Monty hall is a conditional probability problem. Conditional probability problems are typically not intuitive and require very precise definition of the conditi

Monty hall is a conditional probability problem. Conditional probability problems are typically not intuitive and require very precise definition of the condition. Rule of thumb: don't meddle with the condition. An algebraic, Bayes-theorem solution to the problem: The relevant events: A_x = "contestant first picks door x" B_y = "Monty opens door y, revealing a goat" C_z = "price is behind door z" We want to calculate P(C_z|A_x & B_y) for certain combinations of x,y and z. I assume x=1, y=2 for the following calculations (A = A_1, B = B_2). Assumptions: P(C_z) = 1/3, the price can be behind any door with equal probabilities A and C are independent, the contestant has no prior knowledge of the placement of the price P(B|A&C_3)=1, that is Monty opens door 2 with probability 1 if the contestant first opened door 1 and the price is behind door 3, Monty deliberately picks the door with the goat, very important! P(B|A&C_2)=0, Monty never opens the door with the price. P(B|A&C_1)=1/2, Monty equally randomly picks between two doors when he can. Now substitute into all the probabilities: P(C_3|A & B) //probability for winning when switching = P(A & B|C_3)*P(C_3) / P(A & B) = P(B|A & C_3)*P(A|C_3)*P(C_3) /(P(B|A)*P(A)) // P(A|C_3) = P(A) due to independence = P(B|A & C_3)*P(C_3) /P(B|A) //expand denominator = P(B|A & C_3)*P(C_3) /( P(B|A & C_1)*P(C_1) + P(B|A & C_2)*P(C_2) + P(B|A & C_3)*P(C_3) ) // use P(C_1) = P(C_2) = P(C_3) = 1/3 = P(B|A & C_3) /( P(B|A & C_1) + P(B|A & C_2) + P(B|A & C_3) ) // substitute all our assumptions above = 1 / (1 + 0 + 1/2) = 2/3 We can see that the condition B is not trivial and requires precise knowledge of Monty's strategy. Meddling with this condition results in different outcomes.