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$\begingroup$ I like your explanation and how it negates that rule! I did wonder if it was necessary but couldn't prove it. $\endgroup$
– Hand-E-Food
Commented Feb 20 at 8:24
$\begingroup$ Note this generalizes to an n-skull game: suppose you have n-2 skulls labeled 1 through n-2, and two unlabeled skulls. By knowing the parity of the permutation, you can always label the last two skulls and find the bead. $\endgroup$
– Tyler Seacrest
Commented Feb 25 at 7:03
$\begingroup$ A variation of this puzzle fooled Penn & Teller. In that case it was rock, paper, scissors instead of skull with gold tooth, skull with bead, skull with nothing, but the answer was still based on the parity of the number of swaps. $\endgroup$
– Steven Schoch
Commented Mar 2 at 2:35

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