As B Boom wrote, the first pirate can make a proposal that gives him all but 49 (about, depending on the rules) pieces of
All the pirates are eminently logical; the last pirate knows that if he gets a chance to make a proposal, he will get all of the loot:
|…||5th to last||4th to last||3rd to last||2nd to last||last pirate|
|if last proposes:||…||—||—||—||—||all the loot|
If we’re playing where ties are enough, the second to last pirate knows he can get all the loot if he gets a chance to make a proposal [There are many famous woman pirates, but to keep things simple, we will stick to masculine pronouns].
|if 2nd to last:||…||—||—||—||all the loot||0|
What should the third to last pirate propose, should he get a chance? The last pirate expects nothing if the second pirate gets a chance to propose, and so will gladly support the third pirate’s proposal for a single gold piece:
|3rd to last:||…||—||—||almost all||0||1|
How does this continue?
etc. It is clear that the 100th to last pirate (i.e. the first pirate in line) should dole out a single gold piece to the 98th, 96th, etc pirates. The game is odds vs. evens!
What if ties are not enough? Any proposal by the second to last pirate will be vetoed, and the second pirate faces expulsion or even death! His vote, then, can be bought by the third to last pirate for 1 gold piece (if the 2nd only faces expulsion) or 0 gold pieces (which would be better than getting executed!) So this works out as:
Failed proposals = death, ties are not enough
etc. Again the pattern is clear; the 100th to last pirate should give 1 gold piece to the 98th to last, two to the 97th to last, and 1 to the 95th, 93rd ,etc.
Finally if the 2nd to last pirate fears only expulsion, it takes 1 gold piece for the 3rd to last pirate to buy his vote. Things come out a little differently, and we’ll let you finish the table. (Careful! It takes a while for the final pattern to shake out!)
Failed proposals = expulsion, ties are not enough