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.
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.