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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 (talk • contribs)
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 (talk • contribs)
The article currently claims that "Krauss and Wang (2003:10) conjecture that people make the standard assumptions even if they are not explicitly stated." I can't speak for everyone, but I assumed that the host was trying to win against the guest. So if the host knew what was behind each door, and the host hadn't picked the car in the first round, then the host would pick the car before the second round. And if the host did not know, then the host would choose at random. Each of these interpretations is at odds with the "standard assumptions" and each changes the odds and the best strategy for the guest. They obviously can't be talking about all people, since some of us don't make the "standard assumptions," so which people? 18.104.22.168 (talk) 21:36, 2 January 2017 (UTC)
Discussions about the math should be at the /Arguments page
I would know what would happen if another player in the same Monty Hall game had chosen other door (not the one open by the presenter) and given the opportunity to change door also, which would be the probability for him, and the sum of probabilities for both two players?.22.214.171.124 (talk) 03:42, 4 May 2017 (UTC)
Suppose there are two trifles, trifle A being ten times bigger than the trifle B. A ring is hidden in one of them by Mr X. For Mr Y the chance of teh ring being in trifle B is 1/11. Or is it? Does the size really matter? FleischerDan (talk) 14:34, 8 August 2017 (UTC)
Probably belongs on the 'arguments' page (and is an analogy to the first stage only of the MHP) but using the information available to Mr Y, and making the assumption that a ring is smaller than a trifle, it's 1/2. Freddie Orrell 15:05, 8 August 2017 (UTC) Correction: that assumption abut size is superfluous since the ring is successfully hidden, and I should not have included it. Freddie Orrell 22:27, 12 January 2018 (UTC)
But isn't there a FOURTH possibility: The car is behind door No. 1 and the host opens door No.2 NOT door No.3.
That isn't a possibility because the table doesn't make reference to which door is opened. It only makes reference to which door is chosen. In 1/3 of cases you will pick door 1. In some proportion (generally 50%) of those the host opens door 3 and in the remainder door 2. But irrespective which door is opened in the 1/3 of cases where you pick the door with the car behind it you win by staying. In the 2/3 of situations where you pick a door with a goat behind it you lose by staying. SPACKlick (talk) 16:03, 2 October 2017 (UTC)