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Therefore, showing an open door with a goat reveals nothing new about which door has the car.

Two-thirds of the time, our initial pick will be wrong. In that case, Monty cannot randomly pick a door out of the remaining two to reveal. He must pick the door that has a goat. This gives the additional insight into which door has a car.

Conversely, one-third of the time, our initial pick will be correct so Monty can then open any of the two remaining two doors. Switching in that case will then lead us to the wrong door. The good news is, that only happens one-third of the time!

Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50%$, whereas there is definitely a goat behind door $C$.

In particular, let $C_{text{goat}}$ be the event that there's a goat behind door $C$, and $C_{text{revealed}}$ be the event that door $C$ is revealed, and $A_{text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{text{car}}$ and $C_{text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50%$. Thus, one concludes that
$$P(A_{text{car}}|C_{text{revealed}})=P(A_{text{car}})=frac{1}3.$$
The calculation you've done (annotated with a $neq$ sign where things go wrong) is:
$$P(A_{text{car}}|C_{text{revealed}})neq P(A_{text{car}}|C_{text{goat}})=frac{1}2$$

Let me try to guess your birthday. My guess is April $2$nd.

Ignoring leap years, remove $363$ dates that are not your birthday and that are not April $2$nd (because I chose that), so that we are now left with only two options. You offer me to switch to the date that I didn't pick, which I do.

That I was right in my initial guess is unaffected by the dates that you subsequently remove.

The flaw in your thought process is that in the $50:50$ case with only two doors where one is a goat and one is a car, that is the only information you have. In the other case where you have two goat doors and one car door, when you open a goat door, you gain some information by doing this. Thus you appear to have the same $50:50$ situation, but you have some added information.

The idea of assigning $50/50$ probabilities to two alternatives
makes sense only when the two alternatives are truly symmetric.

If you just start with two doors, you know there is a car behind
exactly one, and nothing else has happened concerning those doors,
then $50/50$ probabilities make sense.

But once you have chosen one of three doors and Monty has opened
another door (having been forced by the rules of the game to open
a door that wasn't yours and didn't have the car),
you no longer have a symmetric situation. One of the remaining
doors is an available choice because you already chose it once;
the other is an available choice either (A) because you already
chose the car and Monty got to take one of the "goat" doors away at random,
or (B) because you chose a goat initially and Monty had to take away
the remaining "goat" door, leaving the door with the car.
Case (B) is twice as likely to occur as case (A), not $50/50$,
and which door has the car is now completely a function of whether
you're in case (A) or case (B), so it's also not $50/50$.

The error in your argument is that you don't start with two doors, you start with three.

If you pick one door out of three, you have a one in three chance of having picked the right one ($tfrac{1}{3}$), and a two in three chance of having picked the wrong one ($tfrac{2}{3}$).

If you have picked the wrong door, Monty shows you which door has the car behind it, by opening the other door. You win a car by switching.

Conversely, if you have picked the right door, switching makes you lose the car.

Since there's a $tfrac{2}{3}$ chance of picking the wrong door at first, there's a $tfrac{2}{3}$ of winning the car after switching. The chance of picking the right door is $tfrac{1}{3}$, so there's a $tfrac{1}{3}$ chance of losing the car after switching.

example 1 which shows there would be no advantage in shifting: lets say there are one million doors and I pick No 1. Monty reveals 999998 doors which are empty so now we have only 2 doors, one that I picked and another one. According to the paradox the possibility would be 1/999999 = almost 100% certain that the shift would have advantage. Obviously wrong.

Convinced?

example 2 which shows there would be no advantage in shifting:
Lets say instead of one player we have 2 players and one of them choose door one and another choose door 2 and Monty opens door three. If the argument is true shifting would increase the chance for both players. Obviously not possible.

Convinced?

I'm amazed people still argue this.

Example 1 yes you would have an extremely high chance of winning if you selected the door that was filtered out 999,999 doors. If not, you are basically arguing that Door 1 had a 50/50 shot at winning a contest that had 1 out of 1,000,000 odds. The fact the game show host had to filter the losing 999,998 doors out of the set makes switching advantageous since that door is weighted by the 999,998 losing doors that were filtered out.

Example 2. This example isn't applicable to the stated problem. When you introduce a 2nd player the host no longer can filter 2 doors down to 1 door unless both players choose the same door. If they choose the same door then it's the same classic Monty Hall problem just 2 iterations happening simultaneously. If 2 players select different doors then Monty has to open the 3rd unchosen door. According to you, that is 50/50 right? This is your premise but you're forgetting another option.

What you're forgetting is if neither player selected the correct door. In this case, Monty would have to Open player 1 or 2 door. This would then force that particular player to move as he now knows for a fact he will lose without switching. The player who's door was opened would switch obviously but he now has a 50/50 shot instead of 1/3. The player who's door wasn't opened should switch as this had no impact to him other than the typical Monty hall problem, i.e he now has 2/3 chance.