The Monty Hall Problem - a state-space explanation

There are 3 doors  ?   ?   ?  one of which has a Prize.	So the universe is in one of 3 possible states state1  P   -   -  state2  -   P   -  state3  -   -   P
Now, you choose a door.  ?   ?   ?	Looking inside, we see P in just one of the 3 possible states. state1  P   -   -  state2  -   P   -  state3  -   -   P  If the universe is in state1, we get the P. But if it is in state2 or state3, we don't. Our chances are 1/3 for "P", and 2/3 for "-".
Now "Monty" takes the other two doors state1  P   -   -  state2  -   P   -  state3  -   -   P	and opens one without the prize state1  P   -   -  state2  -   P   -  state3  -   -   P  Now which door do you want, red or green? state1  P   -  state2  -   P  state3  -   P  In green, we get two P's! If the universe is in state2 or state3, we get the P. Only if it is in state1, do we lose. So our chances are 2/3 for "P", and 1/3 for "-".
As you know, a 2/3 chance of getting the Prize is better than a 1/3 chance, so we tell "Monty" we've switched our choice to green.

Notes:
  A friend called to ask about the Monty Hall Problem.
    I liked this explanation, but it didn't work over the phone.
    Not seeing it already online, I wrote this.
  This presentation is similar to this Answer to the Monty Hall Problem,
    but sufficiently different that I went ahead with this one anyway.
   
Doables:

History:
  1998.Oct.15  Created.