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 basic flaw in the reasoning is that it assumes that each of the 12 *possible* outcomes are equally likely -- because without such an assumption "same number of outcomes" does not translate into "equal probability".

In many simple probability situations we can get a plausible "all outcomes are equally likely" assumption from symmetry arguments. But that won't hold here because the group of outcomes where the contestant chose the prize door are fundamentally different from the group of outcomes where the contestant chose a non-prize door; the host have different amount of options in the two cases.

As a simpler illustration that it doesn't always work to assume that all the possibilities are equally likely, consider this different game:

1. First the contestant flips a fair coin.
2. If the coin came out heads, the host rolls a fair die once; otherwise the host keeps rolling the die again and again until he rolls 6.

Here there are *a priori* 12 outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}. But some of them are impossible; if we exclude them there are only 7 outcomes: {H1, H2, H3, H4, H5, H6, T6}. In 6 of the cases the coin flip was head; only one cases was the tails.

*Can we then conclude the odds of flipping heads is 6 to 1?*

Of course not. The coin is still fair, and what happens *after* it is flipped can't influence the probability of coming up tails. So this shows that comparing numbers of outcomes doesn't always give true probabilities, which ruins your Monty Hall analysis.
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 $6\times \dfrac1{18}=\dfrac13$.