Tattletale is correct and the only when with a coherent anaylsis
in this entire thread.

Assume door 1 is chosen
Another way to look at it is, there are really two games being
played. One when the car is beyond is behind door 1 and one when
the car is behind door 2 or door 3

The winning percentage for each door for the second game is
100% 1/3 of the time. The winning percentage for each door
for the first game is 0% 1/6 of the time (if Monty treats
each door the same).

The total winning percentage for either door is then
100% *1/3 + 0% 1/6
-------------------- = 1/3 / 3/6 = 6/9 = 2/3
1/3 + 1/6

1/3 + 1/6