Question: I still do not believe the "correct" solution to the Monty Hall Problem.

Here is my reasoning:
The player can pick from $1$ of $3$ doors.
The prize can be behind $1$ of $3$ doors.
Monty will open $1$ of $3$ doors.

$3 \times 3 \times 3 = 27$ possible sequences of events.

In $15$ of those possible events, Monty either opens the door the player picked or the door with the prize. Since Monty will not do either of those, these $15$ events are removed from the possibilities.

Of the remaining $12$ possibilities, $6$ times the player wins if he stays with his original pick and $6$ times he wins if he switches. It looks to me like the player has the same chance of winning whether he stays or switches.

Could someone please explain the flaw in my reasoning.

Answer: 
The reason that your numbers come out like that is because there are two possible doors for Monty to open if the contestant has already picked the right answer, but only one possible door for Monty to open if the contestant has picked wrongly. In the "two possible doors" case, switching to either of the doors loses, but this only happens in 1/3 of cases, so it is not meaningful to count in this way.

Here is an image. D indicates the door opened, the right side is the contestant's initial choice, and the top indicates which doors Monty can open:

![enter image description here](https://i.stack.imgur.com/tSMuZ.jpg)

If the prize is at A (D=A above), the contestant picks A, then Monty can choose either B or C doors (as marked in green circles). If the contestant picks B when the correct answer is A, then Monty can open only C, and vice-versa if the contestant picks C (the pink circles are Monty's choices)

Assume that the contestant's pick $P$, the prize distribution $D$, and Monty's choice $M$ are uniformly distributed among the possibilities, then
$$
P(M=B|P=A,D=A)=P(M=C|P=A,D=A)=1/2
$$
but
$$
P(M=B|P=B,D=A)=0
$$
and
$$
P(M=C|P=B,D=A)=1
$$
so since the pick and the prize are independent,
$$
\begin{align}
P(M=C,P=B,D=A)&=P(M=C|P=B,D=A) P(P=B) P(D=A)\\&=1\times\frac13\times\frac13\\&={1\over9}
\end{align}
$$
but
$$
P(M=C,P=A,D=A)=P(M=C|P=A,D=A) P(P=A)P(D=A)={1\over18}
$$
In other words the green circles only have probability $1/18$ whereas the pink circles have probability $1/9$.

The multiplicity of your counting just relates to Monty's freedom to choose either B or C when the contestant has the right answer.
Monty does not have three choices, but only one or two. Initially before Monty opens a door the probabilities are as you expect: 

* You choose A, prize behind A: probability $\frac19$
* You choose A, prize behind B: probability $\frac19$
* You choose A, prize behind C: probability $\frac19$
* You choose B, prize behind A: probability $\frac19$
* You choose B, prize behind B: probability $\frac19$
* You choose B, prize behind C: probability $\frac19$
* You choose C, prize behind A: probability $\frac19$
* You choose C, prize behind B: probability $\frac19$
* You choose C, prize behind C: probability $\frac19$

Given this, Monty has to open a door which is not you initial choice and which is not the one with a prize. Sometimes Monty has a choice of door, and sometimes not, leading to $12$ possibilities with varying probabilities. Once Monty has opened a door, you can win or lose by switching and no further chance is involved as there are only two remaining doors. The probabilities become:

* You choose A, prize behind A, Monty opens B: probability $\frac1{18}$, switching loses
* You choose A, prize behind A, Monty opens C: probability $\frac1{18}$, switching loses
* You choose A, prize behind B, Monty opens C: probability $\frac1{9}$, switching wins
* You choose A, prize behind C, Monty opens B: probability $\frac1{9}$, switching wins
* You choose B, prize behind A, Monty opens C: probability $\frac1{9}$, switching wins
* You choose B, prize behind B, Monty opens A: probability $\frac1{18}$, switching loses
* You choose B, prize behind B, Monty opens C: probability $\frac1{18}$, switching loses
* You choose B, prize behind C, Monty opens A: probability $\frac1{9}$, switching wins
* You choose C, prize behind A, Monty opens B: probability $\frac1{9}$, switching wins
* You choose C, prize behind B, Monty opens A: probability $\frac1{9}$, switching wins
* You choose C, prize behind C, Monty opens A: probability $\frac1{18}$, switching loses
* You choose C, prize behind C, Monty opens B: probability $\frac1{18}$, switching loses

So overall switching wins with probability $6\times \dfrac19=\dfrac23$ and loses with probability $6\times \dfrac1{18}=\dfrac13$.