Jon,

As I said earlier in the ammended table, you, like Escamillo are modelling events that don't happen.
If your first choice is correct the house still only lifts one cup once. Your model shows the house lifting both cups once, as two separate events.
Only one of these two events actually occurs. (For simplicity let's give each a probability of 1/2), then;

PRIZE IS IN CUP 1
Choose 1, lift 2, stay on 1 : win + (1/2)
Choose 1, lift 3, stay on 1 : win + (1/2)
Choose 2, lift 3, stay on 2 : loose (1)
Choose 3, lift 2, stay on 3 : loose (1)

PRIZE IS IN CUP 2
Choose 1, lift 3, stay on 1 : loose (1)
Choose 2, lift 1, stay on 2 : win + (1/2)
Choose 2, lift 3, stay on 2 : win + (1/2)
Choose 3, lift 1, stay on 3 : loose (1)

PRIZE IS IN CUP 3
Choose 1, lift 2, stay on 1 : loose (1)
Choose 2, lift 1, stay on 2 : loose (1)
Choose 3, lift 1, stay on 3 : win + (1/2)
Choose 3, lift 2, stay on 3 : win + (1/2)

(again only for sticking as in your example)
The chance to win is (1/2 + 1/2) = 1
in (1/2 + 1/2 +1 +1) = 3
1/3

AndyG