How Can Pirate P1 Maximize His Gold While Ensuring His Survival?

[moshek]

5 pirates P1,P2,P3,P4,P5 are on one boat they just stole 1000 gold coins. They want to share it between themselves. Every one in his tern can suggest how to do it . If more 2 pirates agree with that, this is what a happened, But if 3 or more disagree with him then they throw him to the sea.

What can the first Pirate can suggest so he can have agreement and also the maximum numbers { very many} of coins he can have?

 

[MathematicalPhysicist]

5 pirates P1,P2,P3,P4,P5 are on one boat they just stole 1000 gold coins. They want to share it between themselves. Every one in his tern can suggest how to do it . If more 2 pirates agree with that, this is what a happened, But if 3 or more disagree with him then they throw him to the sea.

What can the first Pirate can suggest so he can have agreement and also the maximum numbers { very many} of coins he can have?

what happens after his turn and let say that the number of pirates has changed too? does the condition stand still although the number of people to disagree with may change?

 

[marcus]

what happens after his turn and let say that the number of pirates has changed too? does the condition stand still although the number of people to disagree with may change?

dear Moshe, we need to know how it is if there are just two pirates

 

[moshek]

My mistake

The pirate that give the suggestion should have more pirate ( include him) that support his suggestion otherwise they throw him to the sea.
So if only 2 stay then the other pirate should agree with him

My question was not pose correctly!
I am sorry for that
Thank you both for your question.

Moshek

p.s : I don't know the answer ( yet)

www.physicsforums.com/showthread.php?t=17243

 

[marcus]

thanks for the link to Organic Science which has a poem about science
mathematics and philosophy (Wittgenstein and fly in klein bottle!)

it is reassuring to find someone who does not draw a sharp division between science and poetry

some people here like to quote an unpleasant anti-poetry saying of PAM Dirac which is not really true


now about the pirates

number them P5, P4, P3, P2, P1
(the opposite of your numbering) so the first pirate to propose a split
is P5

the game is different if there is negotiation and promises
but for simplicity assume there is no bargaining for votes
only the first pirate proposes a split and the others say yes or no


the rule is the proposer gets thrown into the sea unless he gains an absolute majority (more say yes than say no)

if there are two pirates P2 and P1, then P2 must propose a split and P1 will obviously always say no
then it will be 1 versus 1 and P2 will be thrown into the sea
and P1 (the last pirate) will get all the gold coins

so then we must back up to the case of three pirates P3, P2, P1
Now it is interesting because P2 will know that he MUST ACCEPT
the proposition no matter what it is
because if he does not accept and also P1 says no then it will
be down to the two pirate situation and the last pirate will get all the gold.

the last pirate never has any reason to say yes, no matter what the offer is
even if they propose to give him all the gold and give themselves nothing he can still say no if he wants (saying no works in all cases for the last pirate)

So if I were P3, I would propose that my share be 1000 and the other two shares be zero
I know that P1 will say no
but I know that P2 will say yes because he does not want to be thrown into the sea. Life is more important than money.

Now we know that in case of three pirates, the first pirate always gets all the gold. P3 can propose a split (1000, 0, 0) and he will win.

So we can back up to the case of four pirates P4, P3, P2, P1
now P3 will always say no
because if anyone votes with him then P4 goes in the sea and there are 3 pirates and P3 becomes first pirate and gets all the gold


So in case of 4 pirates, P3 will say no and the other two P2 and P1
must say yes, or else they will get nothing
so the first pirate can propose that he keeps everything---maybe he should offer P2 and P1 each one coin just to keep them a little happy
so P4 will propose a split (998, 0, 1, 1)


In case P2, P1------P1 says no and gets all the gold

In case P3, P2, P1------P2 will always say yes (for fear of drowning) so P3 can offer any split he wants, being a pirate he will offer them nothing and keep it all.

In case P4, P3, P2, P1-----P3 will say no because he likes the situation with three pirates, P4 should offer P2 and P1 each one coin and keep the rest
I think P4 should offer a split (998, 0, 1, 1)

The case P5, P4, P3, P2, P1 is the first interesting case
I am thinking that P4 will say no because he likes the situation with four pirates, but P3 does not like that situation because he gets nothing then. So P3 votes yes if he is even offered one coin

the question I ask is how much P5 should offer P2

I think P5 should offer (995, 0, 1, 2, 2) as a split

then P4 will say no (but he would say no anyway)
P3 will say yes because 1 coin is better than the zero he gets in the next case

well you can analyze this further, this is just a first approximation
and no one really knows how pirates think
I think perhaps they think like the CEOs of large corporations
that is, somewhat worse than animals, however it is difficult to
be certain

 

[honestrosewater]

Just some thoughts so far...

number them P5, P4, P3, P2, P1
(the opposite of your numbering) so the first pirate to propose a split
is P5

the game is different if there is negotiation and promises
but for simplicity assume there is no bargaining for votes
only the first pirate proposes a split and the others say yes or no

In what order do they answer? Ex. P5 proposes, then P4 answers yes or no, then P3 answers, etc. Can one pirate pass (p times) and make the others answer first? Can the proposing pirate decide in what order the others answer? etc.? What difference does it make?

the rule is the proposer gets thrown into the sea unless he gains an absolute majority (more say yes than say no)

if there are two pirates P2 and P1, then P2 must propose a split and P1 will obviously always say no
then it will be 1 versus 1 and P2 will be thrown into the sea
and P1 (the last pirate) will get all the gold coins

Assuming pirates are "like the CEOs of large corporations" ;) wouldn't it be "better for business" for P1 to agree with P2 if P2 proposes that P1 get all the gold- instead of disagreeing with him regardless- *if* there is already a "pirate code" saying that if one pirate saves another's life... see where I'm going? That doesn't break the no negotiating rule, does it? Or am I going too far?
I can think of more pros and cons of letting P2 live. Ex, pro: one more pirate for the good guys to chase (less chance of P1 getting caught)... you may say that's one more pirate stealing from good guys (less booty available to P1), but who says pirates can't steal from other pirates? con: P2 can try to steal from P1...
Happy thoughts
Rachel

 

[MathematicalPhysicist]

A Puzzle for Pirates

5 pirates P1,P2,P3,P4,P5 are on one boat they just stole 1000 gold coins. They want to share it between themselves. Every one in his tern can suggest how to do it . If more 2 pirates agree with that, this is what a happened, But if 3 or more disagree with him then they throw him to the sea.

What can the first Pirate can suggest so he can have agreement and also the maximum numbers { very many} of coins he can have?

i knew this puzzle sounded familiar to me, ian stewart has covered this puzzle in his column in scientific american.
the originator of the puzzle is Steven Landsberg (but his version is with 10 pirates and and 100 gold coins), if you want the answer (and to spoile it for yourself) here's it:
http://members.aol.com/istewjoat/sampleMR.html#sampleMR

 

[honestrosewater]

spoil it for yourself?
Nope unintended? ;)
cause they're pirates... er...

 

[moshek]

Dear marcus:

Thank you for your interesting attitude to this problem. It open my mind and i think your solution is correct or atlist must be very close to it. Do you think that for P6 it will be :

(993,0,2,2,2,1)

I am glad to know that Ian stewart is connect the pirate nice problem since i wrote about him in my poem "For mathematics" since in the Epilog to his book the nature of number he wrote about the possibility to create new mathematics which will be more flexible then the one we have today.
well Wittgenstein imagine that kind of new mathematics look like the Klein Bottle.

Honestrosewater you give a nice new point to think about this problem ,
lets think about this point..

Best
Moshek

www.physicsforums.com/showthread.php?t=17243

 

[honestrosewater]

Well, here's ONLY the question from LQG's linked site:

"Ten pirates have got their hands on a hoard of 100 goldpieces, and wish to divide the loot between them. They are democratic pirates, in their own way, and it is their custom to make such divisions in the following manner. The fiercest pirate makes a proposal about the division, and everybody votes --- one vote each including the proposer. If 50% or more are in favour, the proposal passes and is implemented forthwith. Otherwise the proposer is thrown overboard and the procedure is repeated with the next fiercest pirate.

All the pirates enjoy throwing people overboard, but given the choice they prefer hard cash. They dislike being thrown overboard themselves. All pirates are rational, know that the other pirates are rational, know that they know that... and so on. Unlike other puzzles with an apparently similar form (see Monks, blobs, and common knowledge, Scientific American August 1998, 96-97), this one does not hinge upon the revelation of some piece of 'common knowledge'. Any common knowledge is already commonly known. Moreover, no two pirates are equally fierce, so there is a precise 'pecking order' --- and it is known to them all. Finally: gold pieces are indivisible and arrangements to share pieces are not permitted (since no pirate trusts his fellows to stick to such an arrangement). It's every man for himself"

This variation requires only 50% or more, so P1 can't throw P2 overboard if it gets down to just them. But they still don't specify any order to the voting, perhaps it doesn't matter.
Happy thoughts
Rachel

 

[marcus]

...This variation requires only 50% or more, so P1 can't throw P2 overboard if it gets down to just them. But they still don't specify any order to the voting, perhaps it doesn't matter.
Happy thoughts
Rachel

I think the voting must be simultaneous
like everybody writes down their vote on a scrap of parchment, which is folded and put in the Pirate hat
and then taken out and counted

If the rules are changed so it is not an absolute majority but only
50% or more required then it is a whole other problem

Rachel your signature mentioning rosewater reminds me of
something connected to piracy in the Rosewater book by K. V.
the motto "grab much too much or you may get nothing at all"
It was a book where he was trying to understand America from
a human standpoint. Ever read it?

 

[moshek]

Well if 50% is fine and not only more that may change the nice solution of Marcus I wonder what he think about that. I will try my best also.

Thank you
Moshek

 

[moshek]

We may think about the general problem :

Pir(n,c) which mean -

n pirates and c coins to share with.

Best
Moshek

 

[honestrosewater]

If we define each pirate's priorities, in order, as:
1) Avoid getting thrown overboard
2) Maximize your share of gold
3) Maximize number of pirates thrown overboard

When it's between P1 and P2,
P2 must
1) Get P1 to vote yes.
P1's first priority is satisfied,
1) He/she can't be thrown overboard, since he is not making the proposal.
P1 satisfies his 2) and 3) by voting no to anything P2 proposes.
So P2 cannot satisfy his 1).

When it's between P1, P2, and P3,
P3 must
1) Get P2 or P1 to vote yes.
Since P2 must vote yes to satisfy his 1), P3 can propose to keep all the gold for himself and satisfy his 1) and 2).
Note in this case P1 gets no gold, and no one is thrown overboard so doesn't satify his 2) or 3).

When it's between P1, P2, P3, and P4
P4 must
1) Get two of P1, P2, or P3 to vote yes.
P1 will vote yes if offered any gold to satisfy his 2), otherwise he votes no to satisfy his 3).
P2 will vote the same way as P1.
P3 will vote no in all cases
So P4 must offer P1 and P2 some gold to satisfy his 1), and to satisfy his 2) he offers them each 1 gold piece and keeps the rest.

When it's between P5, P4, P3, P2, and P1
P5 must
1) get two of P4, P3, P2, or P1 to vote yes.
P1 will vote yes if offered more than 1 gold piece, otherwise he votes no.
P2 will vote the same way as P1.
P3 will vote yes if offered any gold, otherwise he votes no.
P4 will vote yes if offered more than 998 gold pieces, otherwise he votes no.
So P5 must offer two pirates gold to satisfy his 1), and to satisfy his 2), he offers P3 1 piece and either P2 or P1 2 pieces, and keeps the rest.

So
P5 gets 997 gold pieces
P4 gets 0
P3 gets 1
P2 gets 2 or 0
P1 gets 2 or 0

This is similar to Marcus', but I don't know if this is right. I need to check it, and could clean it up a lot, but 1+N heads are better than 1, so here it is...

 

[honestrosewater]

Marcus, no I haven't read the book. Rosewater has a personal meaning, long story. (But I'm a writer, so if you're intereted, I'd be glad to tell you)
Rachel
P.S. I'm not sure about this part:
When it's between P1, P2, P3, and P4
P4 must
1) Get two of P1, P2, or P3 to vote yes.
P1 will vote yes if offered any gold to satisfy his 2), otherwise he votes no to satisfy his 3).
P2 will vote the same way as P1.
P3 will vote no in all cases
So P4 must offer P1 and P2 some gold to satisfy his 1), and to satisfy his 2) he offers them each 1 gold piece and keeps the rest.

I think P3 might vote yes in some case. I'm working on it.

 

[honestrosewater]

No, I was right the first time, P3 always votes no.

But
"When it's between P5, P4, P3, P2, and P1
P5 must
1) get two of P4, P3, P2, or P1 to vote yes.
P1 will vote yes if offered more than 1 gold piece, otherwise he votes no.
P2 will vote the same way as P1.
P3 will vote yes if offered any gold, otherwise he votes no.
P4 will vote yes if offered more than 998 gold pieces, otherwise he votes no.
So P5 must offer two pirates gold to satisfy his 1), and to satisfy his 2), he offers P3 1 piece and either P2 or P1 2 pieces, and keeps the rest."
I think is wrong.

If P5 offers P3 1 piece of gold, P4 knows that P1 or P2 will vote yes if offered 2 gold pieces, and P4 will get nothing. So P4 will accept 1 gold piece and...

P5 gets 998
P4 gets 1
P3 gets 1
P2 gets 0
P1 gets 0

Yes, I think this is right. It's funny because in the other problem, "the meek inherit the worth" and if this is correct, it's the survival of the fittest ;)
Ha
Rachel

 

[honestrosewater]

Okay, I have checked it, and the solution
P5-998
P4-1
P3-1
P2-0
P1-0
is correct, as far as I can see.
Just two quick notes.
Every pirate will vote for his own proposal to satisfy his 1) and 2).
The pirate making the proposal cannot satisfy his 3) without NOT satisfying his 1) (higher priority), since HE will be the one going overboard ;)
Happy thoughts
Rachel

 

[marcus]

Okay, I have checked it, and the solution
P5-998
P4-1
P3-1
P2-0
P1-0
is correct, as far as I can see.
Just two quick notes.
Every pirate will vote for his own proposal to satisfy his 1) and 2).
The pirate making the proposal cannot satisfy his 3) without NOT satisfying his 1) (higher priority), since HE will be the one going overboard ;)
Happy thoughts
Rachel

as for the analysis it is very nice to see,
brava
happy thoughts to you

but it would be more correct to say
"Every pirate will vote for his/her own proposal ..., since HE/SHE will be the one going overboard"

Every writer should know the history of the famous Irish pirate
Grace O'Malley
who was also the queen of a certain independent island
and who had diplomatic contact with Elizabeth I around 1590 or so IIRC
and was give useless gifts by the other female monarch.
Grace was a tough customer and she and her crew took many
a fat English prize. Therefore the masc. pronoun should not be
used indiscriminately with pirates.

However the analysis matters more so cheers!

 

[honestrosewater]

:) Fair enough, but I at least said he/she once.
Happy thoughts
Rachel

 

[pig]

If P5 offers P3 1 piece of gold, P4 knows that P1 or P2 will vote yes if offered 2 gold pieces, and P4 will get nothing. So P4 will accept 1 gold piece and...

I think this is just a little bit wrong. :)

We can make two different problems here:
a) P5 proposes a division, and then the other pirates vote.
b) P5 makes offers to one pirate at a time, which then votes. Which I believe is the case with your solution.

a:

1) P1 P2

P1 votes no.
P2 dies.

2) P1 P2 P3

P2 doesn't like 1) so he votes yes.
P1 likes 1) so he votes no.
= 0 0 10

3) P1 P2 P3 P4

P3 likes 2) so he votes no.
P2 votes yes if offered at least 1.
P1 votes yes if offered at least 1.
= 1 1 0 8

4) P1 P2 P3 P4 P5

P4 votes yes if offered at least 9.
P3 votes yes if offered at least 1.
P2 votes yes if offered at least 2.
P1 votes yes if offered at least 2.
= 0 2 1 0 7
= 2 0 1 0 7

0 0 1 1 8 wouldn't work here - P4 votes no and gets 8 coins after throwing P5 overboard.

--------------------------------------------------

b:

1) P1 P2

P1 votes no.
P2 dies.

2) P1 P2 P3

P2 doesn't want to die so he votes yes.
= 0 0 10

3) P1 P2 P3 P4

This is where I think it goes a little different. When you said "I think P3 might vote yes in some case.", you were right.

If P4 offers P3 1 gold, he WILL accept it, because he knows that if he refuses, P2 and P1 will accept a coin each and he will get nothing.

So, P4 can offer a coin to any two of P1, P2 and P3 and get his 8 coins.

= 1 1 0 8
= 0 1 1 8
= 1 0 1 8

4) P1 P2 P3 P4 P5

Any of P1, P2 or P3 will accept one coin, because that way they are sure to get it, unlike in 3). Knowing this, P4 will also accept one coin. Therefore, P5 can offer a coin to any two of the other pirates and it will be accepted.

It is possible I made a mistake somewhere so feel free to correct me.. :)

 

[moshek]

Dear honestrosewater:

The one that ask me this nice problem
told me know that the solution is :

997,0,1,0,2

Well i read your nice argument and i believe that you are right !

I will think about your interesting opinion about Zenon paradox.

Best
Moshek

www.physicsforums.com/showthread.php?t=17243

 

[marcus]

:) Fair enough, but I at least said he/she once.

I didn't realize this (I guess I didn't read every post) but I wouldn't put it past you. Not only was it thoughtful to acknowledge the possibility but it provides us with an occasion to remember that brave and rapacious lady pirate, Grace O'Malley. Both Google and the Britannica have details on her, IIRC, in case anyone is interested. Congratulations on the various solutions and Happy thoughts to all.

 

[moshek]

Yes, Congratulations Honestrosewater !

So here are the five :



998 1 1 0 0


I will meet you soon with Zenon !

Best
Moshek

 

[honestrosewater]

Pig,

Let me get some coffee in my system and go through it again. I think you have a point. Oh what a tangled web we weave when we don't set up the rules properly ;)
(This will really annoy me if I was wrong because I checked the P4 propsal like a gazillion times!)
Happy thoughts
Rachel

 

[honestrosewater]

Okay, we had agreed earlier that one pirate makes a public proposal- all voting pirates hear the proposal simultaneously. And all pirates vote simultaneously.
So when P4 makes his proposal, since each pirate votes yes to his own proposal, P4 needs only two other "yes" votes (3/4) to not go overboard. We already agree that P1 and P2 will vote yes only if offered at least one gold coin(1 gc). So we need to find another proposal where P4 gives away 2 gc or less. No pirate will vote yes to P4's proposal if offered 0 gc. So P4 must give 1 gc each to two pirates. Any objections so far?
So,
P1, P2, P3
(1, 1, 0)a
(1, 0, 1)b
(0, 1, 1)c
We know a; b and c are the same: If P1 and P3 are offered 1 gc each, P2 will vote no. P3 knows that P2 will vote no, so P3 will vote no, even though he is offered 1 gc. Because then P4 only has one yes vote (from P1) and so goes overboard and P3 gets all the gold.
So even though P1 and P2 will vote yes if offered 1 gc, P3 is never in the position where accepting 1 gc is best.
Any mistakes so far?

Rachel

I'll write out P5's propsal next.

 

[honestrosewater]

So, between
P4, P3, P2, P1
(8, 0, 1, 1)

P5 has his own yes vote, and needs only two other yes votes (3/5 > 1/2).
No pirate will vote yes if offered 0 gc.
So P5 must offer two pirates n gc.
Look at it from P5's perspective:
Will 2 pirates vote yes for 1 gc each?
P3 will, because he gets 0 gc in the P4 siuation.
So we need to analyze
P4, P3, P2, P1
(1, 1, 0, 0)a
(0, 1, 1, 0)b
(0, 1, 0, 1)c
Any mistakes yet?
Rachel

 

[honestrosewater]

The situation with
P4, P3, P2, P1
(8, 0, 1, 1)

The possible situations when P5 gives away 2 gc
P5, P4, P3, P2, P1
(8, 1, 1, 0, 0)a
(8, 0, 1, 1, 0)b
(8, 0, 1, 0, 1)c

P3 votes yes, P4, P2, P1 vote no in all cases.
So P5 must give away more than 2 gc. (I was wrong)
P3 will still vote yes if offered 1 gc. So P5 will not waste 2 gc on P3.
P5 only needs two yes votes so only needs to offer gc to two pirates.
P4, P3, P2, P1
(2, 1, 0, 0)a
(0, 1, 2, 0)b
(0, 1, 0, 2)c

a) P4, P2, P1 vote no. P3 votes yes.
b) P4, P1 vote no. P3, P2 vote yes.
c) P4, P2 vote no. P3, P1 vote yes.

So I was right the first, first time ;) P5 offers P3 1 gc and offers either P2 or P1 2 gc, keeping 7 for himself.

P5, P4, P3, P2, P1
(7, 0, 1, 2, 0)
(7, 0, 1, 0, 2)

The reason I was wrong is because P4 accepting 1 gc requires a promise- P4 would have to promise P5 that P4 will accept 1 gc if offered. The same goes for similar situations. (Making promises is not allowed)

Yes, pig, you were right. And now I see why. Thanks.

Happy thougths
Rachel

 

[moshek]

So Now 997 for P5 really fit to L. ( who pose me the problem) answer

very good Pig !

Moshek

 

[Pooh]

So I was right the first, first time ;) P5 offers P3 1 gc and offers either P2 or P1 2 gc, keeping 7 for himself.

P5, P4, P3, P2, P1:
1. (7, 0, 1, 2, 0)
2. (7, 0, 1, 0, 2)

I think these are namely the two possible sets of proposals that P5 can offer gaining the maximum amount of coins and avoiding thrown overboard.

But I been thinking! What if they were six pirates! I think this case having namely one possible proposal that satisfies the requirements.

P6, P5, P4, P3, P2, P1:
1. (7, 0, 1, 0, 1, 1)

What about seven pirates! Assuming the above proposal in case of six pirates fits and unique, I think seven pirates will produce namely three possible sets of proposals that satisfy the requirements.

P7, P6, P5, P4, P3, P2, P1:
1. (6, 0, 1, 2, 1, 0, 0)
2. (6, 0, 1, 0, 1, 2, 0)
3. (6, 0, 1, 0, 1, 0, 2)

What about eight! This one is challenging some little! I think this case having namely one unique proposal that satisfies the requirements.

P8, P7, P6, P5, P4, P3, P2, P1:
1. (6, 0, 1, 0, 1, 0, 1, 1)

Shall we talk about nine also! I suppose it must go with namely four possible sets of proposals that satisfy the requirements.

P9, P8, P7, P6, P5, P4, P3, P2, P1:
1. (5, 0, 1, 2, 1, 0, 1, 0, 0)
2. (5, 0, 1, 0, 1, 2, 1, 0, 0)
3. (5, 0, 1, 0, 1, 0, 1, 2, 0)
4. (5, 0, 1, 0, 1, 0, 1, 0, 2)

I can talk about ten pirates too! This case having namely one unique proposal that satisfies the requirements.

P10, P9, P8, P7, P6, P5, P4, P3, P2, P1:
1. (5, 0, 1, 0, 1, 0, 1, 0, 1, 1)

Let me try predicting the only proposal that will be on table from P16!

P16, P15, P14, P13, P12, P11, P10, P9, P8, P7, P6, P5, P4, P3, P2, P1:
1. (2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1)

I think P20 (better call him P20 the poor) is giving up the 10 coins in his only proposal to save his life. In fact P19 giving up the coins too for saving his life. Poor still better than dead! which is the state of P21 no matter what! Glad i am not a pirate, not any more hehe.

If coins are more than 10, say a 100, then P16 is keeping for himself 92 coins in his only proposal! P19 and P20 are no more poor! they are keeping 90 coins! And the best of all, P21 is no more dead! he further is keeping 89 coins! lucky him!

Was I keep constructing mistakes over mistakes! You can discuss my assumption

You can try concluding a general formula for the solution of 'n' pirates with 'm' coins too!

Yes! Happy thoughs!

 

[Pooh]

I think P20 (better call him P20 the poor) is giving up the 10 coins in his only proposal to save his life. In fact P19 giving up the coins too for saving his life. Poor still better than dead! which is the state of P21 no matter what! Glad i am not a pirate, not any more hehe.

If coins are more than 10, say a 100, then P16 is keeping for himself 92 coins in his only proposal! P19 and P20 are no more poor! they are keeping 90 coins! And the best of all, P21 is no more dead! he further is keeping 89 coins! lucky him!

I have said that P20 - in case of 10 coins - has an only proposal, I guess - after some recalculation - that I was wrong, he has got much more than only one way. I said also that P21 is dead! after recalculation I found that I was wrong again! and that P21 has some way - or better say ways - to avoid thrown overboard, but P22 is the dead one. P23 can survive because of this fact!

Thanks!