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Brain teaser time!!

C

chuckywang

Lifer
I saw this on another message board:
Five pirates on an island have one hundred gold coins to split among themselves. They divide the loot as follows: The senior pirate proposes a division, and everyone votes on it. Provided at least half the pirates vote for the proposal, they split the coins that way. If not, they kill the senior pirate and start over. The most senior (surviving) pirate proposes his own division plan, and they vote by the same rules and either divide the loot or kill the senior pirate, as the case may be. The process continues until one plan is accepted. Suppose you are the senior pirate. What division do you propose? (The pirates are all extremely logical and greedy, including you, and all want to live.)
Click to expand...
 
34 to me, 33 to 2nd senior, 33 to 3rd senior?

Something along those lines but you'd need to make it worth it to those two so that they would get the maximum amount if they had voted against and did the same method as this
 
i would give them all the money split between them. Later that night, i would kill them all and take the GOLD!!! ARRRG!! i am a pirate right? 😱
 
say u dont want any of it. they all like that idea, but then everyone else dies because theyre all greedy and kill eachother. u kill the last guy cuz his proposal sucks. u win?

doubt that works, but w/e
 
Oh wait, the 5th pirate is important, he knows he'll get screwed if it comes down to just him and the 4th in charge since the 4th in charge just needs to get his own vote when it's down to two.

hmmm...
 
First pirate proposes 20 each.

Everyone else disagrees, 4 pirates left.

Rinse and repeat until 2 pirates left.
 
The 4th pirate will try to vote no until it is his turn so he can jsut appropriate all to him.
That mean the 5th pirate is more willing to take a deal... not sure if he'll vote for anything though...
 
Originally posted by: hypn0tik
First pirate proposes 20 each.

Everyone else disagrees, 4 pirates left.

Rinse and repeat until 2 pirates left.
Click to expand...

But

"Suppose you are the senior pirate."

😛
 
Originally posted by: YOyoYOhowsDAjello
Originally posted by: hypn0tik
First pirate proposes 20 each.

Everyone else disagrees, 4 pirates left.

Rinse and repeat until 2 pirates left.
Click to expand...

But

"Suppose you are the senior pirate."

😛
Click to expand...

Well, the senior pirate is screwed regardless. If he proposes 0 for himself and split the rest between the others, there's no point to it. If he proposes a lion's share for himself, everyone else will vote no.

Edit: Either way, everyone can get more if they kill the senior pirate.
 
No.1 : 49%
No.2 : 50%
No.5 : 1%

I based this by first thinking of distributing money among the majority each time. Doing it that way No.2 has a chance at 33 peices the first time and 50 pieces the second time before he dies. Assuming the pirates are logical and greedy but not smart, they will not try to give any money to the minority, meaning No.5 has no reason to hope for any money. SO, 50 pieces is reasonable to give no. 2 and 1 piece is good enough for no.5 to buy his vote. i give the rest to me.

No. 4 has a chance at all the money so theres no point in trying to buy his vote. No.3 and No. 2 are the same think in that they both have chance at 50 pieces.

 
Originally posted by: hypn0tik
Originally posted by: YOyoYOhowsDAjello
Originally posted by: hypn0tik
First pirate proposes 20 each.

Everyone else disagrees, 4 pirates left.

Rinse and repeat until 2 pirates left.
Click to expand...

But

"Suppose you are the senior pirate."

😛
Click to expand...

Well, the senior pirate is screwed regardless. If he proposes 0 for himself and split the rest between the others, there's no point to it. If he proposes a lion's share for himself, everyone else will vote no.
Click to expand...

He just needs to get two other pirates to think that his proposed plan gives them more than they would get otherwise.

For example in your scenario, it gets down to two pirates and then the 4th in line says 100 for me 0 for you and then he gets half the votes.

If I propose to give the 5th pirate 1 gold, that might be better than he would expect to get in the long run and he'd vote for my plan.

I'm not sure if there's any way to ensure that he wouldn't wait for another pirate to offer him 2 gold for his vote, but I think there's probably a solution here.
 
Originally posted by: YOyoYOhowsDAjello
Originally posted by: hypn0tik
Originally posted by: YOyoYOhowsDAjello
Originally posted by: hypn0tik
First pirate proposes 20 each.

Everyone else disagrees, 4 pirates left.

Rinse and repeat until 2 pirates left.
Click to expand...

But

"Suppose you are the senior pirate."

😛
Click to expand...

Well, the senior pirate is screwed regardless. If he proposes 0 for himself and split the rest between the others, there's no point to it. If he proposes a lion's share for himself, everyone else will vote no.
Click to expand...

He just needs to get two other pirates to think that his proposed plan gives them more than they would get otherwise.

For example in your scenario, it gets down to two pirates and then the 4th in line says 100 for me 0 for you and then he gets half the votes.

If I propose to give the 5th pirate 1 gold, that might be better than he would expect to get in the long run and he'd vote for my plan.

I'm not sure if there's any way to ensure that he wouldn't wait for another pirate to offer him 2 gold for his vote, but I think there's probably a solution here.
Click to expand...

Hmmmmm. Good point. I may have to rethink this tomorrow.
 
Originally posted by: Mo0o
No.1 : 49%
No.2 : 50%
No.5 : 1%

I based this by first thinking of distributing money among the majority each time. Doing it that way No.2 has a chance at 33 peices the first time and 50 pieces the second time before he dies. Assuming the pirates are logical and greedy but not smart, they will not try to give any money to the minority, meaning No.5 has no reason to hope for any money. SO, 50 pieces is reasonable to give no. 2 and 1 piece is good enough for no.5 to buy his vote. i give the rest to me.

No. 4 has a chance at all the money so theres no point in trying to buy his vote. No.3 and No. 2 are the same think in that they both have chance at 50 pieces.
Click to expand...

Let's say they decide to not vote your plan and it comes down to 4 pirates.

The No.2 pirate just needs to get two votes

He could do

No.2 : 99%
No.5 : 1%

and get the votes he needs.

I get what you're saying about them only giving to the majority, but I thought we were to assume that extremely logical meant they could figure this out.

I'm thinking you need to convince #3 since he thinks #2 will go with that plan.

Maybe

No.1 : 98%
No.3 : 1%
No.5 : 1%
 
Originally posted by: YOyoYOhowsDAjello
Originally posted by: Mo0o
No.1 : 49%
No.2 : 50%
No.5 : 1%

I based this by first thinking of distributing money among the majority each time. Doing it that way No.2 has a chance at 33 peices the first time and 50 pieces the second time before he dies. Assuming the pirates are logical and greedy but not smart, they will not try to give any money to the minority, meaning No.5 has no reason to hope for any money. SO, 50 pieces is reasonable to give no. 2 and 1 piece is good enough for no.5 to buy his vote. i give the rest to me.

No. 4 has a chance at all the money so theres no point in trying to buy his vote. No.3 and No. 2 are the same think in that they both have chance at 50 pieces.
Click to expand...

Let's say they decide to not vote your plan and it comes down to 4 pirates.

The No.2 pirate just needs to get two votes

He could do

No.2 : 99%
No.5 : 1%

and get the votes he needs.

I get what you're saying about them only giving to the majority, but I thought we were to assume that extremely logical meant they could figure this out.

I'm thinking you need to convince #3 since he thinks #2 will go with that plan.

Maybe

No.1 : 98%
No.3 : 1%
No.5 : 1%
Click to expand...

hmm that is true.
 
Ok, I think I have to redo this and find out how much No.5 expects to get at each level.

If it gets down to Three pirates, he knows that 1% is the best he can hope for. That means to ensure that No.2 gets his vote when it comes down to Four pirates, he should offer him 2%.

To beat that offer seems like I would offer him 3% and he knows that will be the best he'll do?

No.3 is going to be ignored by No.2 when there are 4 pirates left, so he'll get nothing if he doesn't vote for your plan, so you give him 1%.

So my new answer is

No. 1 : 96%
No. 3 : 1%
No. 5 : 3%
 
Since they all want to live, 1 and 5 will always vote yes. And since they are all greedy, 4 will always vote no.

For #1 to survive, he has to give #2 more than #2 would receive if there were 4 of them alive (because #2 would have to throw #3 money to vote yes).
 
Originally posted by: YOyoYOhowsDAjelloLet's say they decide to not vote your plan and it comes down to 4 pirates.

The No.2 pirate just needs to get two votes
Click to expand...
That doesn't seem fair though. No halfsies.
 
my brother told me this riddle a long time ago...I think he was asked it at a Job Interview somewhere.....there's some answer that is totally different than what you'd think it'd be...I believe its a trick question of sorts....but I reserve the right to be wrong 😛
 
Originally posted by: MrCodeDude
Originally posted by: YOyoYOhowsDAjelloLet's say they decide to not vote your plan and it comes down to 4 pirates.

The No.2 pirate just needs to get two votes
Click to expand...
That doesn't seem fair though. No halfsies.
Click to expand...

"Provided at least half the pirates vote for the proposal"

4/2 = 2 votes needed
 
Starting from the end, if it's just #4 and #5, #4 will give 100 to himself and 0 to #5 and vote yes ending the game (50%). So #5 will vote to prevent a 2 man scenario as long as he gets more than zero.

So one step back, #3 to #5. #3 will give #5 1 coin (so #5 will support) and #3 gets 99 and #4 gets nothing. So #4 will vote for anything that will prevent it going to a 3 man game as long as he gets more than 0 coins.

One step back #2 to #5. He gives #4 1 coin and himself 99 coins and #3,5 gets nothing. #3 and #5 will vote to prevent getting to a 4 man group as long as they get more than 0 coins.

One step back, #1 to #5. He gives 1 coin to #3, #5 and 98 for himself. End?
 
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