Editing Openai/6933ffb6-9ed8-8010-99bb-a5bc968f2d9b (section)

=== Assistant: Nice one — this is the classic Monty Hall twist where the host doesn't know and just randomly opened a door that happen… ===
Nice one — this is the classic Monty Hall twist where the host doesn't know and just randomly opened a door that happened to be a goat. Short answer:

It doesn’t matter.
* Probability of winning if you stay with Door 1 = 1/2.
* Probability of winning if you switch to Door 3 = 1/2.

Here’s the exact calculation (Bayes-style, compact):

Initial probabilities: car behind Door 1, 2, 3 are each 1/31/31/3.

We observe: host randomly opens one of the two doors you didn’t pick and happens to open Door 2, revealing a goat. Compute the likelihood of that observation under each hypothesis:
* If car is behind Door 1 (prob 1/31/31/3): both Door 2 and 3 have goats, so the host would open Door 2 with probability 1/21/21/2. Contribution = (1/3)⋅(1/2)=1/6(1/3)\cdot(1/2)=1/6(1/3)⋅(1/2)=1/6.
* If car is behind Door 2 (prob 1/31/31/3): Door 2 contains the car, so the host opening Door 2 and showing a goat is impossible. Contribution = 000.
* If car is behind Door 3 (prob 1/31/31/3): Door 2 is a goat, the host opens Door 2 with probability 1/21/21/2. Contribution = (1/3)⋅(1/2)=1/6(1/3)\cdot(1/2)=1/6(1/3)⋅(1/2)=1/6.

Total probability of the observed event = 1/6+0+1/6=1/31/6 + 0 + 1/6 = 1/31/6+0+1/6=1/3.

Posterior probabilities:
* P(car behind Door 1∣obs)=(1/6)/(1/3)=1/2.P(\text{car behind Door 1}\mid\text{obs}) = (1/6)/(1/3)=1/2.P(car behind Door 1∣obs)=(1/6)/(1/3)=1/2.
* P(car behind Door 3∣obs)=(1/6)/(1/3)=1/2.P(\text{car behind Door 3}\mid\text{obs}) = (1/6)/(1/3)=1/2.P(car behind Door 3∣obs)=(1/6)/(1/3)=1/2.

So switching and staying are equally good — each gives a 50% chance of the car.

(Contrast: if the host knows and never opens the car door, switching would give 2/32/32/3 win probability. The difference is the host’s selection process.)