A particularly difficult reasoning problem

After five pirates robbed 100 gold coins, they discussed how to distribute them fairly. They agreed on the distribution principle is:

(1) Draw lots to determine each person's distribution sequence number (1, 2, 3, 4, 5);

(2) Pirates who draw lots. 1 Propose a distribution plan, and then five people will vote. If the plan is agreed by more than half of the people, it will be distributed according to his plan, otherwise 1 will be thrown into the sea to feed sharks;

(3) If 1 is thrown into the sea, No.2 puts forward the allocation plan, and then four people are left to vote. If and only if more than half of the people agree, they will be allocated according to his proposal, otherwise they will be thrown into the sea;

4 and so on.

Assuming that every pirate is extremely intelligent and rational, they can make strict logical reasoning and rationally judge their own gains and losses, that is, they can get the most gold coins on the premise of saving their lives. At the same time, assuming that the results of each round of voting can be implemented smoothly, what distribution scheme should the pirates who have drawn 1 put forward to avoid being thrown into the sea and get more gold coins?

Push from back to front. If 1-3 robbers all feed sharks, there are only No.4 and No.5 left, and No.5 will definitely vote against it, so that No.4 will feed sharks and take all the gold coins. Therefore, No.4 can only save his life by supporting No.3. Knowing this, No.3 will put forward a distribution plan (100,0,0), which will leave all the gold coins to No.4 and No.5, because he knows that No.4 has got nothing, but he will still vote for it. With his own vote, his plan will be passed. However, if No.2 infers the scheme to No.3, it will propose a scheme of (98,0, 1, 1), that is, give up No.3 and give No.4 and No.5 a gold coin each. Since the plan is more favorable to No.4 and No.5 than No.3, they support him and don't want him to be out and assigned by No.3 ... So No.2 took 98 gold coins. However, the scheme of No.2 will be known by 1, and 1 will put forward the scheme of (97,0, 1, 2,0) or (97,0, 1, 0,2), that is, give up No.2 and give No.3 a gold coin at the same time. Because the plan of 1 is better for No.3 and No.4 (or No.5) than No.2, they will vote for 1, plus 1, and the plan of 1 will be passed, and 97 gold coins can be easily put in the bag. This is undoubtedly the scheme that 1 can get the greatest benefit.