I often see people try to "prove" the solution to this problem as the author did, using a large number of trials to approximate the long-term probabilities. Bu

I often see people try to "prove" the solution to this problem as the author did, using a large number of trials to approximate the long-term probabilities. But I don't think this method really proves it, nor does the proof require so much work. Consider that C = car and G = goat. There are three possible configurations of cars and goats, represented by rows of the matrix, and the three doors are represented by the columns: C G G G C G G G C Suppose you chose door #1 (it doesn't matter which door); you then have 1/3 chance of winning: (C) G G (G) C G (G) G C The host reveals another door with a goat. That door is X'd out: (C) X G (G) C X (G) X C Thus, if you stay with your original choice, you retain a 1/3 chance of winning, shown by the set [C, G, G]. If you switch, you have a 2/3 chance of winning shown by the remaining set [C, C, G].