Riddle Me This

Jars

so a woman has 2 children and a homicidal maniac tells her he is going to kill one of them no matter what. which one does she pick

redhead

there actually are circumstances under which pirate #1 (the highest ranked) would vote yes, but that will come up naturally once you understand the right way to approach the riddle.

slym is on the right track

Spudge

Jars wrote: »

so a woman has 2 children and a homicidal maniac tells her he is going to kill one of them no matter what. which one does she pick

Whichever one got the worst grades last semester

Sorry, them's the breaks

Khildith

Jars wrote: »

so a woman has 2 children and a homicidal maniac tells her he is going to kill one of them no matter what. which one does she pick

Wait, in this riddle did he even offer her a choice? Or is it like "I'm gonna kill that one no matter what"

Cause...

Yeah.

BahamutZERO

as far as I can tell #1 (high rank) would only ever want to vote yes to getting 100% of the treasure, because he can't be forced to walk the plank

TheStig

Jars wrote: »

so a woman has 2 children and a homicidal maniac tells her he is going to kill one of them no matter what. which one does she pick

This riddle works better when you tell it to a woman with 2 kids in a secluded location.

SLyM

But if you offer one better than you know what #4 would offer him, he'll accept it because getting anything is better than what #3 gives him.

BahamutZERO

SLyM wrote: »

But if you offer one better than you know what #4 would offer him, he'll accept it because getting anything is better than what #3 gives him.

???
if who offered one better than what, who would accept it?

SLyM

SLyM wrote: »

I think another thing to take into account that if they reach 3 left, 3 could take all of it and 2 would have to agree if he didn't want to die.

This gives #1 and #2 nothing but gives #2 no other option unless he wants to die.

If you can give either of them something better than this and better than what they would get under #4, they should accept it.

E: let's do some clarification.

If 5 offers 1 or 2 something better than what 1 or 2 would get under 4, they should accept it because it's better than what they get under 3 (nothing)

BahamutZERO

stop using so many pronouns and explicitly say which #s you are referring to

Fishman

The solution to the pirate riddle is a bit of a mind fuck, but I won't reveal it because I remember the solution and that's no fun.

101

redhead wrote: »

none of the answers posted so far is correct

edit: oh, and when discussing this in a group, it helps to come up with a convention for what you're going to call the pirates. let's say the lowest ranked pirate is #5 and the highest is #1. helps cut down on confusion when multiple people post answers.

Propose an even split between the lowest three pirates.

MrMonroe

Offer the bulk to the second and third pirates, and a small share for himself. The lowest three pirates then vote in favor of the plan. This is a better deal than the second will get if he waits for the next round, (when he must take an even lower cut in order to make it worthwhile to pirates 3 & 4) and a better deal for the third pirate. The key is to offer the second and third pirates at least their highest possible value, which is, at maximum, 49 to the third pirate and some lesser number to the second which I cannot bother to calculate right now. You win with a 3-2 vote, a small share, and your life.

SLyM

So shouldn't he offer 1 and 2 a pair of coins each, because if they say no they get stuck with 4 who can offer them 1 each?

Wait, this assumes 3 will veto anything until he is the lowest, which probably isn't true.

Shit.

Munkus Beaver

YoSoyTheWalrus wrote: »

50/50 to pirates one and two, then one two and five vote yes to it. #1 and #2 won't vote yes to anything else, they'll just wait until everyone else is dead.

Or you realize that pirates three and four realize the same thing, so you vote to split the money evenly among 3, 4, and 5.

L|ama

YaYa wrote: »

Ubik wrote: »

Poe wrote on both

very good

Straightzi wrote: »

YaYa wrote: »

why is a raven like a writing desk

Poe wrote on both/They both have inky quills.

NO IT'S NOT SUPPOSED TO HAVE A FUCKING SOLUTION GOD DAMMIT THAT IS NOT THE POINT

BahamutZERO

Munkus Beaver wrote: »

YoSoyTheWalrus wrote: »

50/50 to pirates one and two, then one two and five vote yes to it. #1 and #2 won't vote yes to anything else, they'll just wait until everyone else is dead.

Or you realize that pirates three and four realize the same thing, so you vote to split the money evenly among 3, 4, and 5.

if #1 can kill all the other pirates by voting no for everything he will because he'll then get all the money

101

3,4 have to vote for 5's proposal, otherwise they can't win a vote against the other two and die.

so.....
along with my earlier 3-way split.

100% to pirate 5.

SLyM

And if 3 can single-handedly sabotage both 4 and 5 he will because that lets him take it all.

Captain K

yes slym that's totally key!

...but does that make it unsolvable?

The process won't go beyond the round of 3, because of slym's point. Pirate 3 is getting all the coins in that round, and since he knows it, he won't vote for anything in the round of 4 or the round of 5 that doesn't give him all the coins. Knowing his behavior, everyone's proposal will always be that Pirate 3 gets all the coins.

But if all things are equal, and everyone knows that the functional end of the process is the round of 3, won't Pirate 3 just vote against proposals until his round, because he wants to see death?

wait that's it! Pirate 3's vote doesn't actually matter in rounds 4 and 5 because pirates 2 and 4 KNOW he's voting no until his round

pirate 1 always votes no
pirate 3 votes no until round of 3 because he wants to see guys die and then get all 100 coins
pirate 2, without any monetary reason, votes no until round of 3 because he wants to see guys die

so a proposal can't win in the round of 4, since pirates 1 and 3 will vote no in round 4 no matter what, meaning pirate 4's going to vote for whatever pirate 5 suggests--he'll die in the round of 4 otherwise. in the rounds of 3 and 4, pirate 2 won't get any money at all, so he'll vote for any proposal by pirate 5 that gives him money.


Pirate 5 should suggest that Pirate 2 gets 1 coin and that he himself keeps the other 99 coins.


is that it?

101

at the point where it's just 3,2,1 left.

3 is basically fucked.

Munkus Beaver

BahamutZERO wrote: »

Munkus Beaver wrote: »

YoSoyTheWalrus wrote: »

50/50 to pirates one and two, then one two and five vote yes to it. #1 and #2 won't vote yes to anything else, they'll just wait until everyone else is dead.

Or you realize that pirates three and four realize the same thing, so you vote to split the money evenly among 3, 4, and 5.

if #1 can kill all the other pirates by voting no for everything he will because he'll then get all the money

...that's why you give the money to 3, 4, and 5. You don't have to give them anything more than one coin, really, because if you die, then 1 and 2 will vote no on each proposal until they split it 50/50. So anything you propose to 3 and 4 that gives them even one coin will benefit them more than rejecting your plan.

redhead

K, i'm not following all of your reasoning, but although your answer is wrong it seems like you might be thinking about this the right way?

BahamutZERO

How can 3 sabotage 4 and 5? he'd die before they would. #4 just has to vote against it, and then give #5 all the money to stay alive himself. Although then he has no reason to vote against it, because he'd be getting nothing either way...

redhead

also nothing posted so far has been right

SLyM

Munkus Beaver wrote: »

BahamutZERO wrote: »

Munkus Beaver wrote: »

YoSoyTheWalrus wrote: »

50/50 to pirates one and two, then one two and five vote yes to it. #1 and #2 won't vote yes to anything else, they'll just wait until everyone else is dead.

Or you realize that pirates three and four realize the same thing, so you vote to split the money evenly among 3, 4, and 5.

if #1 can kill all the other pirates by voting no for everything he will because he'll then get all the money

...that's why you give the money to 3, 4, and 5. You don't have to give them anything more than one coin, really, because if you die, then 1 and 2 will vote no on each proposal until they split it 50/50. So anything you propose to 3 and 4 that gives them even one coin will benefit them more than rejecting your plan.

1 and 2 won't agree to split it because 1 can just say no, kill 2 and take it all.

101 wrote: »

at the point where it's just 3,2,1 left.

3 is basically fucked.

This is the exact oppisate of true. When 3 are left, 3 says it all goes to him, 2 has to agree or else he goes mano a mano with 1 and loses no matter what he does.

101

I am sticking with basically 5 takes it all.

e:dang

All to 2?

5 votes to save his life, 2 votes to take the money. 4 also votes to save his life - 3,1 will shoot down anything he suggests.

Munkus Beaver

My final answer: 51 coins to pirate 3, 1 to pirate 4, and 49 to pirate 5.

Captain K

okay guys

is 1 the most powerful or the least powerful

Peen

Stale wrote: »

My daughter has many sisters

as many sisters as she has brothers

but each of her brothers

has twice as many sisters as brothers.


how many children do I have?

Spoiler:

3, two daughters and one son.

SLyM

I've been going with 1 the most powerful. A tie mean the plank and the pirates will vote to kill you if they get nothing from voting yes.

BahamutZERO

BahamutZERO wrote: »

How can 3 sabotage 4 and 5? he'd die before they would. #4 just has to vote against it, and then give #5 all the money to stay alive himself. Although then he has no reason to vote against it, because he'd be getting nothing either way...

oh, if they prefer to see death if they get no money, then 3 is definitely fucked if it gets to the round of 3.

redhead

1 is the highest ranked

5 is the lowest ranked and makes the first proposal

SLyM

They prefer to kill you, but they also prefer to stay alive, so 2 will vote with 3 when there are 3 left because it is logically impossible to have 2 people left and have them agree to anything.

BahamutZERO

in the round of 3 #3 would offer 1 coin to #2 and keep the rest, which would be more than #2 would get in the round of 2 since he'd have to give everything to #1.

redhead

the pirates are bloodthirsty and will cause other pirates to die if they can do so without harming their other goals

they're greedier than they are bloodthirsty, and would rather get a single gold coin than cause someone else to die

but more than all of that they have a very strong survival instinct--they'd rather live with 0 coins than die with all 100 (because then what good does the gold do them)

also, because this sometimes gets forgotten when people are trying to solve this, it's important to remember that tied votes result in death. you have to get more than half the pirates to vote for your proposal.

101

All to 2, my final answer.

redhead

not that it seemed like any of you were necessarily misunderstanding any of that, just clarification for general purposes/those following along at home

SLyM

I'm just trying to figure out what would happen if you get voted off. What would 4 do?

He would offer 1 and 2 one coin each, because otherwise they have no incentive not to let him die and let 3 get it all instead, right?

Therefore you have to offer 1 and 2 something better than what 4 would give them, which should be 2 each. This was confirmed as not true, so I assume that there is something I'm missing here.

MrMonroe

101 wrote: »

at the point where it's just 3,2,1 left.

3 is basically fucked.

no, he can give the second to last pirate a 51% share, larger than the one he can get when it's just him and the last pirate, and he should vote for it

read it backwards temporally

if it does get down to just the last two pirates, there essentially has to be a 50/50 split, otherwise you can't get a majority vote, no one dies, no one goes home with money.

if it's the last three, the bottom pirate can offer a 51/49 split to the second to last pirate, a better deal (by one gold piece) than he can get out of the last pirate. Being logical, he will vote for it, and it makes the middle pirate's best possible outcome 49 gold pieces.

if it's the last four, the bottom pirate must get three votes in order to save his life. He has to pay as much or more to two other pirates as their highest possible share to do this. The cheapest way to do that is to pay 50.5 to the second to last pirate and 49.5 to the third. He gets nothing.

In the first round, the lowest ranked pirate can offer 50 to the middle pirate, which he should accept as it's better than anything he can get if he waits, and any amount of money over 0 to the second lowest pirate, as it's better than the nothing he'll be able to dictate in the second round. This gets him two votes in addition to his own.

The lowest pirate, the one proposing the first plan, can take the remainder. (theoretically 49 if he just gives the fourth guy a pittance of one gold piece)

edit: hey look, Munkus got the same answer last page but without typing out the reasoning that rat bastard:

Munkus Beaver wrote: »

My final answer: 51 coins to pirate 3, 1 to pirate 4, and 49 to pirate 5.

with one small arithmetical error