Answer – 98 Coins

The captain says he will take 98 coins, and will give one coin to the third most senior pirate and another coin to the most junior pirate. He then explains his decision in a manner like this…

If there were 2 pirates, pirate 2 being the most senior, he would just vote for himself and that would be 50% of the vote, so he’s obviously going to keep all the money for himself.

If there were 3 pirates, pirate 3 has to convince at least one other person to join in his plan. Pirate 3 would take 99 gold coins and give 1 coin to pirate 1. Pirate 1 knows if he does not vote for pirate 3, then he gets nothing, so obviously is going to vote for this plan.

If there were 4 pirates, pirate 4 would give 1 coin to pirate 2, and pirate 2 knows if he does not vote for pirate 4, then he gets nothing, so obviously is going to vote for this plan.

As there are 5 pirates, pirates 1 & 3 had obviously better vote for the captain, or they face choosing nothing or risking death.

I think the answer should be 25 coins. Divide 100 coins among 4 people and leave 1 with no coin. Now 4 pirates are satisfied. To create a mutiny for anyone  junior in rank, he has to satisfy one more pirate other than the one not getting anything as if he goes against the captain he may have only 1 support in the person who is not getting any money.

The captain divides 100 coins 3 ways by giving himself 33 and two other pirates get 33 each and throws the remaining one coin in the sea.. this way he gets maximum number of coins and enough number of supporters to get his decision approved.

060 can also be the solution……he (captain) divides the share into 5 equal parts and would give 2 pirates their respective shares. so he will keep the share of 3 people.

I think it was be 50 (considering that its a rule that the captain would get the highest number of gold coin)
Let the pirates in order of seniority be A, B, C, D, E
The first captain wld be A. First he would propose that he takes 98 and give 1 each to C and E (as mentioned by john123)
In this case, B would tell E to reject the proposal and he would give him more than 1(so, at least 2). E knows that he would get nothing if he gets A,B and C killed, as D would take 100 for himself and vote for the proposal himself. That would gain him half of the vote and the proposal would be accepted
So, if it comes down to C, he would get a max of 1 coin, as he would choose 1 over 0.
B offers him more than 1. E knows that B is sure to keep his promise as if B offers him 0 coins, there is no way E would accept it and if B offers him 1 coin, he can reject B and move to C’s proposal, where he surely will get 1 coin, and there still a very, very small chance that he gets more than 1.
This can be sure by an example: if a greedy man is given 2 options of boxes, one sure to contain $10 and the other sure to have a minimum of 10, the man would choose the second one as he would not face a loss in comparison to the first one, and would also have a chance to gain a bonus.
So, this shows that to survive, B has to give at least 2 coin to E and would surely do it. C will think all this and come to a conclusion that if B distributes, he would get 0, so it is better to except 1. There is no way E would reject B so B is bound to get accepted and thus he can’t gain more than 1. This secures 1 vote for A
Now thinking all this, A would think that to get E to accept his proposal, A had to give E the max number he can get. In worse case, as captain had to keep the highest number, if B were to secure 1 vote, B would only divide it among him and E, which would be 51 for B and 49 for E. Also, E knows that after A gets killed, B won’t give more than 2 coins as proved above( in C’s proposal).
So, if A offers E 49, E would sure to get 49 and will not reject the proposal for B.
Another example for this, if the greedy man is offered 2 boxes, 1 sure to contain 10 rupees and the other might have less than 10 but a max of 10, he would choose the first box, as he the second box may indulge in a loss in comparison to the first box.
Thus, A would propose 50 coins for himself, 1 for C and 49 for E

96 coins

The problem has a recursive framework, such that the pirates could continue to rebel against the current captain. The ending condition is when only two pirates are left because one of their votes will meet the 50% required vote and no more mutiny can occur. Due to the leadership specification (where each successive group of pirates will lose their highest ranked pirate), these are the possible subsets of pirates throughout the recursive process: {1,2,3,4,5}, {2,3,4,5}, {3,4,5}, {4,5}

As you can see, the ending condition is when only pirates 4 and 5 are left, so lets work backwards through the logic from this point!

{4,5}
Distribution: 4=100, 5=0
Reasoning: If only pirates 4 and 5 are left, pirate 4 can give himself all the coins because his vote will meet the 50% requirement.

{3,4,5}
Distribution: 3=99, 5=1
Reasoning: Pirate 4 will certainly vote against pirate 3 because he always stands to gain far more through mutiny, so pirate 3 needs pirate 5’s vote to avoid mutiny. All he has to do is offer a single coin because it’s better than the zero coins pirate 5 will earn if he rebelled.

{2,3,4,5}
Distribution: 2=98, 5=2
Reasoning: Pirate 2 needs only a single vote to help him avoid mutiny, so he needs only offer pirate 5 a coin more than he’d earn in a mutinous situation.

{1,2,3,4,5}
Distribution: 1=96, 3=1, 5=3
Reasoning: The captain, pirate 1, needs two additional votes to avoid mutiny. He can buy pirate 5’s vote by offering him a coin more than he’d earn in his next best situation. Since pirate 3 will earn zero coins in the next situation, he needs only offer a single coin to earn his vote.

So the captain should give himself 96, the 3rd pirate 1, and the 5th pirate 3.

I had initially guessed 34 based on my intuition, but you can see how wrong that really is!

98