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: 
As another way to look at this, it might be helpful to think of a situation where it would be appropriate to use your calculation.

Suppose the game is played as follows: the player, the game studio, and Monty each secretly choose a door and write their choice on a piece of paper. A judge looks at the papers, and if Monty's choice is the same as the player's or the same as the studio's (i.e., where the prize is kept), then the papers are destroyed and the process is repeated—this will eliminate 15 of the 27 possibilities, as you have noted, and now the player has he same chance of winning whether he keeps his original choice or switches. Your calculation makes sense here.

This is different from the usual Monty Hall problem, however, because Monty normally makes his choice *after* the player and studio have already chosen.

Compare to the coin/die game in Henning Makholm's answer: if the coin and die were tossed simultaneously until you got a legal combination, then the odds of flipping heads would indeed be 6 to 1.
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.