I was recently set this puzzle:
There are five rational pirates, A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.
The Pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.
The Pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates should then vote on whether to accept this distribution; the proposer is able to vote, and has the casting vote in the event of a tie [since this is the right of the proposer]. If the proposed allocation is approved by vote, it happens. If not, the proposer is thrown overboard on the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.
First of all, the pirates want to survive. Secondly, the pirates want to maximize the amount of gold coins they receive and, thirdly, they like throwing other pirates overboard.
Try not to look it up! Takes some quite deep thinking.
A puzzle of pirates and gold
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Parallax
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OK, so I understand the setup, but what is the question?
If it is to predict what will happen, I'll give it a try.
First, if A, B and C get thrown overboard, D will propose all for himself and nothing for E, win the subsequent vote (1 on 1, but he has the tie-breaker) and get all the gold. So D wants A, B and C thrown overboard and will systematically vote against any of their plans.
By contrast, The same outcome leaves E with nothing, so if A and B are thrown overboard he will vote for C, assuming he gets at least 1 gold in that deal.
Consequently, C will vote against A and B so he can propose the following: 99 gold for C and 1 gold for E, a proposal that E is likely to support.
But D knows this! So if B is thrown overboard he'll get nothing instead of having his way! Therefore D will support B as long as the deal leaves him with at least 1 gold. Since B needs only one vote in addition to his own (tie-breaker) then B will propose: 99gold for B and 1 gold for D.
Therefore B will always vote against A.
But if A goes overboard, C and E get nothing, therefore A will offer 98 gold for A, 1 gold for C and 1 gold for E, win the vote because that's the best deal the pirates C and E can hope for and they like gold more than they like throwing other pirates overboard.
Funny thing is, actual pirates (and people) are NOT rational, and if you were pirate A in an economics experiment and proposed that, you'd likely end up thrown overboard.
If it is to predict what will happen, I'll give it a try.
First, if A, B and C get thrown overboard, D will propose all for himself and nothing for E, win the subsequent vote (1 on 1, but he has the tie-breaker) and get all the gold. So D wants A, B and C thrown overboard and will systematically vote against any of their plans.
By contrast, The same outcome leaves E with nothing, so if A and B are thrown overboard he will vote for C, assuming he gets at least 1 gold in that deal.
Consequently, C will vote against A and B so he can propose the following: 99 gold for C and 1 gold for E, a proposal that E is likely to support.
But D knows this! So if B is thrown overboard he'll get nothing instead of having his way! Therefore D will support B as long as the deal leaves him with at least 1 gold. Since B needs only one vote in addition to his own (tie-breaker) then B will propose: 99gold for B and 1 gold for D.
Therefore B will always vote against A.
But if A goes overboard, C and E get nothing, therefore A will offer 98 gold for A, 1 gold for C and 1 gold for E, win the vote because that's the best deal the pirates C and E can hope for and they like gold more than they like throwing other pirates overboard.
Funny thing is, actual pirates (and people) are NOT rational, and if you were pirate A in an economics experiment and proposed that, you'd likely end up thrown overboard.
Ok, without looking at parallax's reply, whatever it is...
The first instinct would be to keep voting against the idea, so unless A or B or C could propose a winning situation then they would get shunted off the boat by the rest.
Except E will then realise that when it comes down to him and D, then D can simply propose keeping all the gold coins for himself and get the casting vote.
So as long as C proposes that E gets one coin, then C can keep all the rest of the coins and D can have nothing.
However, B can do a similar thing, since under that proposal C and D against E and B would be a tie so B gets the casting vote
Therefore, B can propose keeping 98, giving 2 to E and survive, or probably take 99 and propose E gets 1 as before
So, therefore, all A has to do is ensure that either C or D get a coin to get their vote, and E gets a coin.
So A can propose 98 for himself, one coin to E and one to C or D, and when the pirates rationalise it all away then C/D and E will vote yes since they will not come out any better anyway - and C/D would come out worse!
Edit: Looked down at parallax's idea for minimising text. Anyway, need to get back to work, will look at his thoughts later!
The first instinct would be to keep voting against the idea, so unless A or B or C could propose a winning situation then they would get shunted off the boat by the rest.
Except E will then realise that when it comes down to him and D, then D can simply propose keeping all the gold coins for himself and get the casting vote.
So as long as C proposes that E gets one coin, then C can keep all the rest of the coins and D can have nothing.
However, B can do a similar thing, since under that proposal C and D against E and B would be a tie so B gets the casting vote
Therefore, B can propose keeping 98, giving 2 to E and survive, or probably take 99 and propose E gets 1 as before
So, therefore, all A has to do is ensure that either C or D get a coin to get their vote, and E gets a coin.
So A can propose 98 for himself, one coin to E and one to C or D, and when the pirates rationalise it all away then C/D and E will vote yes since they will not come out any better anyway - and C/D would come out worse!
Edit: Looked down at parallax's idea for minimising text. Anyway, need to get back to work, will look at his thoughts later!
Nice puzzle ! ok, here goes:
A wants to survive, he has no chance to throw another pirate overboard, which
leaves him with maximising his gold coin outcome.
All he has to do is get 2 votes. Most likely members to support him are
D and E. (B will vote against him all the time, except for he would get all the coins, which is a bit next to the way. C may speculate on A being thrown overboard ,and then he and B could override E and D in a vote. So C would also vote against A unless he would get very much gold coins, which is unlikely from A. Both B and C have no fear of dying yet and could possibly, in the turn after A's maximise their gold outcome, which makes them vote against A)
So A wants to make sure D and E vote for him.
D and E know, when they don't vote for A, they will most likely end up with nothing(B could , in a tie overrule them) so they have 2 incentives:
see A die or get gold. By the way, D and E will always survive , so it all boils down to seeing A die or getting some gold.
If A proposes 98 Gold for himself and 1 for D and 1 for E he should be save.
MAybe he should make it something like A65 D33 E2 or so...
Thing is , A cannot count on his vote overriding the decision, because it is an
odd number (5 people) he has to make sure his proposal works.
A wants to survive, so he does not look as much on the gold as B or C or D or E.(ascending order B<C<D<E) THat makes E most greedy. D comes next.
Well , I looked it up in wikipedia then :
http://en.wikipedia.org/wiki/Pirate_game
the flaw in my thinking is that B could either call on C or D 's vote by giving each some gold. He would be far better off by giving D gold(less gold more vote ) than giving C gold . THing is, D would get 1 gold either way, so he would probably vote against A and thus see him being thrown overboard while getting a gold coin in B's turn.
A wants to survive, he has no chance to throw another pirate overboard, which
leaves him with maximising his gold coin outcome.
All he has to do is get 2 votes. Most likely members to support him are
D and E. (B will vote against him all the time, except for he would get all the coins, which is a bit next to the way. C may speculate on A being thrown overboard ,and then he and B could override E and D in a vote. So C would also vote against A unless he would get very much gold coins, which is unlikely from A. Both B and C have no fear of dying yet and could possibly, in the turn after A's maximise their gold outcome, which makes them vote against A)
So A wants to make sure D and E vote for him.
D and E know, when they don't vote for A, they will most likely end up with nothing(B could , in a tie overrule them) so they have 2 incentives:
see A die or get gold. By the way, D and E will always survive , so it all boils down to seeing A die or getting some gold.
If A proposes 98 Gold for himself and 1 for D and 1 for E he should be save.
MAybe he should make it something like A65 D33 E2 or so...
Thing is , A cannot count on his vote overriding the decision, because it is an
odd number (5 people) he has to make sure his proposal works.
A wants to survive, so he does not look as much on the gold as B or C or D or E.(ascending order B<C<D<E) THat makes E most greedy. D comes next.
Well , I looked it up in wikipedia then :
http://en.wikipedia.org/wiki/Pirate_game
the flaw in my thinking is that B could either call on C or D 's vote by giving each some gold. He would be far better off by giving D gold(less gold more vote ) than giving C gold . THing is, D would get 1 gold either way, so he would probably vote against A and thus see him being thrown overboard while getting a gold coin in B's turn.
An interesting variation....
A custom dungeon consists of 65 screamers and Lord Chaos. Each denizen, including Lord Chaos, has a salary of one Gor coin. This is because there was a screamers' revolution, and a democracy was founded. Lord Chaos lost all his power, so that he doesn't even have the right to vote.
What Lord Chaos can do once per year is to suggest a redistribution of the salaries. Each creature's salary must be a non-negative integer number of Gor coins, and the salaries must sum to 66. When Lord Chaos has made a suggestion, each denizen except Lord Chaos himself can vote for or against it, or choose not to vote. The suggestion is accepted if the number of votes for it is greater than the number of votes against it.
Each screamer is selfish, and will therefore vote for anything which increases its own salary, and against anything that decreases it. No screamer will bother to vote if it doesn't affect its own salary.
Lord Chaos is not only selfish, but also clever. What salary does he end up with, and how long (i.e how may votes) does it take him to get it?
A custom dungeon consists of 65 screamers and Lord Chaos. Each denizen, including Lord Chaos, has a salary of one Gor coin. This is because there was a screamers' revolution, and a democracy was founded. Lord Chaos lost all his power, so that he doesn't even have the right to vote.
What Lord Chaos can do once per year is to suggest a redistribution of the salaries. Each creature's salary must be a non-negative integer number of Gor coins, and the salaries must sum to 66. When Lord Chaos has made a suggestion, each denizen except Lord Chaos himself can vote for or against it, or choose not to vote. The suggestion is accepted if the number of votes for it is greater than the number of votes against it.
Each screamer is selfish, and will therefore vote for anything which increases its own salary, and against anything that decreases it. No screamer will bother to vote if it doesn't affect its own salary.
Lord Chaos is not only selfish, but also clever. What salary does he end up with, and how long (i.e how may votes) does it take him to get it?
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Parallax
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- Location: Back in New Jersey
Alright, here goes:
First year: 33 screamers get 2 gor coins (33 for), 32 creamers get 0 (32 against), LC get 0. Proposal passes. The 32 disenfranchised screamers will never again get paid and we will not consider them anymore.
Year 2: 17 of the 33 previous screamers get 3 GOR (17 for). 16 get nothing (16 against). LC gets 15. Proposal passes and LC's sacrifice from last year has already paid off tremendously.
Year 3: 9 screamers get 4 GOR each (9 votes for). 8 of 17 get 0 (8 against) 48 screamers are unchanged. LC gets 30 GOR.
Years 4, 5, 6 The process is repeated with 5 against 4 for 5 GOR each, 3 against 2 for 6 GOR each and 2 against 1 for 7 GOR each. LC is earning 52 GOR or year 6.
Year 7: 1 GOR each for 3 previously broke screamers (3 for), 0 for all other creamers (2 against). Approved. LC gets 63 GOR coins and I don't think he can get more than that, ever.
Total: 0 + 15 + 30 + 41 + 48 + 52 + 63 = 249 GOR coins in 7 years and 63 GOR coins every year afterwards. It might be beatable by shunning the big-earners earlier, to maximize early returns at the expense of a longer process.
First year: 33 screamers get 2 gor coins (33 for), 32 creamers get 0 (32 against), LC get 0. Proposal passes. The 32 disenfranchised screamers will never again get paid and we will not consider them anymore.
Year 2: 17 of the 33 previous screamers get 3 GOR (17 for). 16 get nothing (16 against). LC gets 15. Proposal passes and LC's sacrifice from last year has already paid off tremendously.
Year 3: 9 screamers get 4 GOR each (9 votes for). 8 of 17 get 0 (8 against) 48 screamers are unchanged. LC gets 30 GOR.
Years 4, 5, 6 The process is repeated with 5 against 4 for 5 GOR each, 3 against 2 for 6 GOR each and 2 against 1 for 7 GOR each. LC is earning 52 GOR or year 6.
Year 7: 1 GOR each for 3 previously broke screamers (3 for), 0 for all other creamers (2 against). Approved. LC gets 63 GOR coins and I don't think he can get more than that, ever.
Total: 0 + 15 + 30 + 41 + 48 + 52 + 63 = 249 GOR coins in 7 years and 63 GOR coins every year afterwards. It might be beatable by shunning the big-earners earlier, to maximize early returns at the expense of a longer process.
well,
I would say *bleep* votes or years until he earns *iiieek!* gor coins per month.
here is chaos' (or my) math:
first year:
Chaos : 0
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 11220
a screamer gets 0 gor coins,
2 screamers each get two gor coins(from poor screamer and Chaos himself)
second year:
Chaos : 0
2222 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 11000
three screamers get 0 gor coins,
4 screamers get each plus one goins(gor coins) totaling 2 gor coins for each of them.
third year:
Chaos : 0
2222 2222
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1100 00000
screamer with 0 gor coins x 7
8 screamers with 2 gor coins , rest have got 1 gor coin
fourth:
Chaos : 0
2222 2222
2222 2222
1111 1111
1111 1111
1111 1111
1111 1111
0000 0000
0000 00011
0x15, 2x16
fith:
Chaos : 0
2222 2222
2222 2222
2222 2222
2222 2222
0000 0000
0000 0000
0000 0000
0000 00011
0x31, 2x 32 totalling 63, 2 screamers get 1 gor coin still
sixth:
Chaos : 14
3333 3333
3333 3322
2000 0000
0000 0000
0000 0000
0000 0000
0000 0000
0000 00022
Chaos finally gets some coins! 7 screamers get cropped because he maximises 15 screamers to vote against him, with 0x8 he did please 16: 3x14 2x2
seventh:
Chaos : 29
4444 0000
0330 0033
3000 0000
0000 0000
0000 0000
0000 0000
0000 0000
0000 00033
3 get nothing then and he pleases 9 with that move: 5x2 get each 3 , and 4x3 get each then 4 gor coins.that leaves 7 unused . Leave 2 and crop the rest, 5x3 gor coins for chaos in addition
eight year: (i'll leave out the zeroes)
Chaos : 29
4444 3333
333
11 relevant screamers remaining: he needs to please 6:
one 4 gor coin screamer would please 8, he takes remaining 2
plus the 3x4 (12):
Chaos : 31
4444 4440
ninth:
chaos : 41
0005 555, rest 0
he takes 10 , pleases 4, ruins 3
tenth:
chaos:
5555, rest 0 he gives 3 screamers 1 and yields 2x5:now he gets 51
00661
eleventh
0720 yields 4, now he gets 55
twelth form:
0031 yields 5, now he gets 60
thirteenth year:
1002 yields 1 leaves chaos at 61
I would say *bleep* votes or years until he earns *iiieek!* gor coins per month.
here is chaos' (or my) math:
first year:
Chaos : 0
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 11220
a screamer gets 0 gor coins,
2 screamers each get two gor coins(from poor screamer and Chaos himself)
second year:
Chaos : 0
2222 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 11000
three screamers get 0 gor coins,
4 screamers get each plus one goins(gor coins) totaling 2 gor coins for each of them.
third year:
Chaos : 0
2222 2222
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1111 1111
1100 00000
screamer with 0 gor coins x 7
8 screamers with 2 gor coins , rest have got 1 gor coin
fourth:
Chaos : 0
2222 2222
2222 2222
1111 1111
1111 1111
1111 1111
1111 1111
0000 0000
0000 00011
0x15, 2x16
fith:
Chaos : 0
2222 2222
2222 2222
2222 2222
2222 2222
0000 0000
0000 0000
0000 0000
0000 00011
0x31, 2x 32 totalling 63, 2 screamers get 1 gor coin still
sixth:
Chaos : 14
3333 3333
3333 3322
2000 0000
0000 0000
0000 0000
0000 0000
0000 0000
0000 00022
Chaos finally gets some coins! 7 screamers get cropped because he maximises 15 screamers to vote against him, with 0x8 he did please 16: 3x14 2x2
seventh:
Chaos : 29
4444 0000
0330 0033
3000 0000
0000 0000
0000 0000
0000 0000
0000 0000
0000 00033
3 get nothing then and he pleases 9 with that move: 5x2 get each 3 , and 4x3 get each then 4 gor coins.that leaves 7 unused . Leave 2 and crop the rest, 5x3 gor coins for chaos in addition
eight year: (i'll leave out the zeroes)
Chaos : 29
4444 3333
333
11 relevant screamers remaining: he needs to please 6:
one 4 gor coin screamer would please 8, he takes remaining 2
plus the 3x4 (12):
Chaos : 31
4444 4440
ninth:
chaos : 41
0005 555, rest 0
he takes 10 , pleases 4, ruins 3
tenth:
chaos:
5555, rest 0 he gives 3 screamers 1 and yields 2x5:now he gets 51
00661
eleventh
0720 yields 4, now he gets 55
twelth form:
0031 yields 5, now he gets 60
thirteenth year:
1002 yields 1 leaves chaos at 61
Ok, suddenly have a surprise easter weekend away, so not goign to leasuirely work this one out - my idea (Before I check out the others) is:
Each year chaos basically takes one coin from half the screamers with coins from a central core of 64, 32, 16, 8 etc and gives it to the others, skimming any additional coins. To balance the votes, he then gives a scream (or screamers) a coin (the first from his own salary) to ensure creater votes. So the first year he gives his salary to a lone screamer, the second year he gives that screamer another coin, the third year it's cheaper to give two zero coin screamers a coin and steal the three coins fro mthe well off screamer, and so on. Using that logic, it takes him seven votes to remove the core grounds, and an additional year to re-allocate the final welath from his screamer core. Doing quick calculations I got it to 62 coins for him, with 2 coins to one screamer and two screamers with one coin. I probably messed up - let's see parallax's solution now!
Each year chaos basically takes one coin from half the screamers with coins from a central core of 64, 32, 16, 8 etc and gives it to the others, skimming any additional coins. To balance the votes, he then gives a scream (or screamers) a coin (the first from his own salary) to ensure creater votes. So the first year he gives his salary to a lone screamer, the second year he gives that screamer another coin, the third year it's cheaper to give two zero coin screamers a coin and steal the three coins fro mthe well off screamer, and so on. Using that logic, it takes him seven votes to remove the core grounds, and an additional year to re-allocate the final welath from his screamer core. Doing quick calculations I got it to 62 coins for him, with 2 coins to one screamer and two screamers with one coin. I probably messed up - let's see parallax's solution now!