Rick. More documentation as to his 'Ownership' of the article.
Now, after his encouragement to suggest edits, he says this:
Looks like an 'Ownership' issue to me.
Basically, Rick is saying that the Article is already perfect. Therefore, any changes will make it un-perfect. And he knows, because he's written most of it, nominated it for FA, 'sheperded' it through FA, and personally responded to every FA critism himself. Through his 4 years of Ownership-fueled protectionism, he also has granted the existing Article 'tenure'. I wonder if there is such a policy as WPTENURE? Glkanter ( talk) 04:02, 13 February 2009 (UTC)
I'll probably have to document my claim. Here's another good one from Rick:
Thanks, dude. You're too generous. No, really. Glkanter ( talk) 08:01, 13 February 2009 (UTC)
"This page is 495 kilobytes long. It may be helpful to move older discussion into an archive subpage." Plus 7 archive pages dating as far back as 2005.
With my edits to this talk page, I have been accused of violating WP:CIVILITY, and worse. I'm comfortable with everything I've written. As I wrote at the start of this section, it's difficult for a minority voice to be recognized as credible, when it defies the Conventional Wisdom. So, I am trying to accomplish what seems, so far, to be a difficult task. The only tool I have is my ability to post on various Wikipedia pages. Talk about a probability problem! How do I know who to turn to?
So, I post here. I can't post in that pseudo-collegial, 'oh, beg to differ, chum', style that seems so popular. And I can't out-reference or out-WP:Policy Rick. He's got like 10,000 article edits to his name. No, all a punter like me can do is write with common sense and clarity, and try to draw some attention. So I have my own 'style'. Now you know why. Glkanter ( talk) 14:49, 14 February 2009 (UTC)
Let me ask it this way, in the 1/3 of the time when I select the car, how is Monty's motivation or actual actions going to change the fact that the car was my choice? And that by switching, I would give up the car? How is the 'equal goat door constraint' the solution to that question? It isn't. For probability proof purposes, like the most simple 2 or 3 line proof, we never even need to go there. I reject anew Morgan et al.
Glkanter (
talk)
16:13, 14 February 2009 (UTC)
Here's my new proof.
Look at that. No Monty. No Monty's behaviour. Thank you, thank you very much.
Glkanter (
talk) 16:21, 14 February 2009 (UTC)
Glkanter (
talk)
16:37, 14 February 2009 (UTC)
I think this is even better.
No Monty, and now no goats. Glkanter ( talk) 16:53, 14 February 2009 (UTC)
Those of you who reject the existance of Original Research are excused, and free to resume cataloging references to Bigfoot, Unicorns, etc. Glkanter ( talk) 17:17, 14 February 2009 (UTC)
I just realized that any answer other than 'never switch' inproves your liklihood of winning. The worst you can do is not switch, and go 1/3. Do anything else, I'm just thinking 'go random', and yours odds increase, in this case to 50%.
Glkanter (
talk)
23:13, 14 February 2009 (UTC)
I came here from a request on WT:WPM. But the conversation is too long to read. Could each person post a 1-paragraph summary of what she or he thinks are the main issues under discussion? thanks, — Carl ( CBM · talk) 20:30, 8 February 2009 (UTC)
In September or October I used the Monty Hall Problem Article to further my understanding of the puzzle. I found about 1000% more editorial matter than I would have expected. It's roughly 1% useful, 99% BS. Since then, I've been trying to establish why I believe it's 99% BS in order that this be deleted. I have chosen to use Probability Theory as the (only) appropriate method to convice the other editors. I have proferred what I am confident is a valid proof, and have requested that other editors find any errors. This has gone on for 4 or 5 months now. I am not unique in this. There is about 4 years of history attempting to do what I have set out to do. Rick says I'm not addressing the right problem. I insist that I am. Here's the proof. It's longer than it needs to be for clarity:
Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.
There's another dozen points of disagreement. There's no point getting to those until we resolve the question of whether or not the 1st Solution (or something like it) is correct, and that all that other stuff is BS.
But, we're basically at a standstill. Neither Rick or I is going to pursuade the other. I want to see the page improved, finally, and requested that our Conflict get some Escalated Resolution help. Glkanter ( talk) 21:06, 8 February 2009 (UTC)
Glkanter I totalt agree with you the current article is Bull crap! 99% useless and 1% useful.
It does not do a good job communicating a simple and straightforward unconditional solution!
I totaly agree with you and I want you to change it!! I have however started to lose interest because RickBlock is married to the current article and he interpret any attempt of changes as an attack on him. I just wonder why is his name RickBlock by the way?
I thought the whole point was to collaboration?! I am actually a bit insulted by his choice of name!!--
92.41.214.24 (
talk)
22:14, 8 February 2009 (UTC)
I'm still not up to speed on this, but I have a few general comments.
I also have a few questions if none of you objects:
— Carl ( CBM · talk) 23:03, 8 February 2009 (UTC)
@Glkanter: look at what happens in 18 realisations of the game;
I sorted the outcomes for a better overview:
Choice 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 Car 1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3 Open 2 2 3 3 3 3 3 3 2 2 2 2 3 3 3 3 1 1 3 1 1 1 1 1 2 2 2 2 1 1 1 1 1 1 2 2
6 times there was the same situation as in the player's case; here they are:
Choice 1 1 1 1 1 1 Car 1 1 2 2 2 2 Open 3 3 3 3 3 3
The conditional (!) probability to win the car by switching is (surprised?): 4/6 (BTW: the unconditional probability is 24/36, different, but numerically the same!) After all that is allready said, what needs to be said more? If you're not convinced, I guess you don't want to be convinced, or worse. Nijdam ( talk) 23:13, 8 February 2009 (UTC)
It is not the general relativity theory either, yet the above figures explain it all. Please scream, if it make you feel better, I'll go to sleep. Nijdam ( talk) 00:03, 9 February 2009 (UTC)
What would make this article better would be a rearrangement of content that puts like things together, progresses from simple to complex, and uses headings that clearly define the issues. Here is a suggestion:
1 History of the problem (This would not be about the Three Prisoners problem or other “pre-history” problems, but about what happened when the problem appeared in Parade and in follow-up columns. This satisfies the needs of the casual reader.)
2 Explanations clearly consistent with the original solution (This would include 2.1 Name for the first explanation, 2.2 Name for the second explanation, etc.)
3 The common objection (This would be about the difference between asking the more general question whether one should switch, versus the more specific question whether one should switch when, say, door #3 is opened.)
4 Explanations that meet the common objection (This would include 4.1 Name for the first explanation, 4.2 Name for the second explanation, etc. Bayes would be here.)
5 Another objection (This would be about the objection that one does not know if the host is as likely to open one wrong door as another when the player chooses the right door, and the Morgan, et al. solution.)
6 Similar problems (The Three Prisoners problem would be moved into this section.)
7 Theories about the problem
Omit objections that clearly contradict the problem as stated, or put them into a final section 8: Other objections and why they are irrelevant
This proposal, or something like it, meets the needs of both the casual reader who has heard of the problem and wants to know what the deal was, and the reader who is interested in a greater understanding of the issues. Simple314 ( talk) 01:10, 9 February 2009 (UTC)
The Monty Hall Paradox is only a paradox for people who like to compare apples to oranges. If a player were to pick one door out of three, two out of three times that player would be wrong. So a player should always switch if they get a second chance. With respect to the POSSIBLE OUTCOMES of this game, there are more scenarios of a player who did switch won a car. But that is not to say that the act of 'switching' improves probability. The probability of picking one door out of two is ALWAYS 1 to 2. This is true by definition. Trying to say othrwise is simply meaningless, and you can not prove it with a computer simulation. Just as you can not prove that the probability of flipping heads of a fair coin is 1 to 2. Try it, it is only true by definition. —Preceding unsigned comment added by 68.198.159.238 ( talk • contribs)
Total tries | Total times you had the ace |
---|---|
Number of times dealer showed the heart two | and you had the ace |
Number of times dealer showed the diamond two | and you had the ace |
Total tries | Total times you had the ace |
---|---|
60 | 20 (1/3 of total tries) |
Number of times dealer showed the heart two | and you had the ace |
40 (2/3 of total tries) | 20 (half of these tries) |
Number of times dealer showed the diamond two | and you had the ace |
20 (1/3 of total tries) | 0 (all of these) |
Your last paragraph confuses me. Isn't it a premise of the puzzle that the dealer discards a two (reveals a goat) from among his two cards? There's never anything random about this action. It may be random how he chooses which of the two two's, but the fact that he's going to discard a two is a premise. Let's switch back to Monty Hall. Unless the contestant knows something about how Monty chooses 'which goat', he gains no new knowledge when Monty reveals the goat. If he knew something about how Monty chose goats, then that's a new premise or constraint. Which makes it a completely different puzzle.
We have a saying in my business. "If it looks like a duck, quacks like a duck, and smells like a duck, it's a duck."
At the risk of showing my ignorance, I don't think there is a difference between the conditional and unconditional problems. It's already agreed that they have the same statistical probabilities. Now, it looks to me like you're using the exact same simulation that I would use to 'aid in the understanding' of my proposed proof. Since the probabilities are the same, I would suggest that there is no difference in the total knowledge (useful information) on the contestant's part. (That is, the premises and constraints are identical, unlike the 'Monty always chooses the left-most goat-door' variant). Lastly, In order for the two puzzles to be different, yet have the same probabilities, it seems that those probabilities would diverge at some time, yet co-incidentally return to the same values. I don't see that happening. They just both start out at 2/3 and never go anywhere.
Glkanter ( talk) 17:04, 9 February 2009 (UTC)
To summarize: in response to my question, you found it necessary to add a constraint, in order to make it a conditional probability problem. That makes it a different puzzle than the Monty Hall Problem. Further, one would not say that the probabilty for this new puzzle is 2/3. Going back to the Monty Hall Puzzle, since this new constaint (left-most goat-door) is known by the contestant (or else, what's the point?), there are scenarios when he knows exactly where the car is. So the answer would be stated "100% when Monty reveals door #3, 100% when I've chosen door #3 and Monty reveals door #2, etc." So this contestant makes more informed choices than the contestant in the real puzzle. So the contestant will win MORE than 2/3 of the time.
Glkanter ( talk) 17:50, 9 February 2009 (UTC)
Two things. In the specific door 1, door 3 question, why isn't it 2/3? Doesn't the contestant have a 2/3 chance of picking a goat? Then Monty shows the other goat, leaving a car. That's all we know at this point, right? That fits in perfectly with my new and improved, customized Super Proof! (Tip of the hat to a certain MH, who I shall keep out of this.)
The second thing is, I reject the door 1, door 3 literal interpretation. (Not that it makes any difference, as I just proved.) That's not how Probability Theory works. And that's not what makes this an internationally known paradox. Heck, the real Monty didn't even use doors. He used curtains.
Glkanter ( talk) 05:16, 10 February 2009 (UTC) Glkanter ( talk) 06:14, 10 February 2009 (UTC)
The third point is, the 'left-most door variant' is a different puzzle. It adds a new constraint on Monty. Which, in some cases, causes Monty to reveal the exact location of the car. My proof was never intended to solve that problem. I wish you would stop using this as your 'proof' that my proofs are flawed. I think it's this action on your part that I react the most strenuously against.
Glkanter ( talk) 05:24, 10 February 2009 (UTC) Glkanter ( talk) 06:14, 10 February 2009 (UTC)
I've wondered how you would respond when the inevitable happened, and your arguments were proven to be meritless. Now we know. You attack the messenger, and start posting that for Wikipedia purposes, the Truth (being right) is less important than having been published.
I've asked for Wikipedia intervention for days now. I welcome any actions you take to bring closure to this festering sore.
Glkanter ( talk) 18:35, 10 February 2009 (UTC)
There may be a kind of solution that comes close to the "simple intuitive solution", but is really a solution to the problem. Suppose the player is in the situation of having chosen door 1 and being opened door 3. We indicate this situation as [13]. Of course other players may find themselves in [12], [21], [23], [31] or [32]. Together all these 6 situation represent the unconditional one. With the "normal" strategie of the host, there is no fundamental difference between these 6 situation, hence the (conditional) probability of winning the car by switching is the same for each. But the overall probability, the unconditional, is 2/3, as we know from the intuitive (simple) solution. But then all the conditional probabilities must aldo be 2/3. Will this be satisfactory? Nijdam ( talk) 15:38, 10 February 2009 (UTC)
Hey Rick, which "peer reviewed academic sources" did you get this gem from?
That was sentence #1 of the Solutions section until I deleted it last year. It is 180 degrees counter to every bedrock principle upon which Probability Theory rests.
Glkanter ( talk) 21:46, 10 February 2009 (UTC)
I am trying to get agreement on a hypothetical question. Suppose the question posed was this:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door and the host opens another door which has a goat and he does this in a way the does not affect the probability that your original choice was a car. He then says to you, "Do you want to pick the other unopened door?". What is your advice?
I know that this is not the question actually posed but in the above case do you accept that the conditional case is identical to the unconditional one? Martin Hogbin ( talk) 22:09, 10 February 2009 (UTC)
I think that the statement of
Boris Tsirelson confirms the point that I am making.
In that case, why do we not consider all the other actions that occur between the original choice and the decision to swap as conditions? 09:39, 11 February 2009 (UTC)
What I would like to have is an addition to this article of a very simple and convincing explanation of the essential Monty Hall problem.
I believe that, for this purpose, it is acceptable to consider the unconditional problem for the following reasons:
1 The inability of most people to get the right answer, however the problem is formulated, is what makes it notable.
2 The academics who have studied the problem have to some degree (understandably) obfuscated the central problem with unwarranted and unjustified complications, including making it necessarily conditional.
3 Even if the conditional version is considered to be the best one, it is not uncommon in mathematical expositions to start by answering a related, and often simpler, problem and then to explain how this relates to the full blow problem.
The more academic discussion can remain as it is for those who are interested. Martin Hogbin ( talk) 23:09, 10 February 2009 (UTC)
We can start by observing that, if the player switches, he always gets the opposite of his original choice. The are no conditions involved, no probabilities, no dependence on host action - this is a certainty. Obviously of he does not switch he keeps his original choice.
All we have to show now is that the action of the host does not affect the probability the he had originally chosen a car, or to put it another way, that no information is revealed that might indicate the players original choice was.
Let me ask this question, looking at the Parade statement, what is the information that is revealed? Martin Hogbin ( talk) 23:23, 10 February 2009 (UTC)
For clarity in the discussion: I will remove any comment of someone else besides RickBlock and Martin Hogbin given between this header and the corresponding trailer and move it above this header. Also I put this section allways at the end of this page.
@RickBlock. It's sad and frustrating all these people, hardly knowing anything about probability, but yet thinking they just know all the ins and outs of this problem. From now on, I'll only discuss matters with you (as one expert), if necessary, and with MartinHogbin, who seems to me at least responding in an adequate manner. Others may profit by reading the comments, I will no longer respond to them. Also I think it is about time this discussion ends. My suggestion above under "work around" (maybe not so well chosen title) is a sound reasoning, in the line of the "simple solution". It lookes if you crititizised it. It is (seemes?) easy to proof and may be helpfull in understanding the conditional nature as well. Nijdam ( talk) 22:55, 10 February 2009 (UTC)
Like you. I also strongly object the 'unconditional' non-solution, that's the whole point in this discussion. Maybe I didn't express myself clear enough. The "work around" idea is to incorporate the basic idea of this unconditional reasoning as to explain the right conditional explanation. Of course in the 'random goat' strategy, only then there symmetry in the 6 basic cases. And this symmetrie is not a conclusion of the calculation of the condtional probs. The suymmetry leads to equalness of the conditional probs. Hence it can be used ([123] means chosen 1, opend 2 car 3) as follows: for all a,b,c all different:
Hence all conditional probs equal 1/3. But it remain (are) cond. probs.
The 'random goat' strategy is part of the premisses and I think the 'unconditional' non-solution doesn't find support in other cases. or is this the point you are making, that people reason in the same way even when the host has another strategy? Nijdam ( talk) 12:26, 12 February 2009 (UTC)
Okay! Indeed is the problem symmetric only with the 'random goat' strategy. But this strategy is part of the formulation of the problem. The symmetry doesn't stem from the equality of the probs, but from the fact that the problem is then invariant under permutations of the doors. And hence all the conditonal probs must be the same. And this implies they are equal to the unconditional 1/3. But they are of course conditional. So I don't say the 'simple reasoning' is correct, but a kind of 'simple reasoning' may be used. And yes, I know, without the 'random goat' there is no symmetry. And also then it are the conditional probs that had to be considered. Nijdam ( talk) 12:26, 12 February 2009 (UTC)
What is there to be proven? In the 'random goat' strategy there is no reference to any specific door, hence no door plays a different role than another one. Hence all the situations on stage chosen a and opened b are equivalent. Is this original research? I think, this is what the advocates of the 'non-solution' really have in mind wthout being able to formulate it. Nijdam ( talk) 17:07, 12 February 2009 (UTC)
@Martin Hogbin. I repeat the following points:
chosen 1 1 1 1 2 2 2 2 3 3 3 3
car 1 1 2 3 1 2 2 3 1 2 3 3
opened 2 3 3 2 3 1 3 1 2 1 1 2
prob 1 1 2 2 1 2 2 1 2 2 1 1 /18
The columns form the outcomes with their probability (/18 as indicated)
In which situation is the player? That is which event has occurred? Well in the stated problem [13], which is:
chosen 1 1
car 1 2
opened 3 3
prob 1 2 /18
consisting of 2 outcomes, and really a proper part of the sample space. In the second column the car is behind the not chosen door, and the well known conditional (!) probability of winning the car by switching is: (2/18)/(3/18) = 2/3.
Nijdam (
talk)
23:41, 10 February 2009 (UTC)
A am suggesting that the formulation of the problem by Morgan et al introduced unnecessary uncertainties. Here is the Parade statement for reference:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
What does it say about the host offering the choice, it says: 'He then says to you, "Do you want to pick door No. 2?" '. Note that it does not say 'sometimes says to you' or 'may say to you' but 'says to you'. There is no mention in the statement of the problem that the host may not offer you the choice. That possibility is an invention of Morgan.
It says the host, 'opens another door'. It does not give the way that the host decides which door to open. We must therefore take this as random. This is perfectly normal in probability problems, in cases where the method of selection is not given we take it as random, this is not an assumption it is an obligation.
Does the host always reveal a goat. The statement says, 'which is a goat'. It is therefore pure invention to assume that it might not be.
There are an infinite number of perverse actions that could occur in the problem - what if sometimes there are three goats? We do not address these possibilities because they are not part of the stated problem.
Martin Hogbin ( talk) 11:34, 11 February 2009 (UTC)
To support a point that I make above I ask people to answer this question:
A ball is chosen from a bag containing only red and white balls. What is the probability that it will be red?
Please answer the question as posed. Martin Hogbin ( talk) 22:11, 11 February 2009 (UTC)
You're aiming at the answer 1/2. Probability is here to be understood as relative frequncy in all possible realisations of the experiment. The symmetry in the colours brings the answer. Nijdam ( talk) 12:38, 12 February 2009 (UTC)
To the question, as posed, I believe that 0.5 is the correct answer. To say that I am aiming at that answer suggests that ther may be more than one answer. I think there is only one correct answer to my question and that it is 0.5. Are we all agreed on that or not? Martin Hogbin ( talk) 20:26, 12 February 2009 (UTC)
'What is meant by probability'? Surely we are not meant to be discussing that here.
This talk page is actually for talking about changes to the article, not for talking about the ins and outs of the Monty Hall problem or probability theory or the meaning of what a "bag" is. There are really only three legitimate topics for discussion here.
1) Things not in the article that should be added.
2) Things in the article that should be deleted.
3) Things in the article that should be changed.
The 0.999... article has a subpage of its talk page called "Arguments", see Talk:0.999.../Arguments, and a FAQ. I've set up such a structure here, in the hopes that it might be helpful. If anyone wants to move a section they created here to that page, please go right ahead. I'd appreciate everyone's help in moving sections from here to there when it seems appropriate. Thanks. -- Rick Block ( talk) 01:35, 13 February 2009 (UTC)
Yes, I would like to see the paragraphs in the Solutions section that says the first solution offered does not solve the Monty Hall Problem be removed. I believe is it not consistant with what the majority of the published references says.
I would also like to see the references to the minority point of view as espoused by Morgan et al moved out of the Solutions section.
I believe both of the changes are consistant with Wikipedia policy regarding regarding authoritative sources.
Rick, how will we know when it's proper to edit the Article? Would that be when a consensus is reached? Would any editors reserve a veto power?
Glkanter ( talk) 02:19, 13 February 2009 (UTC)
I fell for it. I let him use the old 'rope-a-dope' on me. Next thing you know, instead of argueing about my proof, we'll be argueing about 'who is more mainstream', or 'what is prominant'. We are not arguing about your proof. Your proof is WP:OR and of no relevance, no matter how correct it may be. Please realise that. Leaving out the hyperbole would be helpful. To comment on the principles: no, this is not a democracy, yes all have the same potential to contribute, no Rick doesn't have a veto, no Rick did not say that he had a veto, no the argument that the content has been in place for ages is not a compelling reason to keep it William M. Connolley ( talk) 08:31, 13 February 2009 (UTC)
I guess I'll have to understand and follow Wikipedia protocol if I'm going to be successful in ending Rick's Ownership of this Article. This is from Wikipedia:Ownership of articles:
Rick, as you are aware, I believe you have improperly taken ownership of the Monty Hall Problem article to the detriment of the article. I estimate this has been the case for nearly 4 years. I would like to find a solution to this, in order that I, and others could edit the article, without fear of either your veto, or entering into a reversion war.
I lay out my reasoning beginning with my post of 17:35, 12 February 2009 (UTC), which comes in the section title Conventional Wisdom.
I look forward to your response to this serious issue.
Thank you Glkanter ( talk) 07:36, 13 February 2009 (UTC)
"I have done my level best over many days now to help you understand why the change you seek won't happen" (The Dictator aka Rick Block, 2009) '
This to me implies that "Rick block" actually belives that he can decide by himself what changes should be made to the Monty hall article !!!! Rick Block should decide everything and if someone does anything that "Rick Block" does not like then "Rick Block" is going to mobilize enough support to force such a person out of the game by telling people to revert any changes to the article. Abuse and Bullying on a higher level !! Definitely ownership with capitalized 0 --
92.41.74.138 (
talk)
21:36, 13 February 2009 (UTC)
William M. Connolley if you remove my comment another time (just cause you cant stand the truth) I will remove ALL your comments :-) --
92.41.74.138 (
talk)
21:36, 13 February 2009 (UTC)
There is another source of unnecessary complication introduced by the academic papers on this subject and it is this:
The Parade statement is clearly written from the point of view of the player, 'Suppose you're on a game show...What is your advice?'. The questioner is obviously asking for advice as to what they should do if they were a contestant on the show.
Some academics, however, have reformulated the question from the perspective of some imaginary third party. Someone who may know, for example, that the host has a preference for a particular door. This is something that the player does not know.
As a result of this reformulation, there have been countless arguments on this page over whether it matters if the player knows about the host's preferences. This reformulation is a major obstacle to understanding the solution. Martin Hogbin ( talk) 09:30, 13 February 2009 (UTC)
I have an alternate suggestion regarding the conditional vs. unconditional question that keeps coming up. How about if we add a more comprehensive discussion about this, perhaps like the following (a slightly edited version of an explanation I originally posted above that might have been buried before many people noticed it). If we don't add it to the article, I suggest that it be included in the FAQ linked at the top of this page. The wording may not be quite Featured Article quality, but I think it is close. -- Rick Block ( talk) 05:28, 14 February 2009 (UTC)
Morgan et al. and Gillman claim the Monty Hall problem must be solved as a conditional probability problem. As usually stated, the problem asks about a player who has picked a door and can see which of the remaining two doors the host opens in response, and is then given the opportunity to switch doors. The question pertains not to all players in general, but only to the subset of players who have picked a door (for example door 1) and then have seen the host open a different door (for example door 3). The point concerns the difference between
and
is the unconditional probability of winning by switching. It is roughly what fraction of all players who play the game will win by switching. Gillman suggests this would be the probability of interest in a revised version of the problem where "you need to announce before a door has been opened whether you plan to switch." [italics in the original!]
is the conditional probability of winning by switching given that the player initially picked door 1 and the host has opened door 3. It is roughly what fraction of all players who pick door 1 and see the host open door 3 will win by switching.
These are different questions which may or may not have the same numeric answer. As usually stated, Morgan et al. and Gillman both say the Monty Hall problem asks the conditional question, not the unconditional one.
An unconditional solution such as "a player has a 2/3 chance of initially picking a goat, all of these players who switch win the car, so 2/3 of all players who switch will win the car" correctly answers the unconditional question, but doesn't necessarily say anything about the conditional problem. For example, this solution says the answer for both the Parade version and the Krauss and Wang version of the Monty Hall problem is 2/3. The Morgan and Gillman papers both introduce a different variant, identical to the Krauss and Wang version except instead of
they specify
In this variant, an unconditional solution also correctly answers the unconditional question. The player still has a 2/3 chance of initially picking a goat, and in this case the host must reveal the other goat, so the solution above still applies and 2/3 of all players who switch will win the car. However, in this variant, if the player initially picks door 1 and the host opens door 3 the player knows the car is behind door 2 and has a 100% probability of winning by switching. If the player initially picks door 1 and the host opens door 2, either the player's initial pick was the car (a 1/3 chance) or the car is behind door 3 (also a 1/3 chance). This means, given that the host has opened door 2, the player's chance of winning by switching is 1/2. The point of introducing this variant is to show the difference between the unconditional and conditional questions. In this variant, these questions have different answers exposing the difference between unconditional and conditional solutions.
Using an unconditional solution produces the correct answer for the more exact Monty Hall problem as stated by Krauss and Wang, but this is in effect a coincidence. The problem with the approach is hidden because in this version the conditional and unconditional answers are the same. Applying the same solution to a variant where the unconditional and conditional solutions are different, like the "leftmost door variant", reveals the problem.
This is exactly the same as responding to the question "what is 22?" by saying "2 plus 2 is 4, the answer is 4". This is the right answer, and a true statement, and related to the right reasoning, but not actually right. This wrong reasoning can be exposed by asking, "what is 33?" Similarly, the problem with using an unconditional solution to answer the Monty Hall problem can be exposed by asking about a slightly different variant, where the wrong reasoning produces the wrong answer.
So, how does this work in practice? There's no vote taking, and consensus doesn't necessarily rule the day. But Rick can still change his article, right?
Glkanter (
talk)
10:14, 14 February 2009 (UTC)
Actually I think we now should change the conditional solution to an unconditional one. Concensus has been reached (4 against 1)
Glkanter, Heptalogos, Martin Hogbin and 92.41.236.242 against "Rick Block" which means that the unconditional solution is the winner
What do you want to include in the unconditional solution guys? --
92.41.236.242 (
talk)
12:52, 14 February 2009 (UTC)
To help structure this discussion, I'm adding subsections. Please stay on topic within the subsections. -- Rick Block ( talk) 18:18, 14 February 2009 (UTC)
The reason I am suggesting adding this content is because the basic point made in these two papers is clearly not widely understood. If the article includes a comprehensive discussion of what they say, the wording can be improved by anyone who may be able to express the point better. We should not add the content if the point is insignificant, or mathematically discredited. It is my belief that these two papers, nearly simultaneously published (Morgan et al. was first by 2 months), are the first mathematically rigorous treatments of the Monty Hall problem which makes them crucial to include in this article. I also believe these papers were published specifically in response to the nationwide furor over this problem that followed the Parade articles published in 1990-1991. The Morgan et al. paper is cited 49 times according to Google scholar. The Gillman paper is cited 30 times. As far as I know, the central point they make is not reflected in non-academic literature, but this doesn't mean we shouldn't include what they say. -- Rick Block ( talk) 18:18, 14 February 2009 (UTC)
You started by saying the fully defined problem is directly equivalent to the 3 prisoners problem. This is somehow contradictory to your opposition to accept that it is about conditional probability. The 3 prisoners problem is! Nijdam ( talk) 21:10, 14 February 2009 (UTC)
If you think the suggested wording doesn't fairly summarize what the papers are saying, please say how or where (the analogy at the very end is original, but everything else is directly from the papers). The papers are short, but I could add references to specific paragraphs if that would help. A point they both make, not in the proposed wording above, is that viewing the problem as a conditional probability problem means the Parade version of the problem lacks a condition necessary to justify the 2/3 answer. Specifically, they BOTH mention this version does not say the host picks between two goat doors randomly. BOTH of these papers make this point, and provide a solution where the host's preference for one door or another is left as a variable q. They BOTH show the probability of winning by switching is then in the range of .5 to 1, exactly equal to 1/(1+q). I think this additional point should likely be added as well, but first things first. -- Rick Block ( talk) 18:18, 14 February 2009 (UTC)
Arguing that what these papers say is wrong in a mathematical sense has no bearing on whether the content should be included or not. If you would like to discuss the points they make, please post at /Arguments or Wikipedia:Reference desk/Mathematics. -- Rick Block ( talk) 18:18, 14 February 2009 (UTC)
I think making the two versions of the problem clear in the article might prevent a lot of argument. Martin Hogbin ( talk) 20:34, 14 February 2009 (UTC)
Rick thanks for making the players plural in my recent addition. I agree that is is a better mathematical representation and it is more PC.
I have restored the emphasis on 'always' to indicate that there is no doubt, uncertainty, or probability other than 1 for a swap to change a car to a goat and a goat to a car. I think this needs to be spelled out to the reader so that there are no lingering doubts. I was going to add either a footnote or a picture to make this absolutely clear. What do you think.
I have also put back the probabilities of swappers and non-swappers getting a goat. You may consider this obvious and superfluous but to a newcomer to the problem this may not be so. Martin Hogbin ( talk) 09:05, 14 February 2009 (UTC)
Thinking about this some more, I am not sure that this solution should even be in the lead section, which is meant to be a summary of the article as a whole. Would it be better to make a very brief reference in the lead and move this explanation into the start of the 'solution' section, maybe with pictures? What we have now (and had before) is one simple solution in the lead and a different one in the body. Martin Hogbin ( talk) 17:14, 14 February 2009 (UTC)
Quote: "But if that event cannot in any way possibly affect the outcome then it is a null condition". Martin, which rule says so? Heptalogos ( talk) 22:02, 14 February 2009 (UTC)
I do agree with you, and actually I was hoping for you to say that 'reducing sample space' is not a rule for conditional probability, but 'affecting probabilities' is. Somehow nobody reacts to that statement. Can you please, on the Arguments page? Heptalogos ( talk) 10:50, 16 February 2009 (UTC)
Have you guys had a chance to see my newest solution (it's not a 'proof', after all)? If I'm right, it shows the unconditional solution solves the conditional problem. And, I won't quote anybody here, but if you follow my 'contribs', you will find an editor-of-apparent-authority saying that the solution is a valid unconditional proof of the MHP. All that would remain is to find a published source, which, since it's right, must be out there. But, see for yourself. Make your own judgements... Glkanter ( talk) 12:37, 15 February 2009 (UTC)
This is the suggested unconditional solution. Feal free to make any improvments to the layout etc. This is the best I could do.--
92.41.74.110 (
talk)
21:03, 14 February 2009 (UTC)
There are three doors. Two doors have a goat behind them and one door has a car behind.
(Remember goat is bad and car is good)
1) Choose one door
2) Game host opens one door containing a goat
3) Do you change door?
For example:
Door 1 | Door 2 | Door 3 |
---|---|---|
Goat | Goat | Car |
Now you have two opions: Not switch or Switch
Not switch
pick | show | outcome |
---|---|---|
1 | 2 | lose |
2 | 1 | lose |
3 | 2 or 1 | win |
If you do not switch the probability of winning is 1/3
Switch
pick | show | outcome |
---|---|---|
1 | 2 | win |
2 | 1 | win |
3 | 2 or 1 | lose |
If you switch the probability of winning is 2/3
You should therefore choose to switch doors
92.41.74.110 (
talk)
21:03, 14 February 2009 (UTC)
pick (No. 1) | show (No. 3) | outcome |
---|---|---|
car (1/3) | goat (1) | lose (1/3) |
goat (2/3) | goat (1) | win (2/3) |
It's one thing to mention the "chances" as you called them, another is what there meaning is and where they come from. Hopefully this leads you to finally understand that what you called "chances" here, are the conditional probabilities. How else could door No. 3 have "chance" 1 to reveal a goat, whence the (unconditional) probability of having a goat behind it, is 2/3? Nijdam ( talk) 13:29, 15 February 2009 (UTC)
I don't know what the meaning is of this 'Bert' stuff. May be you're watching way too much Sesam Street, I at least don't see any connection with the discussion. And you may persist in thinking the doors are not labelled, why not, you may even think there are no doors at all. Or 3 doors and 7 little goats and a wolf, or prince William and his girlfriend behind one door and his grand grand uncle behind the curtains. If it makes you feel better, please do. Nijdam ( talk) 17:56, 16 February 2009 (UTC)
Rick went to http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics today to drum up support for his proposed changes.
I followed him by informing them of my allegations of Ownership by Rick. I suggested that today's behaviour alone shows this problem. Glkanter ( talk) 22:21, 14 February 2009 (UTC)
Rick proposed a major addition to the article today. Just hours after about 4 of us were discussing the merits of a 'formal' proposal for deleting stuff. Then he goes to your page to drum up support. What's up with that? Who's stirring the pot? Who's not showing good sense? I know of one highly respected math editor who today described 'the discussion is so turbulent'. And Rick proposes a major addition that he flat-out knows his primary critic is dead-set against? No. That's not normal.
Here's a probability: if you don't look into it, you will not find what I'm describing. However, If you look, you might find I'm right.
Glkanter (
talk)
04:54, 15 February 2009 (UTC)
You might even ask me if I have any documentation to support my claim.
Glkanter (
talk)
05:04, 15 February 2009 (UTC)
As the former leader of the free world once said, 'Fool me once, <pause>, uh, shame on you. Fool me again...won't get fooled again!'
I fell for the 'rope-a-dope' a second time! Turns out in C S's case, my apology was unnecessary. He, at least, had already made up his mind. Hostile mother.
And don't be coy. Tell me why I am 'so wrong you don't even know why you aren't going to garner any consensus.' I'll change my posting techniques if it's futile. Just tell me the rules. Glkanter ( talk) 13:07, 15 February 2009 (UTC)
The article says:
The host opens a known door with probability p, unless the car is behind it ( Morgan et al. 1991). | If the host opens his "usual" door, switching wins with probability 1/(1+p). If the host opens the other remaining door, switching wins with probability p/(1+p). |
The latter formula cannot be correct. In the extreme case, set p=1 and assume that the host does not open his usual door. This can only have been because the car is behind the usual door, so switching in this case wins with certainty. However p/(1+p) is only 50%.
According to my own analysis, the 1/(1+p) is correct, but the other formula should be 1/(2–p). (It is okay for the two probabilities not to add up to 1; conditional probabilities under mutually exclusive conditions have no necessary relation).
Is this an error in the source, in the reporting of it here, or in my understanding of what the behavior is supposed to be? – Henning Makholm – Henning Makholm 04:46, 15 February 2009 (UTC)
Is there a procedure for investigating WP:Ownership allegations? I'll be happy to share everything I can.
Otherwise, read the Conventional Wisdom section. That's where I lay everything out. —Preceding unsigned comment added by Glkanter ( talk • contribs) 06:03, 15 February 2009 (UTC)
Rather than argue about unconditional/conditional issues we could agree, for the moment at least, to continue what is already done in the article. This is what I have done in my changes to the lead.
I think it is agreed by all that if we consider the overall probability of getting a car of players who swap compared with players who stick there is no conditional issue. All that is needed is to talk in terms of players who swap and players who stick (in other words we consider players to be either stickers or swappers). I think it is generally accepted that, for any reasonable interpretation of the problem, stickers have a 1/3 chance of getting a car and swappers a 2/3 chance and that no conditional probability is required to show this.
The two problems with this approach that I see are firstly that I do not believe that it is actually necessary, and secondly that it detracts from the true nature of the problem as a game show in which the player is presented with a choice after a particular action of the host. This makes an overall probability solution justifiably less convincing to some, nevertheless it may suit others.
However, by being careful with our wording it might be possible to present a very simple and clear solution that is acceptable to all. As far as I am concerned the overall solution already given is fine in principle, it just needs improving in terms of words and pictures. Martin Hogbin ( talk) 10:21, 15 February 2009 (UTC)
(edit conflict) I'm late to this party, and the preceding discussion is very long and sprawling. Let me just dump my opinion, more or less from first principles, on what seems to be the matter:
The classical problem, as stated, is one where the plain and orthodox way to turn it into a mathematical question is to ask about a conditional probability. Calculating this conditional probability in the most by-the-book, pedestrian, not-trying-to-be-smart way will yield the correct answer. On the other hand, a mathematically mature reader will be able to see that because of symmetry, the same correct answer can be arrived at with considerably less calculation by imagining that the player makes his choice to swap or not swap in advance; then all one has to do is calculate an old-fashioned non-conditional probability. All of the foregoing is fact, and hopefully not in dispute; otherwise we're worse off than it seems.
Now, different readers will have different trouble with those facts, and we must try to satisfy all of them, even if that means that some readers need to read explanations that are not directly relevant to the understanding which that particular reader finds easiest. The choice between the two approaches is a trade-off between knowing more theory (conditional probabilities), or using less theory but applying more problem-specific cleverness (invoking symmetry). What one prefers is subjective, and not something that is objectively inherent in the problem.
The sticker-swapper argument is both popular and extremely attractive to readers who are capable of seeing that it is a valid approach, so it would be a bad encyclopedia that did not present it at least somewhere. On the other hand, seeing that it is valid is not entirely trivial, and there will be other readers who are more easily convinced by a straightforward conditional probability. So this must also be presented.
However, finding the right answer is not the end-all. The more curious reader will also want to investigate not only what is right, but also how he managed to be wrong when he initially thought it would be 50/50. And in order to discuss that -- an interesting and now famous perceptional problem that this article must address -- I cannot see how that can be discussed without the machinery of conditional probabilities.
What we then have to decide, from an editorial viewpoint, it not whether to present one or the other raw solution to the original problem, but merely which one to present first. Personally I would gravitate towards starting with the conditional probabilities, the one that requires less ad-hoc cleverness, but reasonable arguments for the opposite order can also be made. In the end I think it is more important the both explanations are good than which order they appear in.
Now have I misunderstood everything? – Henning Makholm 17:29, 15 February 2009 (UTC)
The policies say to resolve a dispute, 3rd party input from subject matter experts may help.
So, Rick and I have solicited this input.
Now, I've been criticized for mis-representing that feedback.
So, absent an edit to this talk page from the subject matter expert, how does one incorporate their contributions into the larger discussion?
I think one expert has unambiguously defended my proof. I think another has said he doesn't know what Morgan's problem with the unconditional solution is, so he'll research it further.
But I don't want to put words in people's mouths. I don't need to. It's out there in the edits.
As far as I can tell, the only tool available to me is to cut and paste quotes and links into new edits.
How do I incorporate those views expressed into this discussion, so as to build a consensus? Glkanter ( talk) 15:08, 15 February 2009 (UTC)
The feeling I got from discussing this page with subject matter experts is that they see it as a black hole, which they wish to fervently avoid. Please go to this page to determine for yourself the validity of this statement. http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics
I believe, then, that any efforts to encourage them to post on this page would be counter-productive.
Nobody has answered my question. 3rd party subject matter experts HAVE made comments, suggestions and opinions. How do we incorporate those into this discussion?
We've referred to Boris. I think he has a doctorate in Mathematics. Here's his page. Read the last section. I think it's reasonable to conclude that he's agreeing with me. But you decide! http://en.wikipedia.org/wiki/User_talk:Tsirel Glkanter ( talk) 19:24, 15 February 2009 (UTC)
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
CS, thanks for the explanation. It's all very obvious what you say, and actually that's not what I was looking for. That's why I referred to the Arguments page; here's where I ask the common rule for a problem to be conditional. Most of us are discussing the Krauss and Wang (2003:10) statement of the problem anyway, so there need not to be much discussion about what was meant by Marilyn. Even if so, it's another discussion outside maths. My question is what exactly makes the Krauss and Wang statement a conditional problem, following which rule. Heptalogos ( talk) 22:25, 15 February 2009 (UTC)
I'm trying to eat my humble pie quietly, but I just gotta ask, what do you mean when you say: "By the way, I hope I haven't given the impression that I think the problem must be understood as conditional."? I'm afraid that is what I think you're saying. I've admitted defeat here precisely because I thought you said that it must be. And I'm not arrogant enough to argue math with a math PHD.
Glkanter (
talk)
00:16, 16 February 2009 (UTC)
This may help me understand. You wrote: "If you assume the unconditional problem, this information is not necessary, since as you pointed out, you only had a 1/3 chance to begin with of getting the car, so if you can switch, you should." What is it about the conditional problem that makes that statement not applicable to it's solution as well?. Monty's motivation, behaviour and actions don't affect that 1/3 chance, do they? —Preceding
unsigned comment added by
Glkanter (
talk •
contribs)
01:26, 16 February 2009 (UTC)
I think an emphais on the unconditional solution for the Monty Hall Problem artcile is appropriate.
This would mean de-emphasising dramatically much of the remainder of the existing article. Glkanter ( talk) 17:08, 16 February 2009 (UTC)
File:MontyXXX.jpg-- Pello-500 ( talk) 14:56, 16 February 2009 (UTC)
What is the difference in content between this and the former "Suggested Unconditional Solution"? Heptalogos ( talk) 16:45, 16 February 2009 (UTC)
You choose a door | |||||
---|---|---|---|---|---|
You choose a door with a goat | You choose a door with a goat | You choose a door with a car | |||
You Stick | You Change | You Stick | You Change | You Stick | You change |
You get a goat | You get a car | You get a goat | You get car | You get a car | You get a goat |
If we were to put a simple unconditional solution this might be a better choice, taken from the 'Curious Incident...'
We can assume that the two goats and the car is located as follows: |
---|
Door 1 | Door 2 | Door 3 |
---|---|---|
Goat | Goat | Car |
We now have two opions: Not Switch or Switch |
---|
Not Switch | Switch | ||||
---|---|---|---|---|---|
Pick | Show | Outcome | Pick | Show | Outcome |
1 | 2 | lose | 1 | 2 | win |
2 | 1 | lose | 2 | 1 | win |
3 | 2 or 1 | win | 3 | 2 or 1 | lose |
If we do not switch the probability of winning is 1/3 | If we switch the probability of winning is 2/3 |
---|
We should therefore choose to switch doors if we want to increase our probability of winning the car |
---|
I know that I favour a simple solution but we do need to recognize that this is a featured article and we should tread carefully. The last few changes make the page much worse in my opinion and should be reverted. Let is try to reach a consensus on how to proceed. before we make such dramatic changes. Martin Hogbin ( talk) 23:13, 16 February 2009 (UTC)
Leaving out the door numbers in the wording of the problem, essentially changes the problem and takes away not only its charm, but also changes it to a form it never was intended to. I.e. it no longer would be equivalent to the three prisoners problem, from which it seemes to be derived, but especially it no longer reflects what happens in the Monty Hall show. Nijdam ( talk) 00:31, 17 February 2009 (UTC)
It's been 24 hours since I posted this:
applicable to it's solution as well?. Monty's motivation, behaviour and actions don't affect that 1/3 chance, do they?
Haven't heard a peep back yet.
So, it's over. You know it, and I know it. And I was right. From my very first post.
Rick Block, I commend you, sir. You were led, badly, by someone who you trusted, with good reasons. You were always, always considerate and professional. I hope someday you and I are on the same team.
C S, you are garbage.
I posted some fun stuff on my talk page. Check it out!
http://en.wikipedia.org/wiki/User_talk:Glkanter —Preceding
unsigned comment added by
Glkanter (
talk •
contribs)
03:08, 17 February 2009 (UTC)
So, now that it has been proven that the alleged conditional statements of the MHP can be solved using the unconditional solution, there really is no 'conditional problem'. Just so much noise. Because Monty can't change the past, so it's still a 1/3 chance I picked the car.
Let's focus on improving the Wikipedia reader's experience with this article. —Preceding unsigned comment added by Glkanter ( talk • contribs) 01:39, 17 February 2009 (UTC) Glkanter ( talk) 03:10, 17 February 2009 (UTC)
I can't follow y'all's heated arguments too good, but I tried to add a bit to make it clear that the assertion that the probability of getting a car is still 1/3 after Monty opens a door depends on some unstated assumptions about whether he always does this or not. The logic did not follow without this, and Martin took it out, so I just put a bit to say it depends on some assumptions. Feel free to further clarify, but don't go backwards to the condition of stating a "deduction" that doesn't follow. Dicklyon ( talk) 02:27, 17 February 2009 (UTC)
My statement of overall probability always holds, see Morgan et al. The host always opens a door, it says so in the question. This is not disputed by anyone except you. Martin Hogbin ( talk) 22:40, 17 February 2009 (UTC)
I just realized that a whole new crew of interested editors are joining the fray.
It's been happening for 4 years now, and will continue, with no reason for abating.
So, I'll eventually be gone, but someone new will take my place.
Why not try to reduce the 'heat' before it starts?
What would help is if the Math guys would, in writing, present the consensus 'house' line on the issue. Put it on the talk page. Not like the old banner that was torn down. Something that would pro-actively indicate the other subject matter experts' opinions are in line with the emphasis of the article. —Preceding unsigned comment added by Glkanter ( talk • contribs)
I've revised the new version of the initial solution a bit. I think rearranging the figure so the "host opens Door 3" and "host opens Door 2" cases are adjacent shows what's going on somewhat better. Actually, flipping these so the "host opens Door 2" cases are on the left might be even better (since Door 2 is shown on the left in the figure). -- Rick Block ( talk) 05:14, 17 February 2009 (UTC)
Any mileage in this? It does not rely on host action but does trick the reader.
Initially you have a 1/3 chance of picking a car and a 2/3 chance of picking a goat. After Monty has opened a door revealing a goat you will always get the opposite of your original choice if you switch. Martin Hogbin ( talk) 19:18, 17 February 2009 (UTC)
I only briefly comment on what I over and over read. Namely, that sticking to your (initial) choice, should logically lead to sticking to your chances, i.e. dealing with unaltered probabilities. I guess many of the people defending the "unconditional" solution think along this line, and never even thought it can be otherwise. But they should! Sticking to your choice is one thing, chances do not change, but situations do, and hence new probabilities have to be calculated: conditional ones of course, given the new situaton. Nijdam ( talk) 22:37, 18 February 2009 (UTC)
Section moved to Talk:Monty Hall problem/Arguments#Decision Tree. -- Rick Block ( talk) 14:20, 20 February 2009 (UTC)
Section moved to /Arguments#"Choice" vs. "Guess". -- Rick Block ( talk) 14:41, 24 February 2009 (UTC)