Divide gold coins

The topic is like this: five pirates robbed 100 gold coins, and they decided to divide them: step one, draw lots to decide their numbers (1, 2, 3, 4, 5); Step 2: First, 1 puts forward the distribution plan, and then five people vote. If and only if more than half of the people agree, they will be distributed according to his proposal, otherwise they will be thrown into the sea to feed sharks. Step 3: After the death of 1, the distribution plan will be put forward on the 2nd, and then four people will vote. If and only if more than half of the people agree, they will be distributed according to his proposal, otherwise they will be thrown into the sea to feed sharks; The fourth step, and so on.

The answer is: from the back to the front, if 1-3 robbers all feed sharks, only No.4 and No.5 are left, and No.5 will definitely vote against it and let No.4 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!