Monty Hall Host Policies

Change the host’s policy to see why the same goat reveal can imply different odds.

Monty Hall Host Policies

The Monty Hall problem is a famous probability puzzle based on a television game show. A contestant chooses one of three doors, hoping to win a car hidden behind it. The other two doors conceal goats. After the initial choice, the host opens one of the unchosen doors to reveal a goat. The central question is: should the contestant stay with their original choice or switch to the remaining closed door?

The standard answer is that switching wins with probability 2/3. However, this relies on a specific assumption about the host's policy: the host knows where the car is, must open a goat door, and must choose uniformly between the two goat doors if the contestant initially selected the car. If we change the host's policy, the conditional probability of winning by switching changes drastically.

Worked Example

Let's condition on the specific door that is opened and observe what happens under two different host policies. Assume the prior probability of the car being behind any door is 1/3. Let C be the car door, X be the chosen door, and O be the opened door. Suppose X = 1, and we observe the host open door 3 to reveal a goat (so O = 3 and C ≠ 3).

Informed Host (Standard Monty Hall):If C = 1, the host has two goat doors (2 and 3) and chooses one uniformly. The likelihood of opening door 3 is P(O=3 | C=1, X=1) = 1/2. If C = 2, the host must open the only unchosen goat door, which is door 3. The likelihood is P(O=3 | C=2, X=1) = 1. The posterior probability of staying (C=1) is proportional to 1/3 * 1/2 = 1/6. The posterior probability of switching (C=2) is proportional to 1/3 * 1 = 1/3. Normalizing these gives P(C=1 | O=3) = (1/6) / (1/2) = 1/3, and P(C=2 | O=3) = (1/3) / (1/2) = 2/3.
Uninformed Host (Monty Fall):The host does not know where the car is and simply opens one of the two unchosen doors at random. The likelihood of opening door 3 is 1/2 regardless of where the car is. If C = 1, P(O=3 | C=1, X=1) = 1/2. If C = 2, P(O=3 | C=2, X=1) = 1/2. (If C = 3, the host would reveal the car, but we condition on the fact that a goat was revealed, so C ≠ 3). The posterior probability of staying (C=1) is proportional to 1/3 * 1/2 = 1/6. The posterior probability of switching (C=2) is proportional to 1/3 * 1/2 = 1/6. Normalizing these gives P(C=1 | O=3, goat) = (1/6) / (1/3) = 1/2, and P(C=2 | O=3, goat) = 1/2.

Experiment

You can test these conditions in the Monty Hall Host Policies module.

Set the Host to "Informed".
Set the Chosen door to 1 and the Opened door to 3.
Observe that the Stay probability is 0.333333333333333 and the Switch probability is 0.666666666666667.
Now change the Host to "Uninformed", keeping the doors the same.
Observe that the Stay probability becomes 0.5 and the Switch probability also becomes 0.5. The observation probability (that the host opens door 3 and reveals a goat) drops to 0.333333333333333.

Limitations

The model assumes a uniform prior for the car's location and that the host's choices (when they have multiple options) are uniformly random. In reality, a host might have a bias (e.g., always opening the lowest-numbered door available). Such biases would require a different likelihood function. Additionally, this model only allows for finite, exact evaluation (exact enumeration of the 3 doors). It evaluates invalid observations (e.g., the host opening the door the contestant just chose) by marking them as undefined or invalid.

References

Random Services: The Monty Hall Problem - Provides a detailed mathematical breakdown of the game, including variations on host strategies and their effects on the posterior probabilities.

This source offers comprehensive derivations for the different probabilities associated with varying assumptions about the host's behavior.

Monty Hall Host Policies