The Riddler — Can You Solve The Puzzle Of The Pirate Booty?

Took anoth­er one week hia­tus from The Rid­dler, but am back to this one’s!

The Rid­dler is a week­ly puz­zle series from Oliv­er Roed­er at FiveThir­tyEight that usu­al­ly involves some com­bi­na­tion of math, log­ic and probability.

This weeks FiveThir­tyEight Rid­dler, Can You Solve The Puz­zle Of The Pirate Booty is:

Ten Per­fect­ly Ratio­nal Pirate Logi­cians (PRPLs) find 10 (indi­vis­i­ble) gold pieces and wish to dis­trib­ute the booty among themselves.

The pirates each have a unique rank, from the cap­tain on down. The cap­tain puts forth the first plan to divide up the gold, where­upon the pirates (includ­ing the cap­tain) vote. If at least half the pirates vote for the plan, it is enact­ed, and the gold is dis­trib­uted accord­ing­ly. If the plan gets few­er than half the votes, how­ev­er, the cap­tain is killed, the sec­ond-in-com­mand is pro­mot­ed, and the process starts over. (They’re muti­nous, these PRPLs.)

Pirates always vote by the fol­low­ing rules, with the ear­li­est rule tak­ing prece­dence in a conflict:

1. Self-preser­va­tion: A pirate val­ues his life above all else.
2. Greed: A pirate seeks as much gold as possible.
3. Blood­thirst: Fail­ing a threat to his life or boun­ty, a pirate always votes to kill.

Under this sys­tem, how do the PRPLs divide up their gold?

Extra cred­it: Solve the gen­er­al­ized prob­lem where there are P pirates and G gold pieces.

Unfor­tu­nate­ly, I don’t have any gifs or R sim­u­la­tions this week. Instead I’ll be tak­ing an iter­a­tive, log­i­cal approach.

We start with the case of 1 pirate.

Num­ber of Pirates/Pirate Rank	Rank A (Cap­tain)
1 Pirate	10 Gold Pieces

If there’s only 1 pirate, he gets all the gold. Now at two pirates:

Num­ber of Pirates/Pirate Rank	Rank A (Cap­tain)	Rank B
1 Pirate	10 Gold Pieces
2 Pirates	10 Gold Pieces	0 Gold Pieces

At 2 pirates, Pirate A (the cap­tain) still keeps 10 gold pieces. Pirate B, has no threat to his life (rule 1) & vot­ing to mutiny against Pirate A moves him up to the 1 pirate sce­nario, which means he gets 10 gold pieces (rule 2, pirates seeks as much gold as pos­si­ble). Con­se­quent­ly, Pirate B votes to mutiny, but since Pirate A does not vote to mutiny, the crew lacks the major­i­ty it needs for a mutiny.

In the case of 3 pirates:

Num­ber of Pirates/Pirate Rank	Rank A (Cap­tain)	Rank B	Rank C
1 Pirate	10 Gold Pieces
2 Pirates	10 Gold Pieces	0 Gold Pieces
3 Pirates	9 Gold Pieces	0 Gold Pieces	1 Gold Piece

In the case of 3 pirates, the cap­tain will vote against the mutiny to save his own life (rule 1). Pirate B will always vote for mutiny since mov­ing up to the 2 pirate sce­nario means Pirate B gets 10 gold pieces. The key becomes Pirate C, If pirate C gets just 1 gold piece, he won’t vote for the mutiny since that means the crew moves to the 2 pirate sce­nario (& Pirate C becomes Pirate B), and he won’t get any gold pieces. Pirate A keeps 9 gold pieces, since by rule 2 pirates are greedy, and he only needs to give away 1 gold piece to Pirate C to gar­ner his vote. Con­se­quent­ly, only Pirate B votes for the mutiny, and the crew lacks the major­i­ty to kill the captain.

In the case of 4 pirates:

Num­ber of Pirates/Pirate Rank	Rank A (Cap­tain)	Rank B	Rank C	Rank D
1 Pirate	10 Gold Pieces
2 Pirates	10 Gold Pieces	0 Gold Pieces
3 Pirates	9 Gold Pieces	0 Gold Pieces	1 Gold Piece
4 Pirates	9 Gold Pieces	0 Gold Pieces	1 Gold Piece	0 Gold Pieces

Every­thing from the 3 pirate sce­nario holds true. In this case the cap­tain does­n’t need to give any gold pieces to Pirate D, since Pirate C and him­self already vote against the mutiny, and con­se­quent­ly, only pirate B/D vote for the mutiny, but lack the major­i­ty to enact it.

Now the 5 Pirate Scenario

Num­ber of Pirates/Pirate Rank	Rank A (Cap­tain)	Rank B	Rank C	Rank D	Rank E
1 Pirate	10 Gold Pieces (GP)
2 Pirates	10 Gold Pieces (GP)	0 GP
3 Pirates	9 Gold Pieces (GP)	0 GP	1 GP
4 Pirates	9 Gold Pieces (GP)	0 GP	1 GP	0 GP
5 Pirates	8 Gold Pieces (GP)	0 GP	1 GP	0 GP	1 GP

In the 5 pirate sce­nario, sim­i­lar to the 3 pirate sce­nario, the cap­tain has to bribe Pirate E with 1 gold piece so that he will vote against the mutiny.

Say Pirate E gets 0 pieces of gold. There’s no immdi­ete threat to his own life (rule 1). Killing the cap­tain moves him to the 4 pirate sce­nario where he’s rank D, and still gets no extra gold. So greed isn’t a dri­ving fac­tor (rule 2). How­ev­er, since these pirates are a blood­thirsty bunch (rule 3), Pirate E will vote to mutiny with­out a bribe.

At this point the pat­tern should be clear. Every oth­er pirate after the cap­tain gets 1 gold piece each, and the cap­tain keeps the rest for him­self. So the 10 pirate sce­nario yields:

Num­ber of Pirates/Pirate Rank	Rank A (Cap­tain)	Rank B	Rank C	Rank D	Rank E	Rank F	Rank G	Rank H	Rank I	Rank J
1 Pirate	10 Gold Pieces (GP)
2 Pirates	10 Gold Pieces (GP)	0 GP
3 Pirates	9 Gold Pieces (GP)	0 GP	1 GP
4 Pirates	9 Gold Pieces (GP)	0 GP	1 GP	0 GP
5 Pirates	8 Gold Pieces (GP)	0 GP	1 GP	0 GP	1 GP
6 Pirates	8 Gold Pieces (GP)	0 GP	1 GP	0 GP	1 GP	0 GP
7 Pirates	7 Gold Pieces (GP)	0 GP	1 GP	0 GP	1 GP	0 GP	1 GP
8 Pirates	7 Gold Pieces (GP)	0 GP	1 GP	0 GP	1 GP	0 GP	1 GP	0 GP
9 Pirates	6 Gold Pieces (GP)	0 GP	1 GP	0 GP	1 GP	0 GP	1 GP	0 GP	1 GP
10 Pirates	6 Gold Pieces (GP)	0 GP	1 GP	0 GP	1 GP	0 GP	1 GP	0 GP	1 GP	0 GP

So that solves the orig­i­nal ques­tion of “how do the PRPLs divide up their gold”!

What about the gen­er­al­ized case of P Pirates and G gold pieces. Under the 10 pirate sce­nario, the min­i­mum amount of gold pieces need­ed to pre­vent a mutiny is below:

Num­ber of Pirates/Pirate Rank	Rank A (Cap­tain)	Rank B	Rank C	Rank D	Rank E	Rank F	Rank G	Rank H	Rank I	Rank J
10 Pirates	0 Gold Pieces (GP)	0 GP	1 GP	0 GP	1 GP	0 GP	1 GP	0 GP	1 GP	0 GP

The cap­tain still doles out 1 gold piece each to every oth­er pirate after him (total of 4 pieces), in order to guar­an­tee their vote against mutiny. The cap­tain’s main pri­or­i­ty, per rule 1, is to keep him­self alive, so he does­n’t need any gold for himself.

Say, only 3 gold pieces are avail­able. The cap­tain can no longer bribe Pirate I, and con­se­quent­ly, Pirate I votes for the mutiny, cre­at­ing 6 votes for the mutiny and 4 votes against. So we can see that the amount of gold pieces deter­mine the max­i­mum amount of pirates that are alive. In equa­tion form it looks like;

Pirates Alive ℗  = [Gold Pieces (G) * 2] + 2

The amount of gold the cap­tain gets is:

Gold Cap­tain Gets = Gold Avail­able (G) — [[Num­ber of Pirates ℗ / 2] — 1] 

where the quan­ti­ty [Num­ber of Pirates ℗ / 2] is round­ed up to the near­est itneger

So going back to the ques­tion, the answer to how gold is dis­trib­uted with 10 pirates and 10 pieces of gold:

The cap­tain gets 6 pieces of gold, with every oth­er pirate fol­low­ing the cap­tain get­ting a piece each (4 total pirates). 

And for the gen­er­al case of P pirates and G gold.

There is a mutiny until  [Gold Pieces (G) * 2] + 2 > Pirates Alive. After which every oth­er pirate fol­low­ing the cap­tain gets a gold piece each, with the cap­tain tak­ing the bal­ance of gold pieces.