Now that you have that down, allow the pirates to secretly bribe each other out of their own savings before the vote. If multiple votes occur, previous bribes are added to a pirate's savings. If a pirate is thrown overboard, his remaining personal savings are looted and added to the pool of coins to be divided. The personal savings of each pirate are secret from all the other pirates, but all pirates always have more in savings than the next pirate in order of decreasing seniority.

According to the pirate's code, a bribe is a binding contract, but only to the extent that breaching the terms of the contract results in returning the amount of the bribe after the breach occurs. Otherwise, the bribed pirate is thrown overboard. A pirate may therefore spend the bribe money before reneging, if reneging will still yield enough money to pay back the bribe afterward.

Pirate P[x] has savings s[x], and s[x] > s[x+1].

The degenerate case where n=1 is easy to deduce. P[1] proposes {pool}, votes yes, and wins.

The case where n=2 is also easy. P[1] proposes {pool, 0}, votes yes, and wins.

When n=3, it gets more complicated. P[1] needs one more vote to win. If P[2] is able to propose a split, he will be proposing {0, pool + s[1], 0}. So the cost of P[2]'s vote is at least pool + s[1] + 1, which is normally impossible to achieve for P[1]. But since P[2] does not know how much s[1] is, other than that it is more than s[2], P[1] might be able to risk it. But since P[1] doesn't know s[2], bluffing a lower amount for s[1] is risky, as if the bluff amount is lower than s[2] + 1, then P[2] will immediately know to vote no. The cost of P[3]'s vote is at least 1. P[1] and P[2] will therefore be bidding competitively for P[3]'s vote. P[1] has the choice of offering value to P[3] publicly in the proposal, or secretly as a bribe. Either way, P[1] is likely to propose 0 for P[2].

So if P[1] proposes { pool, 0, 0 }, and bribes P[3] to vote yes with s[1], P[2] might try bribing P[3] with any amount from 1 to s[2] to vote no, with no effect. The net effect is { +pool -s[1], 0, +s[1] }.

Remember also that any bribe P[1] pays to P[2] could be added to a bribe from P[2] to P[3]. If P[1] bribes 1 to P[2] to vote yes, and s[1]-1 to P[3] to vote yes, and P[2] bribes s[2]+1 to P[3] to vote no, then if P[2] and P[3] both vote no, P[1] is thrown overboard for losing the vote, and P[2] is thrown overboard for reneging on a bribe and not having the coin to pay it back, so P[3] gets everything. If P[3] suspects that P[1] may have bribed P[2] to vote yes, and knows that P[2] bribed him to vote no--which would be useless unless P[2] himself intended to vote no--then P[3] may vote no on the possibility of getting pool + s[1] + s[2], which would otherwise be impossible for him. So knowing this, P[1] may be confident that any bribe to P[2] to vote yes would not reasonably be re-bribed to P[3] to vote no.

So P[1] could bribe 1 to P[2] to vote yes, tell P[3] that he had bribed P[2], tell P[2] that he had told P[3], and then propose {pool - 1, 0, 1}. It would be useless for P[2] to bribe P[3] unless he intended to reneg, he can't add P[1]'s bribe to his own, and he knows that s[1] - 1 >= s[2]. With the 1 from the pool, he could not match such a theoretical bribe anyway. So the net effect is now { +pool -2, 1, 1 }.

I'm not exactly sure if that is a stable solution or not.

I think maybe that P[3] might be able to get more by bribing P[1] to vote no, or P[2] to vote yes.