Revisions to Pirates and gold coins

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edited Jan 5, 2021 at 19:12

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^{‡: I prefer to think of the pirates as kind heartedkindhearted people who _care_care for the sharks and don't want them to go hungry.

†: It can be argued that if a solution is known to be unstable, _all_all other pirates will vote no, since it doesn't matter who gets the coins — they won't get to keep them. But for the stable solution, it _is_is necessary to distribute the coins like this.}

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edited Jan 4, 2021 at 16:51

bobble

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$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 1 & 1 & c & \text{yes} \\ \\ 2 & 1 & 0 & \text{no} \\ & 2 & c & \text{yes} & \text{breaks tie} \\ \\ 3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & c-1 & \text{yes} \\ \\ 4 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & 4 & c-1 & \text{yes} & \text{breaks tie} \\ \\ 5 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & 4 & 0 & \text{no} \\ & 5 & c-2 & \text{yes} \\ \\ \text{...} \\ \\ 2c & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n & 1 & \text{yes} & \text{breaks tie} \\ \end{array}$

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
1	1	c	yes
2	1 2	0 c	no yes breaks tie
3	1 2 3	1 0 c-1	yes no yes
4	1 2 3 4	0 1 0 c-1	no yes no yes breaks tie
5	1 2 3 4 5	1 0 1 0 c-2	yes no yes no yes
...
2c	1 2 3 ... n	0 1 0 1	no yes no yes breaks tie

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+1 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} \\ \end{array}$

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+1	1 2 3 ... n-1 n	1 0 1 0 0	yes no yes no yes

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+2 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-2 & 1 & \text{yes} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+2	1 2 3 ... n-2 n-1 n	0 1 0 1 0 0	no yes no yes no yes breaks tie

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+3	1 2 3 ... n-3 n-2 n-1 n	1 0 1 0 0 0 0	yes no yes no no no yes shark bait

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+4 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+4	1 2 3 ... n-4 n-3 n-2 n-1 n	1 0 1 0 0 0 0 0	yes no yes no no no yes potential shark bait yes breaks tie

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+5 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-5 & 1 & \text{yes} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+6 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-6 & 1 & \text{yes} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+7 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-7 & 1 & \text{yes} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+8 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-8 & 1 & \text{yes} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{yes} & \text{potential shark bait} \\ & n-2 & 0 & \text{yes} & \text{potential shark bait} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+5	1 2 3 ... n-5 n-4 n-3 n-2 n-1 n	0 1 0 1 0 0 0 0 0	no yes no yes no no no no yes shark bait
2c+6	1 2 3 ... n-6 n-5 n-4 n-3 n-2 n-1 n	0 1 0 1 0 0 0 0 0 0	no yes no yes no no no no yes yes shark bait
2c+7	1 2 3 ... n-7 n-6 n-5 n-4 n-3 n-2 n-1 n	0 1 0 1 0 0 0 0 0 0 0	no yes no yes no no no no yes yes yes shark bait
2c+8	1 2 3 ... n-8 n-7 n-6 n-5 n-4 n-3 n-2 n-1 n	0 1 0 1 0 0 0 0 0 0 0 0	no yes no yes no no no no yes potential shark bait yes potential shark bait yes potential shark bait yes breaks tie

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+9 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-9 & 0 & \text{no} \\ & n-8 & 0 & \text{no} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+9	1 2 3 ... n-9 n-8 n-7 n-6 n-5 n-4 n-3 n-2 n-1 n	1 0 1 0 0 0 0 0 0 0 0 0 0	yes no yes no no no no no no no no no yes shark bait

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 1 & 1 & c & \text{yes} \\ \\ 2 & 1 & 0 & \text{no} \\ & 2 & c & \text{yes} & \text{breaks tie} \\ \\ 3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & c-1 & \text{yes} \\ \\ 4 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & 4 & c-1 & \text{yes} & \text{breaks tie} \\ \\ 5 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & 4 & 0 & \text{no} \\ & 5 & c-2 & \text{yes} \\ \\ \text{...} \\ \\ 2c & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n & 1 & \text{yes} & \text{breaks tie} \\ \end{array}$

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+1 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} \\ \end{array}$

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+2 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-2 & 1 & \text{yes} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+4 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+5 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-5 & 1 & \text{yes} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+6 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-6 & 1 & \text{yes} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+7 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-7 & 1 & \text{yes} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+8 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-8 & 1 & \text{yes} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{yes} & \text{potential shark bait} \\ & n-2 & 0 & \text{yes} & \text{potential shark bait} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+9 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-9 & 0 & \text{no} \\ & n-8 & 0 & \text{no} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
1	1	c	yes
2	1 2	0 c	no yes breaks tie
3	1 2 3	1 0 c-1	yes no yes
4	1 2 3 4	0 1 0 c-1	no yes no yes breaks tie
5	1 2 3 4 5	1 0 1 0 c-2	yes no yes no yes
...
2c	1 2 3 ... n	0 1 0 1	no yes no yes breaks tie

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+1	1 2 3 ... n-1 n	1 0 1 0 0	yes no yes no yes

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+2	1 2 3 ... n-2 n-1 n	0 1 0 1 0 0	no yes no yes no yes breaks tie

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+3	1 2 3 ... n-3 n-2 n-1 n	1 0 1 0 0 0 0	yes no yes no no no yes shark bait

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+4	1 2 3 ... n-4 n-3 n-2 n-1 n	1 0 1 0 0 0 0 0	yes no yes no no no yes potential shark bait yes breaks tie

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+5	1 2 3 ... n-5 n-4 n-3 n-2 n-1 n	0 1 0 1 0 0 0 0 0	no yes no yes no no no no yes shark bait
2c+6	1 2 3 ... n-6 n-5 n-4 n-3 n-2 n-1 n	0 1 0 1 0 0 0 0 0 0	no yes no yes no no no no yes yes shark bait
2c+7	1 2 3 ... n-7 n-6 n-5 n-4 n-3 n-2 n-1 n	0 1 0 1 0 0 0 0 0 0 0	no yes no yes no no no no yes yes yes shark bait
2c+8	1 2 3 ... n-8 n-7 n-6 n-5 n-4 n-3 n-2 n-1 n	0 1 0 1 0 0 0 0 0 0 0 0	no yes no yes no no no no yes potential shark bait yes potential shark bait yes potential shark bait yes breaks tie

Nr. of pirates	Pirate meekest to fiercest	Nr. of coins	Vote
2c+9	1 2 3 ... n-9 n-8 n-7 n-6 n-5 n-4 n-3 n-2 n-1 n	1 0 1 0 0 0 0 0 0 0 0 0 0	yes no yes no no no no no no no no no yes shark bait

Minor clarifications

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edited Jan 20, 2017 at 15:33

SQB

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Let's put the number of coins at $c$ and see what happens if we increase $n$. We'll number the pirates from meekest ($1$) to fiercest ($n$).

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 1 & 1 & c & \text{yes} \\ \\ 2 & 1 & 0 & \text{no} \\ & 2 & c & \text{yes} & \text{breaks tie} \\ \\ 3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & c-1 & \text{yes} \\ \\ 4 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & 4 & c-1 & \text{yes} & \text{breaks tie} \\ \\ 5 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & 4 & 0 & \text{no} \\ & 5 & c-2 & \text{yes} \\ \\ \text{...} \\ \\ 2c & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n & 1 & \text{yes} & \text{breaks tie} \\ \end{array}$

If we define a solution as stable if nobody gets fed to the sharks, we can say that all solutions so far have been stable. This is because for each $n$, there are more or at least as much (in which case the tie is broken) pirates who would be worse off in the solution for $n-1$.

But now, we're at $n = 2c + 1$. Are we getting ready for carnage? Not quite:

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+1 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} \\ \end{array}$

Here, the fiercest pirate can escape alive, but without gold.

Okay, but now surely the sharks are getting fed? Let's see.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+2 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-2 & 1 & \text{yes} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

The fiercest pirate now broke the tie to stay alive.

How about $n = 2c + 3$ then?

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

Here, finally, at $n = 2c + 3$, we reach our first unstable solution. The fiercest pirate can buy $c$ votes, adds his own vote for a total of $c+1$, but the opposition has $c + 2$ votes.
Note that here, pirate $n-3$ is pirate $n-2$ from the solution before it. Voting $\text{no}$ yields a coin. Pirates $n-2$ and $n-1$ get no gold by voting against this proposal, but as per the third rule, vote to feed the sharks.^‡

But does that mean that from now on, all pirates get their feet wet? Not at all.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+4 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+4 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

Now both the fiercest pirate and the almost-as-fierce pirate are voting for their lives, since the solution with one pirate less is unstable and will see the almost-as-fierce pirate being fed to the sharks. The fiercest pirate breaks the tie and this solution is stable.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+5 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-5 & 1 & \text{yes} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+6 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-6 & 1 & \text{yes} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+7 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-7 & 1 & \text{yes} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+8 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-8 & 1 & \text{yes} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{yes} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+5 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-5 & 1 & \text{yes} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+6 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-6 & 1 & \text{yes} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+7 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-7 & 1 & \text{yes} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+8 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-8 & 1 & \text{yes} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{yes} & \text{potential shark bait} \\ & n-2 & 0 & \text{yes} & \text{potential shark bait} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

And we've found a stable one again.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+9 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-9 & 0 & \text{no} \\ & n-8 & 0 & \text{no} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

And so on.

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.

This yields the following formula for stable solutions: $$ n \leq 2c \lor n = 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$$

The procedure by which to distribute the coins is based on the fact that a pirate can only buy votes if coins are given to the opposite pirates from the next stable solution.
It is as follows:

For $n \leq 2c$ start with givinggive a coin to theevery pirate with the same parity as $n$ and skip every other, with any left over coins going to the fiercest pirate.
For $2c + 2^{z-1} < n \leq 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$ start with the meekest pirate with the opposite parity of $z$ and move up in ferocity.^†

^{‡: I prefer to think of the pirates as kind hearted people who _care_ for the sharks and don't want them to go hungry.
†: 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. But for the stable solution, it _is_ necessary to distribute the coins like this.}

Let's put the number of coins at $c$ and see what happens if we increase $n$. We'll number the pirates from meekest ($1$) to fiercest ($n$).

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 1 & 1 & c & \text{yes} \\ \\ 2 & 1 & 0 & \text{no} \\ & 2 & c & \text{yes} & \text{breaks tie} \\ \\ 3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & c-1 & \text{yes} \\ \\ 4 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & 4 & c-1 & \text{yes} & \text{breaks tie} \\ \\ 5 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & 4 & 0 & \text{no} \\ & 5 & c-2 & \text{yes} \\ \\ \text{...} \\ \\ 2c & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n & 1 & \text{yes} & \text{breaks tie} \\ \end{array}$

If we define a solution as stable if nobody gets fed to the sharks, we can say that all solutions so far have been stable. This is because for each $n$, there are more or at least as much (in which case the tie is broken) pirates who would be worse off in the solution for $n-1$.

But now, we're at $n = 2c + 1$. Are we getting ready for carnage? Not quite:

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+1 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} \\ \end{array}$

Here, the fiercest pirate can escape alive, but without gold.

Okay, but now surely the sharks are getting fed? Let's see.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+2 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-2 & 1 & \text{yes} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

The fiercest pirate now broke the tie to stay alive.

How about $n = 2c + 3$ then?

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

Here, finally, at $n = 2c + 3$, we reach our first unstable solution. The fiercest pirate can buy $c$ votes, adds his own vote for a total of $c+1$, but the opposition has $c + 2$ votes.
Note that here, pirate $n-3$ is pirate $n-2$ from the solution before it. Voting $\text{no}$ yields a coin. Pirates $n-2$ and $n-1$ get no gold by voting against this proposal, but as per the third rule, vote to feed the sharks.^‡

But does that mean that from now on, all pirates get their feet wet? Not at all.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+4 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

Now both the fiercest pirate and the almost-as-fierce pirate are voting for their lives, since the solution with one pirate less is unstable. The fiercest pirate breaks the tie and this solution is stable.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+5 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-5 & 1 & \text{yes} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+6 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-6 & 1 & \text{yes} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+7 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-7 & 1 & \text{yes} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+8 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-8 & 1 & \text{yes} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{yes} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

And we've found a stable one again.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+9 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-9 & 0 & \text{no} \\ & n-8 & 0 & \text{no} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

And so on.

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.

This yields the following formula for stable solutions: $$ n \leq 2c \lor n = 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$$

The procedure is based on the fact that a pirate can only buy votes if coins are given to the opposite pirates from the next stable solution.
It is as follows:

For $n \leq 2c$ start with giving a coin to the pirate with the same parity as $n$ and skip every other pirate.
For $2c + 2^{z-1} < n \leq 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$ start with the pirate with the opposite parity of $z$.^†

^{‡: I prefer to think of the pirates as kind hearted people who _care_ for the sharks and don't want them to go hungry.
†: 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. But for the stable solution, it _is_ necessary to distribute the coins like this.}

Let's put the number of coins at $c$ and see what happens if we increase $n$. We'll number the pirates from meekest ($1$) to fiercest ($n$).

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 1 & 1 & c & \text{yes} \\ \\ 2 & 1 & 0 & \text{no} \\ & 2 & c & \text{yes} & \text{breaks tie} \\ \\ 3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & c-1 & \text{yes} \\ \\ 4 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & 4 & c-1 & \text{yes} & \text{breaks tie} \\ \\ 5 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & 4 & 0 & \text{no} \\ & 5 & c-2 & \text{yes} \\ \\ \text{...} \\ \\ 2c & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n & 1 & \text{yes} & \text{breaks tie} \\ \end{array}$

If we define a solution as stable if nobody gets fed to the sharks, we can say that all solutions so far have been stable. This is because for each $n$, there are more or at least as much (in which case the tie is broken) pirates who would be worse off in the solution for $n-1$.

But now, we're at $n = 2c + 1$. Are we getting ready for carnage? Not quite:

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+1 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} \\ \end{array}$

Here, the fiercest pirate can escape alive, but without gold.

Okay, but now surely the sharks are getting fed? Let's see.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+2 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-2 & 1 & \text{yes} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

The fiercest pirate now broke the tie to stay alive.

How about $n = 2c + 3$ then?

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

Here, finally, at $n = 2c + 3$, we reach our first unstable solution. The fiercest pirate can buy $c$ votes, adds his own vote for a total of $c+1$, but the opposition has $c + 2$ votes.
Note that here, pirate $n-3$ is pirate $n-2$ from the solution before it. Voting $\text{no}$ yields a coin. Pirates $n-2$ and $n-1$ get no gold by voting against this proposal, but as per the third rule, vote to feed the sharks.^‡

But does that mean that from now on, all pirates get their feet wet? Not at all.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+4 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

Now both the fiercest pirate and the almost-as-fierce pirate are voting for their lives, since the solution with one pirate less is unstable and will see the almost-as-fierce pirate being fed to the sharks. The fiercest pirate breaks the tie and this solution is stable.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+5 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-5 & 1 & \text{yes} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+6 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-6 & 1 & \text{yes} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+7 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-7 & 1 & \text{yes} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+8 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-8 & 1 & \text{yes} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{yes} & \text{potential shark bait} \\ & n-2 & 0 & \text{yes} & \text{potential shark bait} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

And we've found a stable one again.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+9 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-9 & 0 & \text{no} \\ & n-8 & 0 & \text{no} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

And so on.

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.

This yields the following formula for stable solutions: $$ n \leq 2c \lor n = 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$$

The procedure by which to distribute the coins is based on the fact that a pirate can only buy votes if coins are given to the opposite pirates from the next stable solution.
It is as follows:

For $n \leq 2c$ give a coin to every pirate with the same parity as $n$, with any left over coins going to the fiercest pirate.
For $2c + 2^{z-1} < n \leq 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$ start with the meekest pirate with the opposite parity of $z$ and move up in ferocity.^†

^{‡: I prefer to think of the pirates as kind hearted people who _care_ for the sharks and don't want them to go hungry.
†: 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. But for the stable solution, it _is_ necessary to distribute the coins like this.}

Changed MathJax delimiter back to $

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edited Jul 2, 2014 at 9:41

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Typo, grammar.

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edited Jun 5, 2014 at 8:08

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Oopsie. Fixed mixed up yes/no

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edited May 21, 2014 at 18:32

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Delimiters yet again

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edited May 21, 2014 at 18:21

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Replaced MathJax delimiter

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edited May 21, 2014 at 17:57

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MathJax! MathJax! MathJax!

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edited May 21, 2014 at 10:50

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created May 20, 2014 at 20:19

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