“
The probability that Monty opens door B if the prize is behind door B is: P(Monty opens B|B) = 0. The probability that Monty opens door B if the prize is behind door C is: P(Monty opens B|C) = 1. We can now calculate the probability that Monty opens door B: P(A) × P(Monty opens B|A) + P(B) × P(Monty opens B|B) + P(C) × P(Monty opens B|C) = . Finally, apply Bayes theorem: P(A|Monty opens B) = . P(C|Monty opens B) = . In plain English: if you happen to choose door A and Monty opens door B to reveal a lemon then the probability of the Bugatti being behind door C is . If you ever find yourself in this
”
”
Stephen Webb (If the Universe Is Teeming with Aliens ... WHERE IS EVERYBODY?: Seventy-Five Solutions to the Fermi Paradox and the Problem of Extraterrestrial Life (Science and Fiction))