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: 
In my opinion the reason why the Monty Hall problem is confusing is because people make the solution unnecessarily complicated. I have a very simple solution that goes like this.

**The player would win by switching doors if and only if (s)he chose the wrong door initially, which happens 2/3 of the time.**
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.