A pirate ship holding five pirates finds a treasure of 1000 golden coins. The treasure must be split among the five pirates. They are, in order of rank: Pirate 1, Pirate 2, Pirate 3, Pirate 4, and Pirate 5.

The pirates have three important characteristics: infinitely smart, infinitely bloodthirsty, and infinitely greedy.

Starting with the lowest ranked Pirate 5, they each must propose a way to split the treasure among them. If the proposal is not accepted by a majority of the pirates, the proposing pirate will be thrown overboard.

What proposal should Pirate 5 make?

ok I think I’ve got something. a few additional details to clarify:

-50% is not a majority and would get the pirate thrown overboard

-pirates are more greedy than bloodthirsty, even for one coin

-pirate 1 can beat pirate 2 if it comes down to them

-a pirate can vote for his own proposal

***WARNING: SPOILER BELOW***

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

I worked it backwards to get to the winning solution:

2 pirates left: if it were down to 2 pirates, pirate 2 would have no choice but to give all the gold to pirate 1. if he proposed anything else, pirate 1 would vote against it, the vote would be 50/50 and pirate 2 would go overboard. arguably pirate 1 would vote against pirate 2 and throw him overboard even if he gave him all the gold just to fulfill his bloodthirsty nature. I’ll proceed under that assumption.

3 pirates left: pirate 3 knows that pirate 2 is screwed no matter what so he will vote for anything p3 proposes. (in this sense, p1 also knows that he is screwed because it will never get down to 2 pirates). so p3 takes all the gold, and p2 votes in favor just to save his life. p1 votes against, but they have a majority.

4 pirates left: p4 needs 3 votes for a majority. p3 will vote against no matter what because he knows that when it gets down to 3 pirates, he will get all the gold. so p4 offers p1 and p2 each one piece of gold and takes 998 for himself, and gives nothing to p3. p1, p2, and p4 vote in favor.

5 pirates left: p5 also only needs 3 votes. as above, p4 will vote against no matter what since he gets 998 if p5 goes overboard. however, p3 knows that he will get nothing if p5 goes overboard so he will vote in favor for just 1 gold piece. p5 only needs one more vote from either p1 or p2. so he gives p1 2 pieces of gold (one piece more than p4 would give him) and gives p2 nothing since he doesn’t need his vote anymore. so p1 votes in favor and p2 against.

solution:

pirate 5 proposes that he take 997, p1 gets 2, p3 gets 1, and the rest

get nothing. p1, p3, and p5 vote in favor. p2 and p4 vote against. the

proposal passes.

thoughts?

Killer job.

Is it possible in your solution that p2 get 2 coins and p1 get nothing? (Basically swapping the two.)

Or, is it possible for p5 to propose that he take 998, p3 gets 1, and p1 gets 1? If this works, he makes out with another coin. Remember, with 5 pirates left, p3 needs to cut a deal pronto or he’s screwed when p4 proposes. If we can find a reason for one of the other guys to accept only 1 coin right here, it seems p5 could make out with 998.

yes, p1 and p2 are interchangeable. but whomever you choose has to get 2 coins because they would each get 1 from p4 anyway, so their bloodthirsty nature would cause them to toss p5 overboard just for the hell of it if the deal wasn’t better.

p3 gets 1 coin because he would get nothing from p4 anyway, so he needs an incentive not to also toss p5 for the hell of it.

so I think 997 is the best p5 can do.

Of course, yes. The bloodthirsty aspect of things does lend itself to an early killing of p5 if he couldn’t cut a better deal. I was thinking they’re infinitely greedy with their spare time, and figured they’d consider cutting a deal right now just to get the thing over with.

great riddle

I had the same riddle, but with the opposite conditions: the highest rank proposes a division of the loot, and a 50/50 vote counts as a passing vote for the proposal.

In that case, you can also take the problem backwards and examine the possibilities.

if 2 are left, the highest of rank will get all the gold.

If 3 are lest, the highest will get everything save one coin that he will give to the lowest of rank so that the middle man gets busted.

And so on.

The 5-pirate scenario ends with everything save 6 coins going to the highest ranking dude, while the p2 gets nothing, the p3 gets 1, p4 gets 2 and p5 gets 3. And that’s cause they’re bloodthirsty, which means that they will not settle for a solution that is “as good as it would have been, had they thrown the p1 overboard”.

Cheers!

If the pirates can make deals during this process than all of these solutions are incorrect. When p5 proposes that he essentially gets all the gold, p1 p2 and p3 realize that their only chance at gold is to agree to split it three ways. The problem with this riddle is that there are an infinite number of plausible solutions. With real pirates, they would all end up dead and the tresure goes to the bottom of the sea.

What if there are even number of pirates. Say 6.

With 6 pirates, you just have to make a better deal than they would get at 5 pirates. The pirates are “infinitely smart” and know that P5’s offer would be best for them if P6 doesn’t make a good offer. P6 will need 3 votes + his own.

P5’s offer:

P1 – 2 or 0 coins

P2 – 0 or 2

P3 – 1

P4 – 0

P5 – 997

P6’s offer would have to be:

P1 – 3 or 0 (interchangeable with P2)

P2 – 0 or 3

P3 – 2

P4 – 1

P5 – 0

P6 – 994

P6 must spend 3 on either P1 or P2, since the *possiblity* of getting 2 from P5 favors their greedy and bloodthirsty aspects.

I don’t think the high ranked pirate should give more than 1 coin to a pirate.

P5’s offer:

P1 – 1

P2 – 0

P3 – 1

P4 – 0

P5 – 998

P6’s offer would have to be:

P1 – 0

P2 – 1

P3 – 0

P4 – 1

P5 – 0

P6 – 996

This is sufficient since, P2 and P4 knows that if they vote against this distribution they might end up getting none as P5 will not give them even one coin. Hence if the pirates are really greedy then they need not give more than one coin to a pirate. Any takers?

I dont understand.. If p5 says that he wants 900+ in gold then why dont all of the other pirates say hell no and do away with him? Why would p1 and p3 or any p for that matter be happy with 1 coin? Maybe im not understanding the terms of the riddle here but if it starts with p5 proposing then no pirate would accept him getting 998 and the rest 1…

Pirate 1 – 50

Pirate 2 – 25

Pirate 3 – 0

Pirate 4 – 0

Pirate 5 – 25

Pirate 2 knows that if Pirate 1 goes overboard, he won’t get any coins.

Pirate 5 can possibly get 100 coins, but that won’t ever happen and I’ll prove that a little further down so he’s content with the 25 coins.

Pirate 1 – Overboard

Pirate 2 – 50

Pirate 3 – 0

Pirate 4 – 0

Pirate 5 – 50

I just threw these numbers out there because there’s no way Pirate 2 could appease more than 1 other Pirate in this situation, so this combination wouldn’t work and Pirate 2 gets thrown overboard.

Pirate 1 – Overboard

Pirate 2 – Overboard

Pirate 3 – 50

Pirate 4 – 50

Pirate 5 – 0

This is why Pirate 5 would take the Pirate 1’s offer of 25 coins because he knows that if it gets down to 3 Pirates, he’ll be the odd man out and get no coins. Pirate 4 says yes to this because he knows if Pirate 3 is thrown overboard, Pirate 5 would reject any of his offers and take all 100 coins to himself. Pirate 3 could theoretically pick 99 coins and give Pirate 4 1 and know that he would accept it because 1 coin would better than the 0 he would get after being rejected by Pirate 5.

+++ I was doing the riddle as if there was 100 coins not 1000, so you can do the math by yourself.

OK, haha, I’ve thought about it and I was wrong and I’ve changed my answer.

Pirate 1 – 50

Pirate 2 – 0

Pirate 3 – 0

Pirate 4 – 25

Pirate 5 – 25

Pirate 2 wouldn’t say yes to my original answer of 25 coins because he knows if Pirate 1 is thrown over, he would then get the 50 coins and the table would be

Pirate 1 – Overboard

Pirate 2 – 50

Pirate 3 – 0

Pirate 4 – 25

Pirate 5 – 25

But Pirate 4 and Pirate 5 were already offered 25 by Pirate 1, so there’s no use in saying no to that and saying yes to this 1, so this option wouldn’t come up. Pirate 4 knows that if Pirate 1 & 2 gets thrown over board, then he’s in a bad situation because has to pick his poison. Pirate 3 could pick 100 coins for himself and then it’d be up to Pirate 4 to decide who he wants to give 100 coins to because he knows if he says no and throws Pirate 3 overboard, then Pirate 5 will end up with the 100 coins by virtue of him rejecting Pirate 4’s proposal. So, if Pirate 3 decides to give himself 99 coins and Pirate 4 1 coin, Pirate 4 would accept this because 1 coin is better than 0.

So, that’s why Pirate 4 & 5 would say yes to Pirate 1’s original offer.

and sorry, I’m leaving a lot of comments, but I remember this riddle as Pirate 1 making the first proposal and so on and so forth, so that’s why my Pirate 5 isn’t the first to make a proposal.

Lets say the highest ranking pirate in P5 and lowest is P1

Starting backward in situation where only

Situation 1: P2 and P1 are left –> P2 has to give everything to P1 otherwise P1 wont agree and hence P2 will be overboard

Situation 2: P3,P2 and P1 are left –> P3 Proposes to keep 999 and give 1 to P2 and nothing to P1 – In this scenario P3 and P2 will agree but P1 won’t but they got more than 50% votes. P2 will agree since if they put P3 overboard he wont get anything not even this 1 based on Situation 1

Situation 3: P4,P3,P2 and P1 are left –> P4 proposes to take 998 coins and give P1 and P2 one coin each and nothing to P3.

P3 wont agree since in his power he will 999 coins but P2 will agree since even when P3 comes he will get only 1 coin and P1 also agrees since with P3 in power P1 wont get anything as Situation 2 above. So P4,P2 and P1 agrees to this solution.

Situation 4: All pirates are there –> P5 proposes to give one coin to P3 and P1 each and nothing to P4 and P2. P1 agrees since in this situation he will get one coin as in Situation 3. Here P2 also agrees since it doesn’t make any difference to him. P4 and P3 wont agree but who cares P5,P2,P1 have agreed and that is more than 50%.

If you consider their blood thirsty nature then P2 and P1 will reject P5 proposal and kill him and then accept to P4’s proposal (Situation 3) since they are getting the same money.

So P5 will be killed and P4 takes 998 coins.

For P5 to survive he has to give 2 coins to P2 and 1 to P1 and take 997. If the question is asked how much max coins a pirate can get then its 998 and that is P4by killing P5. Otherwise the possible division is 997 to P5 and 2 to P2 and 1 to P1

Little modification

Situation 4: All pirates are there –> P5 proposes to give one coin to P3 and P1 each and nothing to P4 and P2. P1 agrees since in this situation he will get one coin as in Situation 3. Here P2 also agrees since it doesn’t make any difference to him. P4 and P3 wont agree but who cares P5,P3,P1 have agreed and that is more than 50%.

So the best solution is P5 will take 998 and P3 and P1 will have 1 coin each