Talk:Monty Hall problem/Arguments

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)

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.

All doors have the same 3 distributions/patterns of prizes, {cgg, gcg, ggc}, so we examine all possible choices for one door for each of the possible games while varying the car location.

When the player 1st choice contains the car, the host is not restricted by rule 2 since in this case rule 1 does the same thing. The host is free to open 2 doors if done in 2 games. That means there are 4 possible games, all played with the same format. Player makes 1st guess, host opens 1 door which is removed from play, player makes 2nd guess.

There is no requirement for the host to choose 1 door for the 1st half of a game and a different door for the 2nd half of that game. A simple misunderstanding of the logical difference between (2 or 3) vs (2 and 3).

Each of the 4 players ( Ann, Bob, Kay, Leo) choose door 1 for their 1st guess.

Player's 1st choice 1 of 3

Ann:1. c in door 1, host opens door 2, door 3 is closed.

Bob:2. c in door 1, host opens door 3, door 2 is closed.

Kay:3. c in door 2, host opens door 3, door 1 is closed.

Leo:4. c in door 3, host opens door 2, door 1 is closed.

Names of players are used to prevent someone from thinking the same player guesses door 1 twice. For the purpose of analysis, all players play 1 game.

Player's 2nd choice {stay, switch}

1. {c, g}

2. {c, g}

3. {g, c}

4. {g, c}

Where is the advantage?

The game show repeatedly offers the same 4 games. Each player still guesses 1 of 3 doors. The fixation on location obscures the role of choices.

Each player choice is independent of all other players.

The player 2nd choice is always c or g, equivalent to a coin toss. There is no way to predict the outcome for a random event like a coin toss. Probability is an historic frequency of occurrence of an event, i.e. after the event, and is not meant to be taken literally. It's called 'a game of chance' since the player doesn't know which of the 4 games the producers are using, and doesn't win anything on the 1st guess, and can be replaced with a coin toss on the 2nd guess. The player is just along for the ride!~phyti 50.75.56.132 (talk) 18:53, 3 November 2025 (UTC)

The advantage is in _how often_ the game show offers each of those 4 games.

What numbers do you get for the table in the comment just above yours?

JumpDiscont (talk) 20:42, 3 November 2025 (UTC)

OK, if you insist your 4 canonical games are equally probable, then out of 300 shows (where the player initially picks door 1) they should each happen about 75 times, right? In which case, the table looks like this:

Canonical game name	Car location	Occurrences (out of 300)	# of times host opens Door 2	# of times host opens Door 3
Ann	Door 1	75	75	0
Bob	Door 1	75	0	75
Kay	Door 2	75	0	75
Leo	Door 3	75	75	0

If this is what you think, then isn't the car behind door 1 150 times (75 "Ann" plus 75 "Bob") out of 300 (probability 1/2) rather than only 100 (probability 1/3)? And behind door 2 or door 3 only 75 times out of 300 (probability 1/4) rather than 100 (probability 1/3)? Doesn't this clearly show your 4 games are not equally probable? -- Rick Block (talk) 23:18, 3 November 2025 (UTC)

The show offers each of 4 games equally for fairness.

The car is behind door 1 150 games.

The car is behind door 2 75 games.

The car is behind door 3 75 games.

The player who chooses door 1 wins a car ½ of games.

The player who chooses door 2 and door 3 wins a goat ½ of games.

When the player chooses the door with the car, that pattern cgg for the 3 doors is played 2 times as game 1 and 2.

That implies door 1 is chosen twice as often as the other doors.

Read the section ‘probability’ in the paper. It refers to frequency of events. If you manipulate the frequency of an event, you will alter its probability.

‘Introducing a bias’ is a polite way of saying ‘rigging a game’.

Refer to ‘1950’s game show rigging’ on Wiki.~phyti ~2025-31466-71 (talk) 19:06, 5 November 2025 (UTC)

Um, what? In the 300 shows we're talking about (where the player has indeed initially picked door 1), the producers had no idea what door the player would initially pick, right? So, to be fair, they must put the car behind each door with probability 1/3 - about 100 times behind each door - and they don't move it after the player picks a door. The initial pick has a 1/3 chance of being correct regardless of which door the player initially chooses, not 1/2 or 1/4. This is where you keep going wrong. Perhaps this slightly rearranged table might be more clear.

Car location	Canonical game name	Occurrences (out of 300)	# of times host opens Door 2	# of times host opens Door 3
Door 1	(unknown until host opens a door)	100	X (turns out to be an "Ann" game)	Y (turns out to be a "Bob" game)
Door 2	Kay	100	0	100
Door 3	Leo	100	100	0

The occurrences where the car is behind door 1 are split between "Ann" and "Bob" games but only after the host opens a door (so X plus Y equals 100), while all the occurrences where the car is behind door 2 are "Kay" games and behind door 3 are "Leo" games. The question is what are X and Y? -- -- Rick Block (talk) 01:22, 6 November 2025 (UTC)

“The initial pick has a 1/3 chance of being correct regardless of which door the player initially chooses,”

The host opens a goat door, so there is no win for the 1st pick.

Now it’s a 2-door game, and the player wins a prize on the 2nd pick.

If the show offers each of the 4 games once, games 1 and 2 allow the host different choices, and he host cannot open 2 different doors in 1 game.

There are 3 doors but 4 host choices.

My example holds the player 1st choice constant.

The host choice depends on the car location (rule 2).

The player 2nd choice is in the following.

1. {c, g}

2. {c, g}

3. {g, c}

4. {g, c}

There is no advantage to always switch.

Since the individual players choices are independent, there are 2^4 =16 possible combinations for (c or g), same as a coin toss. One sequence of choices would be game 1 & 2 stay and game 3 & 4 switch, resulting in cccc, the maximum, all win a car.

The binomial distribution provides all possibilities.

0 1 2 3 4 (length of sequence for c)

1 4 6 4 1 (occurrence of each sequence)

Average is 2 c's for 6/16=3/8.

If the show doesn't maintain equal probability on average for all 4 games, their game is illegal.

~phyti ~2025-31723-14 (talk) 18:38, 6 November 2025 (UTC)

You seem to agree that the player's initial pick has a 1/3 chance of being correct regardless of which door the player initially chooses, right? If so, isn't this the player's chance of winning the car by sticking with his/her initial choice? Again, imagine 300 games where the player's initial choice is door 1. This initial choice has a 1/3 chance of being correct (or the producers are not actually randomizing the car's initial placement). Now imagine all 300 of these players completely ignore what the host does, closing their eyes and covering their ears and repeating "stay, stay, stay". They had a 1/3 chance of being correct on their initial choice, the car has not been moved after this choice (or the producers would be in big trouble!), so about 100 (1/3, not 1/2) of these players will win the car, right? Please don't simply repeat that there are 4 games that must be equally probable. -- Rick Block (talk) 19:35, 6 November 2025 (UTC)

MH game

The game rules as proposed by Whitaker:

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

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

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

A game is

1. p1 (player1st choice).

2. h host opens goat door.

3. p2 (player 2nd choice).

4. host opens player 2nd choice.

The pattern of prizes is constant c g g.

Marilyn Savant

"The first door has a 1/3 chance of winning, but the second door

has a 2/3 chance. The benefits of switching are readily proven by playing through the six games that exhaust all the possibilities."

Error 1.

False. The host removes a goat door and its contribution to probability.

There are not 3 ways to choose 1 of 2 objects.

Error 2.

Savant interprets game 1 as host opening door 2 half the time and door 3 half the time. If 1 each of the 3 games is played once, how do you play half a game?

There are 3 patterns based on 3 locations/doors, but a game includes host choices, which results in 4 games.

The table shows a deficit of 1 car in the p1 column, favoring a switch.

The paper 'Intuition and Game Shows' demonstrates how manipulating the frequencies of game playing can favor staying or switching. It's like a self fulfilling prophecy.~phyti Phyti (talk) 18:18, 7 November 2025 (UTC)

Can we go through this slowly? Do you agree the player's initial pick has a 1/3 chance of being correct regardless of which door the player initially chooses? Yes or no? -- Rick Block (talk) 18:41, 7 November 2025 (UTC)

Reading your comments above perhaps we don't need to go quite so slowly. Above, you say your answer is yes, but "Only in the simple 3-door game, where the player wins what is behind that door."

So, in a modified game where there are 3 doors, behind one of which is a car, and the player gets a single pick, if we do this 300 times and the players all pick door 1, then about 100 of these players will win a car. The leads to the following table of outcomes (for the 300 "shows").

Car location	Occurrences (out of 300)	Player wins the car
Door 1	100	Yes
Door 2	100	No
Door 3	100	No

Are we in agreement so far? Yes or no. -- Rick Block (talk) 01:13, 8 November 2025 (UTC)

MH game

The game rules as proposed by Whitaker:

1. the host cannot open the player's choice until after the player's 2nd choice.

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

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

A game is left to right.

1. player 1st choice.

2. host opens goat door.

3. player 2nd choice.

The pattern of prizes is constant c g g.

A pattern by itself is not a game.

Each game is played once.

The graph shows how 3 doors becomes 4 games.

Ann chooses door 1, host opens door 2, A switches to door 3.

Bob chooses door 1, host opens door 3, B switches to door 2.

Kay chooses door 2, host opens door 3, K switches to door 1.

Leo chooses door 3, host opens door 2, L switches to door 1.

No one wins a prize until after the players 2nd choice.

Half of players (A & B) win a car if they stay.

Half of players (K & L) win a car if they switch.

Thus there are only 4 possible games.

There are NOT 3 ways to select 1 of 2 objects.

Reason.

When the car is behind the players 1st choice, rule 1 also prevents the host from opening a car door, thus allowing opening door 2 and door 3. In the final analysis, the game rules determine the host options and outcomes, not the locations.

Probability of choosing door 1 as the location of the car is 1/3 when there are 3 doors.

Probability of winning a car is 1/2 when there are 2 doors, and depends on choices.

The game is fair providing the same odds for all players of 1/2. It doesn't require any special field of knowledge for the player nor advanced math. Common intuition makes the correct call. The game show problem is better classified as a logic problem in recreational mathematics.

Phyti (talk) 18:11, 10 November 2025 (UTC)

Except in your scenario you have twice as many people picking door 1 then either door 2 or 3. Which makes no sense.

Lets say the car is behind door 1.

You have 3 players: A, B and C, who pick doors 1,2 and 3 respectively.

If they all stay, Exactly one of them will win, 1/3 chance.

In you 4 game scenario, games 3 and 4 are twice as likely as games 1 and 2. 11cookeaw1 (talk) 07:56, 12 November 2025 (UTC)

Was that a "yes" or a "no"? -- Rick Block (talk) 18:54, 10 November 2025 (UTC)

And, to return the favor, I agree with everything you say up to the line that says "Half of players (A & B) win a car if they stay." This is not actually correct. You are asserting this is true because after the host opens a door there are only two choices. But how we get to the two choices matters. If you'll follow through 300 games with me I believe I can help you understand this. -- Rick Block (talk) 19:16, 10 November 2025 (UTC)

“Except in your scenario you have twice as many people picking door 1 then either door 2 or 3. Which makes no sense.”

It does when you consider host choices.

p1 is door 1 with 2 host choices which allows 1 game for

A and 1 game for B.

p2 is door 2 with 1 host choice which allows 1 game for K.

p3 is door 3 with 1 host choice which allows 1 game for L.

“If they all stay, Exactly one of them will win, 1/3 chance.”

No one wins a prize on the 1st pick, that happens on the 2nd pick when there are only 2 doors. One with car, one with goat.

Your fixation on number of locations is preventing you from seeing the effect of host choices, which are regulated by the game rules.

In a game of chance, all the factors affecting the game are beyond the control of the player. In the MH game, the player does not know the car location, so has no information to guide them.~phyti ~2025-32925-06 (talk) 18:46, 12 November 2025 (UTC)

Can we talk about 300 shows and follow through what happens? It seems you prefer to hold the car location constant in your analysis. So, let's say we have 300 shows where the car is always behind door 1. The players have no idea where the car is, so of the 300 shows about how many players would you expect to initially choose each door? I believe you agree it should be about 100 players initially choosing each door. So far, so good? Yes or no, please. -- Rick Block (talk) 20:46, 12 November 2025 (UTC)

No.

Savant’s game 1a & 1b are 2 games, each with different (player, host, player) actions.

You have to define a game.

You can’t play half a game, so there are 4 games.

That’s why 2 players can pick the same door when it hides a car.

That occurs 1 of every 2 games.

First you play those 4 games for the c g g pattern of distribution, for door 1.

Then you shift the numbers on the doors 123,231,312, (permutation) instead of moving prizes.

Which door is picked 1st doesn’t matter.

Assume I wrote a math paper using a sequence of convoluted equations, including an error that went unnoticed, showing 3+4=8.

Biff reads it and writes a computer program, that reproduces the same results.

I complement Biff on his expertise as a programmer, but Biff did not prove my results were correct.~phyti ~2025-32992-84 (talk) 19:25, 13 November 2025 (UTC)

No?? So if we have 300 shows where the car is behind door 1, how many players (who don't know where the car is) initially pick each door? -- Rick Block (talk) 21:16, 13 November 2025 (UTC)

Perhaps you're starting to see there's something not quite right about your analysis. If you are correct and your A, B, K, and L games are equally probable it means out of 300 games there should be about 75 each. Since A and B are both games where the player initially picks door 1, you're saying out of 300 games the player's initial pick is door 1 about 150 times and door 2 or door 3 only 75 times. This makes no sense, and is indeed incorrect. The problem is that the A, B, K, and L games are NOT equally probable.

The probability of 2 events, X and Y, is the probability of X times the probability of Y if X and Y are independent. So, for example, if I roll a 3 sided die and then flip a coin, the probability of any of the 6 possible outcomes (1-heads, 1-tails, 2-heads, 2-tails, 3-heads, 3-tails) is 1/3 times 1/2, or 1/6. Another way to look at this is to figure out that there are 6 possible outcomes and directly conclude they each have a probability of 1/6. In the case where X and Y are independent this is a perfectly fine analysis. In the MH problem, X and Y are not independent. What the host does depends on the player's initial choice. If this were not true, for example if the host flips a coin to decide whether to open door 2 or door 3 (assuming the car is behind door 1), then there would be 6 equally probable outcomes but in two of these (player picks door 2 and host opens door 2, player picks door 3 and host opens door 3) the host is violating the MH rules.

If X and Y are not independent, the probability of X and Y is the probability of X times the probability of Y given that X has occurred (see Chain rule (probability)). In the MH problem, the probability the player initially picks any particular door is 1/3. Assuming the car is behind door 1, the probability the host opens door 2 or door 3 depends on what door the player initially picks. If the player has picked door 1, the probability the host opens door 2 or door 3 is 1/2 (the MH problem doesn't explicitly say this, but it's a natural assumption) so the probability of your A and B games are both 1/3 times 1/2, or 1/6. If the player has picked door 2, the probability the host opens door 3 is 100% so the probability of your K game is 1/3 times 1, or 1/3. Similarly, the probability of your L game is also 1/3.

This leads to the following table for 300 shows (assuming the car is always behind door 1):

Player's initial choice	Occurrences (out of 300)	Number of times host opens door 2	Number of times host opens door 3
Door 1	100	50	50
Door 2	100	0	100
Door 3	100	100	0
Total	300	150	150

If we hold the player's choice constant (player always picks door 1) and randomize the location of the car we end up with the same table but the first column is the car location. The host opens door 2 about 150 times out of 300 and also opens door 3 about 150 times out of 300. If you initially pick door 1 and then the host opens door 3, the car is certainly behind door 1 or door 2 but about 50 times out of 150 it's behind door 1 and about 100 times out of 150 it's behind door 2. Similarly, if you initially pick door 1 and then the host opens door 2, the car is certainly behind door 1 or door 3 but behind door 1 about 50 times and behind door 3 about 100 times. In either case (host opens door 2 or door 3), you're twice as likely to win the car if you switch. This makes sense because your initial pick has a 1/3 chance of being correct. So if you stay with that choice (regardless of what the host does) you should have a 1/3 chance of going home with the car. -- Rick Block (talk) 16:33, 15 November 2025 (UTC)

MH game

The game rules as proposed by Whitaker:

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

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

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

You started with a good explanation in terms of player-host choices.

Then you went off the road, just as Savant did, into the probability swamp.

Probability is a measure of success=s. With w=number of correct choices, and n=number of possible choices, s=w/n. For 3 doors s=1/3, for 2 doors s=1/2.

The player does not win a prize on their 1st choice. After the host opens and removes a goat door, that ghost door does not contribute to the probability of win a car, thus 1/3 is not a factor in the outcomes. The player wins a prize on their 2nd choice of 1 of 2 doors.

A game is

1. p1 (player stays).

2. h host opens goat door.

3. p2 (player switches).

The table shows how the game rules modify the player-host choices from 3*2*1=6, to 4 legal and 2 illegal. The win car ratio s = 1/2 for both stay and switch.

There is no advantage.

Phyti (talk) 16:30, 17 November 2025 (UTC)

Sigh. s=w/n????

Let's try making the results way more obvious. I assume you agree the following is essentially the same as the MH problem. I take 3 cards from a deck of 52 including the ace of spades (like the producers hiding a car behind one of 3 doors in the MH problem). You pick one without looking at it (like the player's initial pick of one of 3 doors in MH). I look at the remaining two and show you one that is not the ace of spades (like the host opening one of the other two doors in MH). Your claim is that since there are only two cards remaining (2 doors in the MH), one of which is the ace of spades (hides the car in MH), the probability you have the ace of spades is 1/2 (w=1, n=2, so s=1/2). Correct?

OK. Now let's do it again. But this time we start with all 52 cards. You pick one. I look at the rest and show you 50 that are not the ace of spades (keeping only one). w is still 1. n is still 2. Do you have a 50/50 chance that your card is the ace of spades? If you think the answer to this is yes, I strongly encourage you to try this, say 20 times and let us know how it works out. I just did this, and the player who picked the initial card ended up with the ace of spades 0 times out of 20. My guess is if do this you'll be lucky to have the ace of spades once. -- Rick Block (talk) 19:21, 17 November 2025 (UTC)

Rick, you're wasting your time. This has been going on for decades (literally). You'll never penetrate the pseudomathematical wall of certainty. EEng 12:09, 17 November 2025 (UTC)

I may be, but I have had successes before. -- Rick Block (talk) 19:21, 17 November 2025 (UTC)

The good news is there are only two outcomes -- success and failure -- and so your chance of success is 1 out of 2, or 50%! So maybe I was being pessimistic. EEng 23:50, 17 November 2025 (UTC)

MH game

The game rules as proposed by Whitaker:

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

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

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

"the probability you have the ace of spades is 1/2 (w=1, n=2, so s=1/2). Correct?"

Yes.

Probability is a measure of success=s. With w=number of correct choices, and n=number of possible choices, s=w/n. For 3 doors s=1/3, for 52 doors s=1/52.

The card ID =1 thru 52. AceSpades is #1. Probability is on the right.

1 thru 52, 1/52

1 thru 51, 1/51

⋮

1 thru 10, 1/10

⋮

1 2 3, 1/3

1 2, 1/2

1, 1/1

As the number of possible choices decreases, s increases.

If we let the host reveal the last non prize card, there is no need for the player to guess!

With 2 cards left:

"Do you have a 50/50 chance that your card is the ace of spades?"

Yes.

There are 2 scenarios with winning set Y {#1} and losing set N {#2 thru #52}

1. Player picks #1 from Y, which is exempt by rule 1 and 2. Host reveals 50 cards from N.

2. Player picks #x from N, which is exempt by rule 1. Host reveals 49 cards from N.

Both scenarios leave #1 and 1 losing card for the players 2nd guess.

At this point, the game is equivalent to a coin toss, with 2 outcomes.

One must be ace of spades and one is not (per the rules).

It is impossible for all those cards removed to have altered any other card.

Removing all the extra cards only makes for a prolonged and boring game.

If you like magic/illusion, watch 'Penn & Teller-fool us' from Las Vegas.

~phyti ~2025-34815-97 (talk) 15:09, 20 November 2025 (UTC)

Since you're still arguing your theories, I assume you haven't actually tried this like I asked you to. Please try it. 20 times should take only a few minutes. Let us know your results. Like I said, when I did it the initial pick ended up being the ace of spades 0 times out of 20 which seems extremely unlikely if it's actually 50/50 as you claim. -- Rick Block (talk) 16:25, 20 November 2025 (UTC)

MH game

Selvin style, with n boxes.

Box 13 contains a check for 1 million dollars, all others are empty.

Player does not know the location of the prize.

Host does know the location of the prize.

Scenario 1.

Player randomly picks box #13.

Host opens n-2 empty boxes, leaving 1 box #x.

Player randomly picks box #13 or box #x.

Scenario 2.

Player randomly picks box #y that is not #13 .

Host opens n-1 empty boxes, leaving 1 box #13.

Player randomly picks box #y or box #13.

Both cases reduce to a choice of 2 boxes, prize or no prize,

which is independent of n.~phyti ~2025-35677-99 (talk) 18:12, 22 November 2025 (UTC)

I elaborated on your correct reply with a detailed example to

Rick ~phyti ~2025-35677-99 (talk) 18:07, 22 November 2025 (UTC)

So, you still haven't tried it!!?? Please try the 52 card version 20 times and tell us your results. How many times does the player end up with the ace of spades and how many times does the host end up with it? -- Rick Block (talk) 18:43, 22 November 2025 (UTC)

When are you going to reconcile yourself to the fact that this guy is never going to look through the telescope? EEng 17:04, 23 November 2025 (UTC)

Explain how you play the game. That could be an issue.~phyti ~2025-35879-23 (talk) 17:32, 23 November 2025 (UTC)

As above. Start with a full deck of 52 cards. "Player" picks one at random and does not look at it. "Host" looks at the remaining 51 cards and shows the player 50 that are not the ace of spades, keeping one (either the ace of spades or a randomly selected one of the 51 cards that is not the ace of spades). Two cards left. One is assuredly the ace of spades. Should the "player" keep the initial card, or switch to the "host's" card? [This is Selvin's rules for n=52]. Do this 20 times and tell us how many times the "player" ends up with the ace of spades and how many times the "host" ends up with it. -- Rick Block (talk) 18:32, 23 November 2025 (UTC)

Sorry to mingle in this conversation but I'd be interested to know if this computer simulation of the MH problem I've just written is helpful: https://github.com/2072/MontyHallProblemSimulation (I think the commented code should be easy to read by non programmers—please tell me if that's the case or not) Inogan (talk) 00:49, 24 November 2025 (UTC)

Actually, I think this simulator is a better choice https://montyhall.io/ . EEng 02:57, 24 November 2025 (UTC)

Sure, if you just want to play the game, but not if you want to intuitively understand why the probabilities are what they are (in my humble biased opinion).

My simulation is only really useful if you try to read its code of course but it does still show what happens probabilistically when 100000 trial games are actually played. Inogan (talk) 03:11, 24 November 2025 (UTC)

MH game with cards.

The 52 cards are numbered on the back 1 thru 52.

Row 1, classifies a set of 1 card containing A (ace of spades),

and a set of 51 Z cards.

Row 2, player 1st randomly guesses card x, followed by the host revealing the number of Z cards leaving 1 Z card for the player's 2nd guess.

Row 3, player chooses between the A card and a Z card for any game.

There are 52 ways to play a game, but only 1 guess to win a prize.

Don't confuse the two experiences/actions.

Phyti (talk) 19:12, 24 November 2025 (UTC)

Use that on the MH game with 3 doors:

The 3 doors are numbered 1 thru 3.

Row 1, classifies a set of 1 door containing C (the car),

and a set of 2 Z doors.

Row 2, player 1st randomly guesses door x, followed by the host opening the number of Z doors leaving 1 Z door for the player's 2nd guess.

Row 3, player chooses between the C door and a Z door for any game.

There are 3 ways to play a game, but only 1 guess to win a prize.

What two experiences/actions are you referring to?

JumpDiscont (talk) 04:29, 25 November 2025 (UTC)

​

​

I figure I should preemptively address one possible reply:

​

The "host cannot open 2 different doors in 1" MH game with 3 doors, but correspondingly, the "host cannot open 2 different" 50-element subsets of {2,3,4,5,...,51,52} in 1 MH game with 52 cards.

[Car is behind door 1, Host opens door 2] ​ and ​ [Car is behind door 1, Host opens door 3] ​ are possibilities for the MH game with 3 doors, and correspondingly, ​ [Player's 1st guess is the ace of spades, Host reveals all but that and the card numbered 51] ​ and ​ [Player's 1st guess is the ace of spades, Host reveals all but that ace and the card numbered 52] ​ are possibilities for the MH game with 52 cards.

​

JumpDiscont (talk) 05:10, 25 November 2025 (UTC)

MH game 3 doors

Z does not = C.

Row 1, classifies a set of 1 door containing C (car),

and a set of 2 Z doors.

Row 2, player 1st randomly guesses door x, followed by the host opening 1 Z door leaving 1 Z door for the player's 2nd guess.

Row 3, player chooses between the A door and a Z door for any game.

There are 4 ways to play a game, but only 1 of 2 guesses to win a prize.

Don't confuse the two experiences/actions.

More on probability.

If we can list all possible games, there is no need to play them many times. For the MH game, the ideal limit is determined by the ratio of (a winning guess) to (all possible guesses). If I said anything different, it was incorrect.

If a coin toss has 2 possible outcomes the ideal probability is 1/2. After a sequence of 10 H, the probability of the next toss equal to T, is still 1/2. A larger number of tosses approaches the ideal limit of 1/2 more closely than a smaller number of tosses.

In reality the ratio is seldom 50/50, but 45/55, 60/40, 91/109, etc. The ideal does not consider any variation where reality does.

Phyti (talk) 17:42, 25 November 2025 (UTC)

You seem to be choosing count-as-multiple-ways (for MH 3 doors) vs count-as-one-way (for MH 52 cards) based on what you think the probabilities are. ​ If you have any other reasoning here, then how do you get that

​

for MH game 3 doors, [Car is behind door 1, Host opens door 2] and [Car is behind door 1, Host opens door 3] are 2 ways to play a game

​

but

​

for MH game 52 cards, [Player's 1st guess is the ace of spades, Host reveals all but that and the card numbered 51] and [Player's 1st guess is the ace of spades, Host reveals all but that ace and the card numbered 52] are not 2 ways to play a game

​

?

​

​

For example, what about this MH game 3 cards?

​

Start with {[Ace of Spades], [2 of spades], [3 of spades]}. "Player" picks one at random and does not look at it. "Host" looks at the remaining 2 cards and shows the player 1 that is not the ace of spades, keeping one (either the ace of spades or a randomly selected one of the 2 cards that is not the ace of spades). Two cards left. One is assuredly the ace of spades. Should the "player" keep the initial card, or switch to the "host's" card?

​

​

regarding probability:

​

For

Toss a coin. ​ If the result is heads, the roll a

6-sided die, else place that die so it shows 1.

, ​ how many possible outcomes are there, and what is the ideal

probability that the above ends with the coin showing tails?

​

​

JumpDiscont (talk) 20:39, 25 November 2025 (UTC)

“You seem to be choosing count-as-multiple-ways (for MH 3 doors) vs count-as-one-way (for MH 52 cards) based on what you think the probabilities are. ​ If you have any other reasoning here, then how do you get that”

The number of sessions (to distinguish from games) for the car game is determined by the number of host choices, which is 2 when the 1st door guess contains the car.

“for MH game 3 doors, [Car is behind door 1, Host opens door 2] and [Car is behind door 1, Host opens door 3] are 2 ways to play a game”

For 52 cards, there are 52 player choices to play a session of the game.

After the host removes all non-car doors except 1, the player is left with a car door and a non-car door for the 2nd guess. This is when the possibility to win a car occurs. This guess is independent of number of doors, and occurs for all 52 sessions.~phyti ~2025-36567-29 (talk) 18:10, 26 November 2025 (UTC)

The car game and the 52-card game don't differ in the ways you refer to:

​

The number of sessions for the 52-card game is determined by the number of host choices, which is 51 when the 1st card guess is the ace of spades. ​ ​ ​ (If that guess is card number 1, then the host can reveal [every Z card except card number 2] or ... or [every Z card except card number 52]. ​ If that guess is a different card number, then there are still 52-1 host choices.)

For 3 doors, there are 3 player choices to play a session of the game.

After the host reveals all non-[ace of spades] cards except 1, the player is left with an ace of spades and a non-[ace of spades] card for the 2nd guess. ​ This is when the possibility to win occurs. ​ This guess is independent of the number of cards, and yet you still seem to agree that the relevant denominator for the 52-card game is 52.

(Your image for it shows ​ ​ ​ 1 game ​ vs ​ 51 games ​ ​ ​ .)

​

For the ​ MH game 3 cards ​ in my previous comment - ​ "Start with {[Ace of Spades], [2 of spades], [3 of spades]}. ... keep the initial card, or switch to the "host's" card?" ​ - do you count [Player's 1st guess is the ace of spades, Host reveals all but that and the 2 of spades] and [Player's 1st guess is the ace of spades, Host reveals all but that and the 3 of spades] as 2 sessions or just 1 session?

​

JumpDiscont (talk) 20:04, 26 November 2025 (UTC)

"(Your image for it shows ​ ​ ​ 1 game ​ vs ​ 51 games ​ ​ ​ .)"

There is 1 session where x is the A card and 51 sessions where x is not the A card.

------------------------------------------------------------

MH game

Host rules:

1.cannot open a door containing c.

2. must offer the player a 2nd selection (which implies host cannot open 1st guess).

The player chosen from the audience is informed by the host there are 3 doors, 1 containing a valuable prize c and 2 containing nothing. Then the host asks the player who doesn't know the location of c to select 1 of the 3 doors. The player guesses door x (A random guess since there is no information to guide the player).

The host does not open door x to verify a win or loss. The host opens an empty door instead.

Fact 1. The probability of the player opening the door containing c is 1/3.

In both the Steve Selvin and Craig Whitaker defined games, there was never a question about fact 1.

Fact 2. The player does not win a prize for the 1st guess. It can only be a misdirection allowing the viewing audience a perception of player involvement in the game.

Fact 3. For the general game, there are 3*2*1= 6 patterns of distribution for 3 distinct prizes for 3 doors. Since the Selvin and Whitaker games are a special case with 2 prizes of equal value (goats) there are only 3 patterns (1, 3, 4) in the table. When door x contains c, the host can open 2 different doors in separate sessions (a sequence of host actions) increasing the number of sessions from 3 to 4.

Fact 4. If the player always guesses door 1 and the host always opens an empty door, the player 2nd guess is always a choice of c or 0, with probability of win c = 1/2.

Fact 5. Door x can be any of the 3 doors since they all have the same patterns.

Fact 6. Winning c is independent of the player's 1st guess.

Fact 7. Comparing the stay doors #(1 and 2) with the remaining switch doors #(3 and 4) shows no advantage.

`phyti Phyti (talk) 16:28, 28 November 2025 (UTC)

Fact 8. Columns 1 and 2 (where staying gets you the car) have, together, probability 1/3. Meanwhile, columns 3 and 4 (where switching gets you the car) have, each, probability 1/3, and together probability 2/3. I look forward eagerly to your response. EEng 20:43, 28 November 2025 (UTC)

Either similarly to or the other way around from your previous comment, the explanation in your most recent comment would lead to the conclusion that for ​ ​ ​ MH game 52 cards ​ , ​ ​ ​ there are at least 51 sessions - rather than just 1 session - where x is the A card. ​ ​ ​ ​ ​ ​ ​ I would've included the 7 Facts for that in this comment, but I worry that me doing so would lead to you yet again giving [reasoning or [an explanation]] that leads to the same choice from [count-as-multiple-sessions vs count-as-one-session] for both [[MH 3 doors] and [MH 52 cards]], rather than you answering my question:

​

For the ​ MH game 3 cards ​ from [2 comments by me] ago ​ - ​ "Start with {[Ace of Spades], [2 of spades], [3 of spades]}. ... keep the initial card, or switch to the "host's" card?" ​ - ​ do you count [Player's 1st guess is the ace of spades, Host reveals all but that and the 2 of spades] and [Player's 1st guess is the ace of spades, Host reveals all but that and the 3 of spades] as 2 sessions or just 1 session?

​

(It could be that you count them as 2 sessions because the host can reveal 2 different cards in separate sessions. ​ It could be that you count them as 1 session because there are 3 player choices to play a session of the game.)

​

JumpDiscont (talk) 23:23, 28 November 2025 (UTC)

MH game

'random' definition:

with no pattern, or means of being predictable, etc...

When Craig Whitaker asked if there was a winning strategy for the MH game, the answer should have been no. Marilyn Savant, who wrote an advice column for Parade magazine said yes, and explained how. Her 1990 response revealed her lack of understanding of basic probability and logic. That's the history.

session 1.

In examining the game, the player does not win a prize with the 1st random guess of 1 of 3 doors. The host opens a losing door, then offers the player a 2nd guess to see if the player wants to stay with their 1st guess or switch to the remaining closed door.

The player does not know the location of the prize for either guess. The win ratio=1/2.

session 2.

To simplify the game, the host informs the contestant there are 2 doors marked H and T, and one contains a prize. The host will toss a coin and the contestant will win

what is behind the door corresponding to the coin. The win ratio=1/2.

Session 2 shows winning the prize does not depend on any ability of the player, but is a random event. The game show is for entertainment. The viewing audience is rooting for the player, who represents the common person. Thus session 1 has more entertainment value for the public.

____________________________

The 1/3 probability is not effective once the host has reduced the doors from 3 to 2/.

There can't be 3 ways to select 1 of 2 things!

_____________________________

here is another table to to consider

Player always picks door 1. For sessions 1 and 2, host has a choice of doors 2 and 3.

For sessions 3 and 4, host must choose 1 door, 2 or 3.

Compare p1 stay to p2 switch. There is no advantage.

~phyti Phyti (talk) 20:09, 1 December 2025 (UTC)

Despite me having shortened the rest of my reply to focus it on the question for you, you still did what I was worried you would do instead of answering that question, so I am getting even-more-quickly to it this time:

​

For the ​ MH game 3 cards ​ from [3 comments by me] ago ​ - ​ "Start with {[Ace of Spades], [2 of spades], [3 of spades]}. ... keep the initial card, or switch to the "host's" card?" ​ - ​ do you count [Player's 1st guess is the ace of spades, Host reveals all but that and the 2 of spades] and [Player's 1st guess is the ace of spades, Host reveals all but that and the 3 of spades] as 2 sessions or just 1 session?

​

If you yet again don't answer the above question, then my next comment here will likely be my last in reply to you regarding Monty Hall, because you would be able to endlessly go on with

​

choose count-as-multiple-sessions vs count-as-one-session according to whichever of those makes w/n equal what you think the probability is

and

the player doesn't know which of the 2 it is, so 1/2

and

there can't be 3 ways to pick 1 of 2

​

, ​ despite you seeming to (as far as I can tell, you also never actually answered the questions for these) have recognized that [the probabilities can be different from 1/2 even when there are exactly 2 possible doors the car can be behind] and [there can be 52 ways for the ace of spades to be 1 of 2 non-revealed cards].

​

JumpDiscont (talk) 21:46, 1 December 2025 (UTC)

A session is a sequence of actions, (player 1st guess, host action, player 2nd guess).

For the MH game, host action is reveal and remove a goat door.

For the multiple card game, the host action is reveal and remove all non prize cards except 1, leaving a prize and a non-prize card for the player 2nd guess. If the host removes 49 non-prize cards, that is equivalent to a 3 card session.

Adding the extra cards seems counter productive. Now you have a probability of 51/52 of picking a losing card.

Specific to the cards, let the ace have a value of 1, and all non-ace cards have a value of 0. Adding 51 '0' cards to the set, then removing 50 '0' cards doesn't alter the value of the '1' card. There is no rule for card removal, like left to right, so we will assume at random.

That means any one of the 50 can be the first or last to be removed, and has no special significance. Their value does not change over time.

Marilyn Savant thought otherwise.

"Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?"

NO, and WHY? Both have survived the removal process and she doesn't know which one is the car. This is the 'gambler's fallacy', an erroneous belief that the probability of a number occurring increases after long periods of absence. Many lottery players believe this, based on the false idea that all numbers within a given range should appear equally often. If a fair coin yields 10 H in 10 tosses, the probability of T for the 11th toss is still 1/2. All tosses are independent events.

The average person doesn't realize a 'random event' can happen any time since it is 'time independent'.

The MH game always ends with the player's 2nd guess of a car door or a goat door.

The players guess is a random event, independent of the external environment.

Winning a car is a random event and why there is no strategy.

Here is a table for 4 cards. The player always picks card 1. The ace 'A' is placed in all possible positions, and the host removes all loser cards except 1. This allows 9 sessions for 1 player choice, thus 9*4=36 possible sessions. Rules are the same, player choice and prize are exempt from host action.

Allowing 2 pairs per card, there would be 105 sessions for 52 cards. Check it for errors

~phyt Phyti (talk) 18:43, 3 December 2025 (UTC)

(I am de-indendinting: ​ Below is my reply.) ​ ​ ​ JumpDiscont (talk) 04:49, 4 December 2025 (UTC)

​

​

​

"A session is a sequence of actions, (player 1st guess, host action, player 2nd guess)."

From MH 3 doors, there are 8 such sequences for [car is behind door 1] -

​

[player's 1st guess is door 3, host opens door 2, player's 2nd guess is the stay door]

[player's 1st guess is door 3, host opens door 2, player's 2nd guess is the switch door]

[player's 1st guess is door 2, host opens door 3, player's 2nd guess is the stay door]

[player's 1st guess is door 2, host opens door 3, player's 2nd guess is the switch door]

[player's 1st guess is door 1, host opens door 3, player's 2nd guess is the stay door]

[player's 1st guess is door 1, host opens door 3, player's 2nd guess is the switch door]

[player's 1st guess is door 1, host opens door 2, player's 2nd guess is the stay door]

[player's 1st guess is door 1, host opens door 2, player's 2nd guess is the switch door]

​

- rather than just 4, so this seems to be a change from both ways you were counting before.

(If your answer is ​ ​ ​ just double all your counts for MH 3 doors ​ , ​ ​ ​ then the followup is when one adds ​ ", player's 2nd guess is the stay door" ​ to the 2-or-1 from the question.)

​

​

"Now you have a probability of 51/52 of picking a losing card."

​

You have a probability of 2/3 of picking a losing door for MH 3 door, which you're saying becomes 1/2 for MH 3 door, even though you seem to think the 51/52 for MH 52 cards dosn't become 1/2 for MH 52 cards.

​

​

"WHY?" ​ ​ ​ If the host does _not_ necessarily avoid opening the players 1st choice, then indeed stay-or-switch doesn't necessarily matter. ​ If instead the host cannot open the door from the players 1st choice, then because there are 1000000 "ways to play a game, but only 1 guess to win a prize."

​

See your comment with the above quote: ​ ​ ​ There is 1 car-door, and 999999 Z doors. ​ Player 1st randomly guesses a door, followed by the host revealing the number of Z doors leaving 1 Z door for the player's 2nd guess.

​

From MH 52 cards, both the player's 1st choice and the 1 Z card left "survived the removal process and" you "don't know which one is the" ace of spades, but you got that there is only 1 session out of 52 in which the player's 1st choice is the ace of spades. ​ The 1 Z card left from 52 cards being the one numbered 7 on the back, corresponds to the 1 non-1st-choice door from 1000000 being door #777,777.

​

You say ​ ​ ​ "The player does not win a prize for the 1st guess." ​ , ​ ​ ​ but that applies to the cards too: ​ ​ ​ ​ ​ ​ ​ The player gets to keep or switch.

​

You said ​ ​ ​ "Don't confuse the two experiences/actions." ​ , ​ ​ ​ but you have yet to give explanation/reasoning for one which doesn't yield an answer for the other that is different from what seems to be your answer to the other.

​

​

Your table for 4 cards does not exempt player choice from host action. ​ (It shows the host removing the player's choice for s in {4,5,6,7,8,9}.) ​ ​ ​ If you fix that, you'll get 6 rows rather than 9, for presumably 24 possible sessions.

​

I'm not sure I know what you mean by ​ ​ ​ "Allowing 2 pairs per card" ​ , ​ ​ ​ but

there is no integer w such that ​ w/105 = 1/(1+51)

and

if I'm correctly guessing what you meant, then [[1 from each pair] and so [52 of 104]] have the player's first guess be the ace of spades, despite you giving 1 vs 51 for MH game 52 cards

.

​

​

JumpDiscont (talk) 04:49, 4 December 2025 (UTC)

MH game

The game rules as proposed by Whitaker:

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

(That would end the game with a prize and deny the player a 2nd guess.)

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

(That would reveal the car location and end the game.)

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

(Relative to Whitaker's question of a winning strategy.)

Car is behind door 1, goats behind doors 2 and 3.

P is player, H is host, P switches.

S1. P guesses door 1, H opens door 2, (rule 1),P guesses door 3.

S2. P guesses door 1, H opens door 3, (rule 1), P guesses door 2.

S3. P guesses door 2, H opens door 3, (rule 1 & rule 2), P guesses door 1.

S4. P guesses door 3, H opens door 2, (rule 1 & rule 2), P guesses door 1.

H does not violate any rules.

There are 4 possible sessions each played once. There are no half sessions.

P wins a car 2/4 staying, and 2/4 switching.

There is no strategy for a game with random guesses!

-------------------------------------------------------------------

It isn't necessary to play the 4 games twice. Just compare P's 1st guess with P's 2nd guess as above.

For 52 cards, 1=A and 51=0. P picks x. Using the game rules, H removes 50 cards leaving x and r. If x=A then r=0. If x=0 then r=A. It's an A or not A guess.

Whether there are doors, boxes, cards, or shells, is irrelevant. The game always reduces to a guess of 1 of 2 things. Win a valuable prize or win a prize of little or no value.

The MH game show was basically simple and the Selvin/Savant explanation was a misinterpretation of the facts, leading to a rigged game. There is no reason for the dramatic 'Nearly everyone doesn't get it' titles, as published by some (with degrees!).

The average intuition person can avoid the expense and time of having to spend a month meditating in the Himalayas to solve this problem.~phyti ~2025-38965-58 (talk) 16:44, 6 December 2025 (UTC)

Do you think ​ keep vs switch ​ matters for ​ ​ ​ MH game 52 cards ​ ?

("Start with a full ... the "host's" card?" ​ from Rick Block)

​

If no, then:

You previously said ​ ​ ​ "There is 1 session where x is

the A card and 51 sessions where x is not the A card." ​ .

Do you now count in a way that gives the same number of sessions

for ​ ​ ​ x is the A card ​ ​ ​ as it gives for ​ ​ ​ x is not the A card ​ ?

​

If yes to the first question in this comment, then:

You are saying stay-vs-switch doesn't matter for 3 doors, and you

seem to think stay-vs-switch doesn't matter for 1000000 doors,

but you accept that ​ keep vs switch ​ matters for 52 cards.

Do you think [the difference between the problem statements

that causes the change between mattering and not mattering]

is [whether the non-switch option at the end is "stay" or "keep"] or

[the number of doors/cards the problem starts with] or something else?

​

​

I'm not yet responding in more detail because, without me knowing your

answers to the above, either [more likely than not most of the response]

or [probably a majority of the response] would be stuff you agree with.

(which of the two for that either-or,

depends on how long the response would be)

​

For example, I was going to lay out the application

of your ​ MH game ​ reasoning to ​ ​ ​ MH game 52 cards ​ ,

but based on ​ ​ ​ "It's an A or not A guess." ​ , ​ ​ ​ maybe

you now think the result of that application is correct.

​

​

JumpDiscont (talk) 03:15, 7 December 2025 (UTC)

MH game

So many believe the 1/3-2/3 probabilities determine the results.

This is why they don't.

There are 3 possible locations, thus 1/3 probability of (p)layer guessing the car location.

The number of sessions to play the game depends on the number of (h)ost actions.

For all possible p1-h-p2 sequences, there are 3*2*1=6 cases.

Rule 1 prohibits h from opening the p1 guess since the player gets a 2nd guess p2 when a prize is verified.

Rule 2 prohibits h from opening the car door, which eliminates sessions 4 & 5 reducing 6 sessions to 4 (8 for stay & switch).

The stay vs switch results on the right show no advantage with a win ratio of 1/2.

Notice when player 1st guess is door 1, they don’t always win the car. To guess the car door does not imply winning the car.

I was only interested in revealing the errors in the Selvin & Savant explanations. That’s done. Maybe someone else will debate your variations of the game.

Phyti (talk) 19:13, 8 December 2025 (UTC)

"I was only ... Selvin & Savant explanations. That's"

only done if either

Given the rules on how the host behaves

(which I believe neither Selvin nor Savant stated, and so,

perhaps neither of them recognized the importance of),

you no longer think their conclusion

(switching is better) is an error.

or

The reason it's done is ​ ​ ​ you won't be replying again ​ .

.

​

Even given the rules on how the host behaves, I was not

arguing in favor of Selvin's or Savant's _explanations_.

I was only arguing that given those rules,

switch is better, 2/3 vs 1/3.

​

JumpDiscont (talk) 20:12, 8 December 2025 (UTC)

When you analyze a problem it helps to know its history.

Craig Whitaker's problem was equivalent to Steve Selvin's problem, just different prizes and containers. Savant responded to Whitaker's question knowing the rules.

"I was only arguing that given those rules,

switch is better, 2/3 vs 1/3"

But that result is caused by manipulation of session frequencies, bias, a rigged game.

Both Selvin and Savant interpreted the host choices literally as [(box/door 2) OR

(box/door 3) but not both, when it should be [(box/door 2) AND (box/door 3)], 1 session for each box/door.

The refined table shows 3 doors with fixed prizes while the player guesses the range of doors. Column w shows prizes for stay. Columns 6-8 show the effect of different frequencies for the 4 sessions.

For f1 s= 2/4 with no difference, for f2 s=1/3 so switch, for f3 s=2/3 so stay.

It's just pure manipulation, and it's illegal.

When the player gets to their 2nd guess when they get a prize, the host asks the player,

"do you want to stay or switch?". A choice of car or goat, 1 of 2 things!

~phyti Phyti (talk) 19:43, 12 December 2025 (UTC)

Here is a different perspective of the MH game.

The player 1st guess is not a prize winning event, but the 2nd guess is.

What then is the purpose of the 1st guess?

It allows the player to select one of the doors in the 2nd guess.

The host selects the remaining door r for the 2nd guess, by eliminating a goat door.

The 4 sessions.

c g g prize

1 2 3 door

--------------

p h r c g

p r h c g

r p h g c

r h p g c

For a fair game all sessions must occur with the same frequency, else it's illegal.

The player with their simple intuition knows there is a car behind one of the doors and a goat behind the other door, since 1 goat has been removed., and 3-1=2.

The player guess is always 1 of 2 things, so s=w/n=1/2. Their is no advantage.~phyti Phyti (talk) 19:48, 12 December 2025 (UTC)

If by "illegal", you're referring to case/statute law

- ​ rather than ​ not valid math/probability ​ - ​ then:

Do you have any citation for either

(a) ​ Unequal session frequencies constitute at least one of manipulation/bias/rigging

​ ​ ​ ​ in the law sense, and as such would make the game/show illegal.

or

(b) ​ Unequal session frequencies would make the game/show illegal for some other reason.

?

​

Note the first part of (a): ​ For cases/statutes about manipulation/bias/rigging, I'd certainly hope that unequal session frequencies _caused by following the game/show rules_ don't constitute such manipulation/bias/rigging, because otherwise

2 players, in each round the players are equally likely to win that round, stop when someone has won 2 rounds

would be illegal, since it makes [goes all 3 rounds] and [ends in 2 rounds] equally likely, whereas there are twice as many sessions for the former as there are for the latter.

​

​

Now, assuming that [following the rules would make the session probabilities unequal] doesn't make [following the rules] illegal: ​ From your table, the next step is figuring out which of f1,f2,f3 correspond to the session frequencies caused by following the rules for Monty Hall.

​

​

"A choice of car or goat, 1 of 2 things", with different probabilities:

​

See my ​ "There are only 2 doors. The ... it's showing 1]." ​ show, and note that although I said the player's 1st guess was the secondary event for MH, that was before we started specifically on ​ ​ ​ Player chooses door 1 ​ . ​ ​ ​ ​ ​ ​ ​ When limiting to ​ ​ ​ Player chooses door 1 ​ , ​ ​ ​ the secondary event is ​ ​ ​ where the car was placed. ​ ​ ​ ​ ​ ​ ​ (This happens before the player chooses a door: ​ Do you think whoever places the car should know that door 1 is special, and so be more likely to put the car behind door 1 than behind door 2?)

​

​

JumpDiscont (talk) 08:29, 13 December 2025 (UTC)

"What then is the purpose of the 1st guess?"

​

​

The purpose is restricting what the host can do.

​

If it didn't - so, the host opens a random goat door even if that's the 1st guess, and one asks about stay-vs-switch in the cases where that's not the 1st guess - then the frequencies would all be equal, because for example,

Prob(car behind 1 and player chooses 1 and host opens 3)

=

Prob(car behind 1 and player chooses 1) *

Prob(host opens 3 given (car behind 1 and player chooses 1))

=

Prob(car behind 1 and player chooses 1) * 1/2

=

Prob(car behind 2 and player chooses 1) * 1/2

=

Prob(car behind 2 and player chooses 1) *

Prob(host opens 3 given (car behind 2 and player chooses 1))

=

Prob(car behind 2 and player chooses 1 and host opens 3)

.

​

The difference, is whether ​ Prob(host opens 3 given (car behind 2 and player chooses 1)) ​ is 1/2 or 1.

(If you'd say that's 1 even without the restriction since we're assuming the host didn't open the 1st guess, then see Bertrand's Boxes for why the prior probability can matter even after a thing happened.)

​

​

​

"For a fair game all sessions must occur with the same frequency, else it's illegal."

​

So, if each one of the 2-or-3 gives exactly 1 winner and is fair,

then standard best-of-three between two players is illegal?

​

​

​

JumpDiscont (talk) 08:29, 13 December 2025 (UTC)

"If by "illegal", you're referring to case/statute law

- ​ rather than ​ not valid math/probability ​ - ​ then:

Do you have any citation for either"

Refer Wiki,"1950s quiz show scandals", resulting in stricter federal regulations of tv gameshows.

"Note the first part of (a): ​ For cases/statutes about manipulation/bias/rigging, I'd certainly hope that unequal session frequencies _caused by following the game/show rules_ don't constitute such manipulation/bias/rigging,"

The gameshow producers would not play the game with any bias!

The 1/3 vs 2/3 results from Selvin's and Savant's misinterpretation of how the game is played.

The MH game is simple and intended for the average contestant to participate. They do not have to be qualified as done on 'Jeopardy'.

Which door contains the car is irrelevant since all doors have the same prizes.

The equal sessions average is for different players over a period of time.

The player, not knowing the car location can only make a random guess for the 1st and 2nd guess.

rule 1. Host cannot open player's 1st guess (would end the game).

rule 2. Host cannot open door with the car.

The host (and their staff) knows the location of the car (to avoid accidentally revealing it and ending the game).

After the player makes the 1st guess, the host per the rules, does not open the 1st guess door. That door becomes 1 of the pair for the 2nd guess. The host opens a goat door and the remaining closed door r becomes the other 1 of the pair for the 2nd guess.

The simple intuition of the player: there are only 2 doors since 3-1=2, and there are only 2 prizes since the host removed 1 goat.

The game involves random (not predictable) events thus there is no system/method of play to form a basis for a strategy. You cannot 'beat the system' as claimed by Savant.

An analogy is a coin toss. An analysis of a history of 1000 tosses reveals a measure (ratio) of (successful call)/(number of tosses) of approx. 1/2. Note the measure of success is an AFTER THE FACT conclusion for a large number of events and not for future individual events.

A coin toss NOW is not predictable, but has a probability of 1/2 per convention.

The 1st guess for the location of the car which offers no prize, is 1 of 3.

The 2nd guess for the location of the car which offers the contents of the door, is 1 of 2.~phyti ~2025-41684-02 (talk) 17:43, 18 December 2025 (UTC)

for reference 28 in the ​ 1950s quiz show scandals ​ article:

I am not finding the Franklin article at the ISSN link, and the doi link shows me only the first page. ​ ​ ​ ​ ​ ​ ​ The only thing relevant on the first page of the Franklin article is ​ ​ ​ "... federal offense for predetermining the outcome of a" ​ , ​ ​ ​ but I'm saying 1/3 vs 2/3, rather than that it's predetermined.

​

for reference 27 in the ​ 1950s quiz show scandals ​ article:

I was assuming the listening/viewing public knew the rules, rather than knowledge of them being special and secret assistance to the contestant. ​ In this case, I would say that following such rules is not engaging "in any artifice or scheme".

​

Are you referring to anything else from the ​ 1950s quiz show scandals ​ article?

​

​

"The gameshow producers would not play the game with any bias!"

​

Indeed, they never played it.

​

​

"... since all doors have the same prizes."

​

I was assuming that [whichever doors have goats don't also have a car] and [whichever door has a car doesn't also have a goat]. ​ Without these assumptions, I think most of what we've been going over would be nonsense.

​

​

"The player, not ... and 2nd guess." ​ ​ ​ and

"The game involves random ... a basis for a strategy." ​ :

​

See my ​ "There are only 2 doors. The ... it's showing 1]." ​ show.

​

​

Do you think the probability of the car being behind the 1st guess would be different if there was no 2nd guess?

​

No-Switch Hall:

1 car and 2 goats are hidden, as in Monty Hall. ​ The player guesses a door, as in Monty Hall. ​ The host opens a door, following the rules we are assuming for Monty Hall. ​ Unlike in Monty Hall, in No-Switch Hall there is no switch-stay choice.

After the above, what is the probability that the car is behind the player's guess?

​

​

JumpDiscont (talk) 22:17, 18 December 2025 (UTC)

"I was assuming the listening/viewing public knew the rules, rather than knowledge of them being special and secret assistance to the contestant. ​ In this case, I would say that following such rules is not engaging "in any artifice or scheme"."

They knew the rules as presented to the public. The deception in the 1950's was all internal, until a few players leaked the details. We currently (2025) have deception in some sports events. It's nothing new.

"Indeed, they never played it."

It was a hypothetical game proposed by Steve Selvin in 1975 to 'American Statistician' and Craig Whitaker in 1990 to Marilyn Savant writing for 'Parade magazine', but the game rules were specified and understood by all involved in the correspondence.

An excerpt from the letter to 'The American Statistician', Aug. 1975.

"Monty Hall wrote and expressed that he was *[not "a student of statistics problems"] but "the big hole in your argument is that once the first box is seen to be emprty, the contestant cannot exchange his box." He continues to say, "Oh and incedentally,** [after one [box] is seen to be empty, his chances are no longer 50/50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so." I could not have said it better myself."]

It's humorous comparing what Monty Hall said at * and later at **. The 2nd reinforced the 1st. He obviously was lacking in basic understanding of probability, along with Selvin.

"Do you think the probability of the car being behind the 1st guess would be different if there was no 2nd guess?"

No. Positioning the car is independent of future player-host actions.

"No-Switch Hall:

1 car and 2 goats are hidden, as in Monty Hall. ​ The player guesses a door, as in Monty Hall. ​ The host opens a door, following the rules we are assuming for Monty Hall. ​ Unlike in Monty Hall, in No-Switch Hall there is no switch-stay choice.

After the above, what is the probability that the car is behind the player's guess?"

If there is no 2nd guess, those rules don't apply and the host opens the player's guess.

The probability of winning a prize equals the probability of car location which is 1/3.

You are playing the simple game, and not the modified game which required those rules.~phyti ~2025-41821-55 (talk) 17:39, 20 December 2025 (UTC)

So, you don't think ​ the probability of the car being behind the 1st guess ​ would be different if there was no 2nd guess, but for No-Switch Hall, you say the probability of winning a prize equals 1/3.

​

For No-Switch Hall, do you think the probability of winning a prize is different from the probability that the car is behind the player's guess?

​

If yes, then you didn't answer my previous comment's question regarding No-Switch Hall.

​

If no, then:

You're getting 1/2 for Monty Hall and 1/3 for No-Switch Hall, but you don't think ​ the probability of the car being behind the 1st guess ​ would be different if there was no 2nd guess. ​ ​ ​ No-Switch Hall was me trying to just remove the 2nd guess from Monty Hall. ​ ​ ​ ​ ​ ​ ​ Do you get a different ​ ​ ​ show description ​ / ​ problem statement ​ ​ ​ for ​ ​ ​ Monty Hall with no 2nd guess ​ ?

​

I am playing simple games so that we are more likely to agree regarding these games, which would allow zooming in on where the disagreement starts.

If you end up saying that ​ ​ ​ whether-or-not there is a 2nd guess ​ ​ ​ affects ​ ​ ​ the probability of the car being behind the 1st guess ​ , ​ ​ ​ then the next step of this zoom-in will be replacing ​ contestant makes the contestant's 1st - and perhaps only - guess ​ with ​ ​ ​ host rolls a 3-sided die and places an arrow pointing towards the door whose number is the result ​ , ​ ​ ​ so that if you say the answer for this die-roll version still depends on whether-or-not there is a guess after the host opens a door, then I can replace 1 contestant with 2 contestants, since the die roll would make the host necessarily able to follow the rules for both.

​

(As far as I can recall, you haven't made this argument, but in case you would've, I am mentioning that ​ necessarily able to follow the rules for both ​ is why ​ ​ ​ "Contestant 1 chooses door 1, Contestant 2 chooses door 2, and Host opens door 3. ​ By symmetry, door 1 and door 2 must be equally likely." ​ ​ ​ doesn't work for Monty Hall:

Following the Monty Hall rules with respect to Contestant 1 is incompatible with following the Monty Hall rules with respect to Contestant 2, and the probabilities depend on how this is resolved.)

​

JumpDiscont (talk) 19:38, 20 December 2025 (UTC)

MY only interest here is the Marilyn Savant response to Craig Whitaker's question of a game strategy.

My examples reveal game manipulation introduces a false winning strategy.

It's not a fair game, and it's illegal.

refence: fcc.gov/general/broadcast-contests

The statutory provision regarding contests is set forth at Section 508 of the Communications Act of 1934, as amended (the ``Act) (47 U.S.C. § 509). Section 508(a) of the Act (47 U.S.C.. § 509(a)) provides that it is unlawful for any person, with intent to deceive the listening or viewing public:

1.To supply to any contestant in a purportedly bona fide contest of intellectual knowledge or intellectual skill any special and secret assistance whereby the outcome of such contest will be in whole or in part prearranged or predetermined.

2.By means of persuasion, bribery, intimidation, or otherwise to induce or cause any contestant in a purportedly bona fide contests of intellectual knowledge or intellectual skill to refrain in any manner from using or displaying his knowledge or skill in such contests, whereby the outcome thereof will be in whole or in part prearranged or predetermined.

3.To engage in any artifice or scheme for the purpose of prearranging or predetermining in whole or in part the outcome of a purportedly bona fide contest of intellectual knowledge, intellectual skill, or chance.

Number 3

~phyti ~2025-42318-15 (talk) 17:50, 22 December 2025 (UTC)

"MY only ... a game strategy."

I haven't looked for the quote by her, but I get the impression that at least part of her response was ​ ​ ​ switch-is-better, 2/3 vs 1/3 ​ . ​ ​ ​ ​ ​ ​ ​ I am not defending anything else from her, and even ​ ​ ​ switch-is-better, 2/3 vs 1/3 ​ , ​ ​ ​ I'm only defending under the rules we've stated for how the host determines which door the host opens. ​ ​ ​ ​ ​ ​ ​ If you're no longer interested in ​ ​ ​ switch-vs-stay under the rules we've stated for how the host determines which door the host opens ​ , ​ ​ ​ then OK.

​

What do you mean by ​ ​ ​ "false winning strategy" ​ ?

I imagine you mean the strategy doesn't actually do better than 1/2, but at this point I'm checking in case you mean something else.

​

for the rest of your comment:

So, even when the rules are known, if they would cause the session probabilities to be non-equal then following them is an ​ ​ ​ "artifice or scheme" ​ ?

​

JumpDiscont (talk) 22:50, 22 December 2025 (UTC)

Look back at the table in my 12-12-2025 post.

The frequencies f1, f2, f3, show the sessions can be manipulated for whatever strategy you want, stay, switch, or do nothing. That’s why her claim to beat the system is false.

She didn’t do the manipulation deliberately, only due to lack of understanding of bsic probability and logic.

A high IQ doesn’t make her infallible.

The game show producers would not play some sessions more than others creating a bias favoring some over others. A fair game offers the same opportunity to all players.

There is nothing wrong with the game rules.~phyti ~2025-42521-06 (talk) 18:32, 23 December 2025 (UTC)

Does ​ [following the rules woluld cause the session frequencies to be not all equal] ​ cause you to regard ​ [following those rules] ​ as manipulation?

If yes, then: ​ ​ ​ Under that definition and the assumptions we've stated, the Monty Hall Problem is manipulated. ​ One can nonetheless ask what probabilities follow from the rules.

If no, then: ​ ​ ​ Most probability prroblems could have manipulation. ​ For a coin flip, the coin could be biased. ​ For a die roll, the die could be weighted. ​ For cards, the shuffler could be cheating.

"There is nothing wrong with the game rules." ​ ​ ​ is a surprising take from you, considering you had switched from ​ ​ ​ [arguing against my claim that following the game rules causes session frequencies f2] ​ ​ ​ to ​ ​ ​ [talk of ​ manipulation/bias/rigging/illegal ] ​ .

JumpDiscont (talk) 23:37, 23 December 2025 (UTC)

The MH games proposed by Selvin and Whitaker were fair games with no deception.

The erroneous interpretations by Selvin and Savant, was the source of manipulation.

The manipulation has nothing to do with following the rules.

This is an excerpt from the latest revision at

https://drive.google.com/file/d/1QJ9KFzLuo5c1AiuXSm2K_xxqrVMH7vBM/view?usp=sharing

A session is playing the game with varying player-host actions.

The format is p1(player 1st guess), h (host removes a door), r (remaining closed door). On the left, the 4 possible sessions are listed for the prize distribution of c g g for 3 doors.

Rules. Host must open a door different from the player's guess

and can only open goat doors.

On the right, are the pairs from p1 and r on the left in terms of prizes, which are the choices for p2, the player's 2nd guess. For this part of the session, the host is only required to verify which prize is won.

The player can guess door 1 with the car for 1/2 the sessions since the host can open doors 2 and 3, whereas sessions 3 and 4 allow opening 1 door.

Even though the player 1st guess is correct, there is no prize for that guess.

The prizes are awarded for the player 2nd guess p2.

All p2 guesses are (c or g) with a win car ratio of 1/2.

If Selvin's elements replace Savant's elements, the results are the same as fig.3.

There is no advantage to switch.

~phyti Phyti (talk) 18:36, 26 December 2025 (UTC)

Does your definition of ​ fair game with no deception ​ require that

(a) ​ all session frequencies are equal

(b) ​ if there is an obvious answer, then that obvious answer is correct

?

​

(c) ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ I am using the assumptions on how the host determines which door the host opens. ​ ​ ​ ​ ​ ​ ​ Do you think I'm using anything else from the ​ ​ ​ "interpretations by Selvin and Savant" ​ ?

​

You didn't address the apparent inconsistency between your answers, so I am making these one-or-two self-contained:

​

​

(d) ​ Just after

1 car and 2 goats are hidden, as in Monty Hall. ​ ​ ​ The player guesses a door, as in Monty Hall. ​ The host opens a door, determined as follows.

The host doesn't open the door the player guessed. ​ The host knows where the car is, and doesn't open that door. ​ If that door is the door the player guessed, then the host chooses uniformly at random which goat door the host opens.

There is no switch-stay choice.

, ​ what is the probability that the car is behind the player's guess?

​

(This question is asking for the probability that ​ ​ ​ the car is behind the player's guess ​ , ​ ​ ​ even if this probability is different from the probability of winning a prize.)

​

​

​

If your answer to (d) is 1/3, then

​

​

(e) ​ Just after

1 car and 2 goats are hidden, as in Monty Hall. ​ ​ ​ The player guesses a door, as in Monty Hall. ​ The host opens a door, determined as follows.

The host doesn't open the door the player guessed. ​ The host knows where the car is, and doesn't open that door. ​ If that door is the door the player guessed, then the host chooses uniformly at random which goat door the host opens.

The player gets to switch or stay, as in Monty Hall.

, ​ what is the probability that the car is behind the player's 1st guess?

​

(This question is asking for the probability that ​ ​ ​ the car is behind the player's 1st guess ​ , ​ ​ ​ even if this probability is different from the probability of winning a prize.)

​

​

.

​

​

​

regarding the excerpt:

​

​

(f) ​ For

The host rolls a 3-sided die, with sides {1,2,3}. ​ If it shows 1, then the host flips a coin, else the host places the coin showing heads. ​ ​ ​ ​ ​ ​ ​ The player guesses ​ ​ ​ "equal to 1" ​ ​ ​ or ​ ​ ​ "greater than 1" ​ . ​ ​ ​ ​ ​ ​ ​ The player wins a car if and only if the player's guess is correct regarding the number shown by the die.

, ​ do you think either guess is better than the other?

​

If your answer to (f) is yes, then apply that excerpt's reasoning to (f):

A session has a die side and a coin side. ​ There are 4 possible sessions. ​ ​ ​ The player can guess ​ "equal to 1" ​ for 1/2 the sessions since the coin can show heads and tails, whereas sessions 3 and 4 allow 1 side for the coin. ​ ​ ​ Even though the die shows 1, there is no prize for that result. ​ The prizes are awarded for the player's guess. ​ ​ ​ All guesses are ​ ( "equal to 1" ​ , ​ "greater than 1" ) ​ with a win car ratio of 1/2. ​ ​ ​ ​ ​ ​ ​ There is no advantage to ​ ​ ​ "greater than 1" ​ .

​

​

​

JumpDiscont (talk) 02:25, 27 December 2025 (UTC)

a. The game show producers are responsible for equal session frequencies.

b. The game rules restrict the host to only open goat doors, so host must know car location.

d. Probability of win car for all 4 possible sessions is 1/2.

e. The player 1st guess is irrelevant, and presents the appearance of player involvement to the viewing audience.

The producers could eliminate the 1st guess, present the player with 2 doors, inform them of car in one and goat in the other. Then ask the player to guess which one. Win ratio = 1/2.

--------------------------------------------------------

f.

Host rolls a die with results (1, 2, 3).

If 1 then coin toss with results (H or T), else host sets value to H.

Player guesses die toss with v = 1 if die value = 1 , else v>1.

For this game, v >1 also equals (not 1), which occurs 2/3 of die rolls.

You are comparing 2 different sets, one with 1 member and one with 2 members.

Fair game would consist of 6 sessions, 3 for H and 3 for T, with win ratio = 1/2.

Host replacing 2 coin tosses with the 2 H-events is manipulation.

Player winning depends on die toss which is 1/3, not on coin toss which is 1/2.

Coin toss is in the future relative to die toss, thus cannot affect the die toss.

--------------------------------------------------------

For clarification, the manipulation of session frequency explains the false strategy proposed by Selvin and Savant.

The game show scandals involved controlling player behavior with prearranged results for the purpose of increasing ratings, and did not alter the game play.~phyti Phyti (talk) 19:31, 30 December 2025 (UTC)

If by ​ ​ ​ "are responsible for" ​ ,

you mean ​ ​ ​ "must ensure" ​ , ​ ​ ​ then:

If the hosts due ensure that, then ​ switch-vs-stay

doesn't matter ​ is correct. ​ ​ ​ However, following the

assumed rules would result in them not doing that.

​

If by ​ ​ ​ "are responsible for" ​ ,

you mean ​ ​ ​ "are the cause of" ​ , ​ ​ ​ then:

Whether-or-not these frequencies are equal is part of our argument.

​

--------------------------------------------------------

​

player's 1st - and perhaps only - guess ​ not equal ​ car location

occurs 2/3 of such guesses

​

"comparing 2 different sets, one with

1 member and one with 2 members" ​ :

Whatever makes ​ 1H , 1T ​ into just 1 element, also makes

( car behind 1 , player picks 1 , host opens 2 )

( car behind 1 , player picks 1 , host opens 3 )

into just 1 element for the

car is behind the player's 1st - and perhaps only - guess

probabilities.

​

car is behind the player's 1st - and perhaps only - guess

depends on that guess which is 1/3,

not on host's door which is 1/2.

​

The host opening a door is in the future relative to both

placement of the car ​ and

the player's 1st - and perhaps only - guess

, ​ thus affects neither the placement nor this guess.

​

--------------------------------------------------------

​

It might feel to you like my questions aren't

leading anywhere - and if you repeatedly

don't explicitly answer the actual questions, and

flip-flop on what answer to them is implied by your reply

, ​ then you indeed could avoid them leading anywhere -

but from the implied

1/2 for ​ the probability that the car

is behind the player's guess ​ from (d)

and ​ ​ ​ yes for (f) ​ , ​ ​ ​ the next

approximate midpoint is the following.

​

(g) ​ For

The host has an otherwise-symmetric disc, with one side labeled 2 and the other side labeled 3. ​ The host rolls a 3-sided die, with sides {1,2,3}. ​ If the die value is 1, then the host flips the disc, else the host places the disc so the label of its shown side is equal to the die value. ​ ​ ​ ​ ​ ​ ​ The player guesses ​ ​ ​ "equal to 1" ​ ​ ​ or ​ ​ ​ "not 1" ​ . ​ ​ ​ ​ ​ ​ ​ The player wins a car if and only if the player's guess is correct regarding the die value.

, ​ do you think either guess is better than the other?

​

(This avoids having an asymmetry between coin sides. ​ On one hand, you might still use what player winning depends on, and - I say correctly - ignore what happens in the future relative to die toss. ​ On the other hand, you might get 4 sessions for this, like you said for d.)

​

JumpDiscont (talk) 00:07, 31 December 2025 (UTC)

MH game

The game is divided into 2 parts. A and B.

A lists the prizes (c, g, g) in the 3 doors 1 to 3, 1 car and 2 goats.

Host must open a door different from the player's guess

and can only open goat doors.

Row 3 lists: d (door ID), and s (session ID).

A session is a sequence of p1 (player 1st door guess), h (host door opened), and r (remaining closed door).

If the player 1st guess contains c, they can play session 1 or 2.

If the player 1st guess contains g, they can play session 3 or 4.

B lists the 2 doors p1 and r in terms of prizes for p2 the players 2nd guess.

The choice is always c or g, with a win c ratio of 1/2.

Each player guesses 1 of 3 doors, and plays 1 of 4 sessions.

~phyti Phyti (talk) 18:59, 3 January 2026 (UTC)

I would suggest you research fcc.gov for broadcast regulations established in 1934 for radio, and amended in 1960 (after the 1950's scandal) for tv.

You can then decide the legality of game shows.

"comparing 2 different sets, one with

1 member and one with 2 members"

Using the 52 card game, there is 1 win session with the AS card, and 51 lose sessions with a blank or not-AS card.

Probability of success is (number of win events)/(number of possible events).

P(win)=1/52, P(lose)=51/52. Probability is a history requiring all possible events.

Without the host participation, a win would be a rare event. But the session isn't complete. The host will remove all blank cards but 1, then ask the player to make a 2nd guess from the 2 remaining cards, AS or blank. The simple intuition of the average person will realize there are only 2 possible endings. The player wins or loses.

If the probability of a correct guess for 3 doors is larger than for 100 doors,

why wouldn't the probability of a correct guess for 2 doors be larger than for 3 doors? I.e. the probability is inversely proportional to number of doors.

This has already been explained. The 1st guess is irrelevant, since there is no prize. Even if it is correct, the player can still lose on the 2nd guess.

g. "do you think either guess is better than the other?"

No. This is the same type of game as you last proposed. The coin or disc is irrelevant since the answer depends on the die toss. There are 3 events, 1 wins, 2 lose. That's a win ratio of 1/3. It's another probability for a random guess.~phyti Phyti (talk) 19:02, 3 January 2026 (UTC)

" ​ ​ ​ g. "do you think either guess is better than the other?"

​

No. This is the same type of game as you last proposed. The coin or disc is irrelevant since the answer depends on the die toss. There are 3 events, 1 wins, 2 lose. That's a win ratio of 1/3. ​ ​ ​ "

​

​

you get a win ratio of 1/3 for the guess ​ ​ ​ "not 1" ​ ?

(That would follow from ​ neither guess is better than the other ​ and one of them having a win ratio of 1/3, but I don't see how you could get fewer wins than loses for the guess ​ ​ ​ "not 1" ​ , ​ ​ ​ so maybe you misread the question.)

​

​

Whatever your reply is, you may want to de-indent it. ​ If you don't, I will probably de-indent mine.

​

​

JumpDiscont (talk) 01:51, 4 January 2026 (UTC)

“The player guesses ​ ​ ​ "equal to 1" ​ ​ ​ or ​ ​ ​ "not 1" ​“

“The player wins a car if and only if the player's guess is correct regarding the die value.”

Case 1. As a set, ‘equal to 1’ has 1 member, {1}.

Case 2. As a set, ‘not 1’ has 2 members, {2, 3}.

Guess case 1, win.

Guess case 2, lose.

Your definition of sets results in a win ratio of ½.

There are 3 possible guesses and only 1 is correct.~phyti Phyti (talk) 17:18, 5 January 2026 (UTC)

"There are 3 possible guesses ..."

"equal to 1" ​ , ​ ​ ​ "not 1" ​ , ​ ​ ​ and what else?

JumpDiscont (talk) 17:28, 5 January 2026 (UTC)

I am deindenting: ​ My followup is below.

JumpDiscont (talk) 00:00, 13 January 2026 (UTC)

If you remain unwilling to actually try the 52-card game - rather than asserting a result from

the chain rule without showing what calculation using the chain rule leads you to that result -

then maybe you'll instead address that for (g), there are only 2 possible guesses:

​

" ​ ​ ​ ​ ​ ​ ​ The player guesses ​ ​ ​ "equal to 1" ​ ​ ​ or ​ ​ ​ "not 1" ​ . ​ ​ ​ ​ ​ ​ ​ "

possible guess number 1: ​ "equal to 1"

possible guess number 2: ​ "not 1"

​

JumpDiscont (talk) 00:00, 13 January 2026 (UTC)

My table and description are playing the game!

The chain rule is not needed.

You can play the game on paper. Phyti (talk) 20:19, 16 January 2026 (UTC)

The contestant/player can visualize 3 possible distributions of the prize w and 2 empty containers on the left.

The probability convention of dividing 1 by the number of locations/containers on the right, is misleading, implying a possible transfer of material between locations.

This is one instance.

Quote from Marilyn Savant response to Craig Whitaker [1]

"Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance."

She transfers the probability from 2 doors to 1 door after 1 door is removed.

That would change the probability table to 1/3w in container 1 and 2/3w in container 2 n and container 3 would be removed.

If any material was transferred, that would have to be an illusion.

Quote from Marilyn Savant response to readers who disagreed [1]

"When you first choose door #1 from three, there’s a 1/3 chance that the prize is behind that one and

[a 2/3 chance that it’s behind one of the others.]

But then the host steps in and gives you a clue. If the prize is behind #2, the host shows you #3, and if the prize is behind #3, the host shows you #2. So when you switch, you win if the prize is behind #2 or #3. You win either way! But if you don’t switch, you win only if the prize is behind door #1."

She divides 1 by the 3 locations resulting in 1/3, 1/3, 1/3.

In the bracketed phrase, she assigns 2/3 to either 1 of the remaining 2 doors which contradicts the 1/3 already assigned to each. After removal of 1 door the 1 for the remaining pair should have been divided by 2. She then neglects the reality that the host can show door 2 and door 3 when the prize is behind door 1, allowing 2 wins in 2 ways to stay.

Following that paragraph she states

"Suppose we pause at that point, and a UFO settles down onto the stage. A little green woman emerges, and the host asks her to point to one of the two unopened doors. The chances that she’ll randomly choose the one with the prize are 1/2, all right." But that’s because she lacks the advantage the original contestant had—the help of the host.

Here she gives the correct answer for choosing 1 of 2 things! There was no advantage. Neither knew the location of the prize when they chose.

Where are the fact checkers? Where is quality control?

[1] Marilyn vos Savant, https://web.archive.org/web/20130121183432/http://marilynvossavant.com

~phyti Phyti (talk) 20:41, 16 January 2026 (UTC)

"[a 2/3 chance that it’s behind one of the others.]"

​

​

Presumably, this was meant as

Prob ​ ( ​ car is behind door 2 ​ or ​ car is behind door 3 ​ ) ​ ​ ​ = ​ ​ ​ 2/3

, ​ rather than as

Prob(car is behind door 2) ​ = ​ 2/3 ​ ​ ​ or ​ ​ ​ Prob(car is behind door 3) ​ = ​ 2/3

.

​

(This is like how there are two elements x of {2,3} such that

x = 2 ​ ​ ​ or ​ ​ ​ x = 3

, ​ even though

there is only one element x of {2,3} such that ​ x = 2

and

there is only one element x of {2,3} such that ​ x = 3

.)

​

​

​

For problem (f) from my 27 December comment, I imagine you wouldn't say you ​ ​ ​ neglected the reality that the coin can show heads and tails when the die shows 1, allowing 2 wins in 2 ways the events described in that problem can go ​ , ​ ​ ​ even though the ​ ​ ​ possibilities ​ - ​ this is reality ​ ​ ​ table for it would have as many rows with ​ ​ ​ die shows 1 ​ ​ ​ as it would have without ​ ​ ​ die shows 1 ​ .

​

I recall you earlier justifying ​ ​ ​ including which door the host opens in your counting ​ ​ ​ with either ​ ​ ​ when the player wins a prize ​ ​ ​ or ​ ​ ​ what wins a prize for the player ​ , ​ ​ ​ but from your much-more-recent comments, you might now think that the possibilities for an intermediate action can need to be ignored even if win-or-lose is not determined until after the intermediate action.

​

​

​

"Here she gives the correct answer for choosing 1 of 2 things!"

​

​

Although she indeed does, see my 15 June comment for why ​ choosing 1 of 2 things ​ is not sufficient to conclude that ​ 1/2 , 1/2 ​ is the correct answer.

​

Also, see the 2nd of my two 13 December comments for why ​ 1/2 , 1/2 ​ is correct for UFO-alien problem but not for the Monty Hall Problem:

For the UFO-alien problem, the ​ Prob(... given ...) ​ probabilities would be tiny rather than 1/2, but there would be no reason to assume that the alien's choice had anything to do with the player's choice or where the car is.

​

​

​

(h) ​ For

​

The host has an otherwise-symmetric disc, with one side labeled 2 and the other side labeled 3. ​ The host rolls a 3-sided die, with sides {1,2,3}. ​ If the die value is 1, then the host flips the disc, else the host places the disc so the label of its shown side is equal to the die value. ​ The player guesses one of the following 2 (and only 2) options.

option 1: ​ "Mars"

option 2: ​ "Neptune"

After the player makes the player's guess of one from among the 2 (and only 2) options, [if the player's guess is "Mars", then [the player wins if and only if the die value is 1]] and [if the player's guess is "Neptune", then [the player wins if and only if the die value is not 1]].

​

, ​ ​ ​ do you think guessing ​ "Neptune" ​ is better than guessing ​ "Mars" ​ ?

​

​

​

JumpDiscont (talk) 07:59, 17 January 2026 (UTC)

"Presumably, this was meant as

Prob ​ ( ​ car is behind door 2 ​ or ​ car is behind door 3 ​ ) ​ ​ ​ = ​ ​ ​ 2/3"

No. She already stated:

"Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance."

She doesn't know the probability depends on number of doors. Transfer of material is magic/illusion.

"For the UFO-alien problem, the ​ Prob(... given ...) ​ probabilities would be tiny rather than 1/2, but there would be no reason to assume that the alien's choice had anything to do with the player's choice or where the car is."

Both players have the same possible guesses NOW, 1 of 2 things. One door contains a car and one door contains a goat. Their guesses are independent of each other and any game history.

"do you think guessing ​ "Neptune" ​ is better than guessing ​ "Mars" ​ ?"

No. The disc orientation does not affect the die toss.

There are 2 results for die≠1, L{2, 3} and 1 for die=1, W{1}.

Win probability = W/ (W+L)=1/3, not 1/2.

This is the source of your mistake. You bet on #3 for a 6 sided die.

There are 6 possible events for a die toss {1, 2, 3, 4, 5, 6}.

There is the win set W={3}. There is the lose set L={1, 2, 4, 5, 6}.

You are comparing 2 different sets of events AS IF they are 2 equal events.

I.e., you win or lose.

The probability is (successful event)/(all possible events). W/(W+L)=1/6.

Thus 1/6 + 5/6 = 1. P(win + lose)=1. There are 2 classes of events, but they typically are not equal in magnitude.~phyti Phyti (talk) 18:36, 21 January 2026 (UTC)