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Another Pack of Prudent Pirates (Posted on 2004-04-01) Difficulty: 2 of 5
The five pirates have found another 500 gold coins and wonder how to split them up. This time they say that unless a majority (more than 1/2) say yes to a plan, the one that proposed that plan will get killed and they will move on to the next plan. The order of plan making starts with 5, then 4, 3, 2, 1.

The pirates will try above all else to stay alive, even if it means accepting no coins. If they will stay alive either way, they would like the most coins possible. Also, the pirates have been on board the ship for a while and are getting tired with each other, so if faced with the decision to reject a plan or keep it, they will reject it if nothing else matters more to them.

What should Pirate 5's offer be?

See The Solution Submitted by Gamer    
Rating: 3.8333 (6 votes)

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Solution re: Dividing the pieces of eight | Comment 2 of 12 |
(In reply to Dividing the pieces of eight by Brian Smith)

I don't completely agree, though my answer is almost the same.  First, remember that the problem states "so if faced with the decision to reject a plan or keep it, they will reject it if nothing else matters MORE to them."  So when you say Pirate 1 would still vote for Pirate 2's plan, I don't agree.

Here's my thought process:

One Pirate: He'll get all 500 coins.

Two Pirates: 2 need's 1's vote, but 2 cannot make 1 happier than the One Pirate case because he can't give 1 more than 500 coins.  Therefore, no matter what 2 offers 1, 1 will reject the plan and 2 will die.

Three Pirates: 3 needs just one more vote in addition to his own.  He knows that 2 will die in the Two Pirate case, so even if 3 gives 2 nothing, 2 will still vote for this plan.  So 3 can keep all 500 coins, 2 will get nothing but will still vote for this plan, and 1 will get nothing and will reject this plan.

Four Pirates: 4 needs two votes in addition to his own.  4 can't make 3 happier, so since he won't get that vote anyway, 4 will offer nothing to 3.  So now 4 needs to make both 2 and 1 happier than the Three Pirate case.  So 4 can keep 498 coins, 3 will get nothing and will reject this plan, 2 will get one coin and vote for this plan, and 1 will get one coin and vote for this plan.

Five Pirates: 5 needs two votes in addition to his own.  If 5 tried to make 4 happier (giving him 499 coins for example) he could only make 3 happier by giving him one coin.  But then 5 gets nothing - there must be something else he can do.  If 5 gives up 4's vote and offers him nothing, then 5 can just concentrate on making two of 3, 2, and 1 happier.  He can easily and cheaply make 3 happier by giving him just one coin, since this is more than he would get in the Four Pirate case.  In order to make either 2 or 1 happier, 5 would need to give one of them 2 coins.

I can't tell if there is a real difference between "offer two coins to Pirate 2, and no coins to Pirate 1" and "offer no coins to Pirate 2, and two coins to Pirate 1."  I would think that it would make a little more sense to offer two to Pirate 2 since in ANY other case he get's either less coins or killed.  Technically Pirate 1 would be happy with 2 coins since that's more than he's get in the Four Pirate case, but there's that chance that the 500 coins possibility is lurking in the back of his head even though he should know he couldn't really get there.  So I wouldn't be confident with buying Pirate 1's vote with 2 coins.

So, I think Pirate 5 should offer a plan of keeping 497 coins, giving no coins to Pirate 4, one coin to Pirate 3, two coins to Pirate 2, and no coins to Pirate 1.

However, I wouldn't argue with someone who thinks Pirate 5 should offer a plan of keeping 497 coins, giving no coins to Pirate 4, one coin to Pirate 3, NO coins to Pirate 2, and TWO coins to Pirate 1.

Later!


  Posted by nikki on 2004-04-01 16:22:30
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