Talk:Monty Hall problem/Arguments

The Monty Hall problem explained here

https://drive.google.com/file/d/18FFyRgOYEXsrHy4IQGpEI9Qhc1mtURvt/view?usp=sharing

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

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

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

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

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

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

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

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

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

The latest version, minor revision.

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

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

for your pdf:

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

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

​

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

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

If yes, then what is a pair of consecutive problems

from that page for which you give different answers?

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

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

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

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

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

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

​

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

, ​ and assumes the show meets that definition.

or

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

​

.

​

​

For

​

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

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

​

,

​

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

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

​

.

​

​

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

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

​

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

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

​

​

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

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

​

​

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

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

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

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

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

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

Simple questions for User:Phyti

Collapsing lengthy section. --Super Goku V (talk) 03:41, 22 January 2026 (UTC)

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.

mh sards

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.

1	2	3		row		1	2	3
C	Z=not C			1		C	Z=not C
x	host opens 1 Z door except #3			2			x	host opens 1 Z door except #2
C		Z		3		C	Z
1 game						2 games

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)

If we can list all possible games, there is no need to play them many times – That only applies if you list all possible games and then analyze them correctly. But if you're not analyzing them correctly, then "playing many times" (i.e. running a simulation) would reveal that fact to you. Unfortunately, you resolutely ignore all entreaties to run the simulation, which allows you to prattle on in blithe ignorance. EEng 18:55, 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.

sessions table

~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

logic table

~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)

Challenge for Phyti

Collapsing lengthy section. --Super Goku V (talk) 03:41, 22 January 2026 (UTC)

It's been two weeks since I've suggested you try the 52 card version and give us your results. Start with 52 cards. Player picks one at random. Host reveals 50 that are not the ace of spades (selecting a random card to keep if necessary). There are 2 cards left - the player has one and the host has one. Record whether the player or the host ends up with the ace of spades. Please do this 20 times and post your results here.

Rick's results: player has the ace of spades 0 out of 20 times, host has it 20 out of 20.

No analysis or other response please. I will hide from view anything posted in response until you post your results. What are Phyti's results? -- Rick Block (talk) 19:32, 11 December 2025 (UTC)

You just make statements, but don' show any proof.

This is the simplest table I can make.

There are 52 cards in a row face down numbered 1 thru 52 on the back. The faces are blank (b) except 1 with an A, the winning card.

A can be in any one of locations 1 thru 52.

There are only 2 possible identities for any card. It is an A-card or a b-card.

The host knows the location of A.

The player randomly guesses location x for the A card.

case 1.

If location x is A, the host removes all b-cards except 1 and x, leaving (A, b).

There are 51 ways for the host to leave a different blank b.

case 2.

If location x is a b-card, the host removes all b-cards except 1, leaving (b, A)

There are 51 ways for the player to choose a b-card.

The number of possible (A, b) pairs equals the number of possible (b, A) pairs.

The player's 2nd guess is always between A and a b-card.

Since there are as many ways to guess A as b, there is no advantage to switch.

The results are the same for all locations

The identity of a card is independent of its location.

The cards remain in their original position (just as the boxes and doors in other game variations).

~phyti Phyti (talk) 18:43, 21 January 2026 (UTC)

nonresponsive analysis and other discussion

I have played it a few times on paper via the tables. The results are the same. The player has to guess which of the 2 remaining cards is the A.

Refer to Marilyn Savant's million door game where there is the player 1st guess and door 777,777 for the 2nd guess. One door remains since it contains the car, the other remains since there has to be 2 doors. All goat doors are identical so it's irrelevant which one. The success ratio is (1 correct guess)/(2 possible guesses) = 1/2.

Show us your way of playing the game that supports your result.~phyi Phyti (talk) 20:35, 16 January 2026 (UTC)

"There are 2 cards left - the player has one and the host has one."

The 2 cards, A and a blank are there for the player to guess which one is the A.

It is NOT the player vs the host. If the host tossed a coin with one face blank and the other face an A, the probability of the A would be 1/2, the same as Heads or Tails! ~2025-41684-02 (talk) 17:50, 18 December 2025 (UTC)

MH game with cards.

The 52 cards are numbered on the back 1 thru 52, and face down in a row.

The ace of spades is 'A' in position 4.

The host's action for a session/(a way to play a game) acts like a filter allowing A and 1 blank card thru for the player's 2nd guess.

Case 1.

Player 'p' 1st pick p1 is card x=4, but wins no prize.

The host 'h' exposes all cards except A and a random 0 card position1, per the rules. The host leaves a 0 card and A for the players 2nd pick p2 when they could win a prize.

Case 2.

Player 'p' 1st pick p1 is card x=51, but wins no prize.

The host 'h' exposes all cards except A and card 51, per the rules.

The host leaves a 0 card and A for the players 2nd pick p2 when they could win a prize.

Conclusion.

The player does not win a prize for picking the correct location of A.

The player can win a prize for picking the A card on the 2nd pick, when it is 1 of 2 cards.

The p2 selection is a random event since the player only knows A is within the set of 2 cards but not which one.

There are 52 possible sessions, each with a probability of ½ to win a prize.

Don't confuse p1 picking the location with p2 possible winning the prize.

MH game with cards.

~phyti

Phyti (talk) 17:56, 15 December 2025 (UTC)

still no having done what Rick Block described?

JumpDiscont (talk) 20:29, 15 December 2025 (UTC)

The many door/box/card case.

There are n cards face down in positions 1 thru n. One card is an ace A, the rest are blank.

Player guesses position x.

If x=A then the host removes n-2 blanks leaving A and 1 blank.

If x≠A then the host removes n-2 blanks leaving A and 1 blank.

The players 2nd guess is 1 of 2 cards, A and 1 blank.

The rules protect the players 1st guess and the A card.

There is NO magic.~phyti Phyti (talk) 19:46, 12 December 2025 (UTC)

You spend so much time arguing about this, surely you can spare 5 or 10 minutes to actually try it. Still waiting for your results (and, again, anything except your results will be hidden from view). -- Rick Block (talk) 02:06, 24 December 2025 (UTC)

Rick's results: player has the ace of spades 0 out of 20 times, host has it 20 out of 20."

The game does not involve who has the ace of spades. ~phyti Phyti (talk) 20:04, 2 January 2026 (UTC)

Isn't it your claim that it's a 50/50 chance who has the ace of spades? How about if we make a bet. You play the host, and I'll play the player. If I guess who has the ace of spades (by staying with my original choice of card or switching to your card), you pay me $10. If I'm wrong I'll pay you $50. Would you take this bet? If so, please take out a deck of 52 cards, pick a random card on my behalf, and reveal 50 cards that are not the ace of spades. I will choose switch every time. Do this 20 times. Would I be paying you money or would you be paying me? How much? -- Rick Block (talk) 21:28, 2 January 2026 (UTC)

"Isn't it your claim that it's a 50/50 chance who has the ace of spades?"

The player has a 50/50 chance of correctly guessing the A (ace of spades) card on their 2nd guess.

There are n-1 cards numbered 2 thru n in a set of '0' cards and 1 card labeled A, face down.

The player guesses card x.

If x=A, the host removes all but 1 0-cards in n-1 sessions, leaving a different 0-card to pair with x.

If x≠A, the host removes all but 1 0-cards in 1 session, leaving A to pair with x.

The host does not have to leave a 0-card since the player has done it, so must leave A.

This session occurs n-1 times.

The list of pairs for the players 2nd guess contains n-1 (A, 0) and n-1 (0, A).

There is no advantage to switch.

An example.

For simplification there are 4 cards in locations 1 to 4. A can be in any of the locations.

The player always guesses location 1, while A is placed in each location.

The host only reveals not-A cards labeled as 0, as indicated by '-' and leaves the A card and the remaining card r.

When the player 1st guess is the A card, the host has more choices of removal, as in the MH game. There are 6 sessions which produce the pairs for the player's 2nd guess p2.

Comparing p1 the stay prizes to r, there is no advantage to switch.

The results are independent of the number of cards.

If you are fixated on the location of the prize and ignore the player-host choices, you won't solve the problem. There is no prize for the 1st guess, only for the 2nd guess.

card table

Phyti (talk) 16:37, 5 January 2026 (UTC)

Why won't you try it 20 times (the player always switches) and tell us your results? -- Rick Block (talk) 17:18, 5 January 2026 (UTC)

There are only 6 possible results to stay and 6 to switch, and no advantage. Playing more sessions will not change the win ratio of 1/2 if all sessions are played with the same frequency.There are no experts, only some people with more experience than others. That's why I don't gamble!~phyti ~2026-14476-1 (talk) 18:47, 7 January 2026 (UTC)

I'm not asking you to try the 3 door version, but rather the 52 card version (which also results in exactly 2 choices left, one of which is the "prize"). For anyone reading along, I assume the reason Phyti is so strongly resisting doing this is because the results make is so obvious that the 2 choices are not equally probable. The actual analysis using Chain rule (probability) here is the following:

1. The probability the player picks the ace of spades AND the host reveals 50 cards that are not the ace of spades is P(player picks ace of spaces) * P(host reveals 50 cards that are not the ace given the player has picked the ace).

2. The probability the player initially picks the ace of spades is 1/52. Everyone understands this.

3. The probability the host reveals 50 cards that are not the ace is 100% (i.e. 1).

4. So the combined probability is (1/52) * 1 = 1/52.

Similarly, the probability the host has the ace is 51/52, and in this case the host also always reveals 50 cards that are not the ace - so the combined probability host has the ace AND host reveals 50 cards that are not the ace is (51/52)*1 = 51/52. If you try this (say, 20 times) you can't help but notice that with 2 cards left, the host nearly always ends up with the ace.

Similarly, in the 3 door MH problem, the probability the player's initial pick is the car is 1/3 and then the host always opens a door revealing a goat (i.e. with probability 1). So the player's chance of winning the car by staying with their initial choice is (1/3)*1 = 1/3. The probability the car is behind one of the doors that is not the player's initial pick is 2/3. In this case, the host also always opens a door revealing a goat (with probability 1), so the probability the player wins the car by switching to the non-picked door is (2/3)*1 = 2/3.

Note that this is not quite the MH problem as typically posed, in which the player is assumed to have picked door 1 and the host has opened door 3. In this event, the probability the car is behind door 1 before the host opens a door is 1/3. The probability the host opens door 3 in this case is 1/2 (the host can open either door 2 or door 3 - and, not generally stated, the host is assumed to have no preference for which door he opens in this case). So, the probability the player picks door 1 and the host opens door 3 is (1/3)*(1/2) = 1/6. If the player initially picks door 1 the probability the car is behind door 2 is 1/3 and in this case the probability the host opens door 3 is 100% (by the rules, the host has no choice but to open door 3). So, if the player has picked door 1 the probability the car is behind door 2 and the host opens door 3 is (1/3) * 1 = 1/3.

You're twice as likely to win the car if you switch.

I'm still waiting for Phyti's results (of the 52 card version). -- Rick Block (talk) 21:45, 7 January 2026 (UTC)

"I'm not asking you to try the 3 door version, but rather the 52 card version (which also results in exactly 2 choices left, one of which is the "prize"). For anyone reading along, I assume the reason Phyti is so strongly resisting doing this is because the results make is so obvious that the 2 choices are not equally probable. The actual analysis using Chain rule (probability) here is the following:"

Rules: host cannot reveal the player 1st guess, and can only reveal 0-cards.

The host knows the location of A.

If the example of 4 cards is correct per the game rules, then n-1 sessions can be extended to 52 cards.

The list of pairs for the players 2nd guess contains 51 (A, 0) and 51 (0, A).

Comparing p1 and r will show there is no advantage to switch.

Your 20 sessions would not be sufficient to prove the win ratio of 1/2.

The player makes a random 1st guess, having no knowledge of A location.

The host randomly reveals the 0-cards, having knowledge of A location.

The player makes a random 2nd guess, having no knowledge of A location.

All actions are independent, thus there is no conditional probability.

If you don’t agree with this explanation, point out the errors.~phyti Phyti (talk) 17:50, 10 January 2026 (UTC)

Your error, and you keep making it, is "All actions are independent, thus there is no conditional probability" (the host is not showing 50 randomly selected cards). Why are you so resistant to trying it 20 times?????? Like I have said, when I did it (with 52 cards) the host ended up with the ace of spades 20 out of 20 times. If this were truly a 50/50 (as you keep claiming) this would be a 1 in a million occurrence. My guess is if you try it the host will end up with the ace of spades at least 19 times. Essentially statistically impossible if it really is a 50/50 chance. Please just try it. Shouldn't take more than 10 minutes. -- Rick Block (talk) 20:14, 10 January 2026 (UTC)

I just put on my user page - https://en.wikipedia.org/wiki/User:JumpDiscont -

more things that may be errors in Phyti's explanation: ​ Whether-or-not they

are errors depends on what Phyti means at various places in that explanation.

JumpDiscont (talk) 22:03, 10 January 2026 (UTC)

1 2 3 ... 50 51 52

Cards are numbered on back and face down spread out in a single row on a table.

Random means unpredictable, and why probability is substituted for lack of knowledge.

There is no method to predict a random event.

The player always guesses card 1, while the position of A is varied from 1 to 52.

The host cannot reveal p1 or A. (Host actions determined by rules.)

This method exhausts all combinations of p1 and A.

case 1. When A is card 1, the host removes a different set of 50 blanks for 51 sessions, leaving p1 and a different blank card r.

occurs 51 times.

case 2. When A is not card 1, the host removes one set of 50 blanks in 1 session, leaving p1 and r.

occurs 51 times.

case 1. A, 0

case 2. 0, A

No advantage to switch.

The set of cards is 51 blanks and 1 A.

The removal of blanks does not alter the remaining cards.

The probability for the location of a card increases as cards are removed.

When there are only 2 cards, they can't have fractional values. It's 0 or 1. Phyti (talk) 15:43, 14 January 2026 (UTC)

If a set of 52 cards, is reduced to C a set of 2 cards, {A, 0} or {0, A} for all 52 sessions, and W is guess the A card, per the ‘chain rule’ P(W|C) =1/2.

The table in Jan 7 shows the card game for 4 cards. The same host removals are done for for 52 cards, 50 per session instead of 2.

Maybe you have a different method of removal that you have not explained. Phyti (talk) 19:54, 12 January 2026 (UTC)

Still haven't tried it yet????? I'll respond further only if you post your results of trying it. -- Rick Block (talk) 22:51, 12 January 2026 (UTC)

I can't try 'it' since I don't know what 'it' is.

Where is your evidence?~phyti Phyti (talk) 15:26, 14 January 2026 (UTC)

1. For the Selvin game, 1 box contains keys, 2 boxes are empty.

After 1 empty box is removed, there are 2 boxes.

2. For the Whitaker game, 1 door contains a car and 2 doors contain goats.

After 1 goat door is removed, there are 2 doors.

3. For the Savant million door version of the Whitaker game, 1 door contains a car and M-1 doors contain goats. After M-2 doors are removed, there are 2 doors, 1 identified as 777,777.

4. For the card version of the Whitaker game, 1 card is A (ace of spades) and 51 cards are blank. After 50 blank cards are removed, there are 2 cards.

The player can only make random guesses.

The game rules allow the host to randomly remove only the losing objects.

possible results:

1. All empty boxes are identical. One box contains keys and one box is empty.

2. All goat doors are identical. One door contains a car and one door contains a goat.

3. All goat doors are identical. One door contains a car and one door contains a goat.

4. All blank cards are identical. One card is A and one card is blank.

Since all losing objects are identical, it's irrelevant which one is paired with the winning object. There are only 2 possible 2nd guesses for each game. One wins and one loses.~phyti Phyti (talk) 15:36, 14 January 2026 (UTC)

"It" is obviously the 52 card version. I'll hide all of this discussion until you post your results. -- Rick Block (talk) 17:03, 14 January 2026 (UTC)

The article page states "It became famous as a question from reader Craig F. Whitaker's letter quoted in (and solved by) Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990". Where is her fact based proof? ~phyti — Preceding unsigned comment added by Phyti (talk • contribs) 18:16, 20 March 2026 (UTC)

Monty Hall game

Archive 1 contains some of the same names as archive 15.

I was not aware of the MH problem until late 2024.

Please do not include me in the group who prolong the debate for years.

Science and the justice system have a common principle. The results depend on evidence. All people don't accept that.

Savant sees IQ tests as measurements of a variety of mental abilities and thinks intelligence entails so many factors that "attempts to measure it are useless".

vos Savant, Marilyn (July 17, 2005). "Ask Marilyn: Are Men Smarter Than Women?". Parade. Archived from the original on October 11, 2007. Retrieved February 25, 2008.

An above average IQ does not imply 'infallibility'. The errors in her response to Craig Whitaker were so basic, a qualified fact checker would have found them immediately.~phyti ~2026-17884-66 (talk) 18:55, 22 March 2026 (UTC)

Monty Hall game

The results for door 1 apply to all doors.

x is door opened and removed from play by the host.

A.

Host question 1. Of the 3 doors {1, 2, 3}, which one contains the car?

setup 1. {c, g, g}

possible guesses:

1= correct, 2= incorrect, 3=incorrect. Success ratio=1/3

There is no prize for this guess.

Marilyn Savant gave the correct probability for question 1 as 1/3.

B.

Host question 2 or 3, but not both.

B1.

Host question 2. Of the 2 doors {1, 2}, which one contains the car?

setup 2. {c, g, x}

possible guesses:

1=win car, 2=win goat. Win car ratio=1/2

B2.

Host question 3. Of the 2 doors {1, 3}, which one contains the car?

setup 3. {c, x, g}

possible guesses:

1=win car, 3=win goat. Win car ratio=1/2

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

What if the player 1st choice was eliminated? We begin at B, with the host opening a goat door. The 1/3, 2/3 probabilities are irrelevant. The player-host actions are the same. The player guesses are random acts. The game rules predetermine the possible host actions and therefore the results for the game.

More facts.

1. Steve Selvin and Marilyn Savant used the same method of playing a game session, when the player guesses the door/box with the prize (door 1). Since the host cannot open 2 doors in one session, the host opens door 2 in half of sessions 1 and door 3 in the other half of sessions 1. That creates a bias which is used as the strategy in Craig Whitaker's inquiry to Marilyn Savant.

2. After investigating game show scandals in the 1950's, the FCC amended the regulations for broadcasting game shows via television about 1960.

The biased frequency of game play would have been classified as a rigged game, illegal if used on any game show after 1960. Of the 5 cases listed, this one would apply.

"4. 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."

The fair game frequency allows an equal opportunity for all players. With no bias, Savant has no strategy to 'beat the system', as she claimed.

reference

[1} The American Statistician, August 1975, Vol. 29, No. 3

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

[3] fcc.gov/general/broadcast-contests

As always, "truth will never be decided by an opinion poll." ~phyti ~2026-18469-04 (talk) 17:09, 24 March 2026 (UTC)

Problem statement is not complete

The italic text giving the problem statement is not complete. In order for the MH problem to have the solution "switching wins 2/3 of the time", the rules of the game must be made clear to the contestant prior to the game starts. That is, the contestant must know that the host will always open a door, no matter the outcome of the contestant's initial guess. If the host only reveals a door when the contestant was initially correct, the a switching strategy looses with 100% probability.

This is a frequent mistake of phrasing the MH problem. The "standard assumptions sorts this out, but I think it could be given already in the formulation. — Preceding unsigned comment added by 2A00:1310:202:3013:0:DDDD:1:5 (talk) 07:42, 24 March 2025 (UTC)

No. If Monty Hall opens a door only when the Contestant's first guess is correct, it's an absurd way to play the game for a TV-show. As soon as Monty Hall opens a door, everyone knows that the Contestant will STICK and win, 100% guaranteed. There will be no suspense. And the claim that Contestants DO NOT KNOW that Monty Hall's behavior, if they pick a goat door, is constrained to open ONLY the other goat door, not any choice between the doors that the Contestant did not pick, is just false. Everyone in the world who has any awareness of this game at all knows that Monty Hall did not, and by the rules of the game COULD not, ever open the door with the car (nor remove it from consideration without opening it) and ask the Contestant to pick or switch. Everyone, Contestant included, knew that if the Contestant's chosen door was the car, ONLY THEN would Monty Hall make a choice and pick ONE of the two OTHER doors (both goats) to open. My recollection is that the door eliminated by Monty Hall from consideration was ALWAYS OPENED. The idea that the door to open would be chosen at random (after the Contestant's first pick was a goat-door) and would therefore sometimes be the car, rendering any further continuation of the game meaningless, is impossible. The idea is that there is some suspense, some unknown until the CONTESTANT makes their last choice, which is ALWAYS the choice that determines the game's outcome.~2026-20619-07 (talk) 20:48, 3 April 2026 (UTC)Christopher Lawrence Simpson

I wrote a recent post at the reference desk that gives a likelihood model addressing how the different shapes of a prior can influence the analysis. The article already does a pretty good job of explaining most of this, although not quite so clearly in a Bayesian idiom. It might be helpful in this discussion (but probably isn't suitable for the article because I freely admit that it is WP:OR as written, although something like it could probably be sourced with enough digging.) Sławomir Biały (talk) 07:04, 4 April 2026 (UTC)

While I think you are actually trying to be helpful, the crux of this problem is not the execution of the math, as a computer program necessarily does flawlessly. The crux of this problem is how you set up the equation from the word problem stated. Marilyn's answer sets up the equation incorrectly, using the correct equation for the first situation incorrectly to model the second choice, a different situation which needs a different equation. No matter how correctly you do the math, using the wrong equation to begin with...solving for the wrong equation...will give you the wrong answer. The program you are using used the equation for the first situation, which is not accurate for the second situation and the question this problem asks. So, when you start with the wrong equation,the program will if course solve for whatever equation you tell it to. The math is not wrong, but the equation you are plugging in to the program isn't the correct equation for the problem. I wrote all of this last year in a different discussion and multiple people debated it, and we ended up proving Marilyn's answer incorrect. Of course, the Moderator of this page wants to keep the charade going, so that whole discussion was deleted :(. I have already published on this topic. :). Happy solving! --AI*girllll ~2026-20861-04 (talk) 19:13, 4 April 2026 (UTC)

Assuming this was the earlier post, it is an interesting thought experiment that illustrates the slipperiness of the problem very nicely, and also why the likelihood model (rather than raw probabilities about "the state of the world") is the "correct" way to understand the problem. I would put it this way: suppose that two contestants vote by secret ballot, and neither one shows their vote to the other. Monty looks at the secret ballots and opens a goat door that is different from both ballots. Each contestant, reasoning only from the information available to them individually, concludes that switching is favorable. But now suppose they reveal their ballots to one another before deciding whether to switch. If they both chose the same door, nothing relevant changes: each still regards switching as favorable. If instead they chose different doors, then they must reassess, because they now know that the two unopened candidate doors were selected symmetrically by the pair. In that case the posterior probabilities of those two doors become equal, so each is assigned probability 1/2! The weird thing that's hard to get though is that, even if they had selected different doors (but didn't share their ballots), they would each (correctly!) assess that switching is favorable, not because the gods-eye state of the world meant that the probability of the other door was 2/3, but because the posterior probability based on each contestant's information favored switching. The reason that both can be 2/3 is that the two posterior probabilities are conditioned on different information, and there is no requirement that the probabilities of switching, conditioned on different information, need to be mutually consistent or add to one.

To make matters even more confusing, if the ballots are also kept secret from Monty, and he just opens a goat door regardless of the contents of the ballots, then he is not conditioning his choice on any available information. Now, obviously if one of the contestants chose the door that Monty opened, they should switch (and the conditional probability of either of the remaining doors containing a car is also obviously 1/2). More strikingly though, even if they didn't choose the Monty door, the conditional probability remains 1/2, and there is no advantage to switching. Likewise, it doesn't matter if the contestants pool their information after the reveal in this case. Sławomir Biały (talk) 20:31, 4 April 2026 (UTC)

Yes... EXACTLY. I used a very similar example in my essay -- they cannot BOTH gain an advantage by switching.. Marilyn's answer is not symmetrical, so it CANNOT be true. It is a very clever charade, much like the 'Where's the Missing Dollar?' word problem. ( https://en.wikipedia.org/wiki/Missing_dollar_riddle ) ~2026-20861-04 (talk) 02:24, 5 April 2026 (UTC)

Marilyn Savant didn't intentionally form an illusion or charade.

Steve Selvin conclusion was the same 15 years earlier.

Both made the same naive assumption that 3 locations implied 3 sessions of game play. Staying with Savant, that meant when the player chose the car door the host could open either one of the remaining 2 goat doors, but not in the same session. Thus her session 1 was played half the time with the host opening goat 1 door and half the time with the host opening goat 2 door.

That means there are only half as many car wins if you stay. The player doesn't know this, but if all sessions are offered with the same frequency on average, anyone tracking the games would detect an advantage for the players that switch doors on the 2nd choice.

A synonym for random is unpredictable, thus you can't predict an event involving random choices. The player can only guess, and the rules predetermine the host actions.

Probability is a property of a set of events, not a property of a single event.

The interpretations of the game are still game rigging under current regulations.

The hypothetical game can be played in a private setting, but can't be broadcast to the public via tv. ~2026-21726-24 (talk) 20:53, 8 April 2026 (UTC)

If you want/have time, check out my essay here:

https://open.substack.com/pub/aigirl334/p/an-open-ltr-to-marilyn?utm_source=share&utm_medium=android&r=56m42g

Happy solving! --AI*girllll ~2026-20861-04 (talk) 02:27, 5 April 2026 (UTC)

I did read you essay. The problem is one of logic, requiring simple math.

Here is the beginning of error detection that leads to running off the road into the swamp of nonsense.

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

Monty Hall game

The question was simple concerning a game strategy. Both people understood the rules for a fair game with no deception. All the “what if’n” does not solve the problem.

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

basic probability

Probability is a measurement from analyzing the history of a large number of events for the frequency of occurrence of a specific event. It is expressed as a ratio of (number of specific events)/(all possible events) or e/u. Its value ranges from 0 (not possible) to 1 (a certainty). It is a substitute for lack of knowledge used for prediction purposes for events that involve many factors (weather) or random events. Possibility is a better factor corresponding more to reality.

the correspondence

Craig Whitaker's question [2]

Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

Marilyn Savant's 1990 first response [2]

Yes; you should switch. [The first door has a 1/3 chance of winning, but the second door has a 2/3 chance.] Here’s a good way to visualize what happened. 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?]

It isn't necessary to go beyond the 1st paragraph of her response to detect errors.

Errors are [enclosed in square brackets].

First response.

1. Whitaker states the player picks door 1, the host opens door 3, leaving 2 closed doors, 1 and 2. Then Whitaker asks his question of strategy. Possible choices equals number of closed doors. The probability e/u must = 1/2.

Keep reading 'error 1' until the light comes on!

2. The host can't open player choice or the car door until after the player 2nd choice. There are M-1 goat doors. Any one can be the 1st or last or anywhere in between for random removal. Non have any special status.

The host has opened M-2 doors leaving 2 closed doors.

The possibility and probability of a goat door containing a car is 0.

Adding M-2 doors with 0 probability to the set, then removing them is redundant, since nothing has changed. The probability e/u must = 1/2.

Keep reading 'error 2' until the light comes on!

If it helps, imagine Whitaker's letter to a common non-celebrity such as a math student.

reference

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

~phyti ~2026-21440-14 (talk) 16:24, 7 April 2026 (UTC)

• Monty looks at the secret ballots and opens a goat door that is different from both ballots – In the case where one ballot is one goat, and the other ballot is the other goat, then Monty can't do what you say he does. So everything after that is pointless. I'm moving this to the arguments page. (And no, I'm not going to debate this.)( EEng 04:29, 5 April 2026 (UTC)

Right, but that is not the point of the counterfactual. I left out that case because including it would have complicated the example without changing the epistemic point. The purpose here was not to propose an alternative game, but to show that small variations in information (including when something is known, and by whom) can completely change the analysis. The most striking features are that if the two secret-ballot contestants pool their information after the reveal, the favorability of switching for both can drop, but if they do not pool their information after the reveal, then each of them correctly assesses that switching is favorable even if they chose different doors! In other words, the posterior relevant to switching is not a fixed gods-eye property of the surviving doors; it depends on the observer’s information set. I have summarized all of the cases in the table below. (Note: in the omitted case where the two ballots are on the two goat doors, no reveal of the stipulated kind is possible. In the later Bayes calculation, that possibility is handled implicitly by assigning zero likelihood to the event $M_{3}$ in those states.)Sławomir Biały (talk) 04:51, 5 April 2026 (UTC)

The real omission in my analysis is not that, but that there is one more piece of information that hasn't been accounted for: in the secret ballots case, where the information isn't pooled after the reveal, do both contestants know that there is another secret ballot in play (or, in a generalized problem, how many secret ballots there might be). The analysis above, while correct on its face, to get to the usual 2/3 implicitly relies on them not knowing that there is a second contestant with a secret ballot! If they do know, then the analysis changes.

Let $C_{1}$ be the event that the car is behind door 1 and $M_{3}$ the event that the car is behind door 3. Suppose that contestant A chooses door 1 and contestant B chooses a door at random, which is unknown to contestant A (but A does know that B has made a definite choice, and that Monty knows both choices). Then Bayes gives, because we have to marginalize over B's choices too:

$P(C_{2}\mid M_{3})={\frac {{\frac {1}{3}}\left[{\frac {1}{3}}P(M_{3}\mid C_{2},B=1)+{\frac {1}{3}}P(M_{3}\mid C_{2},B=2)+{\frac {1}{3}}P(M_{3}\mid C_{2},B=3)\right]}{\sum _{i\in \{1,2,3\}}P(C_{i})\,\sum _{b\in \{1,2,3\}}P(M_{3}\mid C_{i},B=b)\,P(B=b)]}}$

And the priors are

${\begin{array}{c|c|c|c}{\text{Car}}&{\text{Hidden ballot }}B&P(C_{i})\,P(B)&P(M_{3}\mid C_{i},B)\\\hline C_{1}&1&{\frac {1}{9}}&{\frac {1}{2}}\\C_{1}&2&{\frac {1}{9}}&1\\C_{1}&3&{\frac {1}{9}}&0\\\hline C_{2}&1&{\frac {1}{9}}&1\\C_{2}&2&{\frac {1}{9}}&1\\C_{2}&3&{\frac {1}{9}}&0\\\hline C_{3}&1&{\frac {1}{9}}&0\\C_{3}&2&{\frac {1}{9}}&0\\C_{3}&3&{\frac {1}{9}}&0\end{array}}$

Substituting into Bayes, $P(C_{2}|M_{3})=4/7$!!! I.e., switching from $C_{1}$ to $C_{2}$ is still favorable for A, but less so if there is another secret ballot (whose existence A knows about)! The table below has all possibilities. (Note: in the generalized case of n secret ballots whose existence is known, the governing switching posterior is $2^{n}/(2^{n+1}-1)$.)Sławomir Biały (talk) 05:34, 5 April 2026 (UTC)

Secret-ballot variants and whether switching is favorable for contestant A

Monty knows ballots?	A knows B's ballot exists?	Contestants share ballots with each other after reveal?	Relevant ballot pattern	Posterior relevant to switching for A	Is switching favorable for A?	Notes
Yes	No	No	Irrelevant	A assigns probability 2/3 to the other unopened door	Yes	If A is unaware that a second ballot exists, A uses the ordinary one-contestant model.
Yes	Yes	No	Irrelevant	A assigns probability 4/7 to the other unopened door	Yes	A must marginalize over B's hidden ballot, since Monty's action depends on it even though A does not know its value.
Yes	Irrelevant	Yes	Same ballot	A assigns probability 2/3 to the other unopened door	Yes	Pooling does not change anything if both had chosen the same door.
Yes	Irrelevant	Yes	Different ballots	The two surviving candidate doors become 1/2–1/2	No	Once ballots are pooled, the surviving chosen doors are symmetric, so the switching advantage disappears.
No	Irrelevant	No	A chose the door Monty opened	The two remaining unopened doors are 1/2–1/2	No advantage between the two surviving doors	This is an odd variant, since Monty may open A's chosen door.
No	Irrelevant	No	A did not choose the door Monty opened	A's original door and the other surviving unopened door are 1/2–1/2	No	Because Monty did not condition on the ballots, the reveal creates no switching asymmetry.
No	Irrelevant	Yes	A chose the door Monty opened	The two remaining unopened doors are 1/2–1/2	No advantage between the two surviving doors	Pooling the ballots does not change the analysis.
No	Irrelevant	Yes	A did not choose the door Monty opened	A's original door and the other surviving unopened door are 1/2–1/2	No	Pooling still does not matter, since Monty's action was never conditioned on the ballots.

Arbitrary break

Conditional probability: hx is host action and py is player choice.

Game rules.

1. Host cannot open same door as player's choice until after players 2nd choice.

2. Host cannot open the door with the car until after players 2nd choice.

Door 1 2 3

Prize c g g

By rule 1

P(h1|p1)= 0

P(h2|p2)= 0

P(h3|p3)= 0

By rule 2

P(h1|p2)= 0

P(h1|p3)= 0

P(h2|p1)= 1/4

P(h3|p1)= 1/4

P(h3|p2)= 1/4

P(h2|p3)= 1/4.

Game show producers are required to offer each session with the same frequency on average.

Winning a car occurs as a result of choices, not locations.~phyti [[Special:Contributions/~2026-21440-14|

If [the car is always behind door 1] and [your idea is that the player is more likely

to choose door 1 than to choose door 2, either because 1 is the canonical element of

{1,2,3} or because the player has some knowledge that the car is always behind door 1]

, ​ then indeed one can get 1/4 each, by having the player make the player's first

choice be with distribution ​ ​ ​ ​ ​ ​ ​ P(p1) , P(p2) , P(p3) ​ = ​ 1/2 , 1/4 , 1/4 ​ ​ ​ .

​

If one stays with [the car is always behind door 1] but has the player make the player's

first choice with distribution ​ ​ ​ ​ ​ ​ ​ P(p1) , P(p2) , P(p3) ​ = ​ 1/5 , 2/5 , 2/5 ​ ​ ​ ,

then the resulting probabilities will be ​ ​ ​ 1/10 , 1/0 , 2/5 , 2/5 ​ .

​

On the other hand, if

[[where the car goes] and [the player's first choice] are independent]

and at least one of them is uniformly at random, then consider

the following 4 bets placed on what will happen in the game:

​

Alice wins her bet if and only if the player's first choice is the car door.

Bob wins his bet if and only if [the player's first choice is the car door]

and [the host opens the lower-numbered goat door].

Carol wins her bet if and only if [the player's first choice is the car door]

and [the host opens the higher-numbered goat door].

Dave wins his bet if and only if at least one of

"Bob wins his bet" ​ ​ ​ , ​ ​ ​ "Carol wins her bet" ​ ​ ​ happens.

​

JumpDiscont (talk) 23:24, 14 April 2026 (UTC)

Your variations of the game do not match my ideas, and do not alter the evidence of Marilyn Savant's interpretation of the MH game as being illegal.

~phyti ~2026-25276-90 (talk) 17:37, 25 April 2026 (UTC)

~2026-21440-14]] (talk) 16:44, 7 April 2026 (UTC)

Are you regarding

[the player is more likely to choose door 1 than to choose door 2]

as one of my variations of the game?

​

​

If [[the player is more likely to choose door 1 than to choose door 2] matches your ideas], then that seems to be the root of the disagreement:

​

Most people go with [[the door the car starts behind] and [the player's first choice] and independent and uniformly-random] or something close-enough to that, and you seem to take the position that the show

The host flips two fair coins. ​ If they show different sides, then the host flips a third fair coin, else the host doesn't flip the third coin.

would be illegal, because that doesn't maintain equal probability on average for all 6 games.

​

​

If [[the player is more likely to choose door 1 than to choose door 2] does not match your ideas], then I imagine you won't say what you think ​ Alice's , Bob's , Carol's , Dave's ​ win probabilities are, because you recognize that you can't give probabilities which are simultaneously consistent with [[each other] and [what you've been saying so far] and [the probability of the player choosing door 1 is at most the probability of the player choosing door 2]].

​

​

JumpDiscont (talk) 01:38, 26 April 2026 (UTC)

"If [[the player is more likely to choose door 1 than to choose door 2] matches your ideas], then that seems to be the root of the disagreement:"

It does not.

It's the game show producers who are responsible to offer each of the 4 sessions with equal frequency on average.

The player always has a choice of 3 doors.

After A,B,C,D play 1 of the 4 possible sessions games, a 5th player must repeat one of the 4.

You seem to confuse an individual player session with the total set of sessions.~phyti ~2026-26434-72 (talk) 17:17, 30 April 2026 (UTC)

Do you think the game show producers must know ahead of time or otherwise predict - with accuracy

greater than 40% - which of the 3 doors the player will choose? ​ ​ ​ ​ ​ ​ ​ If ​ [no and [the player makes the

player's first choice uniformly at random]] ​ , ​ ​ ​ then what frequency on average can the producers

​

(a) ​ put the car behind door 1

(b) ​ put the car behind door 2

(c) ​ put the car behind door 3

​

to get each of the 4 sessions with equal frequency on average without the producers

[knowing ahead of time or otherwise predicting - with accuracy

greater than 40% - which of the 3 doors the player will choose] ​ ? — Preceding unsigned comment added by JumpDiscont (talk • contribs) 16:03, 1 May 2026 (UTC)

No. The producers cannot predict which door the player will choose. That's why it's a game of chance.

If ​ [no and [the player makes the

player's first choice uniformly at random]] ​ , ​ ​ ​ then what frequency on average can the producers

(a) ​ put the car behind door 1

(b) ​ put the car behind door 2

(c) ​ put the car behind door 3

to get each of the 4 sessions with equal frequency on average without the producers

[knowing ahead of time or otherwise predicting - with accuracy

greater than 40% - which of the 3 doors the player will choose] ​ ?

They can setup each session at random, which will be 1/4 of the time on average.

Placing the car and player choosing a door are independent events.

d: c g g

s 1 2 3  p r

1 p h r  c g

2 p r h  c g

3 r p h  g c

4 r h p  g c

Prize distribution d is c g g, and s is session #.

Player p chooses door, host h opens a door, r is remaining closed door.

The player actions are random.

The host knows the car location allowing randomly opening a goat if they have a choice.

Column 6 is stay prize, column 7 is switch prize.

It clearly shows there is no advantage to switch.

After simulating the sessions for door 1, instead of moving the prizes we move the doors by shifting the numbers, 1 2 3 to 2 3 1 to 3 1 2. Which door contains the car is irrelevant.

Both Steve Selvin and Marilyn Savant assumed there are 3 sessions because there are 3 doors. Their solution of how to open doors 2 and 3 separately was to play their 1st session as half sessions. That is totally unnecessary if there are 4 sessions as shown above. They did not manipulate the frequencies for deception, just lacked experience.

The solution depends on host choices, which depend on game rules, NOT number of doors.

~phyti ~2026-26701-11 (talk) 17:33, 2 May 2026 (UTC)

"The producers cannot predict which door the player will choose", and which of

the 4 possible sessions games happens depends on which door the player chooses -

​

If the car is behind door 1 and the player chooses door 1, then the

session game will be one of ​ ​ ​ session 1 ​ , ​ session 2 ​ ​ ​ from your table.

If the car is behind door 1 and the player chooses door 2 or door 3, then the

session game can't be either of ​ ​ ​ session 1 ​ , ​ session 2 ​ ​ ​ from your table.

​

- so how can the producers ​ "setup each session at random" ​ in your sense?

For example, with what frequency on average would the producers put the car behind door 1?

​

​

"After simulating the sessions for door 1,"

​

You've made tables and asserted that their rows must be

equally likely, but: ​ Have you actually simulated this?

​

(You previously kept declining to simulate, despite

strong pressure from I-forget-the-person's-username.)

​

​

JumpDiscont (talk) 19:44, 2 May 2026 (UTC)

“If the car is behind door 1 and the player chooses door 1, then the

session game will be one of ​ ​ ​ session 1 ​ , ​ session 2 ​ ​ ​ from your table.

If the car is behind door 1 and the player chooses door 2 or door 3, then the

session game can't be either of ​ ​ ​ session 1 ​ , ​ session 2 from your table.”

You are correct!

“- so how can the producers ​ "setup each session at random" ​ in your sense?”

One method, using a computer or mentally pick any whole number n.

Divide n by 4. The remainder can only be (0, 1, 2, 3). Assign those numbers to each session.

“You've made tables and asserted that their rows must be

equally likely, but: ​ Have you actually simulated this?”

Selvin didn’t have a computer in 1975. Savant didn’t use one either. They merely listed what they thought were all possible sessions, then compared wins vs losses.

They both made the same error, concluding there were only 3 sessions based on 3 doors, then adjusting the frequencies to make it work.

All the Savant followers think a computer is somehow infallible. The reality is computers only follow instructions. If you write a program matching her explanation, it will reproduce her erroneous results. It does not prove her explanation is correct, only you can program correctly.

Listing all possible sessions then playing all possible player-host choices is a simulation.

Both you and Rick started in 2009, why haven’t you solved it by now?~phyti ~2026-27056-04 (talk) 17:30, 4 May 2026 (UTC)

That is a fundamental misunderstanding of how computer programs work and what simulations are. You don't even need a computer, you can 'simulate' yourself with a friend and three slips of paper. MrOllie (talk) 17:33, 4 May 2026 (UTC)

​

"One method, using ... to each session."

​

You've given a way for them to choose a session at random,

but not a way for them to setup a session at random.

For example, if the remainder is 3 and the assignment is

0 ​ <-> ​ p h r ​ ​ ​ 1 ​ <-> ​ p r h ​ ​ ​ 2 ​ <-> ​ r p h ​ ​ ​ 3 ​ <-> ​ r h p

, ​ then where do they put the car?

​

​

​

"Listing all possible sessions then playing all possible

player-host choices is a simulation. ​ Both you and Rick

started in 2009, why haven’t you solved it by now?"

​

​

The reasons why we haven't solved your misunderstanding by now are

​

you using words in your own way:

[you say what you did is a "simulation"] ​ and

[you at least were - though maybe you no longer are -

regarding unequal session frequencies as meaning the game is "rigged",

even when them being unequal follows from the rules of the game]

​

and

​

you quite-frequently not answering my questions:

This leads to a lot of time being wasted on parts where [we already agree]

or [our disagreement comes from something more fundamental],

rather than letting me zoom in on where our disagreement starts.

​

.

​

​

​

JumpDiscont (talk) 23:38, 4 May 2026 (UTC)

None of the Savant followers want to admit she was wrong since that would mean they are wrong. The evidence has been presented revealing the errors.

Unless you have contrary evidence, you lose.

Savant presented a table with no computer simulation.

A fair game offers equal opportunity for ALL players. There can't be any bias or favoritism. Her game does not qualify.

If you don't agree, contact the FCC.~phyti ~2026-27440-05 (talk) 16:37, 5 May 2026 (UTC)

You still haven't answered regarding the frequencies of where

the producers put the car, so I am going with, they choose

[where they put it] uniformly at random from among the 3 doors.

​

Under this assumption, my user page now has the evidence.

​

Do you suggest the producers use some other

way of determining where they put the car?

(eg, maybe something involving which session they picked)

​

​

A 2/3 chance of winning for ALL players,

is equal opportunity for ALL players.

​

If you're referring to players who don't know they should switch,

then neither Wheel of Fortune nor Jeopary are fair, because

people know different numbers of phrases and trivia answers.

\begin{sarcasm} ​ I'll contact the FCC right away

to report those two shows. ​ \end{sarcasm}

​

If you mean a _single_-player game being fair/unbiased

requires that win probability equals lose probability, then:

(a) ​ Is that somewhere in the statute?

(b) ​ Have you ever played Minesweeper, or any solitaire card game?

​

​

JumpDiscont (talk) 20:46, 5 May 2026 (UTC)

"You still haven't answered regarding the frequencies of where

the producers put the car, so I am going with, they choose

[where they put it] uniformly at random from among the 3 doors."

[Correct. Any routine could be detected by someone tracking the history of the game.]

Under this assumption, my user page now has the evidence.

"Do you suggest the producers use some other

way of determining where they put the car?"

[All aspects of a 'game of chance' require randomness.

ran·dom [rándəm]

adj

1. without a pattern: done, chosen, or occurring without a specific pattern, plan, or connection

2. lacking regularity: with a pattern or in sizes that are not uniform or regular ]

"A 2/3 chance of winning for ALL players,

is equal opportunity for ALL players."

[Only if all players switch!

The same opportunity means regardless of their choice. Just as it is with a coin toss. It's 1/2 for heads or tails.]

"If you're referring to players who don't know they should switch,

then neither Wheel of Fortune nor Jeopary are fair, because

people know different numbers of phrases and trivia answers.

\begin{sarcasm} ​ I'll contact the FCC right away

to report those two shows. ​ \end{sarcasm}"

[Jeopardy requires screening to qualify as a contestant. It's a true contest of personal mental libraries of historical facts.

Wheel of Fortune allows anyone to play. It's also a true contest of personal familiarity with language.

The first is more difficult with a wide range of subjects.]

"If you mean a _single_-player game being fair/unbiased

requires that win probability equals lose probability, then:

(a) ​ Is that somewhere in the statute?"

[fcc.gov/general/broadcast-contests

"4. 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."]

[There are federal and state regulations for all forms of gambling, including lotteries.]

"(b) ​ Have you ever played Minesweeper, or any solitaire card game?"

[Yes, and a few others.]~phyti ~2026-27738-36 (talk) 16:56, 7 May 2026 (UTC)

This is for your user page.

errors

1. Prob(C=1) ​ = ​ 1/3

[There is no prize for the 1st choice, so 1/3, 2/3 probability is irrelevant.

There is a prize for the 2nd choice when there are 2 closed doors.]

2. Placing the car and player choosing a door are independent events,

Prob(C=1 and P=1) ​ = ​ Prob(C=1) * Prob(P=1) ​ = ​ 1/3 * 1/3 ​ = ​ 1/9

[Therefore there is no connection. If c is in door 1, p can choose any of the 3 doors.

P(p1|c1)=P(p2|c1)=P(p3|c1)=1/3.

3. Similarly, ​ ​ ​ Prob(C=i and P=j and H=k) ​ = ​ 1/9 ​ ​ ​ for each choice of

(i,j,k) from ​ ​ ​ ​ ​ ​ ​ { (1,3,2) , (2,1,3) , (2,3,1) , (3,1,2) , (3,2,1) }

[Where is (1, 2, 3) ?

When does p choose the car door?]

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

The MH game is a dynamic process. The location of prizes is fixed for the entire game. The player can only claim a door for their 2nd choice and the host can only open a goat door until the end of game. Their actions change the state of the game.

d 1 2 3

1 0 1 1

2 0 0 1

3 0 1 0

This is a truth table for the Whitaker game, determining all possible host choices given a player choice. Door ID is 1, 2, 3.

Possible is 1, not possible is 0.

p choices are column 1, h choices are row 1.

Prize distribution is c g g. Applying rule 1 eliminates the diagonal elements (1, 1), (2, 2), (3, 3). Applying rule 2 eliminates host column 1. That leaves 4 possible player-host actions.

The player never knows the car location so their choice is always random.

The host only has choices when the player chooses door 1. Since both reveal goats, they can be considered random. Thus the game rules predetermine the outcomes of each session.

d 1 2 3

1 p h r

2 p r h

3 r p h

4 r h p

The truth table in terms of p-h sessions and r the remaining closed door, for player 1st choice.

Column 1 would be the stay prizes for the 2nd choice. P win c ratio= 2/4.

The r's would be the switch prizes with a win c ratio =2/4.

For a game of chance played with random choices, there are no clues to form a basis for a strategy.~phyti ~2026-27738-36 (talk) 17:05, 7 May 2026 (UTC)

This is the Selvin-Savant version with 3 doors and show how they got their resilts.

d c g g

s 1 2 3 f p r

1 p h r 5 c g restrict 5 sessions

1 p r h 5 c g restrict 5 sessions

2 r p h 10 g c

3 r h p 10 g c

win c 10

win g 20

Prize distribution d is c g g, and s is session #.

Each session is same 3 events:

(p player chooses door, h host opens goat door, r is remaining closed door).

Player p chooses door, host h opens a door, r is remaining closed door.

The player actions are random.

The host knows the car location allowing randomly opening a goat door if they have a choice.

Both Steve Selvin and Marilyn Savant assumed there are 3 sessions because there are 3 doors. Their solution of how the host opens doors 2 and 3 separately was to play their 1st session as half sessions, yet sessions 2 and 3 allow the same host choices.

(Host cannot open player choice nor the car door.)

The example shows each session played with a frequency of 10.

The win c ratio is 10/30 due to the illegal restrictions.

The restrictions are totally unnecessary and their logic is inconsistent. They were not aware of the regulations for tv game shows or they were never considered.

The solution depends on host choices, which depend on game rules, NOT number of doors.~phyti ~2026-27738-36 (talk) 17:11, 7 May 2026 (UTC)

​

"Where is (1, 2, 3) ?"

​

It's from the start of the equality-sequence just

above that sentence: ​ "Prob(C=1 and P=2 and H=3)"

​

​

​

"When does p choose the car door?"

​

​

If you mean the player's first choice:​

That is exactly when ​ ​ ​ C = P ​ .

​

If you mean the player's second choice:​

That is exactly when [[that choice is stay] has the same

truth value as [C=P]]. ​ (i.e., both true or both false)

​

​

​

"Therefore there is no connection."

​

​

If you're referring to something other

than ​ ​ ​ "between C and P" ​ , ​ ​ ​ then what?

If you are referring to ​ ​ ​ "between C and P" ​ , ​ ​ ​ then:​

​

Is this just another way of saying they

are independent? ​ ​ ​ ​ ​ ​ ​ If no, then what else

do you mean by ​ ​ ​ "there is no connection" ​ ?

​

​

​

Given [the player's first choice is uniformly

random and independent of where the car is],

Prob(C=1) ​ = ​ 1/3 ​ ​ ​ indeed is irrelevant, but

when I typed that you hadn't answered regarding the

frequencies of where the producers put the car:​

​

If I had instead gone with

"The producers always put the car behind door 1."

, ​ then you might've said that is

why my reasoning doesn't apply to MH.

("All aspects of a 'game of chance' require randomness.")

​

​

Anyway, below what was already there,

my user page now has the derivation for when

the producers always put the car behind door 1.

​

​

​

You say ​ ​ ​ "The MH game is a dynamic process." ​ ,

but when you gave that in an explanation before and

I asked which other games qualified as such, you

switched phrases instead of answering, and then still

didn't answer regarding the phrase you switched to.

​

Will your use of ​ "dynamic process"

remain functionally equivalent to

"game where the session probabilities are all equal" ​ ,

or are you now willing to reveal, for some other

games I ask about, whether-or-not you regard them as

"dynamic process" ​ es ​ ​ ​ in the relevant sense?

​

​

(The rest of [your reasoning below the 42 hyphens]

also applies to games where you've recognized that

ratio of rows ​ and ​ ratio of how often things happen

are different.)

​

​

​

JumpDiscont (talk) 20:59, 7 May 2026 (UTC)

You are asking the same questions after my posting the same answers.

I don't think the MH game is that complicated. The game is intended to be played by any average person, and without any particular field of knowledge. They simply guess which of 3 doors has a car as a prize. The host then opens a car-less door, then offers the player a 2nd guess of 1 of 2 doors, in case they want to change their 1st guess. A child could play the game.

Since it's a game of chance, there is no strategy for winning more than 1/2 games on average. If there was a strategy of switching on the 2nd choice, potential players would quickly notice and use it. The sponsor would object to the cost of awarding more cars than 1/2 of games played. The viewing audience would get bored, the show would be cancelled.

I can't help you understand.~phyti Phyti (talk) 19:19, 9 May 2026 (UTC)

It is important to note that despite Monty Hall being the namesake of the problem, it is not a description of what happened on this show and was not played for a viewing audience - (not until being featured on Mythbusters decades later, anyway). The version on 'Let's Make a Deal' used different rules. MrOllie (talk) 19:27, 9 May 2026 (UTC)

@ Phyti

​

If you give something you say is an example of

me "asking the same question" after

you "posting the same answer"

- what my question was, and

your post that answered it -

then I will look at your claimed example.

​

If ​ "The host then opens a car-less door"

was enough detail on that part

(so that, when combined with the rest,

Prob(stay wins) ​ and ​ Prob(switch wins)

would be determined)

, ​ then the problem certainly would be simpler.

However, that's not enough detail:

For example, whether-or-not the host

can open the first guess matters.

​

In view of your comment's first sentence,

I am refraining from asking a question here,

and instead pointing out that

_either_ ​ "it's a game of chance" ​ is

just equivalent to your position, _or_

"there is no strategy for winning

more than 1/2 games on average"

does not follow from

"it's a game of chance" ​ .

​

Without me understanding how you sort-of-define

these kinds of phrases (eg, "a game of chance"),

it indeed will be difficult for me to pin

down exactly where you are going wrong.

I suspect the reason you can't help me

understand is, because you think the

phrases on their own are unambiguous,

you don't actually have anything close to

definitions for them, and so you end up using

one meaning for concluding that MH satisfies it,

and another for further inferences regarding MH.

​

JumpDiscont (talk) 22:38, 9 May 2026 (UTC)

phyti is awesome :) . -AI*

girllll AI*girllll (talk) 07:31, 9 May 2026 (UTC)

Thanks.

It's nice to know everyone out there is not thinking the same thing.

Otherwise this would not be worth the effort.~phyti Phyti (talk) 19:17, 9 May 2026 (UTC)

​