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Nine Pirates are splitting 99 gold coins among them. Each of them has already one extra gold coin in their pockets.

Starting from 1st pirate, each will make a proposal on how many coins will be given to each of them.

  • If the majority (including proposer) accept the proposal then the game ends and the split will be done accordingly.

  • If not accepted, the proposer will be eliminated and punished to pay an extra coin to the winner, and the next pirate will follow the same procedure together with the remaining pirates.

  • The winner of the game will be the one who has the highest number of coins at the end!

  • As you guess, pirates are absurdly very smart and supposed to play the game optimally and trying to make as many coins as possible at the end of the game.

So

At the end of the game, how many coins will each of the pirates have?

Note: You may name the pirates as you like, such as A, B, C, or 1st, 2nd, 3rd etc. and the majority means more than 50%.

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  • $\begingroup$ Is the proposal rejected if exactly half of the pirates agree? $\endgroup$
    – Jafe
    Aug 17, 2018 at 16:01
  • $\begingroup$ majority is more than 50%. $\endgroup$
    – Oray
    Aug 17, 2018 at 16:02
  • $\begingroup$ Is this puzzle really different from this one? I mean, besides the number of pirates and gold coins, does it make a difference that the losing pirate gives a coin instead of being thrown overboard? $\endgroup$
    – xhienne
    Aug 17, 2018 at 16:46
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    $\begingroup$ As far as I can tell, the answer is changed by adding the extra coin, number of pirates, and a majority being >50% instead of >=. However, the process for finding the answer is mostly the same. $\endgroup$
    – Bennett Bernardoni
    Aug 17, 2018 at 16:57
  • $\begingroup$ So if the proposal off all pirates get rejected, what happens? Does the first one get to go again ? $\endgroup$
    – Hilmar
    Aug 17, 2018 at 17:08

3 Answers 3

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Updated based of jafe's comment, here is my answer.

A has 93 coins
B has 1 coin
C has 2 coins
Three of D through I will have 3 coins the rest have 1 coin.

Here is the reasoning.

I started with all but one pirate eliminated and worked backwards.

If I is the only pirate he gets 99 coins + 9 extra
For 2 pirates (H and I), I will always say no to get H's extra coin
For 3, G can offer 99 to himself and 0 to H and I. H will accept to keep his extra coin. G and H form a majority.
For 4, F offers in order {97, 0, 1, 1}. G will always say no and F needs 3 for a majority, so he offers 1 to H and I since this is a better outcome for them then the previous case.
For 5, {96, 0, 1, 2/0, 2/0} is offered. 3 votes are needed. G is given 1 which is better than 0 in case 4. One of H or I is given 2 for their vote. The rest are offered 0.
For 6, {94, 0, 1, 2/0, 2/0, 2/0}. Similar reasoning with 4 votes needed. Two of G, H, and I will be given 2 coins. (Updated from 3 due to jafe's comment)
For 7, {94, 0, 1, 2/0, 2/0, 2/0, 2/0}. These are the cheapest 4 votes with 2 needed from F, G, H, and I.
For 8, {92, 0, 1, 2/0, 2/0, 2/0, 2/0, 2/0}. 3 of 5 are given 2 coins.
For 9, {92, 0, 1, 2/0, 2/0, 2/0, 2/0, 2/0, 2/0}. 3 of 6 are given 2 coins.

Since 5 accepted the first proposal everyone keeps their extra coin to give us the answer above.

Here is a table of clarity | A B C D E F G H I 1 | 99 2 | will always be rejected by I 3 | 99 0 0 4 | 97 0 1 1 5 | 96 0 1 2/0 2/0 1 of 2 6 | 94 0 1 2/0 2/0 2/0 2 of 3 7 | 94 0 1 2/0 2/0 2/0 2/0 2 of 4 8 | 92 0 1 2/0 2/0 2/0 2/0 2/0 3 of 5 9 | 92 0 1 2/0 2/0 2/0 2/0 2/0 2/0 3 of 6

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  • $\begingroup$ For 6, I think you can offer 2 to one of the two last pirates. They can't be certain they'll be offered the 2 again when it gets to 5 pirates (they could be offered 0). $\endgroup$
    – Jafe
    Aug 17, 2018 at 17:16
  • $\begingroup$ @jafe Nice catch. This completely changes the answer. $\endgroup$
    – Bennett Bernardoni
    Aug 17, 2018 at 17:34
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Partial, starting from the end...

If there are two pirates remaining, B can reject any proposal A makes and gets all the coins for himself.

From this follows that when there are 3 remaining, A can offer both B and C zero coins, and B has to accept (since otherwise B is heads up with C and forced to lose everything including the extra coin).

If there are 4 pirates, C and D have to accept any proposal of >0 coins. So A can offer B zero coins, C one coin and D one coin.

If there are 5 pirates, C has to accept >0 and D/E have to accept >1. So A can offer C one coin and D (or E) two coins.

Aaand Bennett already figured out the rest :D

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  • $\begingroup$ I feel like the intended solution does not have a choice for who to give coins. But I can't figure out of if there is anything wrong with my reasoning. $\endgroup$
    – Bennett Bernardoni
    Aug 17, 2018 at 16:36
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I've heard a variation of it, where the pirates are all perfect logicians. In that variation, the first pirate was always the captain.

The solution works like this:

The first pirate will always have the most coins. And he only needs to bribe a majority. Let's call him pirate A. Pirate A, needs only to offer 1 coin to pirate C, pirate E, pirate G, and pirate I to win this puzzle. That leaves him with 96 gold coins (he has one in his pocket still). That means, pirate C, E, G, and I will all have 2 gold coins, whereas B, D, F, and H will all only have 1. Here's a video detailing something similar, https://www.youtube.com/watch?v=Mc6VA7Q1vXQ.

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