Puzzling Pirates

Hmm...been some time since I posted. Time for a new puzzle, I first found this one somewhere on the internets:
5 pirates have to divide up a 100 gold treasure they have found. The dividing will be done as follows: the oldest pirate will propose a division of the money among the pirates, and then the pirates vote on whether they accept the proposal or not. If the proposal passes, they divide up the money as the proposal suggested. If the proposal fails, the oldest pirate is killed and the next oldest one gets to make a proposal. This will continue until either a proposal is accepted or only one pirate is left (who takes all the treasure).
The 100 gold pieces cannot be divided in increments any smaller than 1 gold (a proposal cannot have a pirate getting 2.5 gold, for example). The pirates are perfect logicians, and all facts in this puzzle are considered common knowledge. The pirates priorities are: 1. live, 2. get as much money as possible, 3. kill as many other pirates as possible. The pirates ages are ordered, no two of them were born on the same day.
When the pirates all behave as perfect logicians, what is the final division of money among the pirates?

The more pedantic of you will realize that this puzzle is not yet well defined, I have not said how ties are resolved, nor have I been clear on if the proposing pirate gets a vote. I am unclear on both of these on purpose, this is actually 4 puzzles in 1.
1. The proposing pirate gets a vote, and ties result in the motion passing,
2. The proposing pirate does not get a vote, and ties result in the motion passing,
3. The proposing pirate gets a vote, and ties result in the motion failing,
4. The proposing pirate does not get a vote, and ties result in the motion failing.

If my memory is correct, each of these has a different solution, but I can't recall it perfectly. At any rate, they are sort of neat to work out and see the differences in them.

3 comments:

Anonymous said...

If the proposing pirate has a vote:

What if there's only 2 person left? The older person can propose 100 for himself and none for the youngest... put it to vote and it'll be a tie, and the proposal would pass.... Will the younger pirate 'meekly' accept this decision?



If the proposing pirate has no vote:

What if there's only 2 person left?
The older pirates proposal will certainly be rejected by the younger one, as he is the only one who has a vote... Will the older pirate 'meekly' allow himself to be murdered? :)

Kory Stevens said...

You have basically figured out the crux of the puzzle. The pirates will 'meekly' allow exactly the things you said, they have no choice (if you like, there is something there to enforce the rules, maybe a committee of sorts, I just can't find a clean way to fit it into the puzzle).

Now just take what you have found and move upwards to five pirates.

Anonymous said...

i hate puzzles where i know the 'gist' of it..but have to do the 'hard work'.... hahaha

btw, have you ever tried GodTower? (google it)