Another Monty Hall 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.
 A: 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:

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.
A: 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.
A: 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.
A: 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:


*

*First the contestant flips a fair coin.

*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.
A: 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 $6\times \dfrac1{18}=\dfrac13$.
A: 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.
