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5 inteligent pirates find a chest with 100 gold coins. They are trying to decide how to distribute them.

The order of seniority of the pirates is: Pirate 1 is superior to pirate 2, who is superior to pirate 3, who is superior to 4, who is superior to 5.

They have a system of distribution: The most senior pirate proposes a distribution of the coins. The pirates vote on wether to accept the distribution. The proposer is able to vote and, in case of a tie, has the casting vote. If the allocation he has proposed is denied, he is thrown overboard and the next most senior pirate proposes a new distribution.

All the pirates base their decisions on three factors: First of all, they want to survive (duh...). Secondly, they all want to get the most ammount of gold as possible. And thirdly, all the pirates hate each other and would choose to throw another one overboard if it won`t affect the first two factors.

How much gold will each pirate get?



Quem disse que a boca é tua?

Qual é, Dadinho...?

Dadinho é o caralho! Meu nome agora é Zé Pequeno!

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I'll try a reasoning. Pirate 1 is the first to propose, and he wants his proposal accepted, else he'd be killed. He knows pirate 4 and 5 will turn down every proposal, because they are the last to propose and they won't be thrown overboard, so he proposes 33/33/34/0/0 (or something like that), 2 and 3 turn him down and throw him. Then Pirate 2 comes and proposes 50/50/0/0, Pirates 2 and 3 vote yay (3 because he knows it's the maximum he can get), 4 and 5 vote nay, 2 decides yay (because it's a tie)
So in the end:
P1 gets thrown
P2 gets 50g
P3 gets 50g
P4 gets zilch
P5 gets zilch

It's just a theory, probably wrong




zexen_lowe said:

I'll try a reasoning. Pirate 1 is the first to propose, and he wants his proposal accepted, else he'd be killed. He knows pirate 4 and 5 will turn down every proposal, because they are the last to propose and they won't be thrown overboard, so he proposes 33/33/34/0/0 (or something like that), 2 and 3 turn him down and throw him. Then Pirate 2 comes and proposes 50/50/0/0, Pirates 2 and 3 vote yay (3 because he knows it's the maximum he can get), 4 and 5 vote nay, 2 decides yay (because it's a tie)
So in the end:
P1 gets thrown
P2 gets 50g
P3 gets 50g
P4 gets zilch
P5 gets zilch

It's just a theory, probably wrong

Not quite. I can`t tell you why without giving away too much! Just know that P3 wouldn`t agree with that because he can get more...

 



Quem disse que a boca é tua?

Qual é, Dadinho...?

Dadinho é o caralho! Meu nome agora é Zé Pequeno!

Pirate 5 gets everything.



4 ≈ One

If my above answer is incorrect then Pirate 4 and 5 split it 50/50



4 ≈ One

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Pirate 4 gets everything.

Whatever Pirate 1 says, everyone will throw him overboard because It is 1 less person.

Whatever Pirate 2 says, everyone will throw him overboard because It is 1 less person.

Whatever Pirate 3 says, 4 and 5 will throw him overboard because they will always win.

Pirate 4 suggests that he gets everything. Pirate 5 objects, Pirate 4 votes yes. Pirate 4 gets the tie breaker and gets everything.




If you drop a PS3 right on top of a Wii, it would definitely defeat it. Not so sure about the Xbox360. - mancandy
In the past we played games. In the future we watch games. - Forest-Spirit
11/03/09 Desposit: Mod Bribery (RolStoppable)  vg$ 500.00
06/03/09 Purchase: Moderator Privilege  vg$ -50,000.00

Nordlead Jr. Photo/Video Gallery!!! (Video Added 4/19/10)

nordlead said:
Pirate 4 gets everything.

Whatever Pirate 1 says, everyone will throw him overboard because It is 1 less person.

Whatever Pirate 2 says, everyone will throw him overboard because It is 1 less person.

Whatever Pirate 3 says, 4 and 5 will throw him overboard because they will always win.

Pirate 4 suggests that he gets everything. Pirate 5 objects, Pirate 4 votes yes. Pirate 4 gets the tie breaker and gets everything.

 

That was my first guess... but it came to me too quickly so I thought I had to be wrong... so I went with 2 other options XD



4 ≈ One

Dgc1808 said:
nordlead said:
I posted this in another area, but it is one of my favorite problems of all time and no one has solved it yet. A few things to note. Solving this like a proof in algebra makes this a lot easier. Also, I won't post an answer for a few days, as I have to type it up and I'm lazy. There is a short answer, but that is the easy cop out.

1) Start with VG
2) If you possess a string whose last letter is G, you can add an C at the end
3) If you have Vx you can add Vxx to your collection
Example: (VGC => VGCGC)
4) if you have GGG you can replace that with C
5) If you have CC you can drop it
6) Find a way to get VC

 

I was wondering If I can get a PM to this questions answer??

I didn't get as much time as I wanted to ponder on it....

*Searches for answer in thread*..... *sniff*... *sniff*........

 



4 ≈ One

Dgc1808 said:
Dgc1808 said:
nordlead said:
I posted this in another area, but it is one of my favorite problems of all time and no one has solved it yet. A few things to note. Solving this like a proof in algebra makes this a lot easier. Also, I won't post an answer for a few days, as I have to type it up and I'm lazy. There is a short answer, but that is the easy cop out.

1) Start with VG
2) If you possess a string whose last letter is G, you can add an C at the end
3) If you have Vx you can add Vxx to your collection
Example: (VGC => VGCGC)
4) if you have GGG you can replace that with C
5) If you have CC you can drop it
6) Find a way to get VC

 

I was wondering If I can get a PM to this questions answer??

I didn't get as much time as I wanted to ponder on it....

*Searches for answer in thread*..... *sniff*... *sniff*........

 

Sorry, My family came in from out of state, and I didn't know until last night, so this morning I forgot about it and didn't have time. I'll post it soon, I promise.

 




If you drop a PS3 right on top of a Wii, it would definitely defeat it. Not so sure about the Xbox360. - mancandy
In the past we played games. In the future we watch games. - Forest-Spirit
11/03/09 Desposit: Mod Bribery (RolStoppable)  vg$ 500.00
06/03/09 Purchase: Moderator Privilege  vg$ -50,000.00

Nordlead Jr. Photo/Video Gallery!!! (Video Added 4/19/10)

nordlead said:
Dgc1808 said:
Dgc1808 said:
nordlead said:
I posted this in another area, but it is one of my favorite problems of all time and no one has solved it yet. A few things to note. Solving this like a proof in algebra makes this a lot easier. Also, I won't post an answer for a few days, as I have to type it up and I'm lazy. There is a short answer, but that is the easy cop out.

1) Start with VG
2) If you possess a string whose last letter is G, you can add an C at the end
3) If you have Vx you can add Vxx to your collection
Example: (VGC => VGCGC)
4) if you have GGG you can replace that with C
5) If you have CC you can drop it
6) Find a way to get VC

 

I was wondering If I can get a PM to this questions answer??

I didn't get as much time as I wanted to ponder on it....

*Searches for answer in thread*..... *sniff*... *sniff*........

 

Sorry, My family came in from out of state, and I didn't know until last night, so this morning I forgot about it and didn't have time. I'll post it soon, I promise.

 

Ok. I'll just wait.

 



4 ≈ One