g = goat
c = car
x = open door
W = win without change
w = win with change
L = lose without change
l = lose with change.
C = contestant
P = presenter.

Three possibilities (123): ggc gcg cgg

ggc:

C chooses 1
P chooses 2 - gxc (forced choice)
    (i) C sticks - L
    (ii) C twists - w
C chooses 2
P chooses 2 - xgc (forced choice)
    (i) C sticks - L
    (ii) C twists - w
C chooses 3
(a) P chooses 1
    (i) C sticks - W
    (ii) C twists - l
(b) P chooses 2
    (i) C sticks - W
    (ii) C twists - l

Total outcomes for ggc: 8
Total wins if Stick (W): 2
Total wins if Twist (w): 2   

C must either Stick or Twist so the probability of win is 25%.

Repeat for the other two combinations, merely shifting the situation where P has a choice to one of the other positions.

Which is in line with what you might intuit: if the door C chooses is opened at stage 1, the probability would be 1/3. Needing to get two guesses in a row right must reduce the probability.

Now repeat the exercise where all 8 combinations of g and c are permitted.