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, then No.5 will definitely vote against No.4 to feed sharks to keep all the gold bars. Even if No.4 please No.5 to save his life, and put forward such a plan as (0, 100) to let No.5 monopolize the gold bars, No.5 may think it is dangerous to keep No.4 and vote against it to feed it to 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 plan (100,0,0), because he knows that No.4 will unconditionally support him and vote in favor, so adding his own 1 vote will enable him to get100 gold bars.
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 bars. 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 bars 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 bars, and at the same time give No.4 or No.5 2 gold bars, 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 bars can easily fall into the pocket of 1.