[E] Distributing dollars among pirates.

[SURI]

It seems people tired up with the data structure problems.. Here is
one general interesting puzzle.

Six pirates must divide $300 dollars among themselves. The division is
to proceed as follows. The senior pirate proposes a way to divide the
money. Then the pirates vote. If the senior pirate gets at least half
the votes he wins, and that division remains. If he doesn’t, he is
killed and then the next senior-most pirate gets a chance to do the
division.

PART - I : Now you have to tell what will happen and why (i.e. how
many pirates survive and how the division is done)?
[All the pirates are intelligent and the first priority is to stay
alive and the next priority is to get as much money as possible.]

PART - II : Reconsider the pirate problem above, where only one
indivisible dollar is to be divided. Who gets the dollar and how many
are killed?

[Suresh Kumar Maridi]

Does his own vote count while deciding the win?

--
Suresh Kumar Maridi

"it is not our abilities that show what we truly are...... it is our choices"
       - Albus DumbledorE

[SURI]

Not sure, I borrowed it from some site. Found it interesting hence
posted here.

Let us solve it for both cases..

It seems "counting own vote" version is easier than the other then
lets start with this.
But if you disagree then you can start with the case which is easier
for you.

-- SURI

On Jul 15, 9:46 am, Suresh Kumar Maridi <suresh.mar...@gmail.com>
wrote:

[Shyam Prakash Velupula]

Assuming "self" vote counts.

Assuming pirate 6 starts the proceeding, he makes the following offer

1      2     3     4       5       6

$1    $1   $0    $0   $0    $298


With this 1, 2 and 6 would agree for the offer (50%), hence stands valid.

Why 1 & 2 agrees?

Basic case, let's say there are only 2 pirates (1 & 2)

If 2 starts the proceeding, he can take all the money as his vote accounts for 50%. i.e 1 gets nothing
If there are 3 pirates 1 2 3. 2 can exploit the above logic so he will attempt to decline any offer made by 3. But 3 being smart, 3 can offer $1 to 1. pirate 1 also knows the fact that, if he decline this offer, he gets nothing when the turn comes to pirate 2. So he accepts the $1 offer gleefully.

And with 3 or 4 pirates (1, 2, 3, 4), pirate 2 knows that 3 or 4 will make $1 offer to pirate 1 to get 50% or more vote, he will never get a chance. Hence he accepts any offer of $1 from either pirate 5 or 6.

Knowing this, pirate 6 makes the $1 offers to 1 and 2 and claims 50% vote.


Thanks
Shyam Velupula

--
You are limited only by your imagination

[bhargava]

How would the solution change if the pirates are sadistic ?

i.e They would prefer receiving X gold coins and seeing someone executed than just receiving X gold coins. Prefer X gold coins over X-1 gold coins and seeing an execution.

[SURI]

Sadistic !!! May be valid in 1$ case but in case of 300$ they would
prefer having extra dollar than seeing someone executed.

For 300$ Case :
Assuming "self" vote counts.

I got the dollar distribution as

1 2 3 4 5 6

298 0 1 0 1 0

Any other variant solutions for this case ??

-- SURI

[bhargava]

Sadistic without/with self voting, would it differ ?

On Friday, 15 July 2011 13:59:30 UTC+5:30, SURI wrote:

Sadistic !!! May be valid in 1$ case but in case of 300$ they would
prefer having extra dollar than seeing someone executed.

For 300$ Case :
Assuming "self" vote counts.

I got the dollar distribution as
    1    2   3   4   5   6
   298   0   1   0   1   0

Any other variant solutions for this case ??

-- SURI

[bhargava]

sorry, it should be Sadistic/non-sadistic without self voting,