Monty Hall (goat) problem w. elementary events I am trying to solve the Monty Hall problem with elementary events, but I just can't seem to get it right:
I define four sets: 


*

*Actual door ($R$)

*User's first selected door ($S$)

*Moderator pick ($T$) (moderator always opens a door with a goat)

*User's eventual pick ($U$) (never picks revealed goat)


Each of these sets consists of three elements $R = S = T = U = \{ a, b, c \}$, where $a, b, c$ equal the three possible doors.
My base set of events shall be $O = R~x~S~x~T~x~U$. The event $e = (a,~b,~c,~a)$ for example means the car is behind door $a$, the user first picked door $b$, the moderator then opened $c$, and the user then switched to $a$.
Here is a link to this model.
I now define a probability function $P := O \rightarrow [0; 1]$, which assigns 0 to all impossible events (e.g., $(a,~a,~a,~a)$, as the moderator cannot open doors the user picked) and $remaining^{-1}$ to all others. In total I have 24 remaining (valid) events and all of them should be (I would argue) equally likely.
With this model I now define some events:


*

*Winning $W = \{~(r, s, t, u) \in O~|~r = u ~\}$

*Policy 'stay' $P_s = \{~(r, s, t, u) \in O~|~s = u~\}$

*Policy 'change' $P_c = \{~(r, s, t, u) \in O~|~s \neq u~\}$


Lastly, based on these events I would like to compute $P(W~|~P_s)$ as well as $P(W~|~P_c)$. 
I would have expected them to turn out as $1/3$ and $2/3$ respectively; however, both of them are $1/2$ (see model link above and change filter for events).
What am I missing? Am I wrong to model the problem with these elementary events? Is assigning all of them the same probability wrong? (I would argue that since every 'step' (R, S, T, U) happens independently in a model with filtered invalid events this should be fair). 
Update: As @Henry pointed out, my assumption about $T$ being independent is wrong. For example, although in the table $(a, a, c, ?)$ and $(a, b, c, ?)$ have the same number of valid events (2), the probability of $(a, b, c, ?)$ is twice as high (as the moderator is 'forced' into revealing door c in the $(a, b, ?, ?)$ case). $U$ in this case, while technically also dependent, is what one could call pseudoindependent, as in every case where I count it it happens to have two cases, evenly distributed.
 A: You have $3^4=81$ possibilities of which you have filtered out all but $24$.  But these are not equally probable.
What are equally probable are the $3^2=9$ possible combinations of the actual door and the user's first statement.  


*

*In $6$ of the cases they are different and the moderator's door to open is forced: if the user then switches then the user wins while if the user sticks the user loses

*In $3$ of the cases they are the same and the moderator can open either of the other doors: if the user then switches then the user loses while if the user sticks the user wins
A: I believe it's much easier to understand Monty Hall if we think in terms of two possible strategies for the player: "I will stick with my first choice" (STICK) and "I will change doors when offered" (CHANGE). The strategy to be used to play the game has nothing to do with probabilities. The player may decide, in any way he likes, before he plays the game, which strategy he's going to use.
Now, again without talking about probabilities, there are two important facts: under strategy STICK you win if and only if your initial choice is the correct door; and under strategy CHANGE you win the prize if and only if your initial choice is a wrong door.
Hence, if you assign the same probability $1/3$ for the prize to be behind each of the three doors, it follows that under strategy STICK your probability of winning is $1/3$, while under strategy CHANGE your probability of winning the prize is $2/3$. This is why strategy CHANGE is better.
Of course, there is nothing special about the choice of uniform probabilities for the prize being behind each of the three doors. If instead you assign probabilities $a$, $b$, and $c$ (none of which we'll suppose to be equal to $1$), then under strategy STICK you should choose the door with maximum probability, while under strategy CHANGE you should start with the door with the smallest probability. But, again, strategy CHANGE is better because $\max\{a,b,c\}<\max\{1-a,1-b,1-c\}$. Hence, choosing the least probable door and changing when offered to do so is a good way to play the Monty Hall game.
