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.
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