On the pirate question, it helps to think inductively.
Let us use the following notation:
(a,b,c,...)
where a is the numbe of dubloons for pirate 1, b is the number for pirate 2, etc.
If there is only 1 pirate, he takes all the dubloons.
(100)
If there are two pirates, then pirate 2 has to propose a plan. If pirate 2 offers pirate 1 anything less than 100 dubloons, pirate 1 will vote no, kill pirate 2, and keep the dubloons. Therefore, if there are only two pirates, pirate 1 will end up with all the dubloons, and both pirates will remain alive.
(100,0)
If there are three pirates, then pirate 3 needs to propose a plan. He wants to surivive, so he wants to get two votes from the other two pirates. The only such plan he can come up with is to give pirate 1 all the dubloons. If either of the pirates vote no, then they end up in the two-pirates position, where 1 gets everything and 2 gets nothing. Both the pirates will therefore vote yes.
(100,0,0)
If there are four pirates, then pirate 4 needs to get at least two votes. Pirates 2 and 3 would get nothing in the 3-pirate situation, so he doesn't need to offer them anything to get their vote. he can keep the rest of the dubloons to himself.
(0,0,0,100)
In all subsequent numbers of pirates, if there are N pirates, pirate N can keep all the dubloons to themselves as they will have N-2 votes from the pirates would would get nothing anyway if they voted no.
That is what would happen if the pirates vote yes whenever voting yes or no has the same outcome for them. What would happen if they were to vote no is that situation?