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After 2 turns made according to this sequence, Alice and Bob will have a tie. He knows that 30 of the coins are genuine and 70 fake.

You might think that if the sequence of prices doesn’t grow very fast, then using the Thue-Morse sequence is okay. Here is the sequence of prices that I specifically constructed for this purpose: 5,4,4,4,3,3,3,2,2,2,2,1,1,0,0,0. At the end we get a sequence that I decided to call the Fibonacci fair-share sequence. He also knows that all the genuine coins weigh the same and all the fake coins have different weights, and every fake coin is heavier than a genuine coin. He has a balance scale without weights that he can use to compare the weights of two groups with the same number of coins. The strategy is to compare one coin against one coin.

Unfortunately, the puzzle is not available, but my description of it is.

Today let’s look at the third puzzle Derek made for the 2013 Hunt, building on an idea from Tom Yue. Derek tends to write multi-layered puzzles: You think you’ve got the answer, but the answer you’ve got is actually a hint for the next step.

Then the only reasonable way to take turns is ABBBB….

The beauty of the Thue-Morse sequence is that it works very well if there are a lot of items and their consecutive prices form a power function of a small degree k, such as a square or a cube function. A coin collector has 100 coins that look identical.

If both of them have the same value for each piece, then the Thue-Morse sequence might not be good either.