Pirates

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Pirates

Postby DeathRowKitty » Sun Feb 20, 2011 11:58 am

(Note: This puzzle is blatantly stolen from whoever wrote it. I have no idea who that is.)

Five pirates on an expedition for treasure come across 100 gold coins. In order to divide up the 100 coins fairly, the five pirates draw straws in order to determine an ordering of themselves. The pirates then, in order, each propose how the 100 coins should be split up, after which all living pirates vote. If more than half of the pirates (including the pirate proposing this way of splitting the coins) approve, the coins will be divided as that pirate has proposed. If not, the pirate proposing this way of dividing coins is killed and the next pirate makes a proposal. Assume all pirates are perfect logicians (if a logical conclusion exists, they will arrive at it instantaneously). Furthermore, assume their priorities, in order, are:

  1. Surviving
  2. Getting as much treasure as possible
  3. Killing other pirates

That is to say, they will try to survive no matter what, will try to get as much treasure as possible, given that it doesn't kill them, and, given that they will survive and get no less treasure, will always try to kill more pirates. Which pirate's proposal will be accepted and how many gold coins will each receive?
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Re: Pirates

Postby gwtkof » Mon Feb 21, 2011 4:50 am

What happens in the even of a tie?
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Re: Pirates

Postby DeathRowKitty » Mon Feb 21, 2011 9:27 am

A tie isn't more than half. :P
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Re: Pirates

Postby Zapp » Mon Feb 21, 2011 12:32 pm

what I think everybody knows before the votes, but after knowing the order, which led me to the solution:
Spoiler! :
before I start, keep im mind, that the last pirate in the order has the possibility to gain all pieces and his live is guaranteed, therfore he will deny any proposition which leaves him with less than 100 coins. the only possibility to gain the vote of pirate No. 5 is to offer him all coins. every pirate knows that, because they are perfect logicians.
if the first pirate gives all coins to 5 then 2, 3 and 4 will protest, because of their priorities.
if the first pirates equally divides the money between 2 and 3, then he will gain No. 2's vote, because he knows that if it is 2's turn then he will be placed in a similiar situation as 1, because he also needs 3 votes to win, also, No.2 will not be able to get out of his turn with any money, because he will try to satisfy priority 2 for at least 2 pirates, which will in turn satisfy his priority 1.
if it would be No. 3's turn, then would take all the money, and No. 4 will accept it, because No.4 will die on the 4th turn with a 100% probability.
now after looking at all those cases, No. 4 will definitely vote No at turn 1, if he does not obtain 100 coins (on vote 1 and 2 he has their lives in his hands, therefore demanding the maximum amount)
No. 5 will know that he will not gain any money if he is not offered some during turn 1 or 2.he will definitely vote Yes on turn 2 if he is offered money, because that will satisfy all of his priorities.
No 3 will definitely vote no at turn 1 and 2, regardless of the money offered.
knowing all of those conditions, it is obvious, that the only possibility to survive for No 2 is for 1 to win his vote, No 2 will vote Yes on 1. 1 will vote Yes for 1. No 4 will vote Yes, if he ist offered 100 coins. three will never vote yes on 1, 5 will not vote yes if he is not offered money.
also No.4 is only able to gain 100 coins on turn one, therefore he will definitely vote Yes on Turn 1.


my solution:
Spoiler! :
everyone survives and pirate number 4 gets everything
Q.E.D. , or not?
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Re: Pirates

Postby DeathRowKitty » Mon Feb 21, 2011 12:44 pm

Not correct. There are a lot of things in the first spoiler that aren't true.
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Re: Pirates

Postby Phoenix » Mon Feb 21, 2011 1:34 pm

TBH, I thought of something 2 days ago, but I wasn't sure if I dare to post it...
...but I'll try, anyway.

Spoiler! :
95-0-0-3-2


Can share my opinion in case this is true. But I doubt. :oops:
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Re: Pirates

Postby DeathRowKitty » Mon Feb 21, 2011 1:46 pm

Not quite.
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Re: Pirates

Postby Ardilla » Tue Feb 22, 2011 7:03 am

I think that the answer is:
Spoiler! :
96-0-0-2-2


The reasoning is the following:
Spoiler! :
We will start backwards:
1) If the only pirate left is 5, then he will take all the money.
2) Because of 1) if there are 2 pirates left, pirate 5 will vote no, because this way he will get all the money and kill all the other pirates.
3) Because of 2) if there are 3 pirates left, pirate 4 will vote yes, because if this proposal is dismissed he will die. This allows pirate 3 to get all the money.
4) Because of 3) pirates 4 and 5 have no hope in getting any money if this proposal is dismissed, because of that pirate 2 only needs to propose the distribution (0-98-0-1-1) to get the votes of pirates 4 and 5
5) Taking 4) into acount pirate 1 has to offer more money for pirates 4 and 5 to get their vote, giving 2 coins a piece he can ensure their vote, and get the rest.
Spoiler! :
The Spanish Inquisition

Clearly every even integer greater than 2 can be expressed as the sum of two primes.
I have discovered a truly wonderful proof of this proposition, but the signature is too small to contain it.
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Re: Pirates

Postby DeathRowKitty » Tue Feb 22, 2011 12:02 pm

You're really close, but not quite. Your logic was right, but you made a bit of a careless error somewhere.
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Re: Pirates

Postby Ardilla » Tue Feb 22, 2011 12:21 pm

Oh, you are right!
the answer is:
Spoiler! :
97-0-1-2-0
or
97-0-1-0-2

The reasoning:
Spoiler! :
We will start backwards:
1) If the only pirate left is 5, then he will take all the money.
2) Because of 1) if there are 2 pirates left, pirate 5 will vote no, because this way he will get all the money and kill all the other pirates.
3) Because of 2) if there are 3 pirates left, pirate 4 will vote yes, because if this proposal is dismissed he will die. This allows pirate 3 to get all the money.
4) Because of 3) pirates 4 and 5 have no hope in getting any money if this proposal is dismissed, because of that pirate 2 only needs to propose the distribution (0-98-0-1-1) to get the votes of pirates 4 and 5
5) Taking 4) if this proposal is dismissed pirate 3 will receive no money and pirates 4 and 5 will receive 1 coin.
Pirate 2 will not accept the proposal unless he receives 99 or more coins.
Because of all this if pirate 1 gives 1 coin to pirate 3 he will get his vote.
Pirate 1 have to give 2 coins to pirate 4 o 5 to receive one more vote.
With the vote of pirate 3, his own vote, and the vote of either pirate 4 or 5, pirate's 1 proposal will prevail.
Spoiler! :
The Spanish Inquisition

Clearly every even integer greater than 2 can be expressed as the sum of two primes.
I have discovered a truly wonderful proof of this proposition, but the signature is too small to contain it.
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Re: Pirates

Postby DeathRowKitty » Tue Feb 22, 2011 12:24 pm

Correct! :D
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Re: Pirates

Postby xander » Tue Feb 22, 2011 12:26 pm

I've seen a version of this puzzle before, but the voting was handled differently---ties led to the proposal being accepted, rather than rejected. I'm not sure that it makes much difference in the end, however.

Spoiler! :
Let us call the pirates A, B, C, D, and E. A is the first to propose a plan. Working backwards, assume that A, B, and C have all proposed plans and been killed. Pirate D will give all of the money to E. E can veto any plan that D comes up with, so D will choose to give everything to E in the hopes of staying alive.

Suppose that only C, D, and E are left. E knows that he can get 100 gold if C's plan is rejected, so he will vote against any plan that does not give him 100 gold. However, C could give D 1 gold, and keep the other 99. D gets more gold than if it is just between D and E, so he doesn't care that C takes the lion's share. Thus with three pirates, C will keep 99 gold, D will get 1 gold, and E gets nothing.

Suppose that four of the five pirates are left. Pirate B needs the support of two of the remaining pirates. In order to get pirate C's support, he would have to offer up all of the gold, as C is guaranteed 99 gold if B dies, and will happily kill B just to watch him die. Hence we can assume that C will vote against any plan that B proposes. B can get D's support by offering him 2 gold (if he offers just one gold, he votes against B, and gets one gold from C's plan), and B can get E's support by offering him 1 gold (which is one more than he would get under C's plan). Thus with four pirates, B gets 97, C gets nothing, D gets 2, and E gets 1.

Finally, with five pirates, A needs the support of two pirates. He won't get B's support, but he can get C's support for one gold and E's support for 2 gold. Hence the final distribution is:

A - 97 gold
B - 0 gold
C - 1 gold
D - 0 gold
E - 2 gold


xander

EDIT: Bah. Ninja'd while writing. :(
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Re: Pirates

Postby DeathRowKitty » Tue Feb 22, 2011 12:32 pm

Spoiler! :
Your final answer was right (well, you got half of the possible solutions), but you actually made a mistake in the base case of the "induction." Remember: pirates like killing other pirates.
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Re: Pirates

Postby xander » Tue Feb 22, 2011 12:36 pm

DeathRowKitty wrote:
Spoiler! :
Your final answer was right (well, you got half of the possible solutions), but you actually made a mistake in the base case of the "induction." Remember: pirates like killing other pirates.

Spoiler! :
If you are referring to the fact that I suggested that D give all the money to E, it is irrelevant---if D keeps the gold, E kills him; if D gives the money to E, E kills him. I thought I made that point by suggesting that D gives the money to E in hopes of staying alive. Since E prioritizes getting the money above killing D, then maybe E can be bribed. Otherwise, D's actions are irrelevant---either way, D ends up dead, and E gets all of the money.

In the version that I know, this is not the case, because D proposes a plan, and a tie leads to acceptance of the plan. Hence D keeps the money, and E goes poor. This also changes later steps, as B only needs the support of one other pirate, not two.


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Re: Pirates

Postby DeathRowKitty » Tue Feb 22, 2011 12:38 pm

Spoiler! :
If it gets all the way to D, E can keep all the money by killing D or by not killing D, so he automatically votes against D. A pirate's gotta have his priorities. ;)
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Re: Pirates

Postby Phoenix » Tue Feb 22, 2011 1:07 pm

OMG! I can't believe my logic was actually correct!
The only thing I've mistaken is one small little step, so that's why I added extra 1 coin to D and E (and maybe messed up who receives 1 and 2, too, if that already matters). Shame!

P. S. You could have said that I was close... "Not quite." sounds like I messed up a lot. :P
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Re: Pirates

Postby DeathRowKitty » Tue Feb 22, 2011 1:11 pm

I didn't mean for it to sound that way :P

I think I was in a bit of a rush at the time and just gave a quick response
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Re: Pirates

Postby Ardilla » Tue Feb 22, 2011 1:13 pm

The answer is
Spoiler! :
ARRRRRRRRRRRRRRRRRRRR
Spoiler! :
The Spanish Inquisition

Clearly every even integer greater than 2 can be expressed as the sum of two primes.
I have discovered a truly wonderful proof of this proposition, but the signature is too small to contain it.
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Re: Pirates

Postby DeathRowKitty » Tue Feb 22, 2011 1:13 pm

On an unrelated note, your sig still gets me every time xD

Edit: IT JUST GOT ME AGAIN. GAHHH.

You're lucky your sig is completely hilarious or I would be really pissed right now :P
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Re: Pirates

Postby Phoenix » Tue Feb 22, 2011 5:46 pm

DeathRowKitty wrote:I didn't mean for it to sound that way :P

I think I was in a bit of a rush at the time and just gave a quick response


No problem. No feelings hurt! ;)
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