Talk:Monty Hall problem

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50/50

Moved to Talk:Monty Hall problem/Arguments § 50/50

– TheGoatOfSparta (talk) 15:24, 4 December 2024 (UTC)[reply]

Ignorant Monty / Monty Fall - current explanation is incomplete

The table currently describes "Ignorant Monty" solution as "switching wins 50%". However, in this variant, switching and staying are indifferent (when a goat has been revealed by chance by Ignorant Monty) and both in fact win 50%. Suggest that table be updated to state that "switching or staying both win 50%". This is given already in the citation for that Variant, if you read the second page of https://en.wikipedia.org/wiki/Monty_Hall_problem#CITEREFRosenthal2005a 2600:8801:17E2:0:30D0:6149:CBE5:D00B (talk) 17:15, 19 November 2024 (UTC)[reply]

We know the car is behind any of the last two doors 100% of the time, so if the switching one holds it 50% of the time, inevitably the staying one will hold it the other 50%. 190.203.104.150 (talk) 04:54, 11 December 2024 (UTC)[reply]

The completely unnecessary long-winded discussion mainly confuses readers

This article is much more confusing than enlightening.

Instead of straighforwardly explaining the problem and its correct solution, it goes into all manner of alternative theories.

Furthermore, the illustrated explanation contains statements "Probability = 1/6", "Probability = 1/3", "Probability = 1/3", "Probability = 1/6",

without ever stating what these numbers are the probabilities of.

That is very unclear writing.

I hope someone familiar with this subject will fix this.

It needs a better explanation

Gaining more knowledge change conditional probabilities.

P(Door 1 | Door 3 unknown) < P(Door 1 | Not Door 3)

01:10, 8 December 2024 (UTC) Tuntable (talk) 01:10, 8 December 2024 (UTC)[reply]

What needs a better explanation is your post. E Eng 05:36, 8 December 2024 (UTC)[reply]
Not necessarily. Remember there are cases in which P(A|C) = P(A). In those cases we say that A and C are independent. That situation occurs when the condition C reduces both the favorable events and the total events by the same factor, which does not change the proportions; a scaling occurs.

In this specific situation, the revelation of a goat in Door 3 reduces by half the cases in which Door 1 could be incorrect, as it would be wrong if the car were in either Door 2 and Door 3, but Door 3 was ruled out. The problem is that it also reduces by half the cases in which Door 1 could be the winner, because when Door 1 has the car the host is free to reveal either Door 2 or Door 3, so we don't expect that he will always opt for #3. Sometimes he will open #2 and sometimes #3, so the cases in which #1 is the winner and he then opens #2 were also ruled out.

We need to be careful with proportions because they don't necessarily tell if the information was updated or not; we could have updated it while preserving the proportions. To put a more obvious example, just think about a soccer match. Each team starts with the same amount of players: eleven; each starts with 1/2 of the total players on the field. If during the game a player from each team is expelled out, then each team is left with only 10, so despite the total is not the same as in the beginning, each team still has 1/2 of the current total players. 190.203.104.150 (talk) 05:12, 11 December 2024 (UTC)[reply]