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Logic Level 2

Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).

The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go agree with him, the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.

If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.

What is the maximum number of coins the captain can keep without risking his life?


The answer is 98.000.

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4 solutions

Sandeep Bhardwaj
May 15, 2015

Assumptions :

The captain is the pirate 5, the senior most of all the second senior most being pirate 4, the third senior most being pirate 3, the fourth senior most being pirate 2 and the most junior being pirate 1. \boxed{\text{The captain is the pirate 5, the senior most of all} \\ \text{the second senior most being pirate 4,} \\ \text{ the third senior most being pirate 3,} \\ \text{the fourth senior most being pirate 2} \\ \text{and the most junior being pirate 1.}}

The captain says he will take 98 98 coins, and will give one coin to the third most senior pirate and another coin to the most junior pirate. He then explains his decision in a manner like this :

CASE 1: If there were only 2 pirates left. Pirate 2 being the most senior, he would just vote for himself and that would be 50 % 50\% of the vote,and pirate 1 would get nothing. and pirate 2 is obviously going to keep all the money for himself.

CASE 2: If there were only 3 pirates left . Pirate 3 has to convince at least one other person to join in his plan. Pirate 3 would take 99 99 gold coins and give 1 coin to pirate 1. Pirate 1 knows if he does not vote for pirate 3, then he gets nothing (since if he doesn't join the plan, Pirate 3 is killed & CASE 1 comes in action where he doesn't get anything.) So obviously pirate 1 is going to vote for this plan.

CASE 3: If there were 4 pirates left . Pirate 4 would give 1 1 coin to pirate 2, and pirate 2 knows if he does not vote for pirate 4, then he gets nothing (since if he doesn't join the plan Pirate 4 is killed & CASE 2 comes in action where he doesn't get anything). So obviously he is going to vote for this plan.

Finally now there are 5 pirates , so the captain would give one coin each to pirate 1 and pirate 3 since they know that if they don't vote for pirate 5,then they will get nothing (as Pirate 5 will be killed and CASE 3 will come in action). So obviously they are going to vote for this plan.

So, the captain can get maximum of 98 c o i n s \boxed{98 coins} .

But pirate 1 will get one coin even if pirate 5 is killed, so he needs to be paid off with a second coin in order to vote with pirate 5. That brings the maximum coins that the captain can keep to 97, as I see it.

Raffi Kasparian - 3 years ago

If the captain is killed, pirate 4 will win by giving one coin to pirate 2. Thus pirate 1 will not get any coins. Remember, they can all predict the actions of the others because they have infinite intelligence.

Ethan Lukman - 2 years ago

Let's start by naming them from the captain down to the cabin boy ;)

A, B, C, D, & E

Now we have 4 Iterations\cases, the logical approach to solve such problem is to start from the last case back to the first state

Remember the proposal passes by 50% or more so 50% is enough for it and also that every one thinks of what is best for himself only, they don't scheme together nor play it fair (hint their greed drives them)

You will notice that the last state\case is the key to all this

Of course there is a hint missing or let's say it's included but should be emphasized which is "if any pirate finds out that he'll get the same amount of coins whether he agreed or not to the proposal he'll say no"

Mohammad Farhat
Aug 31, 2018

See this TED video

Mr Yovan
Feb 20, 2016

and if there were n n pirates?is there are formula for this?

Have you tried extending the logic to 6, 7, 8, 9, ... pirates? Did any patterns emerge?

Eli Ross Staff - 5 years, 2 months ago

if the number of pirates are odd then --> x gold - (n crews - 1)/2 if the number of pirates are even then --> x gold - (n crews/2 - 1)

Christopher Hurly - 4 years, 10 months ago

Crewman B proposes a mutiny, and as new captain will give each crewmember C, D and E 21 gold, which is more than a fair share divided by five.

The captain in this case is a dead man.

The way B proposes to divvy up the loot, in the face of the Captain's poor choice could be arbitrary, however this is his best bet.

E will never be captain, as E cannot mutiny alone against D.

Likewise, D will never be captain as E will not get any loot if E agrees to mutiny against C.

C is an undesirable captain to both D and E. As long as either D or E gets 1 gold, it wont matter to E either way. If C pays one gold to E, E wont mutiny.

B, on the other hand should pay out to at least 2 crewmembers to avoid mutiny. He could pay out one 33 to one member, 34 to another member, and keep 33 to himself, just to make sure C cant bribe Either D or E with one more gold. It doesnt really matter which...

But he doesnt have to really beat C's maximum bid. D and E know they run the risk of getting nothing if C is ever captain. The best they can hope for is a pittance.

So B splits 63 gold among the rest of the crew, one more than an equal share if the captain had been smarter. He could offer less, and in the face of the captain having walked the plank, he shouldn't. Everyone gets more than the fair share if they hadn't mutinied, everyone gets more than if they agreed with the captain, and B, as the new captain gets 37 gold.

If A realized that dead captains get no gold, he would have paid any other two crewmembers 33 gold, and kept 34 for himself.

Given that everyone knows this, the Captain can only take 34 gold without risking his life.

Joseph Rittenhouse - 3 years, 3 months ago

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