Pirates and Coins

Five pirates discover a treasure chest containing 100 gold coins. Being logical and ruthless pirates, they must decide how to divide the loot. They follow a strict hierarchy based on rank, with Pirate 1 being the most senior down to Pirate 5 being the most junior.

The process works as follows: the most senior pirate proposes a distribution of the coins. All pirates (including the proposer) then vote on whether to accept the proposal. If at least half of the pirates vote in favour, the proposal is accepted and the coins are distributed accordingly. If the proposal is rejected, the proposer is thrown overboard, and the process repeats with the next most senior pirate.

Each pirate’s priorities are, in order of decreasing importance:

survive,
maximise their own coins, and
throw other pirates overboard.

What distribution should Pirate 1 propose to maximise their coins whilst ensuring survival?

Hint

Start by analysing what would happen if there were only 2 pirates left.

Hint

Work backwards from the simplest scenario to understand what each pirate would get in subsequent rounds.

Hint

Pirate 1 needs to secure exactly 3 votes (including their own) to pass the proposal.

Solution

Pirate 1 should propose: 98 coins for themselves, 0 for Pirate 2, 1 for Pirate 3, 0 for Pirate 4, and 1 for Pirate 5.

Let’s work backwards from the simplest case.

2 pirates remaining (Pirates 4 and 5): Pirate 4 proposes 100 for themselves, 0 for Pirate 5. Pirate 4’s vote alone (1 out of 2) constitutes half the votes, so the proposal passes.

3 pirates remaining (Pirates 3, 4, and 5): Pirate 3 needs 2 votes to pass the proposal. Pirate 4 would get 100 coins in the next round, so Pirate 3 cannot win Pirate 4’s vote at any reasonable price. However, Pirate 5 gets nothing in the next round. Pirate 3 can offer Pirate 5 just 1 coin to secure their vote, since 1 coin is better than 0. Pirate 3 proposes 99-0-1, which passes with votes from Pirates 3 and 5.

4 pirates remaining (Pirates 2, 3, 4, and 5): Pirate 2 needs 2 votes. Pirate 3 would get 99 coins in the next round (too expensive to buy). Pirate 4 gets 0 in the next round, so offering 1 coin secures their vote. Pirate 5 gets 1 in the next round, but Pirate 2 can offer them 0 coins since they already have Pirate 4’s vote. Pirate 2 proposes 99-0-1-0, which passes with votes from Pirates 2 and 4.

5 pirates remaining (Pirates 1, 2, 3, 4, and 5): Pirate 1 needs 3 votes (including their own). Pirate 2 gets 99 coins in the next round (too expensive). Pirate 3 gets 0 in the next round, so offering 1 coin wins their vote. Pirate 4 gets 1 in the next round. By priority 3, if Pirate 1 offers exactly 1 coin, Pirate 4 is indifferent and will vote no to throw Pirate 1 overboard. Thus Pirate 4 cannot be won over. Pirate 5 gets 0 in the next round, so offering 1 coin wins their vote.

Pirate 1 proposes 98-0-1-0-1, securing votes from Pirates 1, 3, and 5. This passes with 3 out of 5 votes.

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