The Monty Hall Problem Revisited
- Thread starter raydpratt
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I figured out from my own combinations why switching choices yields a 2/3 advantage:
In the games where the combinatorically expected choice pattern is not allowed, the games are not discarded ("The show must go on."); so, instead, such games are played pursuant to the only remaining choice pattern allowed; and, that results in two more wins per six games played, yielding an average of 4 wins per 6 games, a 2/3 advantage.
In the games where the combinatorically expected choice pattern is not allowed, the games are not discarded ("The show must go on."); so, instead, such games are played pursuant to the only remaining choice pattern allowed; and, that results in two more wins per six games played, yielding an average of 4 wins per 6 games, a 2/3 advantage.
I don't see in your summary that you are using the correct model that the man who reveals the goats behind unchosen doors has actual information about where the goats are.
So there are 9 equal possibilities of where the prize is and what the first chosen door is. Sometimes the host has a choice of two doors and sometimes the choice is forced to one remaining door, but this does not affect the way the total probability of 1 is divided between the 9 choices of prize and first chosen door.
Case I ) With probability 3/9=1/3, the chosen door is where the prize is. The host opens the door behind which is goat A or goat B, so Strategy0 (don't switch) wins in this case and Strategy B (always switch to the unchosen door) loses in this case.
Case II ) With probability 6/9=2/3, the chosen door is not where the prize is. The host opens the door behind which is goat A or goat B. The other goat is behind the chosen door. So Strategy0 (don't switch) loses in this case and Strategy B (always switch to the unchosen door) wins in this case.
So there are 9 equal possibilities of where the prize is and what the first chosen door is. Sometimes the host has a choice of two doors and sometimes the choice is forced to one remaining door, but this does not affect the way the total probability of 1 is divided between the 9 choices of prize and first chosen door.
Case I ) With probability 3/9=1/3, the chosen door is where the prize is. The host opens the door behind which is goat A or goat B, so Strategy0 (don't switch) wins in this case and Strategy B (always switch to the unchosen door) loses in this case.
Case II ) With probability 6/9=2/3, the chosen door is not where the prize is. The host opens the door behind which is goat A or goat B. The other goat is behind the chosen door. So Strategy0 (don't switch) loses in this case and Strategy B (always switch to the unchosen door) wins in this case.
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I woke up earlier than usual this morning, and a simple explanation the Monty Hall problem dawned on me:
You have a 2/3 chance that your first choice will be a goat door; and, if it is a goat door, you will win by switching after Monty Hall eliminates the other goat door.
Said another way, you only have a 1/3 chance of losing by first choosing the grand prize door, having one goat door opened and eliminated, and then switching to the other goat door.
You have a 2/3 chance that your first choice will be a goat door; and, if it is a goat door, you will win by switching after Monty Hall eliminates the other goat door.
Said another way, you only have a 1/3 chance of losing by first choosing the grand prize door, having one goat door opened and eliminated, and then switching to the other goat door.
Hellenologophobia
Registered Senior Member
Isn’t math cool? You can even apply it to choosing a wife. I switched doors and now I have a boat. 
Optimizing Your Wife
Monty Hall: Try It-Play the Game
Great Explanation
Optimizing Your Wife
Monty Hall: Try It-Play the Game
Great Explanation
Beware the "it's so simple" solution: you've thought about this for a long time. Even though the solution now appears so simple and obvious to you, it's likely that someone else who is struggling to understand it will not "get it" from your explanation.
But good for you, though. I'm glad that somebody can change their mind on this forum...
In contrast to your earlier attempts, Ray, I find this to be a wonderful way to analyze the problem.
I only became aware of this problem and the Savant connection in 2024.
What surprised me were professional people who accepted her explanation.
abstract
The game show problem aka Monty Hall problem [1], originated when
Craig Whitaker posed a question of a wining strategy for a modified 3 door game show to Marilyn Savant who wrote articles for Parade magazine.
Her 1990 response was to switch doors when given the option. [2] [3]
The debate of probability of success as 2/3 vs 1/2 has continued until today.
This paper reveals errors in her response.
the game
The logic of the game, with the player never knowing where the car is located until after their 2nd choice, precludes them from defining a basis for a strategy. It's always a random guess. Whitaker's question involved 1 car and 2 goats. Since the goats exist simultaneously, they must have separate identities g1 and g2. The host opening a door to reveal a goat only informs the player of where the car is NOT located. My analysis is more general with Whitaker's description a special case, and excludes any form of deception.
In the graphic, there are 3 distinct prizes, a car 'c' and 2 prizes 'a' and 'b' of lesser value. There are 3!=6 possible arrangements of the prizes per door.
Since all 3 doors must have the same possible arrangements of prizes, it's only necessary to simulate for 1 door, thus the player always chooses door 1. The host is restricted to only open 1 door not containing a car and not chosen by the player, thus door 2 or door 3.
If the host opened 2 doors in one game, the car location is known, the game ends, and the player is denied their option!
A game consists of 3 events:
e1. player chooses 1 of 3 doors, which is door 1 for this example.
e2. host opens 1 of 2 doors in a left to right sequence.
e3. host allows player a 2nd choice of 1 of 2 closed doors (1 or 'op').
Whitaker asks, is there an advantage to stay with door 1 or switch to 'op'.
Without any more detail, e3 determines the ratio of (win a car)/(games played) as 1/2. There are not 3 choices when choosing 1 of 2 doors in the same game!
There are always 2 closed doors remaining, 1 with a car and 1 with no car.
The probability is the same as flipping a coin for H or T, averaging 1/2.
The graphic shows all possible sequences of play.
When door 1 contains 'c', the host has 2 sequences of play vs. 1 sequence when
door 1 does not contain 'c'.
The player wins a car 4/8 games on average, with or without the option.
Failure to recognize the fact of distinct prizes resulting in 2 possible sequences of choices when the car is behind door 1 is the 1st error. Believing that abstract and misleading probabilities have an independent existence is the 2nd error.
Consider that no door contains 1/3 of a prize. Using real world 'possibilities', the values for each door would be a variation of {0, 0, 1}. 1 meaning the door contains a car, and 0 meaning the door does not contain a car. The value for the set of doors is 1 independent of number of doors. I.e. adding more doors for the host to open only makes for a prolonged and boring game.
The key factor is possible choices, not possible locations.
This is also a study comparing abstraction vs reality. The location of the car is fixed and certain for the host. The player's knowledge of its location is uncertain. Thus for the player, probability distributes its location uniformly over all doors.
reference
[1} The American Statistician, August 1975, Vol. 29, No. 3
[2] game show problem, Wikipedia Sep 2024
[3] Marilyn vos Savant, https://web.archive.org/web/20130121183432/http://marilynvossavant.com
What surprised me were professional people who accepted her explanation.
abstract
The game show problem aka Monty Hall problem [1], originated when
Craig Whitaker posed a question of a wining strategy for a modified 3 door game show to Marilyn Savant who wrote articles for Parade magazine.
Her 1990 response was to switch doors when given the option. [2] [3]
The debate of probability of success as 2/3 vs 1/2 has continued until today.
This paper reveals errors in her response.
the game
The logic of the game, with the player never knowing where the car is located until after their 2nd choice, precludes them from defining a basis for a strategy. It's always a random guess. Whitaker's question involved 1 car and 2 goats. Since the goats exist simultaneously, they must have separate identities g1 and g2. The host opening a door to reveal a goat only informs the player of where the car is NOT located. My analysis is more general with Whitaker's description a special case, and excludes any form of deception.
In the graphic, there are 3 distinct prizes, a car 'c' and 2 prizes 'a' and 'b' of lesser value. There are 3!=6 possible arrangements of the prizes per door.
Since all 3 doors must have the same possible arrangements of prizes, it's only necessary to simulate for 1 door, thus the player always chooses door 1. The host is restricted to only open 1 door not containing a car and not chosen by the player, thus door 2 or door 3.
If the host opened 2 doors in one game, the car location is known, the game ends, and the player is denied their option!
A game consists of 3 events:
e1. player chooses 1 of 3 doors, which is door 1 for this example.
e2. host opens 1 of 2 doors in a left to right sequence.
e3. host allows player a 2nd choice of 1 of 2 closed doors (1 or 'op').
Whitaker asks, is there an advantage to stay with door 1 or switch to 'op'.
Without any more detail, e3 determines the ratio of (win a car)/(games played) as 1/2. There are not 3 choices when choosing 1 of 2 doors in the same game!
There are always 2 closed doors remaining, 1 with a car and 1 with no car.
The probability is the same as flipping a coin for H or T, averaging 1/2.
The graphic shows all possible sequences of play.
When door 1 contains 'c', the host has 2 sequences of play vs. 1 sequence when
door 1 does not contain 'c'.
The player wins a car 4/8 games on average, with or without the option.
Failure to recognize the fact of distinct prizes resulting in 2 possible sequences of choices when the car is behind door 1 is the 1st error. Believing that abstract and misleading probabilities have an independent existence is the 2nd error.
Consider that no door contains 1/3 of a prize. Using real world 'possibilities', the values for each door would be a variation of {0, 0, 1}. 1 meaning the door contains a car, and 0 meaning the door does not contain a car. The value for the set of doors is 1 independent of number of doors. I.e. adding more doors for the host to open only makes for a prolonged and boring game.
The key factor is possible choices, not possible locations.
This is also a study comparing abstraction vs reality. The location of the car is fixed and certain for the host. The player's knowledge of its location is uncertain. Thus for the player, probability distributes its location uniformly over all doors.
reference
[1} The American Statistician, August 1975, Vol. 29, No. 3
[2] game show problem, Wikipedia Sep 2024
[3] Marilyn vos Savant, https://web.archive.org/web/20130121183432/http://marilynvossavant.com
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People sure make this hard.
You choose 1 of 3 doors. There's a 1/3 chance you picked the right door, 2/3 chance you picked the wrong door. Monty shows you a goat behind one of the doors you didn't pick... Thing is, you already knew that there was a goat behind one of those doors. He hasn't given you any new information.
There is still a 2/3 chance that the door you originally picked is incorrect.
You choose 1 of 3 doors. There's a 1/3 chance you picked the right door, 2/3 chance you picked the wrong door. Monty shows you a goat behind one of the doors you didn't pick... Thing is, you already knew that there was a goat behind one of those doors. He hasn't given you any new information.
There is still a 2/3 chance that the door you originally picked is incorrect.
DaveC426913
Valued Senior Member
No you didn't.
After you pick a door, but before he opens one, there is still a 1/3 possibility that the goat is behind any of the doors.
Once he opens a door, now you have new information.
It's a matter of people being sure they're right ... when they're not.
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Pinball1970
Valued Senior Member
Outlining the permutations illustrates it. I did not get it first time.
There are 2 goats, 1 car. No matter which door you pick you still know at least one of Monty's doors is a goat.
DaveC426913
Valued Senior Member
Thre graph
Anyway, your solution is going to have to explain the observed effect that says switching is the correct option:
Sorry, my mistake. You get to switch after he opens one of the doors, not before.
Anyway, your solution is going to have to explain the observed effect that says switching is the correct option:
DaveC426913
Valued Senior Member
Nobody is forgetting that. I'm not sure why you think anyone is.
Everybody who has ever tackled this riddle has a crystal clear idea of why their own logic is right, and each of those same everyones has never understood why everyone else doesn't follow their crystal clear logic.
There's an XKCD for that (sort of):
DaveC426913
Valued Senior Member
No. Imagine how that would play out:Everyone agrees you have a 1 out of 3 chance of picking the correct door... Therefore you have a 2 out of 3 chance of having picked the wrong door. So 2 out of 3 times you are better off switching. Seems fairly obvious to me.
Player: "I pick door #1!"
Player: :to self: "Ah but that is 2/3rds likely to be wrong!" I will switch to a different door."
Player: "I switch to door #2!"
Player: :to self: "Ah but that is 2/3rds likely to be wrong!" I will switch to a different door."
Player: "I switch to door #3!"
etc.
If that's not literally what you are suggesting, you'll need to lay it out step-by-step, because - with this riddle more than any other - the devil is in the details.
But here's the converse.Everyone agrees you have a 1 out of 3 chance of picking the correct door... Therefore you have a 2 out of 3 chance of having picked the wrong door. So 2 out of 3 times you are better off switching. Seems fairly obvious to me.
You walk into the room. A previous person has already chosen door 1. Door 3 has been opened and contains a goat. The previous person then leaves.
Now you have two doors to pick from. You never had three doors. Which do YOU choose?
No. I pick door one. Monty eliminates 2 or 3. (He shows you a goat - but you already knew there was a goat behind 1 of Monty's doors.) You now have the illusion of a 50/50 chance.
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