What variable am I missing in order to partition the Monty Hall problem into its fundamental outcomes?

Question

Here are the basic outcomes I have come up with to describe the Monty Hall problem...

Player selects Door A, Monty opens Door A, Player stays, Car is behind Door A = Impossible
Player selects Door A, Monty opens Door A, Player stays, Car is behind Door B = Impossible
Player selects Door A, Monty opens Door A, Player stays, Car is behind Door C = Impossible
Player selects Door A, Monty opens Door A, Player switches, Car is behind Door A = Impossible
Player selects Door A, Monty opens Door A, Player switches, Car is behind Door B = Impossible
Player selects Door A, Monty opens Door A, Player switches, Car is behind Door C = Impossible
Player selects Door A, Monty opens Door B, Player stays, Car is behind Door A = Player wins
Player selects Door A, Monty opens Door B, Player stays, Car is behind Door B = Impossible
Player selects Door A, Monty opens Door B, Player stays, Car is behind Door C = Player loses
Player selects Door A, Monty opens Door B, Player switches, Car is behind Door A = Player loses
Player selects Door A, Monty opens Door B, Player switches, Car is behind Door B = Impossible
Player selects Door A, Monty opens Door B, Player switches, Car is behind Door C = Player wins
Player selects Door A, Monty opens Door C, Player stays, Car is behind Door A = Player wins
Player selects Door A, Monty opens Door C, Player stays, Car is behind Door B = Player loses
Player selects Door A, Monty opens Door C, Player stays, Car is behind Door C = Impossible
Player selects Door A, Monty opens Door C, Player switches, Car is behind Door A = Player loses
Player selects Door A, Monty opens Door C, Player switches, Car is behind Door B = Player wins
Player selects Door A, Monty opens Door C, Player switches, Car is behind Door C = Impossible
Player selects Door B, Monty opens Door A, Player stays, Car is behind Door A = Impossible
Player selects Door B, Monty opens Door A, Player stays, Car is behind Door B = Player wins
Player selects Door B, Monty opens Door A, Player stays, Car is behind Door C = Player loses
Player selects Door B, Monty opens Door A, Player switches, Car is behind Door A = Impossible
Player selects Door B, Monty opens Door A, Player switches, Car is behind Door B = Player loses
Player selects Door B, Monty opens Door A, Player switches, Car is behind Door C = Player wins
Player selects Door B, Monty opens Door B, Player stays, Car is behind Door A = Impossible
Player selects Door B, Monty opens Door B, Player stays, Car is behind Door B = Impossible
Player selects Door B, Monty opens Door B, Player stays, Car is behind Door C = Impossible
Player selects Door B, Monty opens Door B, Player switches, Car is behind Door A = Impossible
Player selects Door B, Monty opens Door B, Player switches, Car is behind Door B = Impossible
Player selects Door B, Monty opens Door B, Player switches, Car is behind Door C = Impossible
Player selects Door B, Monty opens Door C, Player stays, Car is behind Door A = Player loses
Player selects Door B, Monty opens Door C, Player stays, Car is behind Door B = Player wins
Player selects Door B, Monty opens Door C, Player stays, Car is behind Door C = Impossible
Player selects Door B, Monty opens Door C, Player switches, Car is behind Door A = Player wins
Player selects Door B, Monty opens Door C, Player switches, Car is behind Door B = Player loses
Player selects Door B, Monty opens Door C, Player switches, Car is behind Door C = Impossible
Player selects Door C, Monty opens Door A, Player stays, Car is behind Door A = Impossible
Player selects Door C, Monty opens Door A, Player stays, Car is behind Door B = Player loses
Player selects Door C, Monty opens Door A, Player stays, Car is behind Door C = Player wins
Player selects Door C, Monty opens Door A, Player switches, Car is behind Door A = Impossible
Player selects Door C, Monty opens Door A, Player switches, Car is behind Door B = Player wins
Player selects Door C, Monty opens Door A, Player switches, Car is behind Door C = Player loses
Player selects Door C, Monty opens Door B, Player stays, Car is behind Door A = Player loses
Player selects Door C, Monty opens Door B, Player stays, Car is behind Door B = Impossible
Player selects Door C, Monty opens Door B, Player stays, Car is behind Door C = Player wins
Player selects Door C, Monty opens Door B, Player switches, Car is behind Door A = Player wins
Player selects Door C, Monty opens Door B, Player switches, Car is behind Door B = Impossible
Player selects Door C, Monty opens Door B, Player switches, Car is behind Door C = Player loses
Player selects Door C, Monty opens Door C, Player stays, Car is behind Door A = Impossible
Player selects Door C, Monty opens Door C, Player stays, Car is behind Door B = Impossible
Player selects Door C, Monty opens Door C, Player stays, Car is behind Door C = Impossible
Player selects Door C, Monty opens Door C, Player switches, Car is behind Door A = Impossible
Player selects Door C, Monty opens Door C, Player switches, Car is behind Door B = Impossible
Player selects Door C, Monty opens Door C, Player switches, Car is behind Door C = Impossible

If this was a thorough partition into the game's most fundamental outcomes, with each outcome being either equally likely or impossible, then there should be twice as many wins as losses obtained by switching. However, I currently still have a $1:1$ ratio of winners to losers in my partition. What missing variable do I need to include in order to see the correct solution to the game using this approach?

Answer 1

Your model is one where Monty opens a door at random and we observe that by luck the door monty showed was neither the door the player picked (although it could have been) nor the door with the car (although it could have been). Given that what is the probability the car is behind the players door.

The answer is easy to calculate is $\frac 12$. You did nothing wrong. But that is not the Monty Hall problem.

.......

In the Monty Hall problem, Monty does not open the doors equally by probability. Monty is determined that he will and he must open a door that neither the player picked nor the door with the car.

That model would go like this:

Car behind door A; Player picks door A; Monty has choice- weight 1; Monty shows B; player stays. Win.
Car behind door A; Player picks door A; Monty has choice- weight 1; Monty shows C; player stays. Win.
Car behind door A; Player picks door A; Monty has choice- weight 1; Monty shows B; player switches. loose.
Car behind door A; Player picks door A; Monty has choice- weight 1; Monty shows C; player switches. loss.

Car behind door A; Player picks door B; Monty has no choice- weight 2; Monty shows C; player stays. Loose.
Car behind door A; Player picks door B; Monty has no choice- weight 2; Monty shows C; player stays. Loose. Car behind door A; Player picks door B; Monty has no choice- weight 2; Monty shows C; player switches. Win.
Car behind door A; Player picks door B; Monty has no choice- weight 2; Monty shows C; player switches. Win.

(Note these are twice as likely to occur as monty has no choice. So they are weighted twice and listed twice)

Car behind door A; Player picks door C; Monty has no choice- weight 2; Monty shows B; player stays. Loose.
Car behind door A; Player picks door C; Monty has no choice- weight 2; Monty shows B; player stays. Loose. Car behind door A; Player picks door C; Monty has no choice- weight 2; Monty shows B; player switches. Win.
Car behind door A; Player picks door C; Monty has no choice- weight 2; Monty shows B; player switches. Win.

Car behind door B; Player picks door A; Monty has no choice- weight 2; Monty shows C; player stays. lose.
Car behind door B; Player picks door A; Monty has no choice- weight 2; Monty shows C; player stays. loose.
Car behind door B; Player picks door A; Monty has no choice- weight 2; Monty shows C; player switches. Win.
Car behind door B; Player picks door A; Monty has no choice- weight 2; Monty shows C; player switches. Win.

Car behind door B; Player picks door B; Monty has choice- weight 1; Monty shows A; player stays. Win.
Car behind door B; Player picks door B; Monty has choice- weight 1; Monty shows C; player stays. Win.
Car behind door B; Player picks door B; Monty has choice- weight 1; Monty shows A; player switches. loose.
Car behind door B; Player picks door B; Monty has choice- weight 1; Monty shows C; player switches. loose.

Car behind door B; Player picks door C; Monty has no choice- weight 2; Monty shows A; player stays. lose.
Car behind door B; Player picks door C; Monty has no choice- weight 2; Monty shows A; player stays. loose.
Car behind door B; Player picks door C; Monty has no choice- weight 2; Monty shows A; player switches. Win.
Car behind door B; Player picks door C; Monty has no choice- weight 2; Monty shows A; player switches. Win.

Car behind door C; Player picks door A; Monty has no choice- weight 2; Monty shows B; player stays. lose.
Car behind door C; Player picks door A; Monty has no choice- weight 2; Monty shows B; player stays. loose.
Car behind door C; Player picks door A; Monty has no choice- weight 2; Monty shows B; player switches. Win.
Car behind door C; Player picks door A; Monty has no choice- weight 2; Monty shows B; player switches. Win.

Car behind door C; Player picks door B; Monty has no choice- weight 2; Monty shows A; player stays. lose.
Car behind door C; Player picks door B; Monty has no choice- weight 2; Monty shows A; player stays. loose.
Car behind door C; Player picks door B; Monty has no choice- weight 2; Monty shows A; player switches. Win.
Car behind door C; Player picks door B; Monty has no choice- weight 2; Monty shows A; player switches. Win.

Car behind door C; Player picks door C; Monty has choice- weight 1; Monty shows A; player stays. Win.
Car behind door C; Player picks door C; Monty has choice- weight 1; Monty shows B; player stays. Win.
Car behind door C; Player picks door C; Monty has choice- weight 1; Monty shows A; player switches. Loose.
Car behind door C; Player picks door C; Monty has choice- weight 1; Monty shows B; player switches. Loose.

Notice When Staying the player wins 6 out of 18 and loses 12 out of 18. When player switches they win 12 out of 18 and loses 6 out of 18.

Although that is not a very efficient method of proof.

Answer 2

Obviously your list is not a partition into equally likely outcomes of the game, because you have several rows labeled "impossible" which have probability zero of occurring.

So in order to have any hope of showing a partition into equally likely outcomes, you must at least delete all lines that say "impossible".

The question then is whether there is any reason to believe that the remaining "possible" rows are equally likely.

If you count your rows carefully, you should find the following:

• Exactly four "possible" rows in which the player chooses door A and the car is behind door A.
• Exactly two "possible" rows in which the player chooses door A and the car is behind door B.

If every "possible" row were equally likely as any other, this would imply that when the player chooses door A, the car is twice as likely to be behind door A as to be behind door B.

But the placement of the car is independent of the player's choice, so it should be equally likely to be behind either door A or door B after the player chooses door A.

The way to reconcile these facts is to recognize that if the player chooses door A and the car is behind door A, Monty has a choice whether to open door B or door C. In the case where the player chooses A and the car is behind door B, Monty has no choice; he must open door C. The two cases (car behind A or car behind B) have the same probability, but the first case generates twice as many lines of the table due to Monty's increased number of choices. These are just additional lines, not additional probability.

One way to model the game that actually does lead to a list of equally likely events is to assume that in the first case (player chooses A, car is behind A), Monty flips a coin to decide which door to open. In the other case Monty still flips a coin but has to open door C no matter which way the coin comes up. You can add a column for Monty's coin flip, either heads or tails for each possible combination of player door choice and car position; then when Monty has a choice you assign a different door to each coin result, and when Monty has no choice you assign the same door to each coin result.

That's all the randomness you should need to account for. The "Monty opens Door" column is fully determined by the player's first choice, the car position, and the coin toss. You could assume the player also tosses a coin to decide whether to stay or to switch, but all that does is have you work the problem twice, once to decide whether "stay" was to the player's advantage and once to decide whether "switch" was to the player's advantage. Just work out the probabilities for "stay" and whatever they are, "switch" is the opposite.

That's $18$ equally likely outcomes instead of the $54$ unequally likely outcomes you have listed in the question.