![]() 200 satisfied with their coins and 199 currently-victims won't outvote the 400 who want blood (and money). Now, assume one of the top 200 is an idiot and votes for sharks at the critical point. Then the top 200 including the tie-breaking boss joins with approval and with 400+tie votes they outvote these out of bottom 600 who didn't get any gold. Then the leader distributes gold to the bottom 600 at random (1 piece per pirate though).Įach of those on the bottom, who got a gold coin will vote to keep it, because they won't earn more anyway, and have a chance to lose it. The leaders are slaughtered until the total number drops to 800. There's at most 200 pirates who got any gold, and would prefer to keep it, but they have no say in it, for there's another 600 blood-thirsty jerks who believe they can murder the fiercest with full impunity and have nothing to lose (and either gain a chance for gold, or not gain anything). Let's start with the first carnage at N>800. ![]() No optimal strategy exists for the remaining leaders they die. For large N ( > 400) this gets an exponential structure of strings of carnage that can be stopped at certain points: a simple strategy exists for each pirate of fierceness 400+100*2^n (negative n allowed until we start getting fractional pirates). Once the money runs out (N > 400), only the (odd) least fierce pirates get paid, since if they did not, enough would object to kill the These happen to always be the odd-numbered pirates. So we see that the fiercest pirate gives 1 gold to those who would get nothing for N-1. N = 5: 1 needs two votes, so he gives 1g to the two who would get nothing if he were killed, the current 3 and 5. If he is killed, 3 gets nothing (as he becomes 2 in the previous case) N = 4: 1 just needs one other vote (and he breaks the tie by being fiercest). If he is killed, 3 gets nothing, so if he gives 3 1 piece, 3 votes for him, he wins 2-1. ![]() He votes yes, 2 votes no, 1, being fiercest, breaks tie and takes all the gold N = 2: 1 takes all gold (200) for himself. Let's start with small N and use inference: This has the unintuitive effect that if there are more than 400 pirates, the only way for the fiercest to survive is to take none for himself. I believe pirate 1 (fiercest) should give every odd-numbered pirate (3rd fiercest, 5th, etc.) 1 gold, starting from the least-fierce end, and take the remainder (if any) for himself. But for the stable solution, it is necessary to distribute the coins like this. †: It can be argued that if a solution is known to be unstable, all other pirates will vote no, since it doesn't matter who gets the coins - they won't get to keep them. ‡: I prefer to think of the pirates as kindhearted people who care for the sharks and don't want them to go hungry. $$ n \leq 2c \lor n = 2c + 2^z\,|\,z \in \mathbb Z_$ start with the meekest pirate with the opposite parity of $z$ and move up in ferocity. This yields the following formula for stable solutions: When $n > 2c$, we'll see more and more large runs of downvoting pirates, until the number of pirates upvoting to save their lives becomes equal to the number of downvoters and we find a stable solution again.
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