Enter your email address:

June 8, 2010

Five Pirates

5 pirates of different ages have a treasure of 100 gold coins.

On their ship, they decide to split the coins using this scheme:

The oldest pirate proposes how to share the coins, and all pirates
remaining will vote for or against it.

If 50% or more of the pirates vote for it, then the coins will be shared
that way. Otherwise, the pirate proposing the scheme will be thrown
overboard, and the process is repeated with the pirates that remain.

Assuming that all 5 pirates are intelligent, rational, greedy,
and do not wish to die, (and are rather good at math for pirates)
what will happen?


João Marques said...

My solution is:

pirate 1 (P1) is the youngest
pirate 5 (P5) is the oldest

suppose there are 2 pirates left (P1 and P2)
P1 votes "no" and wins the 100 coins.
P2 will never let this happen.

suppose there are 3 pirates left
(P1, P2 and P3)

P2 will always agree with what P3 proposes if he can win something, so P3 can split (0 to P1, 1 to P2 and 99 to P3) and P2 and P3 will vote yes. (P1 doesn't like this)

suppose there are 4 pirates left.
P1 will agree if he can win anything, and P2 will agree if he can win more than 1 coin). P4 will split according to this: 1 to P1, 2 to P2, 0 to P3 and 97 to P4. (P3 is the only who votes "no")

suppose there are 5 pirates left.
P5 will want to make only 2 pirates unhappy, so he can get is proposal pass. He can give 2 coins to P1, and 3 coins to P2 and and he can stay with the other 95 for him.

P1 will vote "yes" and get 2 coins
P2 will vote "yes" and get 3 coins
P3 and P4 will vote "no" and they won't get any coin.
P5 will get 95 coins.

Tiger Woods said...

The question should state, more than 50% vote.

P1: 2
P2: 0
P3: 1
P4: 0
P5: 97

If there are only 2 pirates left then P2 will give all the coins to P1 so that he won't be thrown overboard. If there are 3 pirates then P3 will keep 99 coins and give 1 to P2. If there are four pirates left then P4 will keep 97 coins, give 2 to P2 and 1 to P1.

Anonymous said...

is there a definite answer?

Don said...

P5 eldest......P2, P1 (youngest)
Assuming only two men on deck at the end, is it possible for the youngest (P1) to push the elder guy (P2) overboard? Since we can't expect P2 to be honourable enough to jump overboard if P1 is dissatisfied with his solution, and again, since at least two guys are needed to push a guy overboard - we face a 1:1 solution => 50 gold coins each.

Poss 1. P3, P2, P1 remain
P3- " I can bring P1 over to my side with one more than the amount he will get if I am thrown overboard, i.e., 51. I shall keep 49 for myself."

P3 - 49, P2 - 0, P1 - 51

Poss 2. P4, P3, P2, P1 remain

Now P4 knows if he is thrown aboard, then, according to Poss 1, P2 gets nothing. So he wins P1's allegiance as P3 would do, i.e., with 51 coins and wins over P2 to his side with 1 gold coin. P3 can go to hell.

P4 - 48, P3 - 0, P2 - 1, P1 - 51

Poss 3. P5's gotta stay alive and keep some gold for his rum.

He would think along lines similar to P4's. He needs the support of at least 2 guys.

P5 - 48, P4 - 0, P3 - 0, P2 - 1, P1 - 51

Couple of assumptions -
# 1. Last two men are hopefully evenly matched. That will cause their rationale to overcome greed and share it equally.
# 2. I have assumed the pirates are all guys. Any Elizabeth Swanns out there kindly excuse me.

Matt said...

I'd like to use backwards induction for this problem.

For simplicity I will name the pirates in the position they get to propose a share Pirate 1 being the oldest and Pirate 2 the next in line to choose if Pirate 1 gets thrown overboard.

100 coins between P1 through P5
P5: 100

P4: 100
P5: 0
(P4 votes yes, P5 votes no, still goes through)

P3: 99
P4: 0
P5: 1
(P3 and P5 yes, P4 no, still goes through)

P2: 99
P3: 0
P4: 1
P5: 0
(P2 and P4 vote yes, P3 and P5 no, still goes through with 50% votes)

P1: 98
P2: 0
P3: 1
P4: 0
P5: 1
(P1, P3, and P5 vote yes, P2 and P4 vote no, still goes through)

THUS with the solution, this is what will happen, P1 makes proposal of 98 coins for him, none for P2, 1 for P3, 0 for P4 and 1 for P5 and the offer goes through with 3 out of the 5 pirates voting in favor of the proposal because the 3 in favor are better off than if they let the game progress to the next stage.

Secret Squïrrel said...

I agree with Matt.

Señor Marques errs when he says that (for 3 pirates) P2 will agree if he can win anything. All he needs to do is vote "no" with P1 and he will end up with 100 coins when P3 is thrown overboard. P1 would vote "no" simply because he will otherwise receive nothing - he can't be any worse off and there is a possibility that P2 will give him something (he doesn't know what we know).

Don errs by not realising that the problem stated quite clearly that if 50% or more pirates vote for a method then that is how the coins are shared. Therefore, when there are 2 pirates, the leader voting to keep all of the coins for himself constitutes 50% of the pirates so he is not thrown overboard and keeps all of the coins.

It is bad luck to have a woman onboard a pirate ship so all pirates are men. Pirates with female bits are classified as cabin boys. ;)

Swadhi said...

You have a really great site going!

Anonymous said...

We don't know what happens because we are not there and don't have sufficient information about them.

But lets go down all the way. P5 is the oldest and P0 is the youngest. I assume that their fighting skills are irrelevant to the outcome of this money distribution. The pirates are greedy and intelligent.

The proposal of pirate 5 needs 50% support otherwise he will be thrown overboard, in reality that means the support of three pirates. His proposal might be 32 gold for the three oldest and 2 for the two youngest. With pirate 5 overboard the remaining 4 pirates have increased their chances to more gold.

With 4 pirates left the proposing pirate needs the 2nd oldest pirate accomplish to get the needed 50%. If he can persuade him then he can make any demand, say 45 gold for each and 5 for the two youngest.

But there is the danger of the youngest persuading him to vote against and become first in command. After that a solution might be 46 for the two oldest and 8 for the youngest.
But the bottemline is with the two oldest pirates thrown overboard for the increase of money of the remaining pirates, why should the three youngest pirates keep the rules and trust one another not to be killed for their gold later on the journey? How old is the youngest pirate anyway?

Secret Squïrrel said...

Anonymous (welcome back, Evan), you clearly haven't understood the problem or Matt's explanation.

Pirates an even number "down" from the eldest will always accept a proposal that gives them 1 coin, since they will be offered none if they are an odd number "down". The eldest is greedy and smart, and wishes to maximise his haul. He's not going to offer 32 coins to anybody.

The reason any pirate would "keep to the rules" is irrelevant. You are given that in the statement of the puzzle. Why don't you ask why they chose to become a pirate?

Their actual ages are also irrelevant, save for ordering them accordingly.

Ashish said...

p-1 have 98
p-2 have 1
p-3 have 0
p-4 have 1
p-5 have 0

Christina said...

gr8 site...keep posting

Johnny Come Lately said...

A basic rule is being misinterpreted here. "The oldest pirate proposes how to share the coins, and all pirates remaining will vote for or against it." The proposer does not vote. Therefore, if we call them P5 thru P1 with P5 being the oldest:

P5 needs 2 of 4 votes,
P4 needs 2 of 3 votes,
P3 needs 1 of 2 votes, and
P2 needs P1's vote.

P2 will only get P1's vote for proposing to give all the coins to P1, so if P5, P4 and P3 are thrown overboard then P2 gets nothing. It is thus in P2's interest to vote for any proposal that brings him a single coin.

Knowing this, if P5 and P4 were thrown overboard, P3 could successfully propose to give one coin to P2 and keep the remaining 99 coins. P1, then, has no chance of taking any money if P4 is thrown overboard, so it is in his best interest to vote for any proposal by P4 or P5 that brings him a single coin.

Since P5 and P4 each need two votes to succeed, either could successfully propose to give one coin each to P1 and P2 and keep the remaining 98 coins. That means P4 is out of luck because P5 gets the first opportunity, and the outcome is:

P5: 98 coins
P2: 1 coin
P1: 1 coin

Anonymous said...

Hey.. But the question say the remaining pirates vote.. This means the pirate who propose dun get to vote right?

Bryan said...

P5 will always turn down proposals,since he's the youngest. All P's knows this. P4 knows that if he turns all proposals down until his turn then he will be thrown overboard by p5 so he will accept only p3's proposal,thus turn down p1 and p2. p3 realizes this and turns down all other proposals until his time. P2 is the only one who will vote for p1's proposal,b/c he realizes that when his turn is up everyone will turn it down. Thus p1 is thrown out and p2 also. P3 only needs 1 vote and since he's greedy he keeps all coins to himself since knows p4 will vote for him no matter what.

Anonymous said...

Depends on whether there are sharks in the sea. Cuz if there are the first oldest pirate would not take anything for himself definitely, and split the money among two other pirates only. Then two of them and him would vote for it, so it'd be a 3-2 vote for the oldest pirate.
The two of them would vote for it cuz two pirates sharing the booty is the best possible outcome.
Oldest pirate keeps his ass intact.

But if being thrown overboard is only a mild inconvenience, Im sure he'd try to split some money his way. The rest he splits among the two other older pirates who probably fear being thrown out next, so they'd vote for him.

As the second oldest pirate to vote, you're in as much disadvantage as the first. You'd have to win the vote of two other pirates too. So he'd probably vote for the first pirate even if he took a little money.

But IMO pirates will vote out each other till there's two left. And one will pull out a sword to stab the other.

Anonymous said...

P5 for the sake of self presevartion splits it between p4- 1 coin and p3-99.

P4 realises this is the best deal he's get since he will be the next one to appease the greedy pirates. And probably get 0 coins to save himself.
P3 also realises that this is the best deal he will get.

Anonymous said...


P1 = youngest pirate
P5 = eldest pirate

P1: 2
P2: 0
P3: 1
P4: 0
P5: 97


Working backwards makes it easier to work out:

(two pirates remaining)
P1: 100
P2: 0

P2 must give all his coins to P1, otherwise P1 would vote to have P2 removed, leaving him with the remaining coins. Although this situation leaves him with no coins, it is the only situation that he is able to live.

(three pirates remaining)
P1: 0
P2: 1
P3: 99

P1 will vote to NO unless he is given 100 coins, as voting no will lead to the situation where he gets all coins. In order to be certain that P2 votes yes, P3 must only give 1 coin.

(four pirates remaining)
P1: 1
P2: 2
P3: 0
P4: 97

P1 and P2 know that the three-pirate situation will lead to them receiving less, so they vote YES.

(five pirates remaining)
P1: 2
P2: 0
P3: 1
P4: 0
P5: 97

P1 and P3 know that the four-pirate situation will lead to them receiving less, so they vote YES.