User talk:Gerhardvalentin

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Monty Hall Problem

Just a note informing you that Formal Mediation has begun with 2 mediators. Glkanter (talk) 20:55, 12 August 2010 (UTC)

Thank you!   Gerhardvalentin (talk) 12:55, 14 August 2010 (UTC)

Oh, no! Thank you! I had no idea... Glkanter (talk) 01:36, 15 August 2010 (UTC)

I just undid your entry on the topic, "Should the contestant stick or switch?" I'm not sure whether you were summarizing an opposing viewpoint or not. Would you be able to confirm this?

Also, we are quite serious about the 300 word maximum for this. Writing concise summaries is a way of clarifying thoughts on a topic and, equally important, a way of making mediation more viable. Would you be willing to re-read Will's instructions and try once more? Sunray (talk) 02:00, 15 August 2010 (UTC)

Thank you. Sunray (talk) 15:12, 15 August 2010 (UTC)

Richard tries to cut the crap

Hi Gerardvalentin, I have spent all day "doing my stuff" on the MH mediation page. In an effort to decrease my verbosity I put up some footnotes to some new mediation page contributions by me, on my own talk page. Still struggling with how to do links in wikipedia and how to get notifications when important things are changed. I hope you have time to take a look and do please comment, in whichever way you like. Gill110951 (talk) 13:41, 15 August 2010 (UTC)

Private exploit of private room for interpretation

Richard, Garry: I'm about the words "uniformly at random". The characteristics of the three doors not being further described. Does that mean that a "random distribution" of the three objects behind those three doors seems to be implied? – or not? I think Yes. (Without further private exploit of private room for interpretation).

And again: When I explicitly say: "You pick a door at random, say No. 1", – is the same evidence already strictly implied in the formulation:
"You pick a door, say No. 1", (without the word "random"), as well? – As for me, I would say yes.
As for me, the formulations: those three doors have numbers (1-3) .... you pick a door at random, say No. 1 and: the doors don't have numbers, .... you pick a door at random, let's call it "door No. 1" and: You pick a door, say the leftmost, so let's call it "door No. 1", (without the word "random") all seem to be identical. Or aren't they? Can that lead to a controversy? What say "the sources"?

I think it is most important to clearly understand and fully appreciate and take into account the foundation, the basis on which is talked about. And I guess the formulation given in the MHP evidently "implies" that all of this is done uniformly at random, as well as the opening of one of the two doors by the host, if he has two goats to choose from. (As Marilyn later confirmed that this was the basis of her answer.) Of course anyone is free to "play his own game" and to play what he wants and is free to a priori appoint and specify that "his" host is biased, and to which exact extent his host is biased. But if he does so, in this case, in presenting such variants, it is indispensable to explicitly say so. And, in tracking the trend of "the real world game" :-)), you never may be sure to have "detected" any Host's "door preference" in one hundred million games, if he just clearly "seems" to be biased, but – actually – he isn't biased at all. You - in advance - must have exact knowledge of any host's door preference in the real world, or you have to "appoint" such a given preference for your own game. Please can you give a short comment to my question / POV? What say "the sources"? Would be great. Gerhardvalentin (talk) 17:03, 18 August 2010 (UTC)

Gerhardvalentin, my understanding of this puzzle about a game show seems to be very similar, perhaps identical to your's. And in my naivete', I don't care whether the doors are numbered, or not. Just so long as I can point to each one uniquely (left, center, right) in order to make my selection known. Glkanter (talk) 17:09, 18 August 2010 (UTC)

Wow, thank you for your speedy reply. Yes, you are right: the "door numbers" are not so considerable, my important thing is the implication of "randomness". I learned from Richard: "Either" the distribution of the three objects "or" the door selection of the guest has to be uniformly at random. Either/or (or, at best, both). If it is not guaranteed that the three objects had been distributed uniformly at random, it really is enough to be sure that the guest (she has no info about the actual distribution, whether it is random or not), chooses her door uniformly at random. What do you think? Regards, Gerhardvalentin (talk) 17:20, 18 August 2010 (UTC)

I don't really 'know'. But these are my points:

The puzzle begins: "Suppose you're on a game show..."

I will have no knowledge of how the producer places the car

At that point, with no other requirements or information, I believe it makes no difference which door I chose, and that the likelihood is 2/3 I will choose a goat. Glkanter (talk) 17:29, 18 August 2010 (UTC)

May I say it this way: Even if the car always is known to be behind the same door, but if you make your decision "at random", even then your chance is 2/3 to have picked a goat. Okay? Gerhardvalentin (talk) 17:39, 18 August 2010 (UTC)

I can't guarantee that my choice is 'random'. Maybe I have a 'lucky number'. Will using that lucky number actually affect the odds? I just can't agree with that. I don't think it matters. Glkanter (talk) 17:50, 18 August 2010 (UTC)

And, "Even if the car always is known to be behind the same door..."? Known by whom? Sounds like it's no longer the MHP to me. Glkanter (talk) 17:51, 18 August 2010 (UTC)

Pardon me, Garry, for my unclear wordings, with "Even if the car always is behind the same door, but the guest doesn't now that", I tried to express: "No matter where the car (secretly!) is hidden ..." – it is sufficient that the guest makes her first choice uniformly at random to guarantee a chance of 2/3 to pick a goat. Okay? Gerhardvalentin (talk) 19:59, 18 August 2010 (UTC)

And my argument is that even the 'uniformly at random' portion of your statement is not required. How do the likely outcomes change if the contestant does not select 'uniformly at random'? Glkanter (talk) 21:30, 18 August 2010 (UTC)

Experience: Even given that the guest makes her first selection at random, and even given that the host chooses randomly if he has two goats to show, but without expressly pronouncing that, s.o. will argue "But what if ...". We all know: Marilyn was asked a question (I say "without any malicious trappings"), but s.o. et al. have attacked her that she didn't mention the host's eventual bias, claiming her single solution was faulty. So I guess it must be very clear and unassailable what we are talking about, to avoid turning around. That's what I've learned here. And it is of importance to cite the sources exactly, accounting for which question exactly they are addressing, the intentions and the exact meaning of the remarks, to help avoid misinterpretation. Gerhardvalentin (talk) 23:03, 18 August 2010 (UTC)

Hi. sorry I missed this exchange. Gerhard, you are right in your understanding of what I said. It concerned a mathematical fact. If car is hidden completely at random (ie with equal probabilities 1/3), or if player chooses door completely at random (ie with equal probabilities 1/3), switching will give car with probability 2/3. The mathematics is neutral to what "probability" means. There are two main schools concerning what the word "probability" ought to mean. The Bayesians/subjectivists/epistemologists say that probability measures degree of belief, it is an attribute of the mind of the subject, as he considers some situation or question. That is why Garry doesn't talk about host bias and why he picks a door according to his lucky number. In the story told by vos Savant, our beliefs about the situation would have been identical if the numbers "1", "3" and "2", i.e. in the sequence 1,3,2, had been replaced with any of the other five permutations 1,2,3; 2,1,3; 2,3,1; 3,1,2; 3,2,1. So his subjective probabilities give equal probabilities to the car being anywhere and equal probabilities to either of the host's choices when he has a choice. Equal probabilities means he would bet equal money, either way. It follows that it doesn't matter a damn how he chooses a car. Next, since the probability he hits the car first time is 1/3, the probability he'll get it by switching is 2/3. Because the problem is invariant under permutations of the door numbers, the door numbers are irrelevant. For subjective probabilities, this means that probabilities do not change when adding or deleting specific door numbers to a description of any event. So Garry doesn't condition on the actual door numbers because a priori he knows they are irrelevant. From the subjective probability point of view, his solution is not only correct but also complete. There's no more to be said. There is no strategy which does better than 2/3 because there is no specific situation where it would be beneficial not to switch.

The other people are called frequentists/objectivists/ontologists. They think that probability is "out there" in the real world, independent of any subject's knowledge; and they think you can learn probabilities by doing (or thinking about) many many repetitions. Probabilities are long run relative frequencies.

I say that whatever I think I know about the situation, I will choose my door number by use of a fair randomizer and then switch. Not only will it be the case that in one particular run of the game, I will have subjective probability 2/3 to take the car home with me. It is also the case that if I play the game hundreds of times I'll take the car home 2/3 of the times. Even if Monty Hall is watching me play and trying to trick me and hiding his car and opening doors in all kinds of mean ways.

My present opinion is that there is not more that can be said to a general public about the game. Of course, in a statistics class devoted to conditional probability, it is fun and educative to compute conditional probabilities in different ways, under varying assumptions. In an economics class it is fun and educative to think more about what von Neumann minimax theorem has to say about the problem.

Concerning the present fight on the mediation page, I think we made a lot of interesting progress. I think that the "bad guys" are using the mediation to crush opposition using force of numbers and are not being creative and constructive. It seems they have a dogmatic addiction to a particular point of view. I think wikipedia editors shouldn't push their personal point of view. It also seems to me that we do need a discussion about sources since we have to show that there are reliable sources a-plenty who take a short simple solution to MHP as being correct and complete. We need sources who write about MHP as their primary topic, not sources who use it to illustrate some other point. Carlton is not a good source here. Any author of elementary statistics text books is not a good source. Such authors are not primarily interested in MHP, they are primarily interested in teaching conditional probability, and getting their students to think and learn about conditional probability. Gill110951 (talk) 06:30, 5 September 2010 (UTC)

Yes, you're exactly right. They do not contribute to the correct presentation of the subject of the famous MHP. But, without realizing it, they love to abuse the MHP as their very popular example for students to explain and to teach the practice of conditional probability theory. They should honestly admit this fact but, being real stashed frequentists (!), hide behind wishy-washy faded spurious arguments. Gerhardvalentin (talk) 21:13, 9 September 2010 (UTC)

You're wrong on one item, Gerhardvalentin. They realize it. That's why they're so dismissive of the decision tree derived from Carlton's (and others') simple solutions, and the Combined Doors solution that also has many sources. These solutions demonstrate clearly and unambiguously that they (and their sources) have no valid arguments about the Selvin/Whitaker/vos Savant MHP. Glkanter (talk) 02:03, 10 September 2010 (UTC)

Nijdam's talk page

On June 15, 2010 / 09:52 Nijdam said on his talk page:

Memo

Difference between: "behind one of these doors is a goat" and "behind this door is a goat".Nijdam (talk) 09:52, 15 June 2010 (UTC)

My answer: Before <> after

If the host has no "door preference" and  – in 1/3 of games, having got the chance to choose between two goats –  chooses uniformly at random, as he is supposed to, not irregularly giving away any secret evidence on the actual particular constellation, this secures for each and every game that “we’ve learned absolutely nothing at all to allow us to revise the odds on the door initially chosen by the guest”, and so there of course never can loom any difference at all between "before and after". Not needing any "mathematical proof". The Pws is and remains 2/3, anyway. Unconditionally.

Falk is attesting that such "simple solution" was fully correct if it was based on the presupposition that the host will be choosing one of his two goats uniformly at random, if he has the choice. Only in the absence of such a required presupposition, that means just if it is not specified which one of his two goats the host will show if the guest has chosen the car, you could object:

"Suppose he is biased, ...   (and you know about that bias). He'll always open his preferred door when he can."

So, this cat is out of the bag.

Detecting any "host's bias"

But some conditionalists (e.g. Morgan et al.) prefer to invent a "telltale-host" who could be biased and, by opening of a door, uses then to be flashing additional telltale news and additional information about the respective actual constellation and the actual current secret distribution of the three hidden objects behind the doors. Such a "biased host" will always be providing for some irregular "new condition" in opening of any door showing a goat. As such "host's bias" has to be known in advance (whose knowledge?), the best way to know about such a bias is to leave Monty and to create their own game and to appoint such a given host's door-preference for their own game, and so to always provide for a new "condition" in irregularly opening of a door. Quite easy. Then, by such a biased opening of "one" door those conditionalists indeed will have learned something to permit them, for the unchangeable current constellation, to revise the odds on the door initially chosen by the guest and for the "Probability to win by switching".
If they do appoint an extreme host's door preference e.g., he will be free to always open his preferred door whenever he can. Considering that act, the probability that the car is hidden behind the door initially chosen by the guest suddenly rises from l/3 to l/2, and the probability of the second closed door drops from 2/3 to 1/2. This will happen in 2/3 of all games:

If, in 1/3, he has got two goats to choose from, he can and will open his preferred door, and switching will lose: Pws=0.
If, in 1/3, he has got one goat and the car, the goat being behind his preferred door, he can and will also open his preferred door again, but this time switching will win: Pws=1.

So – whenever this biased host opens his favored door in 2/3 of all games – those conditionalists know that the "conditional probability" of the door originally selected by the guest rises from 1/3 to 1/2 and Pws has dropped from 2/3 to 1/2.   –   For the surplus however:

If, in 1/3, he has got one goat and the car, but the car being behind his preferred door, he will be forced to open his unfavored and avoided door, and switching wins for sure: Pws=1.

Effect: Whenever conditionalists see their host opening his unfavored and avoided door in 1/3 of all games, they know for sure that the "conditional probability" of the door selected by the guest has dropped from 1/3 to zero, and Pws rises from 2/3 to 1, i.e. switching is guaranteed to win. This trick is equivalent to – by opening of one single door – showing the contents of all three doors altogether, to irregularly opening all three doors at once.

That's not the MHP, but conditionalists use to like that, calling it "the heart of the MHP", using "indispensable" mathematical conditional probability theory then for their own variant, to proof and honor their achievement. Pws on average remains unchangeable 2/3, just by telltale info varying around 2/3, from min.=1/2 (twice), to max.=1 (once), and similar scenario if the host is biased to a smaller extent. The results, by additional telltale info, will always vary around 2/3.

Admittedly, although "logically correct", all of that is not the "MHP", it's just a less important side aspect. But some like to ridiculously stylize up such meaningless telltale trivialities to "the main core of the MHP". Although the actual constellation never can be changed by their host, and neither the overall Pws of 2/3. All of those less important side aspects never can be helpful at all to understand and to solve the famous 50:50 paradoxon of the MHP, and we clearly can see that all this ridiculous telltale detour forever will be without any real relevance. Because – in spite of even the most extreme telltale bias of their host – the longest detour just proves what you already knew long before: That the guest should switch indeed, anyway. As per Marilyn vos Savant's advice.  Gerhardvalentin (talk) 17:52, 19 August 2010 (UTC)

About your table

About this table - all 9 lines are equally probable (right?). If the player has initially selected door 1 three lines apply. If, in addition, the host then opens door 3 only two lines still apply (the first one and the 4th one). If these two are equally probable, this table seems to be saying the chances of winning by switching (in this specific case) are 1/2. I know this is not actually true, but it is what your table (and vos Savant's table) seem to be saying. The resolution is that the case where the car is behind door 1 and the player picks door 1 and the host opens door 3 (which is shown, but not individually, as the 1st line) is only half as likely as the case where the car is behind door 1 and the player picks door 1 (which is shown as the 1st line), whereas the case where the car is behind door 2 and the player picks door 1 and the host opens door 3 (which is shown as the 4th line) is just as likely as the case where the car is behind door 2 and the player picks door 1 (which is also shown as the 4th line). Adding and the host opens door 3 changes one of these and not the other, but your table doesn't make this obvious. -- Rick Block (talk) 16:52, 26 August 2010 (UTC)

Thank you, Rick, for your objection. If the player (only in 1/3), has picked the winning door by chance, "say No.1", then sticking will win and switching will loose anytime, independently of which one of his two goats will be shown by the host. The host then can choose evenly and uniformly between his two goats, switching will loose in both cases. Independently of opening door 2 or opening door 3, to stick will win and to switch will loose if the player was lucky to select the car in 1/3. The table shows that. And the table shows that in 2/3 of all possible constellations switching will win and in 1/3 of all possible constellations staying will win. The very advantage of that table is: it helps to avoid misunderstandings and shows that – for the current "Chance to loose" – it is completely irrelevant and meaningless which one of his two doors the host will open. Reason: Because it does not matter. Because it does not matter whether the host opens door 2 or door 3: switching will loose anyway, in both cases. That's a fact. And the table makes this obvious. Please try to make a clearer table that shows all possible constellations, and that shows that it is of no relevance at all, for the current "Chance to loose" – if by chance the player picked the car, which door will be opened by the host, whether door 2 or door 3. He just opens "another door", but never the door chosen by the player: "just another door, say No.3"

Sorry that's not what you like to be my answer, but that's important to emphasize, yes even to underline. Kind Regards, Gerhardvalentin (talk) 18:11, 26 August 2010 (UTC)

Rick, you didn' say anything on that table: No host's bias in all 18 halls, showing all possible situations: Pws when host offers to switch AFTER the host has opened one door showing a goat. Gerhardvalentin (talk) 11:18, 28 August 2010 (UTC)

The rows are no longer obviously equally probable (many are dups). If you're really trying to show all possibilities look at the decision tree in the Grinstead and Snell reference. The case where the car is behind door 1 and the player picks door 1 and the host opens door 3 has probability 1/18 (what makes this 1/18 is the host's choice between door 2 and door 3 being made with probability 1/2). The case where the car is behind door 2 and the player picks door 1 and the host opens door 3 has probability 1/9. -- Rick Block (talk) 15:00, 28 August 2010 (UTC)

So what, Rick? 18 independent games, in 6 games switching will loose and in 12 games switching will win. You can clearly see that they of course are "never obviously equally probable". They never will be, because in every three games one is to loose and two will win.

And in 18 games 12 constellations of course have to be, and therefore are represented identically twice. And you propose the decision tree of Grinstead and Snell? Do you know the purpose for which this decision tree is used by them? To show that the "simple" and the "conditional" approach indeed do differ,

because the so called "conditional approach" deviates from Marilyn vos Savant's version of how the game was played, i.e. in some forbidden "Host's bias" that they take to be given to be known.

Although - as you know - an extreme host's bias will lead to the consequences that, in opening of only one door, he tells the location of the other goat and the car at once. So their decision does not address the question reported by Marilyn vos Savant, who presents an unbiased host, but it addresses a deviant variant. Do you agree? Regards, Gerhardvalentin (talk) 20:48, 28 August 2010 (UTC)

No, I do not agree. Grinstead and Snell's tree shows the outcomes given the host is defined to pick without bias between two goats. The purpose of this tree is to answer the question as it is generally posed, i.e. what is the probability of winning AFTER you've initially selected a door and AFTER the host has opened a door (for example, you picked door 1 and the host opened door 3). They say this is a conditional probability, which is how they use the tree. -- Rick Block (talk) 21:50, 28 August 2010 (UTC)

Thank you, Rick, for your patience. Regards, --Gerhardvalentin (talk) 22:47, 28 August 2010 (UTC)

Morgan's error

Probabilities shown on complete positive distribution of any possible combination / outcome      (No disqualified irresponsible host's blab-out bias. Entire protocol-list, game 1 to game 100 000 000, exists)
Total number of games	Guest selects door	Host opens door	car behind door	# switching wins		# staying wins		%
5 644 754	1	2	1			5 644 754	1/3	5,64 %
10 933 265	1	2	3	10 933 265	2/3			10,93 %
5 555 266	1	3	1			5 555 266	1/3	5,56 %
11 200 220	1	3	2	11 200 220	2/3			11,20 %
5 466 783	2	1	2			5 466 783	1/3	5,47 %
11 199 709	2	1	3	11 199 709	2/3			11,20 %
11 199 762	2	3	1	11 199 762	2/3			11,20 %
5 466 514	2	3	2			5 466 514	1/3	5,47 %
11 200 172	3	1	2	11 200 172	2/3			11,20 %
5 555 418	3	1	3			5 555 418	1/3	5,56 %
10 933 187	3	2	1	10 933 187	2/3			10,93 %
5 644 950	3	2	3			5 644 950	1/3	5,64 %
100 000 000				66 666 315		33 333 685		100,00 %

Same as above, but:         Host is assumed to be biased towards always having opened his door with the largest possible number
Total number of games	Guest selects door	Host opens door	car behind door	# switching wins		# staying wins
11 200 020	1	preferred 3	1			11 200 020	1/2
10 933 297	2	preferred 3	2			10 933 297	1/2
11 200 368	3	preferred 2	3			11 200 368	1/2
11 200 220	1	preferred 3	2	11 200 220	1/2
11 199 762	2	preferred 3	1	11 199 762	1/2
10 933 187	3	preferred 2	1	10 933 187	1/2
10 933 265	1	avoided 2	3	10 933 265	always		never
11 199 709	2	avoided 1	3	11 199 709	always		never
11 200 172	3	avoided 1	2	11 200 172	always		never
100 000 000				66 666 315	2/3	33 333 685	1/3

It's evident:

J.P. Morgan, N.R. Chaganty, R.C. Dahiya and M.J. Doviak: Let's Make a Deal: The Player's Dilemma", 1991:

Morgan's evaluation / opinion that "the six simple explanations F1 - F6" were "false solution", evidently have been based on erroneous foundations and faulty mathematics.

• F1, F3:  Morgan et al. say the player having selected door 1, and the host having opened door 3, the probability that the player wins if she switches can never be 2/3, because "The correct simulation for the conditional problem is of course to examine only those trials where door 3 is opened by the host." – ??? – Haven't they done so? If the player has chosen door 1, the host has opened door 3, and the player then switches, she will win with a probability of 2/3, because in 1/3 of these cases the car will be behind door 1 and the player looses by switching, but in 2/3 the car will be behind door 2 and the player wins by switching.

• F2:  Morgan et al. say that the triples AGG, GAG and GGA are having a probability of 1/3 each, where the triple AGG, for instance, means auto behind door 1, goat behind door 2, and goat behind door 3. This is correct. But Morgan et al. argue for example that the following explanation: "The player choosing door 1 will win in two of these cases if she switches, hence the probability that she wins by switching is 2/3" was incorrect because "GGA is not possible, given door 3 has been opened". ???

• F5: Morgan et al. say that, like F1, the simple solution F5 "is a true statement that answers a different problem. F5 is incorrect, because it does not use the information in the number of the door shown." – In effect however, apart from any "famous host's whistle-blowing tattletale bias", the number of the door shown is completely and totally irrelevant (see 100 million table above). Any "relevance" for the number of the door shown only could arise for the given scenario when math-teachers require that their math-students should explicitly use this number in their conditional probability calculation, for training purpose. Otherwise not.

Do you really disagree with this?

Gerhard, I'm puzzled how you can disagree with the statement "There are multiple reliable sources that present simple solutions without (themselves) distinguishing which of A, B, or C they claim to be addressing" ?

Can you propose a rewording? Eg: "There are multiple reliable sources that present simple solutions without (themselves) explicitly distinguishing which of A, B, or C their solution addresses"? Richard Gill (talk) 08:38, 28 November 2010 (UTC)

Thank you, Richard, I did disagree with this formulation because they may be "said" not to distinguish at all. Gerhardvalentin (talk) 09:36, 28 November 2010 (UTC)

Great! Richard Gill (talk) 21:49, 29 November 2010 (UTC)

Please pardon the intrusion, but 'claim to be' has an undeniable intended negative connotation. The single word 'are' is a perfectly acceptable replacement for those who are not pushing a particular POV.

How about, 'There are numerous reliable sources that present simple solutions to the typical MHP statement [or 1 car and 2 goat puzzle where the contestant chooses door 1 and the host reveals door 3]? Glkanter (talk) 22:02, 29 November 2010 (UTC)

Agreement K&W

Hallo Gerhard, ich habe mein Vorschlag Wikipedia talk:Requests for mediation/Monty Hall problem/Starting points noch ein wenig geaendert, weil es notwendig ist dass Auto und Wahl unabhaengig sind. Ich hoffe du bist noch immer einverstanden. Nijdam (talk) 11:25, 13 December 2010 (UTC)

Danke, Nijdam, für Deine Nachricht, ich bin nach wie vor mit Deiner Formulierung völlig einverstanden. Liebe Grüße an Dich!  Gerhardvalentin (talk) 11:44, 13 December 2010 (UTC)

Your evidence for the Monty Hall arbitration case

Please note that evidence should be only about the conduct of editors. I or someone else will remove anything that is about content as ArbCom will not rule on content, so I'd appreciate it if you'd revise it yourself to focus on conduct and not content. Thanks. On behalf of the Arbitration Committee, --Dougweller (talk) 09:22, 20 February 2011 (UTC)

Thank you, Dougweller, but please consider that the steady firing of the criticized conduct of editors in this very special combat, lasting already for years now, permanently is fanned and foment by the disparate bad antagonism the article presents. Irreconcilably. So in the evidence you HAVE to indicate that antagonism that HAS to be solved. Tank you once more, I'm going to say that in my evidence also. Regards, Gerhardvalentin (talk) 09:51, 20 February 2011 (UTC)

Question for you on the MHP discussion page.

Hello, I have a question for you on the MHP discussion page, concerning your recent revert. Please reply there. glopk (talk) 03:37, 3 April 2011 (UTC)

Thank you, glopk, your question there has meanwhile been answered there.  Gerhardvalentin (talk) 09:29, 3 April 2011 (UTC)

Just wanted to mention, I think you are doing a very good job with your MHP edits. Guy Macon (talk) 03:02, 13 May 2011 (UTC)

Thanks, Guy, for the flowers, I find your efforts beneficial to achieve success for the article. Gerhardvalentin (talk) 13:57, 13 May 2011 (UTC)

Hi! I am pondering whether to re-instate your recently reverted edit on the basis that it was taken out for not following a Point of View that is not supported by consensus. However, I have a problem with the phrase "concretized the actual chances" being too hard for the average reader to understand. Could I get you to rephrase the paragraph using simpler language? No need to put it back in the article; I will do that. Just reword it here. Thanks!

For your reference, here is the deleted paragraph.

"Some sources say that the simple solution disregards the variant that the host, under the game rules, could not be taken to randomly choose which one of his two doors to open if both contain goats and, by his special behavior, could be giving an additional hint on the actual location of the car. Depending on which door he opened, he could have concretized the actual chances for switching. But even in such variant the average chance for winning by switching firmly remains 2/3, so staying can never be better than to switch (see "Criticism of the simple solutions" below). "

Guy Macon (talk) 22:49, 29 May 2011 (UTC)

Thank you, Guymacon, please change it as you like, e.g.: "rectified the actual chances for switching in this special game to a closer actual rate", "closer adustified", substanciated, stated more precisely, updated, upgraded, etc. etc.:

"Some sources say that the simple solution disregards the variant that the host, under the game rules, might not be taken to randomly choose which one of his two doors to open if both contain goats and, by his special behavior, could have been giving some additional hint on the actual location of the car. Depending on which door he just opened in that game, he might have specified more precisely the actual chances for switching to a closer actual rate. But even in such variant the average chance for winning by switching firmly remains 2/3 and staying can never be better than to switch (see "Criticism of the simple solutions" below)."

It is important to additionally mention at that time that, for the MHP, all of this can be of any relevance only under the firm condition that one beforehand is accurately informed about such consistent host's bias, its direction and its degree, otherwise no relevance whatsoever.

So all of that academic "conditionals" just show that you really "could know better" if you just "knew better", i.e. they are of no actual relevance at all to answer the famous question. They are just a reference that you can make allowances for such hypothetical (but never given) respective knowledge about the host's behavior. Interesting in teaching and learning conditional probability theory, yes indeed, but without any impact for answering the famous question.

Thank you once more. Regards, Gerhardvalentin (talk) 05:57, 30 May 2011 (UTC)

A. Gnedin: Total solution

Carlton:
In 1 out of 3 staying wins, and in 2 out of 3 switching wins.

Total solution as per A. Gnedin:
Carlton is right, and the conditional probability of winning, given one's own choice and given the door opened by the host, is completely irrelevant: Choose whatever door at random, and then switch, this will give the car in 2 out of 3 (with probability 2/3).

Conditional probability, given the door numbers, is completely irrelevant. Proven by total solution of A. Gnedin.

[1]

Gerhardvalentin (talk) 21:07, 1 June 2011 (UTC)

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A barnstar for you!

	The Barnstar of Diplomacy
	Gerhardvalentin, you set Newellington straight on the current situation of the Monty Hall Problem and made him feel welcome to be bold and change the article. For doing this peacefully, without getting angry at Newellingtion, a certain Wikipedian has decided to award you this barnstar. Newellington (talk) 14:39, 8 August 2012 (UTC)

Invitation to comment at Monty Hall problem RfC

Because of your previous participation at Monty Hall problem, I am inviting you to comment on the following RfC:

Talk:Monty Hall problem#Conditional or Simple solutions for the Monty Hall problem?

--Guy Macon (talk) 22:18, 6 September 2012 (UTC)

A starbarn for you!

	The Argumentative Wikipedian barnstar
	There is a fine book written by Nobel Prize winning Indian economist Amartya Sen. And I think you deserve this barnstar for the fine, well thought rationales and arguments you present in Wikipedia discussions. Best, Tito Dutta (talk) 15:18, 24 January 2013 (UTC)

Last year, we discussed on another math related question- 1500 years ago, a mathematical genius did some excellent mathematical researches, invented 0 (or started using 0), worked on Trigonometry, Algebra, Astronomy. He also attempted tom find the value of Pi. Our discussion was on his calculation of Pi. Finally the discussion ended without getting any specific answer. Do you have any idea there? You can see the discussion here. If you reply there, please copy the whole section from talk page archive and start a new section at my talk page! --Tito Dutta (talk) 15:27, 24 January 2013 (UTC)

Thank you so much Tito, for having given me your Argumentative Wikipedian barnstar, I very much appreciate this recognition. By the way, your name is familiar to me, as Dutta is the name of a friend of mine and colleague in the same group , we sometimes played chess in the evening, but he won more often than I...
Kind regards, Gerhardvalentin (talk) 16:30, 24 January 2013 (UTC)

Ya, it's Dutta --Tito Dutta (talk) 16:32, 24 January 2013 (UTC)

Tito, thank you for your hint to look at Amartya Sen, his arrow’s paradox etc. As to Aryabhata's way of finding the value of Pi that is much more accurate than 22/7, for me it seems the approach might have been by iteration, as e.g. you easily can get any square root by iteration, in just few steps.

And, maybe in lack of the math. decimal point, he had to find a large number that is accurate enough to give an approximation that, by dividing it by the multiple (the thousandfold or the thousandfold) of a simple digit (the "2" in his case) will give some value that was correct "enough". And I suppose he might have started with some rough value, in his case it eventually could have been at first (3?)*(100+4) (=312), and he just might have tried then to get some suitable approach by varying this value with a special factor.

At first (?) by changing my "3" to some other digit ... finally it was the "8" etc. etc.

I will be trying to see if there is some possible way for me to retrace his approach. Thank you and kind regards, Gerhardvalentin (talk) 17:50, 2 February 2013 (UTC)

Rollback

I have granted rollback rights to your account. After a review of some of your contributions, I believe you can be trusted to use rollback for its intended usage of reverting vandalism, and that you will not abuse it by reverting good-faith edits or to revert-war. For information on rollback, see Wikipedia:New admin school/Rollback and Wikipedia:Rollback feature. If you do not want rollback, contact me and I will remove it. Good luck and thanks. Guerillero | My Talk 03:38, 19 February 2013 (UTC)

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--Guerillero | My Talk 03:38, 19 February 2013 (UTC)

Thank you Guerillero for spontaneously having granted me Rollback and Rewiewing. Gerhardvalentin (talk) 10:29, 19 February 2013 (UTC)

Huh?

What was the purpose of this revert? Boghog (talk) 21:18, 25 April 2013 (UTC)

Thank you Boghog, this obviously was a completely unintended mistype, please pardon the violation. Thank you once more. Gerhardvalentin (talk) 21:29, 25 April 2013 (UTC)

OK, no worries. Thanks for the explanation. Cheers. Boghog (talk) 21:34, 25 April 2013 (UTC)

Helmutzipfel

 Done GiantSnowman 10:17, 16 October 2014 (UTC)

Revert

I'm not sure that this edit that you reverted and gave a warning to the editor for is actually vandalism. Neil916 (Talk) 16:31, 22 October 2014 (UTC)

Thank you, but the name of any subspecies Apogoninae is not even accepted in Species/Synonymy list for the family Apogonidae as currently in FishBase, yet. Regards, Gerhardvalentin (talk) 19:30, 22 October 2014 (UTC)

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