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: 
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 $6\times \dfrac1{18}=\dfrac13$.
If you want a set of outcomes with equal probability,
you need to account somehow for the fact that when the
contestant chooses the prize door on the first guess,
Monty has a choice of which door to open.

This is best modeled as a random variable with two equally likely outcomes.
We might as well assume Monty flips a coin and opens the leftmost possible
door if the coin is heads, the rightmost possible door if the coin is tails.
So:

Prize at A, contestant guesses A, coin is heads $\implies$ open door B.

Prize at A, contestant guesses A, coin is tails $\implies$ open door C.

Prize at A, contestant guesses B, coin is heads $\implies$ open door C.

Prize at A, contestant guesses B, coin is tails $\implies$ open door C.

These are four equally likely events.
The difference with your "solution" is you only counted one
A-B-C event and here there are two A-B-C events
(A-B-heads-C and A-B-tails-C).

If you count all the possible outcomes including the coin flip every time
(even when it isn't "needed") then you have 18 equally likely outcomes;
12 are winners and 6 are losers if you switch.