The accepted standard answer to this question is: Pirate 1 gives No.3 1 gold coins and No.4 or No.5 2 gold coins, and he gets 97 gold coins alone, that is, the distribution scheme is (97,0, 1, 2,0) or (97,0, 1, 0). Now let's look at the following rational analysis:
Let's talk about Pirate No.5 first, because he is the safest and has no risk of being thrown into the sea, so his strategy is also the simplest, that is, if all the people in front are dead, then he can get 100 gold coins by himself.
Next, look at No.4, and his chances of survival depend entirely on the existence of others in front, because if all the pirates from 1 to No.3 feed sharks, No matter what distribution scheme No.4 proposes, No.5 will definitely vote against it and let No.4 feed sharks to keep all the gold coins. Even if No.4 pleases No.5 to save his life and puts forward a plan like (0, 100) to let No.5 monopolize the gold coins, No.5 may think it is dangerous to keep No.4 and vote against it, so that he can feed the sharks. Therefore, rational No.4 should not take such a risk and pin his hope of survival on the random selection of No.5. Only by supporting No.3 can he absolutely guarantee his life.
Look at number three. After the above logical reasoning, he will put forward such a distribution scheme (100,0,0), because he knows that No.4 will unconditionally support him and vote for him, so adding his own 1 vote will make him safely get100 gold coins.
But player 2 also knows the allocation scheme of player 3 through reasoning, so he will propose a scheme of (98,0, 1, 1). Because this scheme is relative to the distribution scheme of No.3, No.4 and No.5 can get at least 1 gold coins. Rational No.4 and No.5 will naturally think that this plan is more beneficial to them, support No.2, and don't want No.2 to go out, so No.3 will be allocated. So number two can get 98 gold coins with a fart.
Unfortunately, One Pirate 1 is not a fuel-efficient lamp. After some reasoning, he also understands the distribution scheme of No.2. The strategy he will take is to give up No.2 and give No.3 1 gold coins, and at the same time give No.4 or No.5 2 gold coins, that is, to propose (97,0, 1 2,0) or (97,0). Because the distribution scheme of 1 can get more benefits for No.3 and No.4 or No.5 than No.2, then they will vote for 1, plus 1' s own 1 ticket, and 97 gold coins can easily fall into the pocket of 1.