One of them thinking games...

[AbsolutDealage]

Originally posted by: GagHalfrunt

Originally posted by: Luagsch
ok... another try.
nr5 will always refuse, cause nomatter what if 1 and 2 are dead he just kills 3 and 4 with his vote and gets 100 diamonds. so the last person that could make a proposal so that everybody lives is 2. 2 could propose 100 for him and 0 for 3,4 and 5. if they refuse 3 and 4 are dead and 5 gets all the diamonds.

cause of this the proposal nr 1 can make is 97 for him, 1 for nr 2,3 and 4 and nothing for nr.5. if they don't agree they will be left with nr.2 's proposal and then 3 and 4 would get nothing if they want to survive.

so nr.1: 97
nr.2: 1
nr.3: 1
nr.4: 1
nr.5: 0

That is so far off it's amazing a human being came up with it.

Actually, I don't think it is that far off.

 

[Luagsch]

Originally posted by: OulOat
Dude, the answer is simple, take all the diamonds and offer everyone else 0 (or offer #2 and #4 one diamond each if they value their life less than a diamond). #2 and #4 will agree. #3 and #5 will disagree. You = profit!!!! I'm da winnAr!!!! I leave it to you to figure out why this combo works.

Hint - #3 and #5 will disagree with every proposal that's not theirs. Why?

doesn't work imo. you need more than 50% of the votes of the other 4 so you'll need 3 people that agree.

 

[marquee]

If there were two people, #2 simply rejects anything #1 says, kills him and keeps his money.

So with 3 people, the first guy just has to get the vote of the second guy, and screw the third. #1 gives 99 to himself and 1 to #2.

With 4 people, #2 is eager to kill #1 to get to the situation where he gets to keep 99. So #1 cannot win #2's vote. He can however win #3's vote by giving him 2 diamonds, and giving 1 diamond to #4. #1 keeps 97

With 5 people, #1 cannot win #2's vote. He only needs two other votes, so the cheapest way for him to do this would be to offer 2 diamonds to #5, and 3 diamonds to #4. This is the best #4 and #5 could do, so they'll accept. Screw #2 and #3, #1 doesn't need their votes anyways.

So in the end, #1 can keep 95 diamonds for himself.

Can anyone pick a flaw in this strategy?

 

[AbsolutDealage]

Reposting my edit from above...:

OK, you have to work backwards.

#5 - will never agree with anyone, because he has no risk, and could end up with all the diamonds

#4 - If there are 2 people left, he will have to give all the diamonds to #5 or he is dead. If there are 3 people left, he will agree with anything above 0, because he knows he gets nothing if #3 dies.

#3 - If there are 3 people left, he will have to split it 0-100-0.

#2 - If there are 4 people left, He dies. If there are 5 people left, he will agree with anything.

So, you essentially have a sure thing vote on 2, and number 3 and 4 will agree with anything above 0. My breakdown is: 98/0/1/1/0.

 

[AbsolutDealage]

Originally posted by: marquee
If there were two people, #2 simply rejects anything #1 says, kills him and keeps his money.

So with 3 people, the first guy just has to get the vote of the second guy, and screw the third. #1 gives 99 to himself and 1 to #2.

With 4 people, #2 is eager to kill #1 to get to the situation where he gets to keep 99. So #1 cannot win #2's vote. He can however win #3's vote by giving him 2 diamonds, and giving 1 diamond to #4. #1 keeps 97

With 5 people, #1 cannot win #2's vote. He only needs two other votes, so the cheapest way for him to do this would be to offer 2 diamonds to #5, and 3 diamonds to #4. This is the best #4 and #5 could do, so they'll accept. Screw #2 and #3, #1 doesn't need their votes anyways.

So in the end, #1 can keep 95 diamonds for himself.

Can anyone pick a flaw in this strategy?

2 diamonds to #5

#5 will never agree. He has absolutely no risk, and it is in his best interest to disagree.

 

[new2AMD]

Originally posted by: Beattie


Question: How does No.1 guy?s proposal can get max number of diamonds and save his life too?

Proposal
2 - 0 - 0 - 49 - 49

#1 has his own vote and gets vote of 4 and 5 for majority.

 

[OulOat]

Originally posted by: Luagsch

Originally posted by: OulOat
Dude, the answer is simple, take all the diamonds and offer everyone else 0 (or offer #2 and #4 one diamond each if they value their life less than a diamond). #2 and #4 will agree. #3 and #5 will disagree. You = profit!!!! I'm da winnAr!!!! I leave it to you to figure out why this combo works.

doesn't work imo. you need more than 50% of the votes of the other 4 so you'll need 3 people that agree.

Uh, you + #2 + #4 = 3. 3/5 > 50%.

Uh, my proposition means you get the all of the diamonds and everyone stays alive. Me = Da Winnar! Poo to the rest of y'all

 

[AbsolutDealage]

Ugh... I keep finding more flaws in every theory I come up with. My head hurts.

 

[Luagsch]

Originally posted by: AbsolutDealage
So, you essentially have a sure thing vote on 2, and number 3 and 4 will agree with anything above 0. My breakdown is: 98/0/1/1/0.

hmm... don't think this works. if 2 and 5 refuse, it's nr2's shot. he will propose 100-0-0-0 from here-on either 3 and 4 agree, or nr5 will always get 100. if they refuse and nr.5 is a sadist he will shoot them nomatter if the give him 100 or not. so nr2 knows his chances of getting all with 3,4 and 5 getting nothing. cause of that nr 1 needs nr5's vote. that's why i got to 97-0-1-1-1.

 

[AbsolutDealage]

Originally posted by: Luagsch

Originally posted by: AbsolutDealage
So, you essentially have a sure thing vote on 2, and number 3 and 4 will agree with anything above 0. My breakdown is: 98/0/1/1/0.

hmm... don't think this works. if 2 and 5 refuse, it's nr2's shot. he will propose 100-0-0-0 from here-on either 3 and 4 agree, or nr5 will always get 100. if they refuse and nr.5 is a sadist he will shoot them nomatter if the give him 100 or not. so nr2 knows his chances of getting all with 3,4 and 5 getting nothing. cause of that nr 1 needs nr5's vote. that's why i got to 97-0-1-1-1.

I still don't see how you will ever get 5's vote. No matter what, if he votes no, it means more diamonds for him. He has no risk of death, so he doesn't care.

 

[marquee]

Originally posted by: AbsolutDealage

Originally posted by: marquee
If there were two people, #2 simply rejects anything #1 says, kills him and keeps his money.

So with 3 people, the first guy just has to get the vote of the second guy, and screw the third. #1 gives 99 to himself and 1 to #2.

With 4 people, #2 is eager to kill #1 to get to the situation where he gets to keep 99. So #1 cannot win #2's vote. He can however win #3's vote by giving him 2 diamonds, and giving 1 diamond to #4. #1 keeps 97

With 5 people, #1 cannot win #2's vote. He only needs two other votes, so the cheapest way for him to do this would be to offer 2 diamonds to #5, and 3 diamonds to #4. This is the best #4 and #5 could do, so they'll accept. Screw #2 and #3, #1 doesn't need their votes anyways.

So in the end, #1 can keep 95 diamonds for himself.

Can anyone pick a flaw in this strategy?

2 diamonds to #5

#5 will never agree. He has absolutely no risk, and it is in his best interest to disagree.

#5 will have to agree to something, if it gets down to 3 people left, he's going to be left with nothing. So wouldnt it be in his best interest to take the diamonds and run?

 

[AbsolutDealage]

#5 will have to agree to something, if it gets down to 3 people left, he's going to be left with nothing. So wouldnt it be in his best interest to take the diamonds and run?

OK, I can see that now... it makes sense.

I just keep thinking that I am missing something big here...

 

[Entity]

Damn, good riddle.

DO NOT CLICK THIS LINK UNLESS YOU WANT TO KNOW HOW TO SOLVE IT.

Re-worded riddle:

(4.) Say that five pirates have one hundred gold coins and they have to divide the coins between themselves. In order of seniority (suppose Pirate 5 is most senior, Pirate 1 is least senior), the most senior pirate proposes a distribution of the coins. He suggets that the pirates all vote and if at least fifty percent accept the proposal, the coins are divided as proposed. Otherwise, the most senior pirate will be executed, and the remaining pirates will start over again with the next senior pirate. Quetsion: What solution does the most senior pirate propose? (Here make the assumption that the pirates are all very intelligent and they are all extremely greedy and that they would prefer not to die. Also "fifty percent" means that three pirates must vote for the proposal when there are five for the proposal to pass; two must vote for the proposal if there are four pirates, etc.)

Beattie, any idea if your riddle might be off in the 50% aspect - kill/no kill aspect?

 

[Luagsch]

Originally posted by: AbsolutDealage

Originally posted by: Luagsch

Originally posted by: AbsolutDealage
So, you essentially have a sure thing vote on 2, and number 3 and 4 will agree with anything above 0. My breakdown is: 98/0/1/1/0.

hmm... don't think this works. if 2 and 5 refuse, it's nr2's shot. he will propose 100-0-0-0 from here-on either 3 and 4 agree, or nr5 will always get 100. if they refuse and nr.5 is a sadist he will shoot them nomatter if the give him 100 or not. so nr2 knows his chances of getting all with 3,4 and 5 getting nothing. cause of that nr 1 needs nr5's vote. that's why i got to 97-0-1-1-1.

I still don't see how you will ever get 5's vote. No matter what, if he votes no, it means more diamonds for him. He has no risk of death, so he doesn't care.

the problem is nr2 proposal. he will say 100-0-0-0. if 3 and 4 disagree they are dead (or propose all the money to nr 5 so where would be the point there).
anyhow. we agree that 3 and 4 are more than happy with 1 diamond. now. nr5 knows this. so if 3 and 4 accept nr2 proposal he will get nothing either. cause of that he will be happy with the 1 diamond nr.1 offers him.

 

[marquee]

Originally posted by: Entity
Damn, good riddle.

DO NOT CLICK THIS LINK UNLESS YOU WANT TO KNOW HOW TO SOLVE IT.

Re-worded riddle:

(4.) Say that five pirates have one hundred gold coins and they have to divide the coins between themselves. In order of seniority (suppose Pirate 5 is most senior, Pirate 1 is least senior), the most senior pirate proposes a distribution of the coins. He suggets that the pirates all vote and if at least fifty percent accept the proposal, the coins are divided as proposed. Otherwise, the most senior pirate will be executed, and the remaining pirates will start over again with the next senior pirate. Quetsion: What solution does the most senior pirate propose? (Here make the assumption that the pirates are all very intelligent and they are all extremely greedy and that they would prefer not to die. Also "fifty percent" means that three pirates must vote for the proposal when there are five for the proposal to pass; two must vote for the proposal if there are four pirates, etc.)

There is a difference with the original riddle posted and the riddle you posted. The original riddle says you gotta get more than 50%, the one you posted says you just need at least 50%. Makes a slight difference, consider the case of 2 robbers/pirates and you'll see what I mean.

 

[Entity]

Originally posted by: marquee

Originally posted by: Entity
Damn, good riddle.

DO NOT CLICK THIS LINK UNLESS YOU WANT TO KNOW HOW TO SOLVE IT.

Re-worded riddle:

(4.) Say that five pirates have one hundred gold coins and they have to divide the coins between themselves. In order of seniority (suppose Pirate 5 is most senior, Pirate 1 is least senior), the most senior pirate proposes a distribution of the coins. He suggets that the pirates all vote and if at least fifty percent accept the proposal, the coins are divided as proposed. Otherwise, the most senior pirate will be executed, and the remaining pirates will start over again with the next senior pirate. Quetsion: What solution does the most senior pirate propose? (Here make the assumption that the pirates are all very intelligent and they are all extremely greedy and that they would prefer not to die. Also "fifty percent" means that three pirates must vote for the proposal when there are five for the proposal to pass; two must vote for the proposal if there are four pirates, etc.)

There is a difference with the original riddle posted and the riddle you posted. The original riddle says you gotta get more than 50%, the one you posted says you just need at least 50%. Makes a slight difference, consider the case of 3 robbers/pirates and you'll see what I mean.

I know. I'm thinking the OP's translation may be off; I could be wrong, but it seems that the original riddle (which was difficult to understand) might be the one I posted.

Rob

 

[OulOat]

Originally posted by: Entity
Damn, good riddle.

DO NOT CLICK THIS LINK UNLESS YOU WANT TO KNOW HOW TO SOLVE IT.

Re-worded riddle:

(4.) Say that five pirates have one hundred gold coins and they have to divide the coins between themselves. In order of seniority (suppose Pirate 5 is most senior, Pirate 1 is least senior), the most senior pirate proposes a distribution of the coins. He suggets that the pirates all vote and if at least fifty percent accept the proposal, the coins are divided as proposed. Otherwise, the most senior pirate will be executed, and the remaining pirates will start over again with the next senior pirate. Quetsion: What solution does the most senior pirate propose? (Here make the assumption that the pirates are all very intelligent and they are all extremely greedy and that they would prefer not to die. Also "fifty percent" means that three pirates must vote for the proposal when there are five for the proposal to pass; two must vote for the proposal if there are four pirates, etc.)

So I was right. HAR HAR HAR. But I would think they would value their lives more than a diamond, so you could keep all the diamonds. Actually even if they don't, there is a way to get 99 diamonds. Because each thief is equally smart, thief 5 should know that thief 3 will screw him over. Thus, he would vote for the proposal if thief 1 offers him one diamond. Of course, thief 4 would vote for even if he didn't get any since he's screwed anyway. Thus, thief 1 would walk away with 99 instead of 98. Me = profit!

 

[new2AMD]

Originally posted by: Luagsch

Originally posted by: AbsolutDealage

Originally posted by: Luagsch

Originally posted by: AbsolutDealage
So, you essentially have a sure thing vote on 2, and number 3 and 4 will agree with anything above 0. My breakdown is: 98/0/1/1/0.

hmm... don't think this works. if 2 and 5 refuse, it's nr2's shot. he will propose 100-0-0-0 from here-on either 3 and 4 agree, or nr5 will always get 100. if they refuse and nr.5 is a sadist he will shoot them nomatter if the give him 100 or not. so nr2 knows his chances of getting all with 3,4 and 5 getting nothing. cause of that nr 1 needs nr5's vote. that's why i got to 97-0-1-1-1.

I still don't see how you will ever get 5's vote. No matter what, if he votes no, it means more diamonds for him. He has no risk of death, so he doesn't care.

the problem is nr2 proposal. he will say 100-0-0-0. if 3 and 4 disagree they are dead (or propose all the money to nr 5 so where would be the point there).
anyhow. we agree that 3 and 4 are more than happy with 1 diamond. now. nr5 knows this. so if 3 and 4 accept nr2 proposal he will get nothing either. cause of that he will be happy with the 1 diamond nr.1 offers him.

What does nr stand for?

 

[GagHalfrunt]

Hmmm, I overlooked that in addition to #2 being a death seat #4 is also a death seat. Anyone making any proposal from either slot is dead, no chance of success. That gives #1 and #3 leverage over #4 to accept almost any deal.

If 1 offers 97-1-0-2-0:

1: agrees, he's cool.
2: agrees, he gets a diamond and life
3: disagrees. irrelevant
4: agrees, gets 2 diamonds and life
5: disagrees, irrelevant

Here's the logic: If #1 is killed off #2 dies too as he can make do acceptable offer. He's forced to take one diamond and like it. #3 is voting no, but it doesn't matter. #4 is the key. He knows that if #1 dies #2 dies too, that leaves only 3-4-5 left. At that point #4 is in trouble because he must agree to any offer by #3. If #3 dies #4 knows he's doomed, that gives #3 the leverage to force a 99-1-0 deal on #4. #4 is left with a choice between gettind 2 diamonds on #1's deal or getting 1 on #3's deal.

 

[Luagsch]

Originally posted by: new2AMD

Originally posted by: Luagsch

Originally posted by: AbsolutDealage

Originally posted by: Luagsch

Originally posted by: AbsolutDealage
So, you essentially have a sure thing vote on 2, and number 3 and 4 will agree with anything above 0. My breakdown is: 98/0/1/1/0.

hmm... don't think this works. if 2 and 5 refuse, it's nr2's shot. he will propose 100-0-0-0 from here-on either 3 and 4 agree, or nr5 will always get 100. if they refuse and nr.5 is a sadist he will shoot them nomatter if the give him 100 or not. so nr2 knows his chances of getting all with 3,4 and 5 getting nothing. cause of that nr 1 needs nr5's vote. that's why i got to 97-0-1-1-1.

I still don't see how you will ever get 5's vote. No matter what, if he votes no, it means more diamonds for him. He has no risk of death, so he doesn't care.

the problem is nr2 proposal. he will say 100-0-0-0. if 3 and 4 disagree they are dead (or propose all the money to nr 5 so where would be the point there).
anyhow. we agree that 3 and 4 are more than happy with 1 diamond. now. nr5 knows this. so if 3 and 4 accept nr2 proposal he will get nothing either. cause of that he will be happy with the 1 diamond nr.1 offers him.

I havent understood any of your proposals. What deos nr stand for?

LMAO, sh17... i've got that from german.... nr. stands for number here. don't know if it is right in english. maybe its No. 1 etc...

 

[step-dawg]

wrong!! i think the question is a little flawed.

 

[gotsmack]

Simple game theory.

#1 = 98
conservative number. he can get 99 or 100 diamonds depending on if he wants to risk it it #2 and #4

#2 = 1
is dead unless he offers #3 99 or 100 diamonds

#3 = 0
will always disagree when offered <99 or <100 because when it gets to him, he can propose #4 gets 1 or 0 and 99 or all for himself

#4 = 1
even if he offers #5 all the diamonds he has 1/2 chance of living

#5 = 0
will always disagree until his turn if offered <100, 50% chance of saying yes if offered all diamonds in round 4

 

[Beattie]

I dont know if any of you noticed, but I had to correct the question. It's different than it was originally.

 

[new2AMD]

Originally posted by: Luagsch

Originally posted by: new2AMD

Originally posted by: Luagsch

Originally posted by: AbsolutDealage

Originally posted by: Luagsch

Originally posted by: AbsolutDealage
So, you essentially have a sure thing vote on 2, and number 3 and 4 will agree with anything above 0. My breakdown is: 98/0/1/1/0.

hmm... don't think this works. if 2 and 5 refuse, it's nr2's shot. he will propose 100-0-0-0 from here-on either 3 and 4 agree, or nr5 will always get 100. if they refuse and nr.5 is a sadist he will shoot them nomatter if the give him 100 or not. so nr2 knows his chances of getting all with 3,4 and 5 getting nothing. cause of that nr 1 needs nr5's vote. that's why i got to 97-0-1-1-1.

I still don't see how you will ever get 5's vote. No matter what, if he votes no, it means more diamonds for him. He has no risk of death, so he doesn't care.

the problem is nr2 proposal. he will say 100-0-0-0. if 3 and 4 disagree they are dead (or propose all the money to nr 5 so where would be the point there).
anyhow. we agree that 3 and 4 are more than happy with 1 diamond. now. nr5 knows this. so if 3 and 4 accept nr2 proposal he will get nothing either. cause of that he will be happy with the 1 diamond nr.1 offers him.

I havent understood any of your proposals. What deos nr stand for?

LMAO, sh17... i've got that from german.... nr. stands for number here. don't know if it is right in english. maybe its No. 1 etc...

ahhh got ya...number

 

[Luagsch]

hmmm now it's getting confusing as we seem to speak of different riddles.

this is how i understood the riddle: if you make a proposition you'll need more than 50% of the other votes (your own vote doesn't count).

how did you guys understand that part?