You should To prove, let us first do the following


The Sleeping Beauty problem itself is a famous problem in the philosophy of probability, and obviously we aren't going to resolve it here. Fortunately, the question here is more concrete, so let's just do it. Of course, you need not say the same thing every time. Conclusion:


This does it: With the 'finishes when have covered all squares' clarification.


The absolute minimum is First some observations: The actual strategy: Next Proof:


To solve the problem, at least I can cross And the strategy is: For improvement, if


User hexomino already figured out the puzzle, and managed to actually find the very complicated path that was exactly how I came up with the game. To recap: The game itself is a lot easier to play than that, though, so I'm posting this self-answer to show how. First, it's very useful to note that in a given board position, every move is uniquely defined by ...


I typed this up in a hurry, there may be errors.


No, it's not possible: If you start by weighing 3 vs 3 and it comes out equal, you only have two weighings to find the fake coin (of unknown weight) among six. In weighing 2, if you have more than three potential fakes on the scale, and it comes out unbalanced, then the remaining weighing (with only three potential results) will not allow you to pick the ...

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