All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
A Pack of Prudent Pirates (Posted on 2002-08-19) Difficulty: 3 of 5
After a long season of plunder, a pirate team of five Prudent Pirates has amassed a booty of 500 golden coins. Before they part their ways, the five decide to divide the treasure.

They that they will each propose a division strategy in order of their seniority: first the oldest pirate will propose the strategy for the division of coins. All five will then vote on it, and if at least half vote "Yes", the strategy will be used to divide the coins. If the majority rejects the plan however, the oldest pirate will be killed, and the whole process will be repeated with the remaining pirates, with the second oldest proposing his strategy.

Since all the pirates are very prudent, each one will want to claim as many coins for himself without getting killed. Given this, how many coins will each of the pirates (5 - 1, with 5 being the oldest) get, and why? What strategy will the oldest pirate propose?

See The Solution Submitted by levik    
Rating: 4.3750 (16 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
How I'd solve it | Comment 16 of 17 |

Ok, A is oldest, E is youngest.

A would split like so:

A=167, B=0, C=166, D=0, E=166

My reasoning is:  To GUARANTEE (or as close as I can possibly get) not being killed I want to split it evenly.  I only need a majority of votes so 3 have to vote for including me.  I need to buy two votes.  I choose E, because he's getting nothing any other way.  He gets nothing if its D and E.  He gets 1 max if its C,D,E because D won't vote for C, because if it fails D takes it all.  C knows E will get nothing if the vote fails and its only D,E.  So, C could offer him 1.  So, its seems C and D are in dominant positions as if either is the oldest remaining they get all or all -1.  But if it gets to B he only needs one other for a draw and he would split with E.  So, B and D are the most dominant.  It would be best for A to work with the two in the least dominant positions.  I would give them every possible reason to want to accept my proposal by sharing it evenly with them.

Possible problems with this idea:

1.  The two may be a little greedier than me and want to take more than an even split at which point they vote down my proposal, but given their positions, it seems reason has the best chance of winning out in A's favor if the split is as I suggest.

The problem with this whole logic puzzle, is that it brings psychology into the situation.  You can't just use Math to solve it.  I'm sure it could be written so as to maximize the math angle and eliminate the psychology, but if you put people into purely logical problems, you can't necessarily determine the "absolute" solution as people aren't 100% logical creatures.  It is a very interesting puzzle to think about nonetheless. 

This was actually played out a while back on a reality show.  Don't remember the name off the top of my head.  Didn't see how it ended, but they all screwed it up anyway.  They could have walked out 10 seconds after the show started if they were smart people.  Duh, all agree to give the money to one person who signs a contract with all members that he has to split the money evenly with everyone after the show.  Not tough.  I guess I don't know if that was against the rules or not, but that's what I would have done.  I would never have voted to give one person all the money otherwise.  Anyway, it was interesting.


  Posted by Michael Rudolph on 2007-05-10 17:25:20
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information