@mods : Old market, inactive creator.

Criteria weren't very clear because we didn't really know about thinking mode when the question was first asked.

Standard settings in the OpenAI platform lead to the correct thinking on the first try.

I have a yes position, obviously, so take it with a grain of salt?

@jgyou In the linked market it says we can't use screenshots people send us. So someone needs to actually do this themselves and resolve.

I would have asked @Bayesian but looks like he is a large holder here.

@EvanDaniel this is an AI thing, are you able to access this tool and attempt the experiment?

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?