Talk:Monty Hall problem/Arguments

50/50

Moved from Talk:Monty Hall problem § 50/50

 – TheGoatOfSparta (talk) 15:28, 4 December 2024 (UTC)

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)

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)

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)

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)

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)

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) — Preceding unsigned comment added by TheGoatOfSparta (talk • contribs)

When the player gets their 2nd guess, it is 1 of 2 doors. 1 contains a car and 1 contains a goat. There are not 3 ways to choose 1 of 2 objects.The probability can only be 1/2.

There cannot be more possible guesses than doors.For details ref:

https://drive.google.com/file/d/1WmNEa9CFm-hiRBFC8ooQLO6YHK7_osYy/view?usp=sharing

~ Phyti (talk) 16:06, 13 June 2025 (UTC)

<sigh> EEng 21:02, 13 June 2025 (UTC)

@Phyti ​ ​ ​ Consider this show:

There are only 2 doors. The host rolls a 6-sided die. If the die lands showing 1, then the host puts a car behind door 1 and a goat behind door 2, else the host [puts a car behind door 2 and a goat behind door 1 and moves the die so it's showing 1].

After the host does as above, what is the probability that the car is behind door 1?

JumpDiscont (talk) 02:01, 15 June 2025 (UTC)

If there are 2 doors, 1 with a car and 1 with a goat, the possibility/probability of guessing the car door is 1/2. The die is irrelevant. You could simply toss a coin with the same results.

The player represents the general public having an opportunity to enrich themselves with very little effort, and increases the viewing audience interest.

The supporters of Marilyn Savant can't accept she made errors in her interpretation of the MH game. That would mean they also made the same errors.

A high IQ doesn't imply you are above average for all forms of knowledge. Phyti (talk) 16:31, 16 June 2025 (UTC)

The range for where I imagine we will switch

from disagreeing to agreeing, is now small-enough

for me to give multiple questions in parallel:

​

Are your answers to questions Q0,Q1,...,Q5,Q6

from https://paste.ee/p/fm5fnuqk all the same?

If yes, then what is that answer?

If no, then what is a pair of consecutive such

questions for which your answers are different?

​

(I imagine there is at most one such pair, but if there

is more than one such pair then just give any such pair:

In particular, if you find such a pair then there is

no need to consider the other questions from the link.)

​

JumpDiscont (talk) 02:41, 17 June 2025 (UTC)

​

To maybe make things easier for you,

"Are your answers to questions Q0,Q1,...,Q5,Q6

...

... are different?"

should've been

​

"Are your answers to questions Q0 and Q6

from https://paste.ee/p/fm5fnuqk the same?

If yes, then what is that answer?

If no, then what is a pair of consecutive questions

from among [Q0,Q1,...,Q5,Q6 in that link]

for which your answers are different?"

​

, ​ since if your answers to Q0 and Q6

are the same then I don't care about your

answers to the other five from the link.

​

JumpDiscont (talk) 03:12, 17 June 2025 (UTC)

If anyone is interested, I have published an article on this, explaining step by step why and how Marilyn's argument is incorrect, and why and how the odds once 1 of the 3 doors is removed is truly 50/50... link below :

https://substack.com/home/post/p-157123989?source=queue AI*girllll (talk) 19:25, 18 June 2025 (UTC)

​

​:: @Phyti: ​ My questions to you still apply.

(and I don't know why WP keeps the above line

less-indented and with two visible colons)

​

The rest of this is ​ @AI*girllll .

​

​

Even if that does the first part, it doesn't do the second:

After it finishes attacking the conditional probability approach to the MHP,

it switches to stating the answer is 1/2 each, without giving why or how.

​

​

My initial guess it that you'd say either something like

"The nonzero numbers are equal, so they condense to the same probability."

or something like ​ ​ ​ "The nonzero numbers are equal and conditional probability

doesn't apply, so they condense to the same probability." ​ .

​

​

In the former case, my followup guess is that you will

not allow zooming in on where between the MHP and

​

a situation I created which is such that

[

[you think it has nothing to do with the original problem]

and [I claim it is a situation in which equal nonzero

numbers condense to different probabilities] and

[you [agree or otherwise don't post disagreement] with that claim]

]

​

you will start agreeing with me that the resulting

probabilities for the two doors are different.

​

​

In the latter case, my followup guess is that you will

not allow zooming in on where between the MHP and

​

a situation I created which is such that

[

[you think it has nothing to do with the original problem] and

[I claim it is a situation in which conditional probability applies

and the resulting probabilities for the two doors are different] and

[you [agree or otherwise don't post disagreement] with that claim]

]

​

you will start agreeing with me that the resulting

probabilities for the two doors are different.

​

​

If my initial guess is correct and the relevant

one of my followup guesses is also correct, then

​

I will point that out once

and

I might point it out again if you bring up that your

argument(s) here is/are ​ ​ ​ not ​ / ​ no longer ​ ​ ​ being opposed

but

I will otherwise stop replying to you, because by not revealing

what probability your approach leads to in situations

that test it, you'd be able to just keep on with roughly

1/3,1/3 ​ becomes ​ 1/2,1/2 , ​ the 2 are equally likely

and/or ​ ​ ​ ​ conditional probability does

not apply here, the 2 are equally likely

​

.

​

​

(However, I would still reply to Phyti, because

at least so far, Phyti has been willing to allow

zooming in on where the disagreement ends.)

​

​

JumpDiscont (talk) 07:34, 19 June 2025 (UTC)

You show the probability of 'win a car' depending on a secondary event (die toss), or conditional probability.

The problem, the method of distribution of prizes for the doors is only known to the host and their staff. It provides no useful information to the player.

The statistics apply to many games, and not to individual games.

The MH game in question did not involve a secondary event, and only a single game played by 1 person. The player would only know there is 1 car behind 1 of 2 doors.

For the general case, there are 3*2*1=6 possible patterns/arrangements of 3 distinct prizes for 3 doors, with c the car.

1 2 3

a b c

a c b

b c a

b a c

c a b

c b a

When c is behind the player's 1st guess door 1, the host has 2 choices, otherwise they have 1 choice.

The patterns are the same for all doors, any prize appears behind any door 1/3 of the time. The player has a 1/3 probability to win any prize on the 1st guess.

The player 1st guess door is not opened to verify the prize. Instead the host opens a non car door, and offers the player a 2nd guess, and it is always 1 of 2 doors.

Using the game rules, there are only 4 possible outcomes resulting from the combination of player and host choices in terms of prizes, and not doors. (shown as a graph Oct. 2024).

The sequence is p=1st guess, h=host choice, r=remaining closed door.

p h r

----

a b c

b a c

c a b

c b a

Goats 1 and 2 can replace a and b for a specific game.

The 4 games can be played twice, once with stay and once with switch, or compare column p to column r.

There is no advantage.

~ Phyti (talk) 16:02, 25 June 2025 (UTC)

regarding "The problem, the ... to the player." and "The player would only know there is 1 car behind 1 of 2 doors.": ​ If you don't assume the player knows the rules of the game, then 1/2 can easily be correct.

Are you arguing for 1/2 even when the rules include

(a) ​ The host can't open the door the contestant chose.

(b) ​ The host knows where the car is, and can't open that door.

(c) ​ If the contestant chose the door with the car, then the host chooses 50/50 which other door the host opens.

and the player knows the rules include ​ ​ ​ (a) and (b) and (c) ​ ?

If no, then I think we don't disagree. ​ If instead yes, then:

​

The player's 1st guess - choosing 1 of the 3 doors - is the secondary event for the MH game in question: ​ If that guess is the car door, then the host opening a door results in the car still being behind the player's 1st guess, else the host opening a door results in the car being behind the remaining door.

"there are only 4 ..., and not doors.", but these 4 are not equally probable: ​ ​ ​ See my ​ "Flip a nickel. ​ If ... nickel is showing heads?" ​ question, and note that for the MH game, what the host chooses later does not alter the historical results of the player's 1st guess.

​

JumpDiscont (talk) 17:19, 25 June 2025 (UTC)

Get a room, guys! AI*girllll (talk) 01:41, 26 June 2025 (UTC)

(a) ​ The host can't open the door the contestant chose.

(b) ​ The host knows where the car is, and can't open that door.

(c) ​ If the contestant chose the door with the car, then the host chooses 50/50 which other door the host opens.

Assume the player chooses door 1.

If (c), the host opens door 2 for game 3,

and opens door 3 for game 4.

Marilyn Savant didn’t know how to resolve this situation.

The host cannot open 2 doors in a game to avoid revealing the car location.

She makes this case different by switching the doors, which produces a bias in number of games, and -1 goat for stay and +1 car for switch. It’s an apparent advantage of her own making.

The game is simple and does not require any complex mathematical analysis. Phyti (talk) 18:10, 27 June 2025 (UTC)

For

"Flip a coin. If it shows heads, then place a 6-sided die so it shows 1, else roll that die.

What is the probability that, after doing the above, the coin is showing heads?"

, ​ do you get 7 games

(coin,die) ​ : ​ (heads,1) , (tails,1) , (tails,2) , (tails,3) , (tails,4) , (tails,5) , (tails,6)

because the die can't land more than one way in a game?

If yes, then do you get 1/7 for the probability that after the [flip followed by [place or roll]], the coin is showing heads?

You previously indicated 1/2 instead of 1/7 for this, with the explanation

"The coin is flipped once, the result is H or T, each with probability of 1/2.

What happens to the die later does not alter the historical results of the coin."

, ​ but for the MH game,

The player chooses a door once, the result is car or goat, with probabilities 1/3,2/3 respectively.

What the host chooses later does not alter the historical results of the player's choice. ​ ​ ​ .

You also said the MH "game is a dynamic process with changing factors as it progresses", but as far as I can recall, you neither answered whether you think my "The host rolls a 3-sided ... is showing 1." is a dynamic game nor gave any other indication of what you mean by "dynamic game" here.

(If your answer for "the probability that after the [flip followed by [place or roll]], the coin is showing heads?" is 1/2, then "goes from 1 game out of 3 to 2 games out of 4" is _not_ enough, because [flip followed by [place or roll]] goes from 1 game out of 2 to 1 game out of 7.)

JumpDiscont (talk) 21:01, 27 June 2025 (UTC)

• Listen, when you guys are done, can you turn out the lights and lock the door on your way out? EEng 19:23, 25 June 2025 (UTC)

Why no love for goats?

Has anyone ever questioned the assumption that the contestant would prefer a car to a goat? I'm actually seriously asking this. While I don't plan to spend hours going through all of the archives, I have never seen anyone call attention to assumptions as to what the contestant prefers (and this would also apply to variations of the Monty Hall problem). I've seen people bring up the issue of the presenter's motivation, but the contestant's motivation seems to be taken as a given in various formulations of the problem. CAVincent (talk) 04:37, 14 June 2025 (UTC)

A car has a higher monetary value than a goat, so even if the contestant's goal were to maximise their ownership of goats, it is in their interest to try for the car. MartinPoulter (talk) 11:01, 17 June 2025 (UTC)

But what if you belong to a primitivist sect that sees all things modern as evil, so that possessing a car even temporarily renders you subject to excommunication and condemnation to eternal hellfire? EEng 22:22, 18 June 2025 (UTC)

Or maybe you just want to literally get Monty Hall's goat. I certainly understand that most hypothetical contestants would prefer the car, but it strikes me as an unexamined assumption that all contestants would. At the very least, I would think that this preference should also be explicit as being one of the parameters of the problem. CAVincent (talk) 05:22, 19 June 2025 (UTC)

==The Monty Hall problem explained here==https://drive.google.com/file/d/18FFyRgOYEXsrHy4IQGpEI9Qhc1mtURvt/view?usp=sharing

~phyti — Preceding unsigned comment added by 108.176.89.195 (talk) 21:11, 21 September 2025 (UTC) Corrected typo.~phyti

One could say that [making what I called the crucial assumptions] _is_ producing "a biased or rigged game", but given those three assumptions as rules, the "game manipulation" from the above document is _by the host_: ​ ​ ​ If the car is behind door 3, then the host has probability 1 of opening door 2. ​ If the car is behind door 2, then the host has probability 1 of opening door 3. ​ If the car is behind door 1, then the host has probbility .5 of opening door 2 and probability .5 of opening door 3.

Contrast this with Monty Fall, where [which door Monty opens] accounts for _at most one of_ [[where the car is], [which door the contestant chose]]. ​ For Monty Fall, the relative frequences would either all be .5 or all be 1/3, unless you rescale to make them all 1.

JumpDiscont (talk) 02:51, 7 October 2025 (UTC)

When the car is behind door 1, and that is the players 1st choice, the host can open door 2 and door 3, but in separate games. you can't play 1/2 a game. Savant made a wrong assumption that the host opened door 2 in 1/2 her game 1 and door 3 in the other half.

There is no reason for not playing each game with the same frequency. Her manipulation allowed her switch strategy, but now it's not a fair game.

A fair game offers all players the same opportunity to win the car via a random guess.

As a game of chance,there is no basis for a strategy.

`~phyti 108.176.89.195 (talk) 16:32, 9 October 2025 (UTC)

The latest version, minor revision.

https://drive.google.com/file/d/1Q43qXVEWyy12sd1wnZVIA3c-5L16yGos/view?usp=sharing

~phyti Phyti (talk) 18:04, 9 October 2025 (UTC)

for your pdf:

Fig.3 is your version of the game with the car being behind door 2 only half as

often as the car is behind door 1. ​ This introduces a bias in favor of staying.

​

Are your answers to problem 0 and problem 16 from my user page

- https://en.wikipedia.org/wiki/User:JumpDiscont - 1/2 and 1/3 in that order?

If yes, then what is a pair of consecutive problems

from that page for which you give different answers?

JumpDiscont (talk) 00:49, 14 October 2025 (UTC)

This should be the last revision, since it includes Selvin's paper, showing both he and Savant manipulated the game frequency to beat the system.

https://drive.google.com/file/d/1x7XJAAKJw6yAJ6kzIQI5JcWtvA1qa1U2/view?usp=sharing

~ Phyti (talk) 18:47, 22 October 2025 (UTC)

"Comparison of the stay results in 1st choice and switch

results in 2nd choice" ​ only ​ "show no advantage" ​ if either

​

One _defines_ ​ "fair game" ​ as ​ all lines are equally important

, ​ and assumes the show meets that definition.

or

One ignores that the 4 lines are not all equally important.

​

.

​

​

For

​

The host rolls a 6-sided die. ​ If the result is in {1,2,3,4}, then

the host places a coin showing heads, else the host flips a coin.

​

,

​

Do you get that there are 8 possible sequences of host actions?

Do you get that at the end, ​ ​ ​ die shows 5 ​ ​ ​ has an advantage over ​ ​ ​ die shows 4 ​ ?

​

.

​

​

If the die-coin game I just described is not fair, then under what

I called the crucial assumptions, the MH game is _also_ not fair:

​

The host behaves deterministically for 2 of the 3 prize locations,

but makes a random choice for 1 of the 3 prize locations.

​

​

If your explanation is that the MH game is a ​ ​ ​ "dynamic game" ​ , ​ ​ ​

then see my ​ "You also said ..." ​ sentence in the 50/50 section.

​

​

JumpDiscont (talk) 22:34, 22 October 2025 (UTC)

The purpose of my paper is to restore confidence in intuition for the common person vs the movement by some to interpret the MH game as beyond their ability to understand. The fans of Marilyn Savant accepted her explanation based on her celebrity status as a person with a high IQ, while the many rejected it based on their experience. The paper shows exactly how it was done. You can propose many variations of the MH game, but they are not the game in question which was understood by both Whitaker and Savant. Phyti (talk) 17:57, 24 October 2025 (UTC)

Then you're doing the common person a disservice, because "best to switch" is indeed the correct solution. And that's not because Marilyn vos Savant has fans but because clear thinking often disagrees with intuition -- the very same reason that bridges are built, and satellites put into space, according to clear thinking and not according to the intuition of amateurs and crackpots. Please remember to turn out the lights and lock the door behind you when you're done. Good night. EEng 17:15, 25 October 2025 (UTC)

P.S. Looks like I used the "turn out the lights and lock the door on your way out" line already on this page -- but apparently the message iosn't getting across.

Are you labeling the 1000's who disagreed with her as amateurs and crackpots? That would be a bold statement, considering they use statistics on a regular basis.~phyti 108.176.89.195 (talk) 17:57, 26 October 2025 (UTC)

Plenty of people "use statistics on a regular basis" without understanding it -- the discipline of "statistics" -- or them -- the "statistics" that result from analyzing raw data. (Anyway, getting the right answer to this problem requires "statistics" about as much as doing your taxes requires "mathematics" i.e. it doesn't.) Such people only become crackpots when they keep arguing for years and years and years that they're right and experts who do such stuff for a living are wrong. EEng 00:11, 27 October 2025 (UTC)

Simple questions for User:Phyti

Am I correct that you believe the following?

Given the player has initially selected door 1, there are 4 possible "games" and these are equally likely:

1) The car is behind door 1 and the host opens door 2

2) The car is behind door 1 and the host opens door 3

3) The car is behind door 2 and the host opens door 3

4) The car is behind door 3 and the host opens door 2

If this is what you believe, then out of (say) 300 times the player initially chooses door 1 I assume you would expect each of these (equally probable) "games" to occur about 75 times.

Is this what you believe?

But if this is true, then isn't the car behind door 1 150 out of 300 times and behind either door 2 or door 3 only 75 out of 300 times? Isn't there a basic assumption that the car is equally likely to be behind any door, so should be behind door 1 (or door 2 or door 3) 100 out of 300 times? How do you explain this? -- Rick Block (talk) 00:37, 28 October 2025 (UTC)

The players guess does not affect the location of the car.~phyti 50.75.56.132 (talk) 20:21, 28 October 2025 (UTC)

The car is behind 1 0f 3 doors, if you consider the location only.

The game involves host-player choices.

When the car is behind the player 1st guess, the host has 2 choices.

The host can only open 1 door per game, thus 2 games, not 2 half games as thought by Selvin and Savant

Now there are 4 games, each consisting of player 1st guess, host opening a door, player 2nd guess. There are more guesses than locations, and that determines win or lose.

You can hold the car in 1 door and vary the guess for the player, or hold the player

to 1 door and vary the car location.

In your example, the player guesses door 1 150 times and doors 2 and 3 75 times.

That compensates for the loss of car wins in the half games with the manipulated frequencies.

The player’s guess does not affect the location of the car.~phyti 50.75.56.132 (talk) 20:22, 28 October 2025 (UTC)

@phyti :

​

You said ​ ​ ​ "the player guesses door 1 150 times and doors 2 and 3 75 times." ​,

but Rick Block says ​ ​ ​ "Given the player has initially selected door 1"

and ​ ​ ​ "out of (say) 300 times the player initially chooses door 1" ​ .

​

Do you nonetheless stay with

"the player guesses door 1 150 times and doors 2 and 3 75 times."

as your answer to Rick Block?

​

JumpDiscont (talk) 20:55, 28 October 2025 (UTC)

Remember what Shaw is said to have said about wrestling with a pig? Why bother? EEng 22:07, 28 October 2025 (UTC)

Contribute something constructive.

Point out what you think are errors in the paper.~phyti 50.75.56.132 (talk) 16:07, 30 October 2025 (UTC)

A point of clarification. The game show producers are obligated to present each game with the same frequency. 300 games = 75 rounds of 4 games each. Each player guesses door 1 2x which = 150. This is a history for 300 players.

MH game rules as proposed by Whitaker.

1. the host cannot open the door from the players 1st guess.

2. the host cannot open a door containing a car.

3. the host must offer the player a 2nd guess.

table of prize won

e 1 2 3 p1 p2

1 c x _ c g

2 c _ x c g

3 _ g x g c

4 _ x g g c

e is game number

car c is behind door 1

player guess left to right, c or g

host opens door x

_ door is not chosen

p1 is player 1st guess (stay)

p2 is player 2nd guess(switch)

no difference for p1 or p2 50.75.56.132 (talk) 16:00, 29 October 2025 (UTC)

To illustrate your mistake better, imagine another scenario where you have to work three times a week: Fridays, Saturdays and Sundays. Every Friday you must go to a department called A, every Saturday you must go to a department called B, while on Sundays sometimes you go to A and sometimes to B, like alternating each week:

... Fridays ---> A, A, A, A, A, A ...

... Saturdays -> B, B, B, B, B, B ...

... Sundays ---> A, B, A, B, A, B, ...

... But the fact that on Sundays you can go to two different departments while on the others only to one is not going to make the weeks have twice as many Sundays as Fridays or Saturdays (obviously). Every week will still have 1 of each. So what's really going to happen is that you will end up going to A less times on Sundays than on Fridays, and to B less times on Sundays than on Saturdays.

But if you were going to list the different types of days you can have, you would get four:

... 1) A Friday that you go to department A.

... 2) A Saturday that you go to department B.

... 3) A Sunday that you go to department A.

... 4) A Sunday that you go to department B.

It's just that the last two possible types of days are repeated half as often as the first two.

That's what occurs in Monty Hall. To think that because the host can make two possible revelations when your door has the car duplicates the amount of times you have the car in your door is an equivalent mistake to saying that the weeks will start having two Sundays in the example above.

You can also compare it to having a deck of collecting cards, where some are repeated twice or more and of others you only have one. If you were to randomly draw a card, the probability of getting a specific result would not only depend on the number of different types of cards, but also on how many there are of each type. Those types that are repeated more are more likely to be drawn. EGPRC (talk) 01:12, 31 October 2025 (UTC)

My paper shows the biased results 1/3 vs 2/3 are a consequence of manipulation of game frequency. You can change the frequencies to favor one choice over another. I am NOT saying they deliberately rigged the game. Their conclusion was a combination of lack of understanding basic probability and logic, and experience.

In the simplest terms:

Player 1^st guess. The probability of the car in the set of 3 doors is1(certainty).

The probability of the car in any 1 door is 1/3.

Player 2^nd guess. The probability of the car in the set of 2 doors is1(certainty).

The probability of the car in any 1 door is 1/2.

The state of the game changes as it progresses. 50.75.56.132 (talk) 16:21, 29 October 2025 (UTC)

@phyti - please think about 300 games. I assume you agree the car is placed randomly and is not moved, so in these 300 games the car is behind door 1 (or door 2 or door 3) in about 100 of these games. With me so far?

Now, let's say the player initially chooses door 1 (in all 300 games). The car doesn't move. In only 100 of the 300 games is the car behind door 1. If the player closes his/her eyes at this point and shouts "stay, stay, stay" paying no attention to what the host does (opens door 2 or door 3), in how many pf these 300 games will the player end up with the car? [hint: 100]

On the other hand, if the player initially chooses door 1, closes his/her eyes and shouts "switch, switch, switch" paying no attention to what the host does (opens door 2 or door 3), in how many of these 300 games will the player end up with the car? [hint: 200 because there are 300 games and staying only wins in 100 of them so switching must win in the rest]

This analysis, resulting in a 2-1 advantage for switching is essentially what vos Savant offered.

Before we continue, can I ask if you agree with this so far? If not, please tell me how this analysis can be incorrect. The salient points are: the car is placed randomly (so behind door 1 about 100 times out of 300), the player initially choosing door 1 doesn't make the car move, so choosing door 1 and staying with this choice (ignoring which door the host opens) means winning 100 times out of 300. The only other option is switching, so switching must win 200 times out of 300. Note that this isn't exactly the problem as described in which the player is deciding to switch after seeing which door the host opens, but this is often easier to understand. We can get to what happens with an eyes-open player next.

Rick Block (talk)

@Phyti - you say above that out of 300 games where the player has initially picked door 1 (before the host has opened a door) you'd expect the car to be behind each door 100 times. On the other hand, you also say the 4 "games" listed at the beginning of this section (the 4 possible conditions in which the host has opened door 2 or door 3) are equally probable meaning out of these same 300 games the car is behind door 1 150 times, and behind doors 2 or 3 only 75 times. Both of these cannot be true. I truly believe I can help you understand this conundrum if you'll let me. Your choice. -- Rick Block (talk) 00:58, 30 October 2025 (UTC)

MH game

The game rules as proposed by Whitaker:

1. the host cannot open the door from the players 1st guess.

2. the host cannot open a door containing a car.

3. the host must offer the player a 2nd guess.

Prize car is behind door 1.

Player's 1st choice 1 of 3,

door 1, host opens door 2, door 3 is closed.

door 1, host opens door 3, door 2 is closed.

door 2, host opens door 3, door 1 is closed.

door 3, host opens door 2, door 1 is closed.

Player's 2nd choice 1 of 2,

{1, 3} {1, 2} {2, 1} {3, 1}, in terms of doors,

{c, g} {c, g} {g, c} {g, c}, in terms of prizes.

Each player choice is independent of all other choices, different players and games.

The player 2nd choice is always {c, g}, c or g equivalent to a coin toss. There is no way to predict the outcome, the result is after the event. Probability is an historic frequency of occurrence of an event.

Basic intuition of the common person still works!~phyti 50.75.56.132 (talk) 16:04, 30 October 2025 (UTC)

Those cases you are counting are not equally likely to occur. Provided that the player's door is only the winner 1/3 of the time, then the two possible revelations that the host can make once the player has in fact picked the car happen 1/3 * 1/2 = 1/6 of the time each. I mean, those two possible revelations DIVIDE that 1/3 in two halves, they do not duplicate them.

This is better seen in the long run. Suppose you played a lot of times, always picking door #1. Every door would tend to be correct with the same frequency, but for every two games that door #1 has the prize, the host will tend to open door #2 one time and #3 one time, on average. In contrast, for every two times that door #2 has the car, he will be forced to reveal #3 in both, so twice as often as when the car is in #1, and similarly, for every two times that #3 has the car, he will be forced to reveal #2 in both.

As you see, each type of revelation is repeated more times when the winner is any of the others than when the winner is yours (again, because yours are shared between the two possible revelations while the other aren't shared), and that's why switching wins more often. EGPRC (talk) 00:58, 31 October 2025 (UTC)

“then the two possible revelations that the host can make once the player has in fact picked the car happen 1/3 * 1/2 = 1/6 of the time each.”

False. The player wins no prize on their 1st guess, only on their 2nd guess, when there are 2 doors. There is never a simple 3-door game played. Whitaker’s rule 3, host offers player a 2nd guess. You have to know what game you are playing.

Game 1, When player picks door with car, that frequency doubles with the host opening door 2 and door 3 in separate games, exactly as they do for games 3 & 4. Games are played consistently the same format.

The game rules restrict the host choices, which affects the player guesses.

For the 2nd guess, the player can reason one door has a car and the other a goat. The 2nd guess is independent of the 1st guess, there is no connection.

The player not knowing the car location, can only make a RANDOM guess.

Look up ‘random’ to learn its meaning. No pattern or method of prediction.

When the football game ends in a tie, who plays 1st is decided by a coin toss., since it is free from any outside influence.

When the fans accepted her explanation, no one asked how or why it worked.

My paper does.

Once the viewing audience learned the supposed strategy. future players would take advantage winning 2x as many cars. The sponsor would object to the cost, and the show would have a short shelf life!~phyti 50.75.56.132 (talk) 17:02, 1 November 2025 (UTC)

Please think about 300 games. I believe you agree the car should be behind each door about 100 times (1/3 each door). If all 300 players pick door 1 and stay with that pick regardless of what the host does, how many of these players win the car? -- Rick Block (talk) 16:59, 30 October 2025 (UTC)

There is no need to play many games.

The post of Oct 30 shows 4 different players playing all 4 possible games.

Each player has the same opportunity to win, 1 of 2 doors.~phyti 50.75.56.132 (talk) 17:14, 1 November 2025 (UTC)

Asserting the 4 different "possible games" are equally probably does not make it so. The host's choice of door to open is restricted by the player's initial choice, so the door the host opens is not independent (if it were, the host would accidentally open the door that is the player's first choice about 1/3 of the time). Following through what happens with 300 games is quite instructive. Can you fill out the following table (thinking about 300 games where the player's initial choice is door 1)?

Car location	Occurrences (out of 300)	# of times host opens Door 2	# of times host opens Door 3
Door 1	100	??	??
Door 2	100	??	??
Door 3	100	??	??

Each row corresponds to a possible position of the car (1/3 chance any door, so 100 times out of 300). The two ?? in each row are the number of times out of the 100 the car is behind that door that the host opens Door 2 or Door 3. The ?? in each row therefore must add up to 100. Once you fill in these numbers we can talk again. -- Rick Block (talk) 18:13, 1 November 2025 (UTC)