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Cake day: November 14th, 2026

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  • Here’s the way I see it. At the start, there are three scenarios. The person is equally likely to be on boat 1, boat 2, or boat 3. Say you pick boat 1, you have a 1/3rd chance of being correct. Then, the goat revealer comes along. They only reveal a boat with a goat that is not the boat you’ve picked. Which means they’re never able to eliminate the possibility of the person being on your chosen boat, but can do it for the other two. When boat 2 is revealed to have a goat, the possibility of a person being on boat 2 is eliminated, and the possibility of a person being on boat 3 is not eliminated when it could have been. The possibility of the person being on boat 1 remains unaffected. The 2/3rds initial chance of the person being in another boat remains unaffected, so the probability pools into the remaining unpicked boat. At least, that is how it works in the monty hall problem.

    It’s different if the goat could be revealed in any boat (including the one you picked) or if the goat is not always revealed, or so on. The goat reveal has to have other information behind it for it to not collapse the possibilities to 50/50.