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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 ...


I'm not entirely sure how to single out the exact ball here as the description given seems to imply that I have four groups of three balls and only three times to use the scale in order to determine the ball. Based on that, I can use standard logic to figure out which group the different ball is in. Let's say that our groups are labeled A, B, C, and D, and ...


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:


Then, Additionally,


However Why? so whatever Bob can arrange these cards in the best case scenario, Let's make it more complex; so even it is 0 to 100, Ann will


Code to find the pattern: (The third number printed out on each line is the winning move if the player to move is in a winning position)


We have a winning strategy for: The first move is:

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