Math Is Fun Forum

With two pirates remaining, the elder one could assign himself all of the gold and his vote would count as the 50% needed to "ratify" that vote. That would leave 2 pirates alive, and the older of the two with all of the gold.

Unless "remaining" means OTHER THAN THE ELDEST, which isn't specified, "remaining" would mean "all of the living pirates, including the eldest."

The youngest pirate has no real incentive to vote against anyone, because no matter how many eliminations happen, when it gets down to 2, he misses out entirely. He can't die, and he CAN end up with no money. He should vote FOR any proposal that give him even a single coin.

This is the actual reverse order:

Oldest pirate = 1
Second oldest = 2
etc

With two left, the older one proposes 100 / 0 (oldest to youngest) and it passes with 50% because of his vote.

However, 3 doesn't want to die, so he would have proposed 99 / 0 / 1 (oldest to youngest) which he and 5 would have passed by 66.6%, because the youngest knows he will get NO money if he votes against.

However, 2 doesn't want to die, so he would have proposed 99 / 0 / 0 / 1 (oldest to youngest) which he and the youngest would pass by 50% because of the same reasons stated above.

However, 1 doesn't want to die, so he would have proposed 98 / 0 / 1 / 0 / 1 (oldest to youngest), which would have passed. 2 votes against it because if the vote fails, he knows he will get 99 coins. 3 votes for it because he knows that if it fails, he the next vote will pass and he will receive nothing. 4 votes against it because he votes against everything knowing that if it comes down to 2 he gets everything. 5 votes for it because he knows that if votes continue to fail, he will be left with nothing. 98 / 0 / 0 / 1 / 1 would also work because 4 would realize that if the vote fails, he could be stuck with nothing on the next vote.

JMHO