Logicians Are Crazy

Time for the solution to the pirates problem from last time. The comments section last time gave away the essentials of the solution, so odds are nobody is going to be surprised by anything in the answers here, but whatever.

I guess I'll solve the four problems in the order I gave them in. For notation, the pirates will be labeled in order of ages as 1,2,3,4, and 5, with 5 being the oldest. First:
The proposing pirate gets a vote, and ties result in the motion passing.

Alright, so if only pirates 1 and 2 are left, then pirate 2 can propose to take all the gold, and the motion will pass. Thus, if 1,2,3 are all left, pirate 3 can offer just 1 gold to pirate 1 and pirate 1 will accept. So then, for 1,2,3,4, pirate 4 can offer 1 gold to pirate 2 and it will pass. Finally, with 1,2,3,4,5 all there, pirate five simply offers 1 gold to pirates 3 and 1.
Pirate 5 gets 98 gold, pirate 3 and pirate 5 both get 1, motion passing 3-2.

Inductioning this up is fairly easy, and will stop when the proposing pirate runs out of gold (Inductioning is a word, right?). Moving on to the next case:
The proposing pirate does not get a vote, and ties result in the motion passing.

Now, if there are only 2 pirates left, pirate 2 is dead no matter what he proposes (as pirate 1 would rather kill him and take all the money than just take all the money. Thus, pirate 2 will accept anything pirate 3 proposes, so pirate 3 will take all the gold himself if 3 pirates are left. With 4 pirates left, pirate 4 can propose 1 gold to pirates 1 and 2 to get their support. In the 5 pirate case, pirate 5 will have to offer 1 gold to pirate 3 to get his support, and will have to offer 2 gold to either pirate 1 or 2, it does not matter which.
Pirate 5 gets 97 gold, pirate 3 gets 1 gold and one of pirates 1 or 2 get 2 gold, the other getting zero, motion passing 2-2.

This is a bit harder to induction up, since at 6 pirates, pirates 1 and 2 do not know who pirate 5 will choose if he gets the chance. Next:
The proposing pirate gets a vote, and ties result in the motion failing.

This problem actually comes out exactly the same as the last one, as if there are 2N pirates left in this case, then N votes of support besides the proposing pirate are needed to pass, just like in the last one (as only 2N-1 votes are actually cast in the last one). If there are 2N+1 pirates left, then N votes of support are needed outside of the proposing pirate, just like the last one.
Pirate 5 gets 97 gold, pirate 3 gets 1 gold and one of pirates 1 or 2 get 2 gold, the other getting zero, motion passing 3-2.

Again, hard to induction up (induction is a verb, right?).
Final case:
The proposing pirate does not get a vote, and ties result in the motion failing.

This one is a bit different, now if there are 2 pirates left, pirate 2 is just dead as before. If 3 pirates are left, pirate 3 is just dead though, as pirate 1 will vote down anything still. So with 4 pirates left, pirates 2 and 3 will support anything, as they are dead otherwise. Finally, with 5 pirates, pirate 5 will offer 1 gold to pirates 1, 2, and 3 to get their support.
Pirate 5 gets 97 gold, pirate 1, 2, and 3 each get 1, motion passing 3-1.

This will induction up to 6 pirates, but no further because at 6 the pirate must choose if he offers 2 gold to two of {1,2,3}, but he does not care which ones.

In conclusion, the oldest pirate wins, alot. Also, logicians are crazy.

1 comment:

edison said...

You've completely ruined the romantic image about pirates that people have in their mind...

There goes the Pirates of Carribean franchise...