The problem that both sides of the spectrum are having is consolidating two different factors into one problem.
Thus we have people on both sides using two completely different processes to explain something where the problem lies in the ability of the problem itself to align fractional mathematics and boolean logic.
To illustrate that both sets of people are both wrong and right in different ways:
The explanation of the issue is that the contestant is more likely to have chosen something “other than a car” originally. This is true. However this statement concludes that the real loss is in the initial choice. This is true.
What is lost in translation however is that you were never under any circumstances going to be allowed to pick “Monty’s” door. The problem itself has to follow its own rules which state that “Monty” is entitled to a door which is never going to be a door that you pick.
Mathematicians are using a choice which exists only under the assumption of the contestant assuming he can pick each door.
The contestant can pick (1) goat door - it does not matter which, because “Monty” is already assigned to the other (1) variable.
The contestant has thus the choice of “goat” or “car” because “Monty” will always be a placeholder for one of those goat doors.
You can never choose “Monty’s Door” and therein lies the problem with the fraction being represented as n/3 wherein n = 1 or 2.
To label this:
Monty’s Door = can only be assigned to him
Initial Choice = contestant choice
Switch Choice = contestant choice
How many choices do we really have available? It appears to be 3 if we include “Monty’s Door” however that is a constant not a variable. Monty will always own (1) goat, with the door being the “placeholder” for that goat. The goat he owns is independent of the door. He will always have a goat behind his door, rendering it impossible for you to “choose” that particular goat.
The problem here is simple:
With (3) doors and (1) prize you will always have a 1/3 chance of car and 2/3 chance of a goat.
Monty will always have a goat door so that door is not available for choice. Monty existing and owning a door in this entire problem has changed the denominator. The denominator has the illusion of being 3 so mathematicians are using the right operations and operands but the wrong denomination, due to the assumption that “Monty’s Door” was ever to be considered in the pool of options.
You only ever had two choices:
Stay or Switch
One will always be a goat and one will always be a car.
From those two axioms, the conclusion should be evident.
The choice was always:
Car on initial
Car on switch
Monty’s door always holds one goat. Its a choice you can never make.
Further:
The mental confusion arises when assigning fixed identifiers to rotating objects. There is never a door 1, 2, or 3 or A, B, or C. We assigned them identifiers to fix their positions, however “Monty” will never be assigned to A, B, or C because he has to go where a goat goes, not with a specific door.
Using fixed identifiers changes the way we view the problem. A variable cannot be a constant and a constant cannot be a variable. The two are boolean opposites.
The problem lies in information transference, not in mathematics. In a scenario without Monty, you have n/3, however if Monty is assigned to one door fulltime, you can never pick his door, regardless of the identifier (A,B,C) which means your probability can only be expressed in n/2 because his door will never be an option in Stage 2.
The very construct of this problem points to how things change in each stage. The problem both groups are having is not realizing that the conditions of Monty being there have impacted the original idea of (3) choices. Without Monty you have (3) choices. At the event in which he exists you only have (2) choices. You cannot choose his goat door even though you cannot presently put a fixed identifier on it.
It makes no difference whether (1) goat is in China and the (2) is in France. Revealing geography is not the same as revealing existence. We already knew both goats existed, so Monty gives useless information. When he opens the door, he is merely displaying to observers that he has chosen a Chinese or French goat. You were always going to be left with the remaining goat or a Car. You never had the option to pick because he would have already chosen the specific goat before ever presenting the option to you. This is irrelevant though because even from Stage 1 we are assuming Monty will be assigned a goat. You cannot assign his goat (though so far he is unidentified in Stage 1) as an option for us to pick.
Based on the actual Monty Hall scenario, even in the beginning you have a choice of (3) doors, but a choice of only (2) prizes - A car and Monty’s least favorite goat (language, name or other identifier to be determined in Stage 2)
Also, he will always be gifting you the entirety of his door’s probability. As with any removal, the contents are distributed evenly to the “stay” and “switch” doors.
You didn’t need to see him do anything to know this. The key point is that he is actually removing both the denominator and the enumerator from the fraction. Your options are (1) car and (1) unidentified goat.
Monty states: "You cannot choose from my pool of options because then I’ll just switch to the other goat in the door you didn’t pick. Regardless, I win because I get a goat either way, which is my function. You can have the leftover doors, but you would have always had a goat and a car option regardless of which language your goat ends up speaking.
Further, if another individual was playing and would receive Monty’s leftovers and yours upon your final decision, then he wins wherever you lose. If you always choose to switch, his prize swaps also, even though its not by choice. If he was originally going to win, now you’ve swapped and made him a loser. If you swap every time, his prize swaps every time and he has also been impacted by variable change…that is to say he should make the same assumption you are making. One person loses from another person’s gain. The leftover door is the other person playing with you. You switch every time so it switches every time to align with your rejected prize.
The door and yourself cannot both have a 66.6 percent chance of a car. But you never had a 66.6 chance, in fact Monty had 100% of a goat which means you could never get “his” goat, “Monty’s Goat”, so you never had (3) options for prizes, only (3) options for doors in Stage 1. Stage 2 shifts the focus by reducing the number of doors to the number of “types” of prizes (car or goat). And that is where the math gets confused. Performing numerical operations alone without separating objects from their assigned placeholders.