5 logical pirates find a 100 coin chest

[Bob-omb]

they are trying to figure out how to split the coins.

However all of the pirates are different rank in that Pirate A is great than Pirate B and so on, such that Pirate E is the lowest rank.

they have decided that let the most senior pirate decide how to split the coins, then all the pirates will take a vote. if at least half the pirates agree with the decision then that is how they split, if more than half disagrees, the most senior pirate is thrown overboard and the next senior pirate decides how to split the gold.

assuming all of the pirates wants to maximize the gold they get without being thrown overboard, how would the gold be split and why?

 

[BobG]

Spoiler

A - 60%; b - 0%; C - 20%; d - 0%; E - 20%

Reasoning: If A, B, and C are tossed overboard, the game reduces to a dictator game between D & E with D getting 100%.

But before getting to that point, you have to have a game between C, D, & E. D will vote to reject no matter what, so the C, D, & E game is really an ultimatum game between C & E. Being human pirates, E will reject any offer below 20% (the tipping point is actually about 16%, but acceptance at exactly the tipping point would be far from certain). Below that percentage, the psychological rewards (and corresponding physiological rewards) of punishing C's unfairness outweigh the monetary rewards, even if C getting 0 money were the only punishment (personally, I think adding in the reward of tossing C overboard would increase the threshold for rejection, but no one has done that type of experiment).

Before getting to the CDE game, you have to have a 4 player B, C, D, & E game. C will surely reject and D will surely receive 0 for rejecting. E gets no advantage either way. The fact that D would suffer for rejecting while E would not essentially reduces this to a 2-player ultimatum game between B and D with 20% once again being the "unfairness" threshhold.

But, before getting to the BCDE game, a five player game has to be played between A, B, C, D, & E. Finally, you have a different game - a 3 player ultimatum game between A, C, and E (since C and E would suffer if the game went to 2-player game between B & D). You would think this would change the dynamics but, surprisingly, a 3 player ultimatum game requiring unanimous consent runs very consistent with the 2-player version. A keeping 6 times as much as C and E is the tipping point, but A would want to keep less than 4 times as much as C and E to be certain he wouldn't get tossed overboard - especially since the act of tossing a pirate overboard is pleasing in itself and would add to the natural tendency to reject unfair offers.

http://ethesis.helsinki.fi/julkaisut/eri/hecer/disc/140/ultimatu.pdf

(Not quite the answer that "pure logic" dictates, but surprisingly similar once allowances for human psychology are made.)

There is a major exception to the "unfairness" threshhold. If the responder can leave comments on an index card that the proposer may or may not read, the responder will accept a more unfair offer. Of course, that experiment didn't allow the responder to use profanity in their responses and that would never work for pirates. (But it does strengthen my feeling that the opportunity to throw a pirate overboard if they reject the offer would increase how much the responder would have to receive. There's a definite psychological reward for punishing unfairness and that psychological reward is worth some money.)

http://www.andrew.cmu.edu/user/exiao/ef.pdf

There's also another exception. If the responder has to state the minimum offer he'll accept before any offer is made, the threshold for rejection is higher - possibly because it's hard to imagine how unfair people can be until it actually happens? And when confronted by the reality of unfairness, people become more accepting of it?

 

[RichardM]

This is not a whole puzzle. There are at least next tree questions:

1. How many pirates is not able split 100 coin chest ? Found smallest number !

2. How many pirate is able split 100 coin chest, when this number is bigger than in question 1 ?

3. 524488 pirates and 100 coin chest. Is able to split coins ?

Answers:
[color="#black"]1. 203 pirates
2. 204 pirates (100 coin for 1,3,5...199 and pirates 203 and 204 will be vote yes)
3. YES (100 coin for 1,3,5...199 and pirates 262245-524488 will be vote yes)[/COLOR]