Even greedier solution to 63. pirates
How about this: The three least senior pirates all say in unison "I demand to get all 100 gold coins" (can't get any greedier than that...). The most senior pirate will not be able to fulfill that request for more than one of them, thus whatever proposal he has, it will be voted down and he dies.
The next most senior pirate now absolutely needs one vote. The threesome stick to their demands, and so he has to select at random one of them and give him all the coins to save his own life.
Monday, January 03, 2005
That breaks more of the problem than just how greedy the pirates are. However you could claim pirates 1,2,3 realise this stratigy gives them a chance of more gold with no risk of death, so I'll go with it, because that makes them betting men.
In the two pirate case, pirate two gets 100 gold.
In the three pirate case, pirate 3 cannot bribe pirate 2 because pirate 2 can have him killed and still get all the gold. Pirate 1 knows that pirate 3 values his life more than gold and informs him that the only way to get his vote is all 100 gold pieces.
In the four pirate case, pirate 4 cannot give the gold to pirate 1 because pirate 1 can watch him die and get the gold anyway, so would have to give all the gold to 2 or 3 since they both realize that pirate 4 values his life and will bargin away all the gold for it. While each ones expected return is only 50 golds (50% probability of being picked in pirate 4's plan) they are both willing to bet for the full 100.
In the five pirate case, pirate 5 cannot get vote 2 or 3 because 2 and 3 are willing to bet on the four person case. But now pirate 1 should NOT be willing to move to the four person case because he wouldn't get anything. An intellegent pirate would then realise that he has to spilt the money with pirate 4. You can still claim that pirate 5 would have to give up all the money to save his life, but now your going to have to come up with more of your wierd logic to decide what the spilt is.
Pirate 1 cannot just keep his 100 golds demand. There is no way to make it logical, because he cannot get it if we start with 5 pirates. Ever. Which makes for a pretty bad bet.
Friday, January 14, 2005
This idea does not work ...
What happens if pirate 4 does not give 100 gold to one of pirates 1, 2 or 3? He dies, but then ...?
Pirate 3 has 100 gold to distribute ... He will give 1 gold to pirate 1 and keep 99 gold himself. If pirate 1 does not vote for this plan, he will get NOTHING! So, pirate 1 and 3 both vote for this plan ... Pirate 2 gets nothing.
So, back to pirate 4 case ...
He keeps 99 himself and gives 1gold to pirate 2. Why does pirate 2 vote for this plan? Because otherwise he gets NOTHING!
Of course this is based on the given information - the pirates are extremely greedy and dont want to die. The only way for pirate 2 to get any money is to vote for pirate 4's plan. He would never make an agreement with pirates 1&3 ... if they were choosing, he would ge nothing!
Wednesday, January 26, 2005
What do you mean the idea doesn't work? Lots of greedy people gamble. The point is that intellegent people gamble BETTER. So the question becomes "how do the pirates gamble?" With the original solution, the pirates arn't willing to gamble, they take the 100% likelyhood of success and get 1 gold piece. However, Joakim Rosqvist argued that with P1,P2,P3 changing thier stratigies, they could raise their expected return, 1/3 chance of getting 100 gold = 33 golds expected return. I was just pointing out that the probabilities weren't equal.
Thursday, January 27, 2005
Fog Creek Home