A classic logic puzzle involves a crew of five perfectly rational pirates attempting to divide 100 gold coins under a strict voting system. Their priorities: survival first, personal wealth second, and finally, maximizing the number of crewmates overboard if indifferent.

The Rules

  1. The most senior pirate proposes a split.
  2. All remaining pirates vote.
  3. A proposal passes with at least 50% (rounding up in case of odd pirates).
  4. If rejected, the proposer dies, and the next most senior pirate makes a new proposal.

Let’s label the pirates from A (most senior) to E (least senior).

Working Backwards

To understand what proposal pirate A should make, we analyze from the smallest group upward.

1 Pirate Left: Pirate E

Only E remains. He keeps all 100 coins. No vote needed.

2 Pirates Left: Pirates D and E

Pirate D must get one vote (his own) to pass a proposal. He gives 100 coins to himself, 0 to E.

Split:
D: 100, E: 0

3 Pirates: C, D, E

C needs 2 votes. He has his own. He can buy E’s vote with 1 coin.

Split:
C: 99, D: 0, E: 1

Why this works:

  • D would prefer C to die and revert to the 2-pirate case where he gets everything.
  • E gets 0 in the 2-pirate case, so 1 coin is better. He votes yes.

4 Pirates: B, C, D, E

B needs 2 votes beyond his own. He gives 1 coin to D (who would get 0 if B dies) to secure his support.

Split:
B: 99, C: 0, D: 1, E: 0

  • C would prefer B dead, hoping for 99 coins in the 3-pirate scenario.
  • D prefers 1 over 0.
  • E gets nothing either way, but prefers B dead, so votes no.
  • B gets yes from himself and D.

5 Pirates: A, B, C, D, E

A needs 3 votes. He gives 1 coin to C and 1 to E (both get 0 in the 4-pirate case), securing their votes.

Split:
A: 98, B: 0, C: 1, D: 0, E: 1

  • B would get 99 if A dies; votes no.
  • C gets 0 if A dies; votes yes.
  • D gets 1 if A dies; votes no.
  • E gets 0 if A dies; votes yes.

With votes from A, C, and E, the proposal passes.

Final Distribution

A: 98 coins
B: 0 coins
C: 1 coin
D: 0 coins
E: 1 coin

Reference