Pirates. (1 Viewer)

[seanieg89]

Five hundred pirates stumble across a chest filled with 100 pieces of gold. Being in a hierarchical society, the pirates have ranks #1 to #500. The pirate king (#1) decides on a way of distributing the gold. Eg 50 for him, 50 for #2.

That night, the pirates all take a vote on whether this distribution is acceptable. If >1/2 of the pirates vote AGAINST the distribution, then the lead pirate is killed and the pirates all move one rank up. The process repeats the next day.

The pirates are all quite adept mathematicians/logicians, and their priorities are:

1) Survival
2) Getting as much money as possible
3) Killing

in descending order.

The process repeats the next day with the new pirate king. Etc.



How many pirates remain when a distribution is agreed upon, and who gets what?

 

[ExcelHSC]

One person will survive, not definite though =/

 

[seanieg89]

Sorry, not quite.

This question is actually rather hard. It is all logical though. Try seeing what happens for small numbers of pirates (so their decisions are simple) and build up.

 

[Thief]

One person will survive, not definite though =/

Well it can't be one person, because once there's two pirates left there will be 50/50 on the vote.

 

[seanieg89]

Exactly.

 

[Sy123]

Does each person, have to have an integer number of gold, or can it be decimals (grammar?)?

 

[seanieg89]

Integers.

 

[barbernator]

they decided to invest the money in a commonwealth netbank saver account and reap that reward of 4.5% interest.

 

[seanieg89]

they decided to invest the money in a commonwealth netbank saver account and reap that reward of 4.5% interest.

Alas, despite their mathematical inclinations, these pirates are not particularly thrifty...and they prefer cold hard gold.

 

[barbernator]

wouldn't the answer just be 199 pirates and the top 100 get 1 piece of gold each?

scratch that.

 

[Carrotsticks]

lol'd hard at barbernator's option.

they decided to invest the money in a commonwealth netbank saver account and reap that reward of 4.5% interest.

 

[seanieg89]

wouldn't the answer just be 199 pirates and the top 100 get 1 piece of gold each?

Not quite.

 

[Sy123]

Ok, Ill take a stab at it, for the first 500 pirates, there will be an overwhelming majority against the distribution, because they didnt get any money. As in, even if lead pirate gave everyone he could 1 piece of gold, then there still will be 400 left unsatisfied. Therefore that night, the lead pirate will get killed. This process continues until there are 200 pirates. When there are 200 pirates, the lead pirate will give himself 1 piece of gold, and will give 1 piece of gold for every rank until 100, this will give him theoretically, 50:50 outcome of votes. Which denies the lead pirate from being killed. (Since it requires to be greater than 50%). However, since the third priority of the 100th pirate kicks in, he would love to see the lead pirate being killed, and because he is in no danger, he decides to vote against the distribution.

Then this process continues, until there are 100 pirates left, and there are still 100 pieces of gold. Being an adept mathematician the lead pirate will choose to give everyone 1 gold piece. However the last 50 pirates, the one with rank 51, will NOT vote against the distribution, since if he does (due to his priority of killing), he will forecast that the process will continue until there are only 2 pirates left, and that he will indeed eventually get killed. Therefore the voting stands at 51:49 always.

And there will be only 100 pirates left.

EDIT: Might be something wrong with my logic, I dont know.

 

[barbernator]

Ok, Ill take a stab at it, for the first 500 pirates, there will be an overwhelming majority against the distribution, because they didnt get any money. As in, even if lead pirate gave everyone he could 1 piece of gold, then there still will be 400 left unsatisfied. Therefore that night, the lead pirate will get killed. This process continues until there are 200 pirates. When there are 200 pirates, the lead pirate will give himself 1 piece of gold, and will give 1 piece of gold for every rank until 100, this will give him theoretically, 50:50 outcome of votes. Which denies the lead pirate from being killed. (Since it requires to be greater than 50%). However, since the third priority of the 100th pirate kicks in, he would love to see the lead pirate being killed, and because he is in no danger, he decides to vote against the distribution.

Then this process continues, until there are 100 pirates left, and there are still 100 pieces of gold. Being an adept mathematician the lead pirate will choose to give everyone 1 gold piece. However the last 50 pirates, the one with rank 51, will NOT vote against the distribution, since if he does (due to his priority of killing), he will forecast that the process will continue until there are only 2 pirates left, and that he will indeed eventually get killed. Therefore the voting stands at 51:49 always.

And there will be only 100 pirates left.

EDIT: Might be something wrong with my logic, I dont know.

hmmm, would that pirate not strategise that if he voted against there he would be fucked?

 

[seanieg89]

Ok, Ill take a stab at it, for the first 500 pirates, there will be an overwhelming majority against the distribution, because they didnt get any money. As in, even if lead pirate gave everyone he could 1 piece of gold, then there still will be 400 left unsatisfied. Therefore that night, the lead pirate will get killed. This process continues until there are 200 pirates. When there are 200 pirates, the lead pirate will give himself 1 piece of gold, and will give 1 piece of gold for every rank until 100, this will give him theoretically, 50:50 outcome of votes. Which denies the lead pirate from being killed. (Since it requires to be greater than 50%). However, since the third priority of the 100th pirate kicks in, he would love to see the lead pirate being killed, and because he is in no danger, he decides to vote against the distribution.

Then this process continues, until there are 100 pirates left, and there are still 100 pieces of gold. Being an adept mathematician the lead pirate will choose to give everyone 1 gold piece. However the last 50 pirates, the one with rank 51, will NOT vote against the distribution, since if he does (due to his priority of killing), he will forecast that the process will continue until there are only 2 pirates left, and that he will indeed eventually get killed. Therefore the voting stands at 51:49 always.

And there will be only 100 pirates left.

EDIT: Might be something wrong with my logic, I dont know.

Not quite, but I feel it might take quite a while to explain why. Try playing with toy situations where there are a small number of gold pieces and a small number of pirates to get a feel for things. The answer is definitely larger than 100.

 

[seanieg89]

(I might leave this one up for a day or so before I answer it, as I stress: it IS quite hard/counter-intuitive.)

Try the monk riddle just below this one in the extracurricular section for something that is easier to solve quickly.

 

[barbernator]

Not quite, but I feel it might take quite a while to explain why. Try playing with toy situations where there are a small number of gold pieces and a small number of pirates to get a feel for things. The answer is definitely larger than 100.

real life meetup. 10:30, bring knives and money. lets test this.

 

[deswa1]

This is a big coincidence but I just heard this exact riddle yesterday- albeit with less pirates. I won't spoil the answer though -> its interesting.

 

[seanieg89]

real life meetup. 10:30, bring knives and money. lets test this.

Shotgun not being starting pirate king.

 

[DAFUQ]

Spoiler

is it 200pirates remaining? cause every pirate king keeps dieing as he gets voted out until theres 200 pirates. Then pirate king shares 1gold to himself and 99 other tory pirates, which is 100. This allows up to 100 other noob pirates to go against the pirate king, without this king being condemned. So theres 200 pirates remaining