[Guest]

There are 5 pirates that need to divide a treasure of 100 gold pieces:

-All 5 pirates are PERFECTLY RATIONAL and PERFECTLY SELFISH

The way it goes: Pirate 1 will propose a division method, then there will be a vote- a majority vote means the division will be accepted. However a minority vote would mean that pirate 1 would be made to walk the plank and would die. Thus leaving 4 pirates. Then it would come upon Pirate 2 to make a proposal and so on.

A tie in the vote means the vote is passed.

The thing to remember is all pirates are perfectly rational and perfectly selfish.

How will the treasure be divided and why?

[akaslickster]

the first guy was to greedy and walked the plank and drowned. The second decided not to be greedy and stay alive so he proposed that they each will get 25 gold pieces and they lived happily ever after.???

[Guest]

the first guy was to greedy and walked the plank and drowned. The second decided not to be greedy and stay alive so he proposed that they each will get 25 gold pieces and they lived happily ever after.???

That is not the answer. There is a logical conclusion to the problem.

[Guest]

i think there was a similiar riddle posted somewhere here.

[Guest]

I'm going to assume that the proposing pirate can vote for himself and that when there is a tie and a vote is "passed" the proposal is accepted

the first pirate will be voted to walk the plank no matter how he divides the treasure, cuz the rest of the pirates will get more money with one less pirate.

likewise for the second pirate and the third pirate.

then the fourth pirate will propose all the money to himself and vote for himself. the vote will end in a tie.

Edited July 17, 2008 by rjsghk107

[Guest]

I'm going to assume that the proposing pirate can vote for himself and that when there is a tie and a vote is "passed" the proposal is accepted

the first pirate will be voted to walk the plank no matter how he divides the treasure, cuz the rest of the pirates will get more money with one less pirate.

likewise for the second pirate and the third pirate.

then the fourth pirate will propose all the money to himself and vote for himself. the vote will end in a tie.

Your assumption is right- the proposing pirate can vote.

Your answer is exactly what came to my mind initially. However, it is not the correct answer. The riddle requires some deep thinking.

[Guest]

No other takers- I will post a hint in some time.

[Guest]

Seeing as No.5 knows that he will get nothing if it gets to No.4, and No.3 knows he will get nothing & killed if he proposes that he takes it all, then No.3 should propose an even split between himself and No.5, which should leave 2:1 votes against No.4. No.2 could possibly cut a deal with No.4 (knowing that Nos 3&5 will screw No.4 over) and take a tie on votes to pass it.

However, knowing this, No.1 could propose a 3-way split between himself and Nos 3&5, which will get voted for as Nos 3&5 both realise that 2&4 will split the treasure.

So, all pirates survive, and the treasure is split 3 ways between Pirates 1, 3 and 5.

Spluh

[Guest]

what happens in the event of a tie vote?

[Guest]

No matter what the first guy proposes, there will be no vote from the rest of them. (Reason, no vote = death sentence for the one who proposes). The next two guy will also face the similar fate.

Remaining two guys....what's next... it will be a tie. 100 golds is being splitted equally among the remaining two.

[Guest]

what happens in the event of a tie vote?

A tie wins the vote

[Guest]

Seeing as No.5 knows that he will get nothing if it gets to No.4, and No.3 knows he will get nothing & killed if he proposes that he takes it all, then No.3 should propose an even split between himself and No.5, which should leave 2:1 votes against No.4. No.2 could possibly cut a deal with No.4 (knowing that Nos 3&5 will screw No.4 over) and take a tie on votes to pass it.

However, knowing this, No.1 could propose a 3-way split between himself and Nos 3&5, which will get voted for as Nos 3&5 both realise that 2&4 will split the treasure.

So, all pirates survive, and the treasure is split 3 ways between Pirates 1, 3 and 5.

Spluh

You are on the right track. what is the division?

[kingofpain]

Working backwards: If it comes down to 2 pirates (Pirates 1, 2 and 3 dead), P4 will propose 100 coins for himself and P5 will get none.Therefore, P5 will support P3 if P3 proposes 99 coins for himself and 1 coin for P5 and 0 coins for P4. This in turn implies that P4 will support P2 if P2 proposes 99 for himself and 1 coin for P4. Therefore, all that P1 has to propose is 1 coin for P5, 1 coin for P3 and 98 coins for himself.

Cheers!

--

Vig

[Guest]

No matter what the first guy proposes, there will be no vote from the rest of them. (Reason, no vote = death sentence for the one who proposes). The next two guy will also face the similar fate.

Remaining two guys....what's next... it will be a tie. 100 golds is being splitted equally among the remaining two.

If a tie wins the vote, why would they split it equally...when 2 pirates r left P4 would vote he gets everything, tie the vote and win.

Even so this answer is incorrect. keep trying.

[Guest]

Think backward induction

I'm going to assume that the proposing pirate can vote for himself and that when there is a tie and a vote is "passed" the proposal is accepted

the first pirate will be voted to walk the plank no matter how he divides the treasure, cuz the rest of the pirates will get more money with one less pirate.

likewise for the second pirate and the third pirate.

then the fourth pirate will propose all the money to himself and vote for himself. the vote will end in a tie.

[Guest]

You are on the right track. what is the division?

Well Seeing as 3 & 5 are better off with anything more than 0, No 1 can offer 98 to himself and 1 each to 3 & 5

[Guest]

Cheers!

--

Vig

Working backwards: If it comes down to 2 pirates (Pirates 1, 2 and 3 dead), P4 will propose 100 coins for himself and P5 will get none.Therefore, P5 will support P3 if P3 proposes 99 coins for himself and 1 coin for P5 and 0 coins for P4. This in turn implies that P4 will support P2 if P2 proposes 99 for himself and 1 coin for P4. Therefore, all that P1 has to propose is 1 coin for P5, 1 coin for P3 and 98 coins for himself.

Spluh and kingofpain- Congratulations. You have both figured it out. Since P3 and P5 along with the rest of the pirates are perfectly rational, they will happily accept 1 piece each determining it to be the best possible outcome for them given the situation.

[Guest]

This is one of my all time favorite riddles. Kindly rate it if you found it interesting.

[Guest]

[spoiler= ...this is probably a way too simple solution for it to be even in the vicinity of correct, but hey, worth a shot]Pirate 1 suggests, that Pirates 1,2 and 3 divide the money equally amongst the. 1 + 2 + 3 concur, ergo, the majority.

[Guest]

[spoiler= ...this is probably a way too simple solution for it to be even in the vicinity of correct, but hey, worth a shot]Pirate 1 suggests, that Pirates 1,2 and 3 divide the money equally amongst the. 1 + 2 + 3 concur, ergo, the majority.

I just realized, that the correct answer has already been posted... sorry.

I'm new! - Don't kill me...

[Guest]

Aren't you missing something?

Pirate 1 and 3 have the extra burden of dying should they not get their vote, while Pirate 5 will only lose out on the coins. By the rules, only if your vote is not passed, you walk the plank, since 5 will never get to vote, he is safe.

I would then suggest that Pirate 1 vote all of the coins to Pirate 5.

[Guest]

uh duh.... rock paper sciccors

Edited July 17, 2008 by Mityily*Hk

[Guest]

the fourth pirate to suggest gets all of the money

first vote - the Proposer votes yes teh remaining pirates vote no (more loot for them)

2nd vote goes 1 against three (once again selfish pirats more loot for them)

3rd vote goes 1 against two

4th vote goes 1 against 1 so the proposal is accepted. the pirate who proposed teh vote said that he gets all of teh money

[Yoruichi-san]

Uhhh...sorry to spoil the party, but this riddle has been done before and then some...search for "pirates" there's Pirates 1, 2, (by magicalcheese) and 3 (by yours truly ;P)...I'm also currently kinda on/off working on Pirates 4...

[unreality]

http://brainden.com/forum/index.php?showtopic=3169

OP: search for "pirate +gold" and you get the pirate gold puzzles (which are great puzzles )