Talk:Monty Hall problem

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Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show?[edit]

I do not think this is correct. Monty Hall made clear that nothing like the Monty Hall Problem, as it is widely understood, could ever have happened on the game show. Martin Hogbin (talk) 14:34, 29 December 2015 (UTC)

And how does this conflict with "most statements of the problem do not match the rules of the actual game show" (which, BTW, is sourced)? -- Rick Block (talk) 17:35, 31 December 2015 (UTC)
By starting with, 'Most statements of the problem..', the wording suggests that there was a real game that was similar to the MHP. We know from Monty Hall's words that nothing like the well known problem could have happened on the TV show. Martin Hogbin (talk) 16:09, 1 January 2016 (UTC)
It's true that the game show in the Monty Hall Problem has rules different from those of Let's Make a Deal. The real Monty Hall varied his moves, to keep the show confusing for contestants and fun for viewers. In the MHP, the host follows a strict algorithm.
The distinction between the hypothetical and real game shows isn't very important to the MHP. What is important is that the statement of the problem stipulate that the hypothetical host, unlike his real-life counterpart, ALWAYS reveals one of the unchosen doors as a loser. Marilyn Vos Savant, for example, failed to include this in her initial discussion of the MHP. TypoBoy (talk) 20:38, 14 March 2016 (UTC)

FourthKind Solution: Monty Hall problem Solved and Confusion Explained[edit]

Discussion not focused on improving the article on the basis of verifiable info from reliable sources. Do not reopen this, and as suggested, consider going to the arguments page. I, JethroBT drop me a line 23:25, 16 February 2016 (UTC)

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

FourthKind Solution: Monty Hall problem Solved and Confusion Explained


There are 2 games being played here on Wikipedia and the media in general. Let’s call them:

1: Monty Hall Game

2: 1/3-2/3 Model Trap Game


Explanation for Monty Hall Game


The Paradox

The question posed in the Monty Hall problem on Wikipedia: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a)


Standard Assumptions

1. The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).

2. The host must always open a door to reveal a goat and never the car.

3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

(Note there are 3)


Answer to the Paradox

Question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a)

Answer: No (with the 3 Standard Assumptions, choose at random)


Not So Simple Solution

If the contestant adheres to all of the 3 Standard Assumptions, the contestant should answer randomly unless the contestant is really good at reading Monty’s “tells”. Toss a coin in your head, use a pocket random generator, etc. The contestant doesn’t even really have to decide anyway. If the contestant can’t decide, guess, that is a random choice itself. A random choice, without reason. It is a 50/50 probability at the real moment of choice (the contestant answering Standard Assumption #3), that is the real moment of consequence when the probabilities become fixed. The moment when win or lose are real possibilities with unchangeable outcomes. Your final answer. There is no consequence in choosing a door initially when the probability for the contestant winning is 0% or losing 0%, the game show producers winning 0% or losing 0%. Why even consider this in the Monty Hall Game? The answer is because it has great consideration in the 1/3-2/3 Model Trap Game. This is the “connection” some mathematicians make between the Monty Hall Game and the 1/3-2/3 Model Trap Game. It is not important to the outcome of the Monty Hall Game probabilities. It is important to the outcome of the 1/3-2/3 Model Trap Game probabilities to explain how the contestant makes an error (losing 2/3 of the time) by “always-staying” in the 1/3-2/3 Model Trap Game. This is why. In the Monty Hall Game the initial door choice is merely used as a “reference point” whether you changed your mind (switch) or not (stay) later in the game. It must be considered in the 1/3-2/3 Model Trap Game. In the 1/3-2/3 Model Trap Game this is a necessary & fixed 4th condition.


After all, the game is not decided when choosing an initial door in the Monty Hall Game but the game is decided at that point in the 1/3-2/3 Model Trap Game. It is 1 of only 2 possible avenues of choice to take in 1/3-2/3 Model Trap Game. In the Monty Hall Game the initial door chosen is merely a reference point. In the 1/3-2/3 Model Trap Game it is described in 1 of only 2 possible answers, always-stay or always-switch, which the player must always choose from (to prove the probability calculations/model). That’s important! That is the difference. The Monty Hall Game has no repeated fixed answer, after all it is a one-time, 2 door, 2 choice, 50/50 event. The reason contestants fear choosing the “wrong door”. Maybe they missed some math advantage? Enter the 1/3-2/3 Model Trap Game.


In the Monty Hall Game the real game from the contestant’s and Monty’s perspective begins after Monty’s 2nd question (via Standard Assumption #3). The game for the contestant (not the viewer) is not after selecting a door of no consequence (sorry Monty). After all the game always continues via Standard Assumption #3. The game never stops after the initial door is chosen by the contestant and the initial door choice has no consequence to the outcome other than being used as a reference point. It is merely used as showmanship for the show (sorry Monty). Down to 2 doors. Win 50% or Lose 50%. Who has the advantage you or Monty. Neither, so the correct answer to the question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a) is “no”.


The Monty Hall Game For the contestant is a one-time event. The contestant gets 1 choice in their life (assuming Monty doesn’t ask them back). No advantage for Monty, no advantage for the contestant. The choice is 50/50 (2 doors left, 2 possible choices WARNING: switch or stay will result in only a 1/2 advantage).


1/3-2/3 Model Trap Game In the 1/3-2/3 Model Trap Game the game continues for as long as you wish. Play always-stay or always-switch for as long as you want. Answer: yes (with the 4 Standard Assumptions, choose the always-switch answer this will result in a 2/3 advantage and the always-stay answer will result in a 1/3 advantage: WARNING: choosing a random-only answer will result in only a 1/2 advantage) It is not the Monty Hall Game. Even Monty stated as such.


Explanation for 1/3-2/3 Model Trap Game


The Paradox Question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a) Answer: Yes (with the 4 Standard Assumptions, choose always-switch answer this gives the player a 2/3 probability of winning choosing always-stay answer will result in only a 1/3 probability of winning)


Standard Assumptions

1. The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).

2. The host must always open a door to reveal a goat and never the car.

3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

4. The contestant must always-stay or always-switch (an improbable event unless the contestant has been on a previous show)

(Note there are 4)


Answer to the Paradox

Question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a) Answer: yes (with the 4 Standard Assumptions, choose the always-switch answer this will result in a 2/3 advantage: WARNING: choosing a random-only answer will result in only a 1/2 advantage)


Simple Solution

If you adhere to all of the 4 standard assumptions and you always-switch, you win 2/3 of the time. If you always-stay you win 1/3 of the time. If you choose a random answer when asked the important question of whether to stay or switch (cheat) you will win 50% of the time playing this model. How unusual (explanation later)!


Confusion Explained: Mixing the 2 different games

You have noticed by now there are 2 different games competing for referencing solutions. 2 different games competing for rules (Standard Assumptions). 2 different games explaining how to win (Simple Solutions). Which is the real game? FourthKind proposes that the Monty Hall Game is the real Deal (oops, thanks Monty). Monty Hall even claims it is the real game. But that is boring and too simple. 50/50, who needs a mathematician for that?


Comparing the correct answers of the 2 games to the original question posed

The original question posed: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a)


Monty Hall Game: no (choose randomly for an even chance)

1/3-2/3 Model Trap Game: yes (“always-switch” choice gives the player a 2/3 probability advantage


Possible Solutions to the Confusion

1: A 4th Standard Assumption such as the one I proposed needs to be added to the 3 Standard Assumptions of the Monty Hall Game before trying to explain the Monty Hall Game using the 1/3-2/3 Model Trap Game.

2: Split the connection of the Monty Hall Game and 1/3-2/3 Model Trap Game entirely (retract the “solutions” to the Monty Hall Game)

3: Start a 1/3-2/3 Model Trap Game page in Wikipedia.

FourthKind FourthKind (talk) 14:02, 16 February 2016 (UTC)



Every time a constant chooses to stick the probability is 2/3 of winning, every time a constant switches the probability is 1/3 of winning. A strategy of randomly choosing to stick, or switch, is statistically less successful than always sticking and more successful than always switching.
Also, a reference for "The contestant must always switch or always stay with their original door choice when offered by the host. (an improbable event unless the contestant has been on a previous show)" would be needed, as Wikipedia aims for verifiability rather than truth - see WP:NOTTRUTH. Jonpatterns (talk) 15:53, 14 February 2016 (UTC)
FourthKind, the Monty Hall problem is infamous for fooling nearly everybody and you are one of those that it has fooled. This page in not for discussing why wrong solutions are wrong or for helping people understand the problem and the correct solution. If you want to do this I sggest that you go back to the arguments page. Martin Hogbin (talk) 16:09, 14 February 2016 (UTC)


Martin Hogbin thank you for commenting on my posting, I truly appreciate it and with all due respect thank you for also helping me prove my points.


You state: "Every time a constant chooses to stick the probability is 2/3 of winning, every time a constant switches the probability is 1/3 of winning"


My Point: I don't see the words "every time" in any of the 3 Standard Assumptions. There needs to be a 4th Standard Assumption for the 1/3-2/3 statistical models to be a valid mathematical argument. This new assumption or method of choosing is not stated before the mathematical arguments are proposed. They are continually referenced as if there are the only 2 possible avenues available to the contestant, not taking into account a 3rd possible avenue of “always-random”. This 3rd avenue is always overlooked and never addressed by any of the 1/3-2/3 models. The 4th Standard Assumption being the one I proposed. The contestant must always switch or always stay with their original door choice when offered by the host.


The Confusion problem with the public: That they are told by some mathematicians that they have a 1/3 chance of winning from the beginning. Before the contestant answers Monty's 2nd question: Do you want to stay or switch? the probability is a 50/50 chance. That is why I believe the public is horrified to be shown such a disadvantage that is often described as not "intuitive".


The confusion being: The problem is explained mathematically with an added condition of "every time" they stay and "every time" they switch. A flaw of reasoning by proving a mathematical model which will ONLY works with this added condition after the 3 Standard Assumptions are explained. This condition incidentally is not in any the 3 Standard Assumptions. Ask yourself who added this extra condition? The answer is the mathematicians who explain the problem using the 1/3-2/3 models. Mathematicians who use this model need this 4th condition to prove Monty has an edge with all the 1/3-2/3 models. There is currently no such condition under the 3 Standard Assumptions. This 4th Assumption needs to be added, or the 1/3-2/3 models to be refined. This is root of the confusion. That is why the public is horrified to be "proven" they were at such a disadvantage. The public is given an explained with a "proof" that is often referred as not "intuitive" and most people fall into a "fool's trap" by staying (which in reality, later in the show (when they both share the “discarded” set) a simple 50/50 chance choice is presented. They can’t figure out why Monty has such an edge (2/3 vs. contestants 1/3). Monty's edge is created by mathematicians and needs a 4th Assumption to be proven accurately and taken seriously. Covertly slipping it in at a later date, unbeknownst to the public undermines the credibility of mathematics in general. Also, mathematicians on YouTube and on the Wikipedia page for the Monty Hall problem do not explain the concept that the "discarded 999,998 choices" in the 1 million-door explanation. Mathematicians fail to explain that these "discarded choices" (also relevant in the 3 doors problem) shown by the host are now a set of "shared choices" by both the contestant and Monty. A set that is shared by both the contestant and Monty now makes the probability a 50/50 one as both sides have 999,999 chances out of 1 million. The 999,998 chances are now exposed to both. The contestant is not stuck with one-in-a-million chance of success as described at the beginning by the mathematician. The 50/50 probability at the end is presented as 1 vs. 1,000,000. Which would you choose? The power of math, or I should say the power of some mathematicians with the 1/3-2/3 models.


Also, I hope the statement: "Wikipedia aims for verifiability rather than truth" is a company slogan and not to be taken seriously. As for me, truth by democracy is not my cup of tea.

Honestly and sincerely thank you again. FourthKind FourthKind (talk) 12:04, 15 February 2016 (UTC)

@FourthKind: It doesn't matter how many times the content plays the game or only once the probability to stick is always 2/3 and switch always 1/3. The success rate over multiple goes is in fact a different question. There are no shared choices, see assumption two '2.The host must always open a door to reveal a goat and never the car.' Or in the many door question 'the host must always open all the remaining doors bar one, only revealing goats'.
Okay, now consider I'm wrong and you are right. Wikipedia will need a reference to back up your claim about the fourth required assumption, such as a peer reviewed paper. Its outside the scope of Wikipedia to do mathematical proofs. The scope of Wikipedia is to write an encyclopedia about proofs that have already been published. Jonpatterns (talk) 13:36, 15 February 2016 (UTC)

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Can we please keep this kind of mathematical or philosophical discussion to the arguments page. Martin Hogbin (talk) 14:55, 16 February 2016 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

To what "standard" do we refer?[edit]

The article currently has a section named "standard assumptions". That's a misnomer; no standards body promulgates a list of statements whose truth can be assumed when analyzing mathematical problems.

This quirky bit of lingo seems to be a sneaky way of excusing people who omit important parts of the problem statement. Marilyn Vos Savant made this error in her initial description of the problem; she described the host opening a non-winning door, but failed, crucially, to state that the host always does this. A reader new to the problem can be forgiven for failing to guess at this rule which, after all, the real Monty Hall did not follow.

This section should be named something more like "stipulations", and shouldn't try to do double duty by both stating the stipulations and subtly claiming that their statement shouldn't be necessary. TypoBoy (talk) 03:10, 15 March 2016 (UTC)

Per the sources in the standard version of this problem all of these are usually assumed even if not stated, you can see from MSV's answer that she assumed the host always opened the door even though she didn't state it. Relation to the original show is irrelevant as the problem isn't based on the show. SPACKlick (talk) 14:49, 15 March 2016 (UTC)
What's not clear is whether she realized at the time that the assumption mattered. (She clearly knew that by the time of the second column, but at the time of the first column, I don't see any evidence beyond her later say-so.) --Trovatore (talk) 17:39, 15 March 2016 (UTC)
All of that may seem correct, but take a closer look. She did not just say that "the host opens just one of the two other doors", but (let's forget about the door #numbers) she wrote indeed: "...and the host, who knows what’s behind the doors, opens another door which has a goat. So IMO indeed it's necessary to take a closer look. Kind regards, Gerhardvalentin (talk) 12:19, 2 May 2016 (UTC)

This section is problematic. The MHP can be solved without the assumption that the host always reveals a goat. For example, the host may sometimes reveal a goat and sometimes not; we are presented with a game in which he does. Is he doing so to help us win the car as we chose a goat first (the Angelic Monty variant), or to discourage us from winning it as we chose the car first (Monty from Hell)? In the absence of any other information, the principle of indifference suggests we have equal chance of either, supporting the 2/3 solution. Anyway, surely the key assumption is that a car is more advantageous than a goat, without which the solution cannot be verified.Freddie Orrell 20:54, 14 June 2016 (UTC) — Preceding unsigned comment added by Freddie Orrell (talkcontribs)

You are wrong. If the host doesn't specifically show a goat, the chance to win by switching to the second closed door is not 2/3. Have a look to university of California, San Diego: "Monty Does Not Know Version", and the "Explanation of the game". If the host did show a goat just by chance, the winning rate is not 2/3, but only 1/2. Gerhardvalentin (talk) 10:48, 21 June 2016 (UTC)
Since I said 'we are presented with a game in which he does', I was not discussing the probabilities associated with not showing a goat. The MHP contains the phrase 'and the host, who knows what's behind the doors', which would be superfluous were he to show a goat just by chance; we may therefore assume he is acting deliberately.Freddie Orrell 09:56, 22 June 2016 (UTC) — Preceding unsigned comment added by Freddie Orrell (talkcontribs)

What was the date of inception for the "Monty Hall problem" in Wikipedia?[edit]

Wikipedia talk pages are for discussion of article improvements, not general discussion

FourthKind FourthKind (talk) 11:56, 21 April 2016 (UTC)

February 1975, as it says in the first sentence.SPACKlick (talk) 12:09, 21 April 2016 (UTC)

If you're referring to the creation date of this article, you can find it at the history page, all the way down to the oldest edits. It was created on by AxelBoldt at 23:29, 22 September 2001 with a reference from a New York Times article in July 21, 1991. Diego (talk) 12:42, 21 April 2016 (UTC)