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:

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

There goes the Pirates of Carribean franchise...

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