Viewing the Monty Hall Dilemma through Bayes' Theorem
Explanation
As is well known, in the Monty Hall problem it is advantageous to switch regardless of where the prize actually is. Setting aside whether one accepts this as a fact, there are people who do not grasp the Monty Hall game intuitively or who are uncomfortable with algebraic expressions.
For convenience, imagine you are the player and you have chosen door 1.
At this point we have no information about the prize, so each door should be considered equally likely. Now suppose the host opens door 3. The host chooses which door to open based on the player’s choice and the actual location of the prize:
- If the prize is actually behind door 3, the host must not open door 3, so the probability that door 3 is opened is $0$.
- If the prize is actually behind door 2, the player chose door 1 and the host cannot open door 2, so the probability that door 3 is opened is $1$.
- If the prize is actually behind door 1, the host may choose either door 2 or door 3 freely, so the probability that door 3 is opened is $1/2$.
Recall that the event of a door being opened is not a random accident. If the host had simply chosen a door at random without any information and happened to pick an empty door, that would be equivalent to starting the game with only two doors from the outset. The host’s opening of a door is not coincidental but a hint that removes a definite wrong option after observing the player’s action. Naturally, the player is in a state with more information than a blind random choice, and it is to be expected that one of the two strategies—staying or switching—will be advantageous.
Since the host’s choice updates the situation, compute the posterior probabilities using Bayes’ theorem. Let the probability that the prize is behind door 1 if you stay be $p_1$, and the probability that the prize is behind door 2 if you switch be $p_2$. $$ p_{1} = {{ 1/3 \cdot 1/2 } \over {1/3 \cdot 1/2 + 1/3 \cdot 1 + 1/3 \cdot 0}} = {{1} \over {3}} $$
$$
p_{2} = {{ 1/3 \cdot 1 } \over {1/3 \cdot 1/2 + 1/3 \cdot 1 + 1/3 \cdot 0}} = {{2} \over {3}}
$$
Some people prefer concrete computation to the intuitive remark “in simple terms, probability is being concentrated.” Using Bayes’ theorem and enumerating cases yields the same result purely by calculation, without requiring convoluted intuition.
Marilyn’s answer
