well the non thinking version gets this wrong: Ooooh — you’ve posed a subtle twist on the Monty Hall problem! Let’s think carefully step by step, because the host’s ignorance really matters here.

Classic Monty Hall (host knows):

• Car is equally likely behind doors 1, 2, or 3.

• You pick door 1.

• The host, knowing where the car is, always reveals a goat.

• Probability car is behind door 1: 1/3.

• Probability car is behind the other unopened door: 2/3.
  → So you should switch.

Your scenario (host ignorant):

• You pick door 1.

• The host randomly opens a door — not by design, just by accident.

• He happens to reveal a goat behind door 3.

Now, because his action was not conditional on avoiding the car, the information you gain is weaker. Let’s calculate.

Step 1: Initial probabilities

• P(car behind door 1) = 1/3

• P(car behind door 2) = 1/3

• P(car behind door 3) = 1/3

Step 2: Condition on what happened

We observed: host opened door 3 → goat.

• If car was behind door 1: host could have opened door 2 or 3. He opened door 3, which has probability 1/2.

• If car was behind door 2: host could have opened door 1 or 3. But he can’t open door 1 (your choice). So the only door he can accidentally open is 3 → probability 1.

• If car was behind door 3: impossible (since door 3 revealed goat). Probability = 0.

Step 3: Apply Bayes’ rule

Relative likelihoods:

• Door 1 case: (1/3) × (1/2) = 1/6

• Door 2 case: (1/3) × (1) = 1/3

• Door 3 case: 0

Normalise: total = 1/6 + 1/3 = 1/2.

• P(car behind door 1 | observed) = (1/6) ÷ (1/2) = 1/3

• P(car behind door 2 | observed) = (1/3) ÷ (1/2) = 2/3

🎉 Answer:

Yes, it’s still to your advantage to switch! Even with the host ignorant, the act of accidentally revealing a goat changes the probabilities in your favour. Door 2 now has a 2/3 chance of hiding the car.

Would you like me to also show what happens if the host had instead accidentally revealed the car?