here we go again
My solution (I’ve tabbed each step because I’ve made it very detailed - you have been warned - answer is underlined in the final tab):
[tab]The only pirate who has no risk of dying (b) is the youngest pirate.
In the most extreme case, where he was the only one left, he would get the maximum of 100 coins all for himself, fulfilling (a) perfectly as well as (b).
In the next most extreme case, with 2 pirates left, the youngest would be the only voter, resulting in either 100% vote in favour of the proposal made by the oldest or 0%, which would get the oldest thrown overboard.
So to survive at all, the oldest has to offer an acceptable proposal for the youngest, who has nothing to lose. The only proposal that the youngest will accept is all 100 coins, or he will vote against it, be the only pirate left, and get all 100 coins anyway, due to (a).
So due to (b), the oldest of the two would offer 100 coins to the other pirate and have to put up with nothing for himself.[/tab]
[tab]With 3 pirates left, there are two voters. The oldest only has to satisfy 1 of them to get “at least 50%” of the vote, so he has to work out which one to satisfy and how much will satisfy them. The other doesn’t need to be offered anything, in order to maximise how much the other 2 get, as with (a) - it’s fine if just one of them votes against the proposal.
He can actually offer just 1 coin to the 2nd youngest, and nothing to the youngest, and take 99 for himself.
It is in the interests of the 2nd youngest pirate to vote for the oldest pirate of 3 to be kept alive (accepting his proposal) as long as the proposal involves 1 coin or more for him, because even accepting 1 coin would be better than the 0 that he would get if the oldest pirate of 3 is thrown overboard (as in the scenario above with only the 2 youngest pirates). To ensure the life of the oldest of 3, the 2nd youngest needs to either be sure that the youngest will vote for him (in order for him to get at least 50% of the vote) whether he himself votes for the proposal or not, or he needs to vote for him himself (in order to give at least 50% of the vote) whether the youngest follows suit or not.
As pointed out, the youngest is ensured (b) and at least 0 coins, so only has to worry about getting as many more coins as he can more than 0, as with (a). He will have worked out the dilemma of the 2nd youngest, and that he’s in danger of getting 0 coins if he votes for any proposal. His concern is to get at least 1 coin, and the best scenario is that both vote against whatever the proposal is, so he can secure 100 coins - but that’s only a possibility if he can be sure that the 2nd youngest will vote against the proposal, then he can too, and he’s in the money.
At this point, he can’t actually be sure that the 2nd youngest will vote against the proposal if 0 coins are offered to him, because he’ll get 0 coins whether he votes for or against. Needing to be sure in order to secure his 100 coins by also voting against any proposal, this possibility goes out the window. So he will also accept anything more than 0 coins, even 1 coin, in order to escape the 0 coins he’s in danger of being offered.
Having worked out this, the oldest of 3 could actually offer either both of them 1, or either one of them 1 and the other 0. The former option only leaves him with 98 instead of 99 as in the latter 2 options. He can be more sure of his proposal being accepted with at least 1 vote if he offers the 1 to the 2nd youngest who is in the worse position - having nothing to lose.[/tab]
[tab]With 4 pirates left, there are 3 voters. The eldest needs to please 2 of these 3 to get at least 50% of their vote.
The 2nd youngest would do better to ensure the survival of the oldest of 4 pirates if he offers him 2 or more coins, else if he is thrown overboard, he’s in danger of getting 1 or less (as in the above scenario with only 3 pirates). Being offered 2 coins or more will get the oldest pirate 1 of the 2 votes he needs to stay alive.
Assuming the 2nd youngest is offered 2 coins, this would leave a maximum of 98 for the 3rd youngest, which is less than he would get if the oldest of 4 pirates was thrown overboard. It’s in his best interest to vote against the eldest. He can ensure this if the eldest offers less than 2 coins to the 2nd youngest, by being the 2nd vote against any proposal, or at least if he can be sure that the youngest will vote against the oldest pirate whether 2 coins are going to the 2nd youngest or not. The youngest is going to vote for whatever he is offered, as long as it is more than the 0 he’s going to get if the oldest of 4 pirates is thrown overboard.
So if the oldest of 4 proposes 2 to the 2nd youngest, 1 to the youngest, and nothing to the 3rd youngest, he can take 97 for himself and still have his proposal accepted.
However, since the 3rd youngest will have worked this out, he knows he is in danger of getting nothing, so will also accept as little as 1 coin (or more). The oldest of 4 will deduce this and question giving 2 to the 2nd youngest. If he offers him nothing, losing his vote, can he guarantee the 2 votes of the other two? No, because if the other two work out that the 2nd youngest won’t vote for the proposal, the 3rd youngest stands to get 98 coins if he also votes against the proposal and gets the oldest of 4 thrown overboard, which he will prefer.
So the oldest of 4 has to give 2 or more to the 2nd youngest, and 1 can go to the youngest and nothing to the 3rd youngest. This is because the youngest will fear the possibility of the oldest of 4 pirates being thrown overboard and getting 0 coins. He is loss averse and the 2nd youngest will be swayed by the gain of 1 coin. The 3rd youngest will vote against but this won’t matter with 2/3 of the vote passing the proposal. Without giving the 1 coin to the youngest, the 3rd youngest has a chance of his vote being combined with the vote of the youngest to move them over to the 3 pirate scenario where the 3rd youngest gets 98 coins. So the 1 coin to the youngest secures that this possibility doesn’t happen. The oldest of 4 gets his 97 coins.[/tab]
[tab]Now the Grande Finale:
With all 5 pirates, there are 4 voters, and so only 2 of these 4 need to be satisfied.
According to 4 pirate scenario above, the pirates most easily pleased will be the 3rd youngest who would stand to get no coins, and the youngest who will get only 1 coin. Offer the youngest 2, the 3rd youngest 1, and they’ll have to accept. The other 2 are going to vote against getting nothing, but that doesn’t affect the proposal of 2 coins to the youngest, 1 coin to the 3rd youngest, nothing for the other two, and the remaining 97 coins for the oldest pirate. It will still pass with at least 50% of the vote in favour.
Perhaps interestingly, this may all change considering the possibility of finding coins in future. This might turn the pirates against each other, especially the youngest one against the others since he will never be thrown overboard if they keep the same rules, and especially since the 2nd oldest and the 2nd youngest are going to want more than nothing in future, and their primary target is going to be the oldest pirate who stands to rip them off in future in a similar way, and he’s the first in line to be thrown overboard.[/tab]