this post was submitted on 03 May 2025
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Every time I see this I spend 10 minutes reading the solution explanation in detail and still end up not gaining any intuitive insights. I'm sure the "switch" probability is better but I always have more questions.
If the host knows where the car is and picks a door that doesn't have the car, I understand that that is technically gained information compared to the starting state in which the initial door was chosen, but why does that information matter? As a counterfactual, why doesn't it remain just as theoretically possible that the unpicked/switch-to door was the initial chosen switch-from door and therefore there is an equal likelihood that both doors contain the car?
That's what keeps stumping me.
If the host were to pick a door randomly, there would be 6 equally likely possibilities.
First, you pick either Goat 1, Goat 2, or Car. In the first case (1/3 chance), the host picks either Goat 2 or Car. In the second (1/3 chance), the host picks either Goat 1 or Car. In the last (1/3 chance), the host picks either Goat 1 or Goat 2.
Out of these 6 possibilities, two of them result in the host revealing a car, which would end the game early. Eliminating those two possibilities, so the host always reveals a goat, leaves 4 possibilities. This is the "new information" that is used by the host.
In the first case (1/3 chance), switching gives you the car. In the second case (1/3 chance), switching gives you the car. In the last case (1/3 chance), switching gives you a goat.
Thank you! This did it.
Breaking down the how the blocks of potential outcomes stemming from the host's choice, and how the reveal of the goat in two choices leaves only the car as a possible outcome for those two choices, was the key information here that filled in the blanks for me.