I enjoy the weekly brain teasers from FiveThirtyEight’s puzzle master, Oliver Roeder. Here’s an intriguing challenge from this week:

The Puzzle Of The Pirate Booty:

Ten Perfectly Rational Pirate Logicians (PRPLs) stumble upon 10 gold pieces and need to devise a fair way to split the treasure among themselves.

Each pirate holds a distinct rank, starting from the captain downwards. The captain proposes an initial distribution plan, and all pirates (including the captain) vote on it. If over half of the pirates support the plan, it is put into action; otherwise, the captain is eliminated, the next in line takes charge, and the process begins anew. (These PRPLs are quite rebellious.)

The pirates follow these voting rules:

1. Self-preservation: A pirate values their life above all else.
2. Greed: A pirate desires maximum gold.
3. Bloodthirst: Absent a threat to their life or treasure, a pirate chooses to eliminate another.

Given these guidelines, how do the PRPLs divide the gold?

Extra credit: Solve the problem with P pirates and G gold pieces.

___________

I believe I’ve cracked it—tap below to uncover and read my solution:

[expand title=”SOLUTION“]

See it from the perspective of the lowest-ranked pirate and progress upwards.

The bottom-ranked pirate (let’s name him P10, with the captain being P1) would ideally want all others eliminated to seize the full 10 gold pieces once he becomes captain, signaling all others are gone.

P9 also favors eliminating those above him. If only he and P10 remain, he can take all 10 pieces for himself while denying P10 any. The vote would end 1-1, ensuring his safety. Thus, there’s no threat to his life.

P8 grasps this and comprehends the potential outcomes when only three pirates remain, and he’s captain:

P9: 10 pieces
P10: 0 pieces

In this scenario, P9 aborts P8’s plan to avoid receiving nothing. Yet, P10 has nothing to lose and votes to kill per rule #3. To secure P10’s vote, P8 should offer him a single piece. P10’s greed prompts him to vote for the plan to gain something before P8 is eliminated, leaving P8 with nine gold pieces. Hence, P8’s allocation:

P8: 9
P9: 0
P10: 1

If four pirates remain, with P7 as captain, he considers all these factors, recognizing he needs just one favorable vote from the three others to survive. His assignment:

P7: 9
P8: 0
P9: 1
P10: 0

Despite knowing that P8 and P10 will oppose, P7 wins P9’s vote, as P9 prefers having one piece over none. P7’s plan is accepted.

This strategy continues up the hierarchy. In every case, the top-ranked pirate can avoid elimination by allotting a piece to those receiving none if he dies:

For the final five pirates, the division would be:

P6: 8
P7: 0
P8: 1
P9: 0
P10: 1

With six pirates left:

P5: 8
P6: 0
P7: 1
P8: 0
P9: 1
P10: 0

With seven pirates remaining:

P4: 7
P5: 0
P6: 1
P7: 0
P8: 1
P9: 0
P10: 1

If eight pirates are left:

P3: 7
P4: 0
P5: 1
P6: 0
P7: 1
P8: 0
P9: 1
P10: 0

With nine pirates remaining:

P2: 6
P3: 0
P4: 1
P5: 0
P6: 1
P7: 0
P8: 1
P9: 0
P10: 1

Ultimately, no one dies because the original captain splits the gold as follows:

P1: 6
P2: 0
P3: 1
P4: 0
P5: 1
P6: 0
P7: 1
P8: 0
P9: 1
P10: 0

Gathering votes from P3, P5, P7, P9, and himself leads to a 5-5 split, enacting his plan.

The solution is: 6, 0, 1, 0, 1, 0, 1, 0, 1, 0