With two pirates, even if the second pirate offers the first pirate all the money, the second pirate’s gonna die.
So with three pirates, the second pirate will vote for anything where she stays alive. So the third pirate can propose to keep all the money and the second pirate will vote for it.
With four pirates, the fourth pirate needs three votes, and no matter what, the third pirate won’t be the deciding vote, because if the fourth pirate’s allocation is voted down, he gets all the money plus one death, which is one death better than the fourth pirate can offer.
But if the fourth pirate dies, the first and second get nothing, so all she needs to offer them is one gold piece each. I’ll write this (1, 1, 0, x-2).
With five pirates, the fifth pirate needs three votes, so he needs to pay off the third pirate with one gold piece and either the first or second with two: (0, 2, 1, 0, x-3) or (2, 0, 1, 0, x-3).
Here it gets tricky, because the first and second don’t know for sure whether they’ll get 0 or 2 if the sixth pirate dies.
With six pirates, the sixth pirate needs four votes, so she has to pay off three other pirates. The fourth pirate can be paid off with 1 for sure. The third pirate can be bought off with 2 for sure. But the first and second may be optimistic sadistic pirates who think they’ll be chosen for a payoff, or realistic sadistic pirates who think the choices will be randomly made.
If they’re realistic, they both expect 1 gold piece on average, plus one death, if the sixth pirate dies. So the sixth pirate will need to offer one of them 2.
So it’s either (2, 0, 2, 1, 0, x-5) or (0, 2, 2, 1, 0, x-5).