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An extension of this question.

As @Jared stated in his answer the solution is:

we assume that the head pirate chooses between multiple possible proposals that maximize his profit by rewarding seniority, then a proposal of $(G-1-\lfloor\frac{n}{2}\rfloor,0,1,2,0,1,0,1,\ldots,\frac{1+(-1)^n}{2})$ will be accepted. The proof is similar to the induction above.

For 9 pirates this gives $\left(G-5,0,1,2,0,1,0,1,0\right)$

The pirate captain has worked this out and knocks the lock off the looted chest to be confronted by a measly 3 gold doubloons!

Is he a dead man?

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1 Answer 1

Lets work our way up and start with just 4 pirates. No problem, a proposal of $(1,0,1,1)$ works based on the answer to the question linked above.

  • 5 pirates: $(0,2,1,0,0)$ works since the two second mates both get more than they would have otherwise.
  • 6 pirates: $(0,1,0,0,1,1)$ works, since again, all pirates who wouldn't die get more money than they would have.
  • 7 pirates: $(0,1,0,1,1,0,0)$ works by the same argument.
  • 8 pirates: This guy is dead. No allotment of coins will allow him to survive. There are 4 pirates getting 0 coins at 7, so there aren't enough coins around for the captain to bribe them.
  • 9 pirates: $(0,0,0,1,0,0,1,1)$ works. The first-mate will vote in favor, because he doesn't want to die. The ones getting coins will also vote in favor, because they would get 0 coins if the captain and first-mate die.
  • 10 pirates: The captain is once again a dead man. The first- and second-mate both know they will survive at 9 pirates, so they can vote to kill, unless the captain gives them each a coin, but there aren't enough coins to go around then - at best the captain can secure 4 votes in favor. Not a clear majority.
  • 11 pirates: The captain and first mate will vote in favor, otherwise they die, plus three other pirates who each get a coin. The vote of 5 to 6 however means no one lives.
  • 12 pirates: Almost! Now, the captain, first-mate and second-mate all vote in favor, otherwise they die, plus another three pirates who each get coins, but with a vote of 6:6, the captain doesn't have a clear majority, so he dies.
  • 13 pirates: AH! The Captain can live once again with $(0,0,0,0,1,1,1,0,0,0,0,0,0)$ or similar. The captain and first three mates all vote for, since otherwise they die. The other three pirates would have received 0 coins at 9 pirates remaining. But now they each get 1 coin again, so they live.

The next number that allows the captain to live is 21 when pirates 14-20 all want to live, plus himself, and he can give 3 other pirates who would have received 0 coins at 13 pirates which gives a total of 11 pirates for, 10 pirates against.

For $n\ge 9$, the number of pirates which allow the captain to live are: $9,13,21,69,133,...$ $=2^{k}+5$

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