Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**arrrr****Member**- Registered: 2011-11-18
- Posts: 1

So let's firstly set out some basics.

1. The most important rule - no one wishes to die - so the first person must set such a divide so he will live and the rest of the crew will be "happy"

2. According to the first rule - it is better to get nothing than to die

3. last rule but still important pirates are inteligent

Also you can take notice of three things

A. the youngest pirate cannot die in this case

B. the second youngest(4th) will always vote Nay so he deserves nothing (till its his turn to divide)

C. Person wich sets up the deal wont vote against himself

Lets make also clear 1-oldest (...) 5- youngest

So the question is what will happen (other version says how much gold can the oldest get and still live)

so accepting the question what will happen ill say:

the youngest pirate gets he will end up only with one coin at best so he changes his fate

the oldest pirate really doesnt wants to die so he knows that if 98:0:1:0:1 wont succeed he eventually will end with no gold nor life

So the oldest proposes 0:1:1:0:98 oldest get nothing second 1 coin third 1 coin youngest 98

the logic is as follows :

if the youngest would be unhappy with getting only one coin His vote would be the most decisive

so if the oldest says he splits 98:0:1:0:1 and the youngest would say no becouse he thinks he deserves more - oldest will die

0:1:1:0:98

1 pirate(oldest) gets nothing and wont vote for himself to be thrown out of board

2nd pirate has to accept the one coin deal and also has to the third pirate

Becouse

if the second pirate wouldnt agree on such deal and lets say the third wouldnt agree either + the 4th - captain goes down

But

Now the 2nd pirate has to divide so he has to set 0:0:0:100 himself wont vote nay also the youngest has 100 so he wont vote nay

if it would be 1:0:0:99 the second could eventualy die and make the third pirate think about his life

so the lets say second said 99:0:0:1 or 1:0:0:99 and died

Now the third pirate Has Only This choice - do i want to die or do i want to live? by die i assume 99:0:1 or 1:0:99 by live = 0:0:100

I think we should accept that the third's pirates choice/step should be the last becouse if the third pirate Knows that when last two pirates died becouse the youngest was unhappy it wouldnt be too Intelligent to make the same mistake the third time so he has to set 0:0:100 wich going backward reverses to 0:1:1:0:98

and 0:1:1:0:98

oldest gets nothing but he is still alive and is double sure he has 50% or more votes even if the youngest would say no the second and third would have to say Yes

second has to accept 1 coin or in the next turn he gets nothing or dies

third - same as above

and 4th gets nothing couse he would always vote no

I think i made myself clear

Offline

**Engenuelino****Member**- Registered: 2011-07-11
- Posts: 8

Have tried to register and did not find an email that was to be sent to me. Have tried to reregister, but get a message that my username or email is already being used. What do I do?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,593

Hi;

Sign up with another email address.

Please do not post any spam. No links to any other sites other than a free site. are allowed. That means no pay homework sites, porn sites, pirated movies, pharmaceuticals, no advertisements of any kind. Nothing that can cause a problem for the forum. All those will be deleted.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline