Pirate's Dilemna

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heuristically_alone
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Pirate's Dilemna

Postby heuristically_alone » Mon Oct 03, 2016 6:58 am UTC

There are 5 rational pirates, A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.

The pirates have a strict order of seniority: A is senior to B, who is senior to C, who is senior to D, who is senior to E.

The pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution. In case of a tie vote, the proposer has the casting vote. If the distribution is accepted, the coins are disbursed and the transaction ends. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.

Pirates base their decisions on three factors with priorities in this order: First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins each receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal. The pirates do not trust each other, and will neither make nor honor any promises between pirates apart from a proposed distribution plan that gives a whole number of gold coins to each pirate.

How much gold can pirate A distribute to each pirate to ensure that he receives the maximum gold possible (without dying)?
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Sabrar
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Re: Pirate's Dilemna

Postby Sabrar » Mon Oct 03, 2016 7:09 am UTC


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heuristically_alone
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Re: Pirate's Dilemna

Postby heuristically_alone » Mon Oct 03, 2016 7:24 am UTC

Not too surprised to see it done before, but couldn't find it myself. Still curious to see everyone's responses, so don't cheat and click on Sabrar's link!
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taemyr
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Re: Pirate's Dilemna

Postby taemyr » Tue Oct 04, 2016 12:49 pm UTC

Question leaves open the assumption of how a pirate would compare some reward in the current scheme with the potential, but not certainty, for a higher reward by a later pirate.

Spoiler:
If there is only one pirate he gets 100 gold.

If there are two pirates one pirate will be thrown overboard, and the other will get 100 gold.

If there are three Pirates the captain gets 100 gold.

If there are four pirates the captain gets a choice. He gets 99 gold in either case, and one of the two least senior pirates gets one gold.

This creates a problem for the five captain case.

If a pirate prefers a sure one gold over throwing the captain overboard and possibly getting a gold coin from the new captain, or if it's predictable which pirate gets the coin in the four pirate case, the captain gets 98 gold, the third senior pirate gets one, and one of the two least senior gets one.

If the pirate prefers taking his chance with the new captain, and it's not predictable who will get the coin, the captain will only get 97 gold, third senior gets 1, and one of the two least senior gets two.

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Soupspoon
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Re: Pirate's Dilemna

Postby Soupspoon » Tue Oct 04, 2016 1:14 pm UTC

taemyr wrote:
Spoiler:
If there are two pirates one pirate will be thrown overboard, and the other will get 100 gold.
AIUI, no.

heuristically_alone wrote:... In case of a tie vote, the proposer has the casting vote. ...

Which also therefor changes the assumptions for greater numbers.

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Re: Pirate's Dilemna

Postby PeteP » Tue Oct 04, 2016 2:24 pm UTC

Spoiler:
If there are two pirates the senior one will give himself all money the vote is a tie so the senior one wins via tie breaker=> There will never be only one pirate and the least senior one doesn't want to reach the stage of having only two left. But the two least senior ones don't have to worry about dying.

At 3 pirates the least senior pirate will thus vote for any plan where he gets at least 1 gold coin (assuming no punishment behavior for bad deals occurs just optimization for a single game), together with the proposer that is a win. And the proposer gets 99 coins. (The second least senior would want the vote to fail except if he gets 100 so he gets nothing.)

At 4 pirates. Well with tie breaker it only takes 2 to win. The least senior one knows he will get one if this fails so 2 coins should suffice. 98 coins for the scond most senior one. 0 for the two in the middle.

At five pirates the second and third least senior ones know they get nothing if it comes to 4 so they get 1 each. The senior one gets 98.


Does that make sense?

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Re: Pirate's Dilemna

Postby measure » Tue Oct 04, 2016 5:24 pm UTC

Spoiler:
PeteP wrote:At 4 pirates. Well with tie breaker it only takes 2 to win. The least senior one knows he will get one if this fails so 2 coins should suffice. 98 coins for the scond most senior one. 0 for the two in the middle.

In the 4-pirate case it is cheaper to buy the vote of the 2nd lowest pirate with 1gold leaving 99 for the captain.

This changes the 5-pirate solution as well.

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Re: Pirate's Dilemna

Postby taemyr » Mon Oct 10, 2016 8:26 am UTC

Soupspoon wrote:AIUI, no.


Not sure how I missed that the senior has deciding vote in the two pirate case.

measure wrote:
Spoiler:
This changes the 5-pirate solution as well.

Spoiler:
Not a lot though, it's still the case that the two pirates that get 0 in the 4-pirate case now gets 1.

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Re: Pirate's Dilemna

Postby Schrödinger's Wolves » Sun Mar 26, 2017 4:38 pm UTC

My solution:
Spoiler:
Pirate E gets all 100 coins if alone. However pirate D only needsvhis own vote to survive against E so he will keep all the coins if he is the oldest alive. Pirate C must therefore pay E 1 coin in order to get his vote if C is the oldest alive. Pirate B must pay D 1 coin to get his vote.(Pirate D is the only one who gets nothing from pirate C.) Thus since pirate A needs 3 votes they will get their own and pay off the cheapest two namely C and E who don't get anything from B. Pirate A thus gets 98 coins all to himself and isn't thrown overboard.

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Re: Pirate's Dilemna

Postby djangochained » Tue Mar 28, 2017 1:25 pm UTC

10 Coins, then everybody else would get 90/4=22.5 coins


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