Chapter 10: Brainteasers 387
other words, he should make his allocation so that it is
preferred to the next scenario by sufficient numbers of
pirates to ensure that he gets a favourable vote.
Pirate 3 allocates 1000 to himself and nothing to the
others. Obviously Pirate 3 will vote for this. And so will
Pirate 2, if he votes against in the hope of getting some
loot he will find himself in the two-pirate situation...in
which case he could easily end up over the side.
Pirate 3 Pirate 2 Pirate 1
0 1000
1000 0 0Now to four pirates. Pirate number 3 is not going to vote
for anything number 4 says because he wants Pirate 4
in the deep. So there’s no point in giving him any share
at all. Pirates 2 and 1 will vote for anything better than
the zero they’d get from the three-pirate scenario, so he
gives them one each and 998 to himself.
Pirate 4 Pirate 3 Pirate 2 Pirate 1
1000 0
1000 0 0
998011With five pirates similar logic applies. Pirate 4 gets zero.
Then Pirate 5 needs to get two votes from the remaining
three pirates. What is the cheapest way of doing this?
He gives one to Pirate 3 and two to either of Pirates 2
and 1. Pirate 5 gets the remaining 997.
Pirate 5 Pirate 4 Pirate 3 Pirate 2 Pirate 1
1000 0
1000 0 0
998011
997 0 1 2 / 0 0 / 2Pirate 6 needs four votes to ensure survival, his own
plus three others. He’ll never get Pirate 5 so he needs
three votes from Pirates 4, 3, 2 and 1. Pirate 4 is cheap,