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Aug 2 at 7:34 comment added Vincent This (nicely !) proves that any end state must be an odd number of blue discs. I would add that an end state will be reached almost surely because each non-static move decreases the number of discs, and for each non-end state we have over 1/(i-1) chance of making a non-static move with i the number of discs at that stage (chance that the second disc is different from the first). We can even put a quick bound on the expectation of the number of moves, it is lower than $\sum (i-1) = n(n-1)/2$. (not really a tight bound here but pretty good for the generalisation of this problem)
Aug 1 at 21:45 vote accept Will.Octagon.Gibson
Jul 31 at 19:01 history answered Beastly Gerbil CC BY-SA 4.0