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Former featured articleMonty Hall problem is a former featured article. Please see the links under Article milestones below for its original nomination page (for older articles, check the nomination archive) and why it was removed.
Main Page trophyThis article appeared on Wikipedia's Main Page as Today's featured article on July 23, 2005.
Article milestones
DateProcessResult
May 3, 2005Peer reviewReviewed
June 25, 2005Featured article candidatePromoted
January 29, 2007Featured article reviewKept
May 18, 2008Featured article reviewKept
June 13, 2011Featured article reviewDemoted
Current status: Former featured article

Any real mathematicians ?

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This article is lovely insofar as it illustrates so well confusion. It should be listed under "Magical Thinking", or "Cognitive bias", and probably "Vain attempts". I will not even begin to correct it, nor explain (I know my WP); just let possible startled readers know that yes, it is a lot of gobbledigook, some of it by people who are perfectly aware and enjoy it.Environnement2100 (talk) 10:55, 8 April 2023 (UTC)[reply]

Do you know your WP enough to know that there are at least 2 articles with Argument pages? This article is one of them. You may have been referring to something other than WP:RS, but my understanding is that on the arguments page, WP:RS only comes in if the conversation-starter brings up changing the article, rather than to shut down other conversations. JumpDiscont (talk) 22:46, 13 April 2023 (UTC)[reply]
Genuinely curious what another article with an Argument page is, can't find any info on Google. 174.44.112.170 (talk) 07:25, 21 March 2024 (UTC)[reply]
I know of Talk:Cantor's diagonal argument/Arguments. - Jochen Burghardt (talk) 12:27, 21 March 2024 (UTC)[reply]
You can search intitle:"Arguments" on Wikipedia and then filter by the talk namespace.
Avessa (talk) 12:31, 21 March 2024 (UTC)[reply]
for anyone who's more curious ​ ^_^ : ​ ​ ​ The 0.999... Arguments page is the one - other than _this_ page - that I knew about before I saw Jochen's reply. ​ ​ ​ ​ ​ ​ ​ JumpDiscont (talk) 07:28, 17 April 2024 (UTC)[reply]
What on earth are any of you talking about? What, concretely, is wrong with the article? Brusquedandelion (talk) 07:37, 21 March 2024 (UTC)[reply]

50/50

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Monty Hall

Prize behind door 1.

Choose door 1. Shown door 2. Swap Result lose.

Chose door 1. Shown door 3. Swap Result lose.

Choose door 1. Shown door 2. No swap. Result win.

Choose door 1. Shown door 3. No swap. Result win.

2 wins and 2 losses from 4 possibilities when you choose the correct door.

Choose door 2. Shown door 3. Swap. Result win.

Choose door 2. Shown door 3. No swap. Result lose.

1 win and 1 loss from 2 possibilities when you choose the wrong door.

50/50 chance. 213.128.242.112 (talk) 18:59, 1 July 2023 (UTC)[reply]

Amazing, that table recapitulates the subject to the point. All difficult aspects of the matter are summarized on a small plot. In a fantastic way, you present complex issues clearly.
Though your result is only possible under special conditions concerning the host's behavior e.g. the "lazy host" who stands next to the object of attention and only wants to open door 3 if possible to avoid long distances. In that case, among many, there is indeed a 50/50 chance to win the trophy.
When we follow all the standard assumptions which are listed in the article and additionally assume that we are dealing with a "balanced" host that means the host's intention to open the one or the other door is completely random (50/50), then the following table would apply.
I added the host's intention to the table: "as wanted" means that the host opens the door he originally wanted to, "door x wanted" means that the host actually wanted to open door x. If he can decide between two "goat doors" he is free to choose one of them and does so randomly with a probability of 50%. If there is the "prize door" and a "goat door" left and he actually wanted to open the "goat door" he must open it, which coincides his intention. If there is the "prize door" and a "goat door" left and he actually wanted to open the "prize door" he must open the "goat door" nevertheless.
But these are two seperate cases that need to be taken into account and this fact affects the probability values of our overall calculation.
Monty Hall
Prize behind door 1.
Choose door 1. Shown door 2 (as wanted). No swap. Result win.
Choose door 3. Shown door 2 (as wanted). No swap. Result lose.
Choose door 3. Shown door 2 (door 1 wanted). No swap. Result lose.
Choose door 1. Shown door 3 (as wanted). No swap. Result win.
Choose door 2. Shown door 3 (as wanted). No swap. Result lose.
Choose door 2. Shown door 3 (door 1 wanted). No swap. Result lose.
2 wins and 4 losses from 6 possibilities.
33.33 chance to win if player decides not to swap.
Choose door 1. Shown door 2 (as wanted). Swap. Result lose.
Choose door 3. Shown door 2 (as wanted). Swap. Result win.
Choose door 3. Shown door 2 (door 1 wanted). Swap. Result win.
Choose door 1. Shown door 3 (as wanted). Swap. Result lose.
Choose door 2. Shown door 3 (as wanted). Swap. Result win.
Choose door 2. Shown door 3 (door 1 wanted). Swap. Result win.
4 wins and 2 losses from 6 possibilities.
66.66 chance to win if player decides to swap. 188.106.91.33 (talk) 10:25, 16 July 2023 (UTC)[reply]
The problem is not well defined. If it is reformulated that the host does not tell which door s/he opens (but this door contains goat)- then you are right. If the problem is formulated in dubious way (as it is) by saying that the host opens, say, door 3 - then, the 50/50 guy could be absolutely right! It is just badly formulated problem - which happens a lot with probability problems! 130.88.75.80 (talk) 14:20, 17 May 2024 (UTC)[reply]
I wrote a whole essay here about 4 months ago, articulating EXACTLY why the Monty Hall problem is based on an illusion. I spelled it out and went through it step by step. Mr.JumpDiscont here DELETED everything I said because he couldn't find the flaw in my logic. He couldn't find the flaw because there wasn't one. The odds ARE actually 50/50. The intro to the actual problem is used to throw the readers focus off onto extraneous information which is not a part of the actual equation.
It's basically a magic trick, used to fool ppl who can't break it down.
Boo on Mr. JumpDiscont for deleting valid commentary and any argument which disagrees and disproves his page 😂😂😅 AI*girllll (talk) 18:32, 12 June 2024 (UTC)[reply]
Oh, wait... my bad...it HASN'T been deleted, it's just on the 'Arguements' page. Go check it out --'Monty Hall 33/66 is based on an illusion '. AI*girllll (talk) 18:35, 12 June 2024 (UTC)[reply]
In addition to the - it's on the Arguments page - part,
It wasn't me who moved your essay there.
and
I have edited this _talk_ page, but as far as I can tell, I've never edited the _article_ page.
and
You have not replied to my response to your essay. (in the section you mentioned of the Arguments page)
.
(Even if you think the lower part of my response has nothing to reply to, there's still:
Do you get 50/50 even under what I called the crucial assumptions, or instead get 50/50 on the basis that those 3 assumptions don't all hold in the real world?)
JumpDiscont (talk) 21:52, 5 July 2024 (UTC)[reply]

Addition to the 'Solutions by simulation' section?

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Can I propose an addition to the 'Solutions by simulation' section - a simplified Python script to give the following result (approximately)

Overall success rate given a stick or switch strategy
win_rate(monty_knows=True, will_switch=True, give_overall_result=True) = 0.6641
win_rate(monty_knows=True, will_switch=False, give_overall_result=True) = 0.3349
win_rate(monty_knows=False, will_switch=True, give_overall_result=True) = 0.3359
win_rate(monty_knows=False, will_switch=False, give_overall_result=True) = 0.3254
Success rate at the point where the contestant has a choice to stick/switch
win_rate(monty_knows=True, will_switch=True, give_overall_result=False) = 0.6685
win_rate(monty_knows=True, will_switch=False, give_overall_result=False) = 0.3337
win_rate(monty_knows=False, will_switch=True, give_overall_result=False) = 0.5004
win_rate(monty_knows=False, will_switch=False, give_overall_result=False) = 0.5002

The Python code is as close to pseudo code as I could make it.


""" Monty Hall probability calculator. """

import random  # Random number generator module

def random_door(except_doors):
    """ Return a random door number which in not in the list given. """
    while True:
        n_doors = 3
        door = random.randrange(1, n_doors + 1)   # Return 1, 2 or 3
        if door not in except_doors:
            return door

def win_rate(
        monty_knows: bool,
        will_switch: bool,
        give_overall_result: bool) -> float:
    """
    monty_knows:
      True if Monty knows which door has the prize, False otherwise.
    will_switch:
      True if, given a choice, the contestant will switch doors.
    give_overall_result:
      True if we want the probability of winning with a given strategy
      False to give the probability of winning from the point when a stick or
            switch choice is made.
    Returns: Probability of contestant winning given the test criteria above.
    """
    # Note: Doors are numbered 1, 2, 3...
    runs = 0                # The number of valid games
    wins = 0                # The number of times the contestant has won so far
    while runs < 10000:
        runs = runs + 1     # Increment the run count (may get decremented later)
        prize_door = random_door(except_doors=[])   # Door with the prize
        choose_door = random_door(except_doors=[])  # Door chosen by the contestant
        if monty_knows:
            # Monty knows which door has the prize.
            # He chooses to open the door without a prize
            second_door = random_door(except_doors=[choose_door, prize_door])
            # The contestant must choose to switch or stick
            if will_switch:
                choose_door = random_door(except_doors=[choose_door, second_door])
        else:
            # Monty does not know which door has the prize
            # He chooses one of the other doors to open
            second_door = random_door(except_doors=[choose_door])
            if second_door == prize_door:
                # Monty has opened the second door and revealed the prize, so
                # the contestant loses.
                if not give_overall_result:
                    # We want to give the probability of winning from the point
                    # of being given a choice, so we discard this scenario from
                    # the results.
                    runs = runs - 1
            else:
                # The second door does not have the prize.
                # The contestant must choose to switch or stick
                if will_switch:
                    choose_door = random_door(except_doors=[choose_door, second_door])
        # Has the contestant won?
        if choose_door == prize_door:
            wins = wins + 1
    return wins / runs

if __name__ == '__main__':
    print("Overall success rate given a stick or switch strategy")
    print(f"{win_rate(monty_knows=True, will_switch=True, give_overall_result=True) = }")
    print(f"{win_rate(monty_knows=True, will_switch=False, give_overall_result=True) = }")
    print(f"{win_rate(monty_knows=False, will_switch=True, give_overall_result=True) = }")
    print(f"{win_rate(monty_knows=False, will_switch=False, give_overall_result=True) = }")
    print("Success rate at the point where the contestant has a choice to stick/switch")
    print(f"{win_rate(monty_knows=True, will_switch=True, give_overall_result=False) = }")
    print(f"{win_rate(monty_knows=True, will_switch=False, give_overall_result=False) = }")
    print(f"{win_rate(monty_knows=False, will_switch=True, give_overall_result=False) = }")
    print(f"{win_rate(monty_knows=False, will_switch=False, give_overall_result=False) = }") 

92.239.201.60 (talk) 09:02, 7 June 2024 (UTC)[reply]

This article, like most articles, is better off without code samples, which are meaningless to many readers. The procedure for simulating with cards is sufficient to make the point and can be understood and implemented by anyone. MrOllie (talk) 13:07, 7 June 2024 (UTC)[reply]

Ignorant Monty / Monty Fall - current explanation is incomplete

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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]