Mixing
Pirate's Game: What if proposer can't vote
Clash Royale CLAN TAG#URR8PPP
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I had read the Pirate's game
With other number of pirates and coins we can think the same way
But I supposed maximum number of coins that pirate A might get if plan still accepted when get a half of votes but the proposer can't vote for his plan ?
The problem will be :
There are 5 pirates, they must decide how to distribute 100 gold coins
among them. The pirates have seniority levels, the senior-most is A,
then B, then C, then D, and finally the junior-most is E.
Rules of distribution are:
- The most senior pirate proposes a distribution of coins.
- All pirates, who not proposes the distribution, will vote on whether to accept the distribution.
- If the distribution is accepted, the coins are disbursed and the game ends.
- If not, the proposer is thrown and dies, and the next most senior pirate makes a new proposal to begin the system again.
- In case of a tie vote the distribution will be accepted
Rules every pirates follows.
- Every pirate wants to survive and doesn't believe in anyone else
- Given survival, each pirate wants to maximize the number of gold coins he receives.
puzzle game-theory
edited Jul 18 at 9:04
joriki
164k10180328
asked Jul 18 at 3:59
Hâm Zon
284
 |Â
show 4 more comments
up vote
5
down vote
favorite
I had read the Pirate's game
With other number of pirates and coins we can think the same way
But I supposed maximum number of coins that pirate A might get if plan still accepted when get a half of votes but the proposer can't vote for his plan ?
The problem will be :
There are 5 pirates, they must decide how to distribute 100 gold coins
among them. The pirates have seniority levels, the senior-most is A,
then B, then C, then D, and finally the junior-most is E.
Rules of distribution are:
- The most senior pirate proposes a distribution of coins.
- All pirates, who not proposes the distribution, will vote on whether to accept the distribution.
- If the distribution is accepted, the coins are disbursed and the game ends.
- If not, the proposer is thrown and dies, and the next most senior pirate makes a new proposal to begin the system again.
- In case of a tie vote the distribution will be accepted
Rules every pirates follows.
- Every pirate wants to survive and doesn't believe in anyone else
- Given survival, each pirate wants to maximize the number of gold coins he receives.
puzzle game-theory
edited Jul 18 at 9:04
joriki
164k10180328
asked Jul 18 at 3:59
Hâm Zon
284
What happens in case of a tie?
– Hagen von Eitzen
Jul 18 at 4:08
The plan of A still accepted when he has 2/4 vote "yes" from B C D E (A can't vote for his plan)
– Hâm Zon
Jul 18 at 4:10
Thank you. I've edited it
– Hâm Zon
Jul 18 at 4:23
1
@joriki, "casting vote" is a phrase meaning "tie-breaking vote".
– Gerry Myerson
Jul 18 at 9:30
1
@GerryMyerson: Thanks, I should have checked -- I'd never heard that before.
– joriki
Jul 18 at 9:45
 |Â
show 4 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I had read the Pirate's game
With other number of pirates and coins we can think the same way
But I supposed maximum number of coins that pirate A might get if plan still accepted when get a half of votes but the proposer can't vote for his plan ?
The problem will be :
There are 5 pirates, they must decide how to distribute 100 gold coins
among them. The pirates have seniority levels, the senior-most is A,
then B, then C, then D, and finally the junior-most is E.
Rules of distribution are:
- The most senior pirate proposes a distribution of coins.
- All pirates, who not proposes the distribution, will vote on whether to accept the distribution.
- If the distribution is accepted, the coins are disbursed and the game ends.
- If not, the proposer is thrown and dies, and the next most senior pirate makes a new proposal to begin the system again.
- In case of a tie vote the distribution will be accepted
Rules every pirates follows.
- Every pirate wants to survive and doesn't believe in anyone else
- Given survival, each pirate wants to maximize the number of gold coins he receives.
puzzle game-theory
edited Jul 18 at 9:04
joriki
164k10180328
asked Jul 18 at 3:59
Hâm Zon
284
I had read the Pirate's game
With other number of pirates and coins we can think the same way
But I supposed maximum number of coins that pirate A might get if plan still accepted when get a half of votes but the proposer can't vote for his plan ?
The problem will be :
There are 5 pirates, they must decide how to distribute 100 gold coins
among them. The pirates have seniority levels, the senior-most is A,
then B, then C, then D, and finally the junior-most is E.
Rules of distribution are:
- The most senior pirate proposes a distribution of coins.
- All pirates, who not proposes the distribution, will vote on whether to accept the distribution.
- If the distribution is accepted, the coins are disbursed and the game ends.
- If not, the proposer is thrown and dies, and the next most senior pirate makes a new proposal to begin the system again.
- In case of a tie vote the distribution will be accepted
Rules every pirates follows.
- Every pirate wants to survive and doesn't believe in anyone else
- Given survival, each pirate wants to maximize the number of gold coins he receives.
puzzle game-theory
edited Jul 18 at 9:04
joriki
164k10180328
asked Jul 18 at 3:59
Hâm Zon
284
edited Jul 18 at 9:04
joriki
164k10180328
edited Jul 18 at 9:04
joriki
164k10180328
edited Jul 18 at 9:04
joriki
164k10180328
164k10180328
asked Jul 18 at 3:59
Hâm Zon
284
asked Jul 18 at 3:59
Hâm Zon
284
asked Jul 18 at 3:59
Hâm Zon
284
284
What happens in case of a tie?
– Hagen von Eitzen
Jul 18 at 4:08
The plan of A still accepted when he has 2/4 vote "yes" from B C D E (A can't vote for his plan)
– Hâm Zon
Jul 18 at 4:10
Thank you. I've edited it
– Hâm Zon
Jul 18 at 4:23
1
@joriki, "casting vote" is a phrase meaning "tie-breaking vote".
– Gerry Myerson
Jul 18 at 9:30
1
@GerryMyerson: Thanks, I should have checked -- I'd never heard that before.
– joriki
Jul 18 at 9:45
 |Â
show 4 more comments
What happens in case of a tie?
– Hagen von Eitzen
Jul 18 at 4:08
The plan of A still accepted when he has 2/4 vote "yes" from B C D E (A can't vote for his plan)
– Hâm Zon
Jul 18 at 4:10
Thank you. I've edited it
– Hâm Zon
Jul 18 at 4:23
1
@joriki, "casting vote" is a phrase meaning "tie-breaking vote".
– Gerry Myerson
Jul 18 at 9:30
1
@GerryMyerson: Thanks, I should have checked -- I'd never heard that before.
– joriki
Jul 18 at 9:45
What happens in case of a tie?
– Hagen von Eitzen
Jul 18 at 4:08
What happens in case of a tie?
– Hagen von Eitzen
Jul 18 at 4:08
The plan of A still accepted when he has 2/4 vote "yes" from B C D E (A can't vote for his plan)
– Hâm Zon
Jul 18 at 4:10
The plan of A still accepted when he has 2/4 vote "yes" from B C D E (A can't vote for his plan)
– Hâm Zon
Jul 18 at 4:10
Thank you. I've edited it
– Hâm Zon
Jul 18 at 4:23
Thank you. I've edited it
– Hâm Zon
Jul 18 at 4:23
1
1
@joriki, "casting vote" is a phrase meaning "tie-breaking vote".
– Gerry Myerson
Jul 18 at 9:30
@joriki, "casting vote" is a phrase meaning "tie-breaking vote".
– Gerry Myerson
Jul 18 at 9:30
1
1
@GerryMyerson: Thanks, I should have checked -- I'd never heard that before.
– joriki
Jul 18 at 9:45
@GerryMyerson: Thanks, I should have checked -- I'd never heard that before.
– joriki
Jul 18 at 9:45
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
As an example, let's consider the case with $5$ pirates ($A$, $B$, $C$, $D$, and $E$) and $100$ coins:
If only $D$ and $E$ are left, then the distribution is $(0,100)$. If $D$ offers anything other than all the coins to $E$, then $E$ votes against the proposal and takes all $100$ for himself/herself. Even with the distribution $(0,100)$, pirate $E$ can vote against the plan if he/she chooses (perhaps due to personality conflicts, etc).
If $C$, $D$, and $E$ are left, then the distribution is $(99,1,0)$. $D$ will vote for this plan as it's better than getting $0$.
If $B$, $C$, $D$, and $E$ are left, then the distribution is $(97,0,2,1)$. In this case, $B$ needs two votes to survive. $B$ won't get $C$'s vote unless $B$ offers $C$ more than $99$, which isn't going to happen. To guarantee the other two votes, $B$ offers one more to $D$ and $E$ than what they would get without $B$.
If all pirates are alive, then the distribution is $(97,0,1,0,2)$. In this case, $A$ needs two votes and isn't going to get $B$'s vote without offering more than $97$, which isn't going to happen. Therefore, $A$ offers a little extra gold to two who would lose out (and get the smallest amounts) if $A$ were killed.
Just for fun, here are the next few cases:
With $6$ pirates, $(96,0,1,2,1,0)$
With $7$ pirates, $(96,0,1,2,0,0,1)$ or $(96,0,1,0,0,2,1)$
With $8$ pirates, things get a little interesting. Since there are two options for $7$, $(96,0,1,0,1,1,1,0)$ might be enough because the pirates who get $2$ in the case of $7$ might not want to risk getting $0$.
With $9$ pirates, any of $(95,0,1,2,1,0,0,0,1)$, $(95,0,1,0,1,2,0,0,1)$, $(95,0,1,0,1,0,2,0,1)$, or $(95,0,1,0,1,0,0,2,1)$ should get enough votes.
With $10$ pirates, $(95,0,1,0,1,0,1,1,1,0)$ should be a good distribution.
The number of coins doesn't really matter, just as long as there are enough coins to distribute nicely. If there are more pirates than half the coins, things might get quite interesting. Likely there will be a lot of killing in those cases to bring the number of pirates down.
In these types problems, sometimes the pirates are defined to be "bloodthirsty." In this case, the pirates choose to kill whenever their two votes would give the same result. This would mean that in the first case (only $D$ and $E$ left, $E$ would always vote to kill $D$).
answered Jul 18 at 9:48
Michael Burr
25.6k13262
add a comment |Â
up vote
0
down vote
Before a definitive answer can be given, two questions need to be answered:
If the pirates are indifferent to a proposal, will they accept it or will they kill the head pirate?
If the head pirate could equally well pay multiple pirates, which will be paid? Might the gold, perhaps, be distributed randomly?
Different answers to these questions will yield different results.
As an example of how to analyse this game, I am going to consider what I take to be the best interpretation: the pirates might kill, and the head pirate, when given a choice, will pay the highest-ranked pirates first. I will call the lowest-ranked pirate #1, the second-lowest #2, and so on.
2: Doesn't matter, pirate #1 gets 100.
3: (100, 0, 0) (pirate #2 may die if this proposal doesn't go through, and will therefore accept it.)
4: (98, 0, 1, 1)
5: (97, 0, 1, 2, 0) (#1 or #2 could have been offered 2 coins, but #2 is ranked higher.)
6: (96, 0, 1, 2, 0, 1)
7: (96, 0, 1, 2, 0, 1, 0)
8: (95, 0, 1, 2, 0, 1, 0, 1)
9: (95, 0, 1, 2, 0, 1, 0, 1, 0)
10: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1)
11: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1, 0)
...
196: (1, 0, 1, 2, 0, 1, 0, 1, ... 0, 1)
197: (1, 0, 1, 2, 0, 1, 0, 1, ... 1, 0)
198: (0, 2, 1, 0, 0, 1, 0, 1, ... 0, 1)
199: (0, 1, 0, 2, 1, 1, 0, 1, ... 1, 0)
200: (0, 1, 2, 1, 0, 0, 0, 1, ... 0, 1) (Pirate #200 can only afford 99 votes, and is killed.)
201: (0, 0, 2, 0, 0, 1, 1, 1, 0, ... 1, 0) (Pirate #201 can only afford 99 votes, but survives because #200 accepts the proposal in order to stay alive.)
202: (0, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 1) (Pirate #202 can only afford 100 votes, and is killed.)
203: (0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 0, 1, 0) (Pirate #203 gets #202's vote for free, and survives 101-101.)
204: (0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0) (Pirate #204 gets no free votes and is killed.)
205: (0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0) (Pirate #205 gets #204's vote for free, but still loses 101-102.)
206: (0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0, 0, 0) (Pirate #206 gets #204 and #205, but loses 102-103.)
207: (0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0, 0, 0) (Pirate #207 get #204-206, and survives 103-103.)
Pirates #208-214 die, but #215 survives 107-107.
Pirates #216-230 die, but #231 survives 115-115.
Pirates #232-262 die, but #263 survives 131-131.
There is a pattern.
Each pirate is separated from the previous one by a power of two. In general, pirate $199+2^n$ survives.
There are also patterns in the distribution of the coins, such that it is possible to find, for any pirate, what they would get with x number of pirates.
In general, for x number of pirates and G of coins, the pirates will run out of coins at $x=2G-2$, and from 2G onwards, only pirates equal to $2G-1+2^n$ will survive, where n is a natural number 1, 2, 3, ...
answered Aug 12 at 4:33
Josh P
63
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
As an example, let's consider the case with $5$ pirates ($A$, $B$, $C$, $D$, and $E$) and $100$ coins:
If only $D$ and $E$ are left, then the distribution is $(0,100)$. If $D$ offers anything other than all the coins to $E$, then $E$ votes against the proposal and takes all $100$ for himself/herself. Even with the distribution $(0,100)$, pirate $E$ can vote against the plan if he/she chooses (perhaps due to personality conflicts, etc).
If $C$, $D$, and $E$ are left, then the distribution is $(99,1,0)$. $D$ will vote for this plan as it's better than getting $0$.
If $B$, $C$, $D$, and $E$ are left, then the distribution is $(97,0,2,1)$. In this case, $B$ needs two votes to survive. $B$ won't get $C$'s vote unless $B$ offers $C$ more than $99$, which isn't going to happen. To guarantee the other two votes, $B$ offers one more to $D$ and $E$ than what they would get without $B$.
If all pirates are alive, then the distribution is $(97,0,1,0,2)$. In this case, $A$ needs two votes and isn't going to get $B$'s vote without offering more than $97$, which isn't going to happen. Therefore, $A$ offers a little extra gold to two who would lose out (and get the smallest amounts) if $A$ were killed.
Just for fun, here are the next few cases:
With $6$ pirates, $(96,0,1,2,1,0)$
With $7$ pirates, $(96,0,1,2,0,0,1)$ or $(96,0,1,0,0,2,1)$
With $8$ pirates, things get a little interesting. Since there are two options for $7$, $(96,0,1,0,1,1,1,0)$ might be enough because the pirates who get $2$ in the case of $7$ might not want to risk getting $0$.
With $9$ pirates, any of $(95,0,1,2,1,0,0,0,1)$, $(95,0,1,0,1,2,0,0,1)$, $(95,0,1,0,1,0,2,0,1)$, or $(95,0,1,0,1,0,0,2,1)$ should get enough votes.
With $10$ pirates, $(95,0,1,0,1,0,1,1,1,0)$ should be a good distribution.
The number of coins doesn't really matter, just as long as there are enough coins to distribute nicely. If there are more pirates than half the coins, things might get quite interesting. Likely there will be a lot of killing in those cases to bring the number of pirates down.
In these types problems, sometimes the pirates are defined to be "bloodthirsty." In this case, the pirates choose to kill whenever their two votes would give the same result. This would mean that in the first case (only $D$ and $E$ left, $E$ would always vote to kill $D$).
answered Jul 18 at 9:48
Michael Burr
25.6k13262
add a comment |Â
up vote
3
down vote
accepted
As an example, let's consider the case with $5$ pirates ($A$, $B$, $C$, $D$, and $E$) and $100$ coins:
If only $D$ and $E$ are left, then the distribution is $(0,100)$. If $D$ offers anything other than all the coins to $E$, then $E$ votes against the proposal and takes all $100$ for himself/herself. Even with the distribution $(0,100)$, pirate $E$ can vote against the plan if he/she chooses (perhaps due to personality conflicts, etc).
If $C$, $D$, and $E$ are left, then the distribution is $(99,1,0)$. $D$ will vote for this plan as it's better than getting $0$.
If $B$, $C$, $D$, and $E$ are left, then the distribution is $(97,0,2,1)$. In this case, $B$ needs two votes to survive. $B$ won't get $C$'s vote unless $B$ offers $C$ more than $99$, which isn't going to happen. To guarantee the other two votes, $B$ offers one more to $D$ and $E$ than what they would get without $B$.
If all pirates are alive, then the distribution is $(97,0,1,0,2)$. In this case, $A$ needs two votes and isn't going to get $B$'s vote without offering more than $97$, which isn't going to happen. Therefore, $A$ offers a little extra gold to two who would lose out (and get the smallest amounts) if $A$ were killed.
Just for fun, here are the next few cases:
With $6$ pirates, $(96,0,1,2,1,0)$
With $7$ pirates, $(96,0,1,2,0,0,1)$ or $(96,0,1,0,0,2,1)$
With $8$ pirates, things get a little interesting. Since there are two options for $7$, $(96,0,1,0,1,1,1,0)$ might be enough because the pirates who get $2$ in the case of $7$ might not want to risk getting $0$.
With $9$ pirates, any of $(95,0,1,2,1,0,0,0,1)$, $(95,0,1,0,1,2,0,0,1)$, $(95,0,1,0,1,0,2,0,1)$, or $(95,0,1,0,1,0,0,2,1)$ should get enough votes.
With $10$ pirates, $(95,0,1,0,1,0,1,1,1,0)$ should be a good distribution.
The number of coins doesn't really matter, just as long as there are enough coins to distribute nicely. If there are more pirates than half the coins, things might get quite interesting. Likely there will be a lot of killing in those cases to bring the number of pirates down.
In these types problems, sometimes the pirates are defined to be "bloodthirsty." In this case, the pirates choose to kill whenever their two votes would give the same result. This would mean that in the first case (only $D$ and $E$ left, $E$ would always vote to kill $D$).
answered Jul 18 at 9:48
Michael Burr
25.6k13262
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
As an example, let's consider the case with $5$ pirates ($A$, $B$, $C$, $D$, and $E$) and $100$ coins:
If only $D$ and $E$ are left, then the distribution is $(0,100)$. If $D$ offers anything other than all the coins to $E$, then $E$ votes against the proposal and takes all $100$ for himself/herself. Even with the distribution $(0,100)$, pirate $E$ can vote against the plan if he/she chooses (perhaps due to personality conflicts, etc).
If $C$, $D$, and $E$ are left, then the distribution is $(99,1,0)$. $D$ will vote for this plan as it's better than getting $0$.
If $B$, $C$, $D$, and $E$ are left, then the distribution is $(97,0,2,1)$. In this case, $B$ needs two votes to survive. $B$ won't get $C$'s vote unless $B$ offers $C$ more than $99$, which isn't going to happen. To guarantee the other two votes, $B$ offers one more to $D$ and $E$ than what they would get without $B$.
If all pirates are alive, then the distribution is $(97,0,1,0,2)$. In this case, $A$ needs two votes and isn't going to get $B$'s vote without offering more than $97$, which isn't going to happen. Therefore, $A$ offers a little extra gold to two who would lose out (and get the smallest amounts) if $A$ were killed.
Just for fun, here are the next few cases:
With $6$ pirates, $(96,0,1,2,1,0)$
With $7$ pirates, $(96,0,1,2,0,0,1)$ or $(96,0,1,0,0,2,1)$
With $8$ pirates, things get a little interesting. Since there are two options for $7$, $(96,0,1,0,1,1,1,0)$ might be enough because the pirates who get $2$ in the case of $7$ might not want to risk getting $0$.
With $9$ pirates, any of $(95,0,1,2,1,0,0,0,1)$, $(95,0,1,0,1,2,0,0,1)$, $(95,0,1,0,1,0,2,0,1)$, or $(95,0,1,0,1,0,0,2,1)$ should get enough votes.
With $10$ pirates, $(95,0,1,0,1,0,1,1,1,0)$ should be a good distribution.
The number of coins doesn't really matter, just as long as there are enough coins to distribute nicely. If there are more pirates than half the coins, things might get quite interesting. Likely there will be a lot of killing in those cases to bring the number of pirates down.
In these types problems, sometimes the pirates are defined to be "bloodthirsty." In this case, the pirates choose to kill whenever their two votes would give the same result. This would mean that in the first case (only $D$ and $E$ left, $E$ would always vote to kill $D$).
answered Jul 18 at 9:48
Michael Burr
25.6k13262
As an example, let's consider the case with $5$ pirates ($A$, $B$, $C$, $D$, and $E$) and $100$ coins:
If only $D$ and $E$ are left, then the distribution is $(0,100)$. If $D$ offers anything other than all the coins to $E$, then $E$ votes against the proposal and takes all $100$ for himself/herself. Even with the distribution $(0,100)$, pirate $E$ can vote against the plan if he/she chooses (perhaps due to personality conflicts, etc).
If $C$, $D$, and $E$ are left, then the distribution is $(99,1,0)$. $D$ will vote for this plan as it's better than getting $0$.
If $B$, $C$, $D$, and $E$ are left, then the distribution is $(97,0,2,1)$. In this case, $B$ needs two votes to survive. $B$ won't get $C$'s vote unless $B$ offers $C$ more than $99$, which isn't going to happen. To guarantee the other two votes, $B$ offers one more to $D$ and $E$ than what they would get without $B$.
If all pirates are alive, then the distribution is $(97,0,1,0,2)$. In this case, $A$ needs two votes and isn't going to get $B$'s vote without offering more than $97$, which isn't going to happen. Therefore, $A$ offers a little extra gold to two who would lose out (and get the smallest amounts) if $A$ were killed.
Just for fun, here are the next few cases:
With $6$ pirates, $(96,0,1,2,1,0)$
With $7$ pirates, $(96,0,1,2,0,0,1)$ or $(96,0,1,0,0,2,1)$
With $8$ pirates, things get a little interesting. Since there are two options for $7$, $(96,0,1,0,1,1,1,0)$ might be enough because the pirates who get $2$ in the case of $7$ might not want to risk getting $0$.
With $9$ pirates, any of $(95,0,1,2,1,0,0,0,1)$, $(95,0,1,0,1,2,0,0,1)$, $(95,0,1,0,1,0,2,0,1)$, or $(95,0,1,0,1,0,0,2,1)$ should get enough votes.
With $10$ pirates, $(95,0,1,0,1,0,1,1,1,0)$ should be a good distribution.
The number of coins doesn't really matter, just as long as there are enough coins to distribute nicely. If there are more pirates than half the coins, things might get quite interesting. Likely there will be a lot of killing in those cases to bring the number of pirates down.
In these types problems, sometimes the pirates are defined to be "bloodthirsty." In this case, the pirates choose to kill whenever their two votes would give the same result. This would mean that in the first case (only $D$ and $E$ left, $E$ would always vote to kill $D$).
answered Jul 18 at 9:48
Michael Burr
25.6k13262
edited Jul 18 at 11:08
answered Jul 18 at 9:48
Michael Burr
25.6k13262
answered Jul 18 at 9:48
Michael Burr
25.6k13262
answered Jul 18 at 9:48
Michael Burr
25.6k13262
25.6k13262
add a comment |Â
add a comment |Â
up vote
0
down vote
Before a definitive answer can be given, two questions need to be answered:
If the pirates are indifferent to a proposal, will they accept it or will they kill the head pirate?
If the head pirate could equally well pay multiple pirates, which will be paid? Might the gold, perhaps, be distributed randomly?
Different answers to these questions will yield different results.
As an example of how to analyse this game, I am going to consider what I take to be the best interpretation: the pirates might kill, and the head pirate, when given a choice, will pay the highest-ranked pirates first. I will call the lowest-ranked pirate #1, the second-lowest #2, and so on.
2: Doesn't matter, pirate #1 gets 100.
3: (100, 0, 0) (pirate #2 may die if this proposal doesn't go through, and will therefore accept it.)
4: (98, 0, 1, 1)
5: (97, 0, 1, 2, 0) (#1 or #2 could have been offered 2 coins, but #2 is ranked higher.)
6: (96, 0, 1, 2, 0, 1)
7: (96, 0, 1, 2, 0, 1, 0)
8: (95, 0, 1, 2, 0, 1, 0, 1)
9: (95, 0, 1, 2, 0, 1, 0, 1, 0)
10: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1)
11: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1, 0)
...
196: (1, 0, 1, 2, 0, 1, 0, 1, ... 0, 1)
197: (1, 0, 1, 2, 0, 1, 0, 1, ... 1, 0)
198: (0, 2, 1, 0, 0, 1, 0, 1, ... 0, 1)
199: (0, 1, 0, 2, 1, 1, 0, 1, ... 1, 0)
200: (0, 1, 2, 1, 0, 0, 0, 1, ... 0, 1) (Pirate #200 can only afford 99 votes, and is killed.)
201: (0, 0, 2, 0, 0, 1, 1, 1, 0, ... 1, 0) (Pirate #201 can only afford 99 votes, but survives because #200 accepts the proposal in order to stay alive.)
202: (0, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 1) (Pirate #202 can only afford 100 votes, and is killed.)
203: (0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 0, 1, 0) (Pirate #203 gets #202's vote for free, and survives 101-101.)
204: (0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0) (Pirate #204 gets no free votes and is killed.)
205: (0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0) (Pirate #205 gets #204's vote for free, but still loses 101-102.)
206: (0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0, 0, 0) (Pirate #206 gets #204 and #205, but loses 102-103.)
207: (0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0, 0, 0) (Pirate #207 get #204-206, and survives 103-103.)
Pirates #208-214 die, but #215 survives 107-107.
Pirates #216-230 die, but #231 survives 115-115.
Pirates #232-262 die, but #263 survives 131-131.
There is a pattern.
Each pirate is separated from the previous one by a power of two. In general, pirate $199+2^n$ survives.
There are also patterns in the distribution of the coins, such that it is possible to find, for any pirate, what they would get with x number of pirates.
In general, for x number of pirates and G of coins, the pirates will run out of coins at $x=2G-2$, and from 2G onwards, only pirates equal to $2G-1+2^n$ will survive, where n is a natural number 1, 2, 3, ...
answered Aug 12 at 4:33
Josh P
63
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Before a definitive answer can be given, two questions need to be answered:
If the pirates are indifferent to a proposal, will they accept it or will they kill the head pirate?
If the head pirate could equally well pay multiple pirates, which will be paid? Might the gold, perhaps, be distributed randomly?
Different answers to these questions will yield different results.
As an example of how to analyse this game, I am going to consider what I take to be the best interpretation: the pirates might kill, and the head pirate, when given a choice, will pay the highest-ranked pirates first. I will call the lowest-ranked pirate #1, the second-lowest #2, and so on.
2: Doesn't matter, pirate #1 gets 100.
3: (100, 0, 0) (pirate #2 may die if this proposal doesn't go through, and will therefore accept it.)
4: (98, 0, 1, 1)
5: (97, 0, 1, 2, 0) (#1 or #2 could have been offered 2 coins, but #2 is ranked higher.)
6: (96, 0, 1, 2, 0, 1)
7: (96, 0, 1, 2, 0, 1, 0)
8: (95, 0, 1, 2, 0, 1, 0, 1)
9: (95, 0, 1, 2, 0, 1, 0, 1, 0)
10: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1)
11: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1, 0)
...
196: (1, 0, 1, 2, 0, 1, 0, 1, ... 0, 1)
197: (1, 0, 1, 2, 0, 1, 0, 1, ... 1, 0)
198: (0, 2, 1, 0, 0, 1, 0, 1, ... 0, 1)
199: (0, 1, 0, 2, 1, 1, 0, 1, ... 1, 0)
200: (0, 1, 2, 1, 0, 0, 0, 1, ... 0, 1) (Pirate #200 can only afford 99 votes, and is killed.)
201: (0, 0, 2, 0, 0, 1, 1, 1, 0, ... 1, 0) (Pirate #201 can only afford 99 votes, but survives because #200 accepts the proposal in order to stay alive.)
202: (0, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 1) (Pirate #202 can only afford 100 votes, and is killed.)
203: (0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 0, 1, 0) (Pirate #203 gets #202's vote for free, and survives 101-101.)
204: (0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0) (Pirate #204 gets no free votes and is killed.)
205: (0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0) (Pirate #205 gets #204's vote for free, but still loses 101-102.)
206: (0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0, 0, 0) (Pirate #206 gets #204 and #205, but loses 102-103.)
207: (0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0, 0, 0) (Pirate #207 get #204-206, and survives 103-103.)
Pirates #208-214 die, but #215 survives 107-107.
Pirates #216-230 die, but #231 survives 115-115.
Pirates #232-262 die, but #263 survives 131-131.
There is a pattern.
Each pirate is separated from the previous one by a power of two. In general, pirate $199+2^n$ survives.
There are also patterns in the distribution of the coins, such that it is possible to find, for any pirate, what they would get with x number of pirates.
In general, for x number of pirates and G of coins, the pirates will run out of coins at $x=2G-2$, and from 2G onwards, only pirates equal to $2G-1+2^n$ will survive, where n is a natural number 1, 2, 3, ...
answered Aug 12 at 4:33
Josh P
63
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Before a definitive answer can be given, two questions need to be answered:
If the pirates are indifferent to a proposal, will they accept it or will they kill the head pirate?
If the head pirate could equally well pay multiple pirates, which will be paid? Might the gold, perhaps, be distributed randomly?
Different answers to these questions will yield different results.
As an example of how to analyse this game, I am going to consider what I take to be the best interpretation: the pirates might kill, and the head pirate, when given a choice, will pay the highest-ranked pirates first. I will call the lowest-ranked pirate #1, the second-lowest #2, and so on.
2: Doesn't matter, pirate #1 gets 100.
3: (100, 0, 0) (pirate #2 may die if this proposal doesn't go through, and will therefore accept it.)
4: (98, 0, 1, 1)
5: (97, 0, 1, 2, 0) (#1 or #2 could have been offered 2 coins, but #2 is ranked higher.)
6: (96, 0, 1, 2, 0, 1)
7: (96, 0, 1, 2, 0, 1, 0)
8: (95, 0, 1, 2, 0, 1, 0, 1)
9: (95, 0, 1, 2, 0, 1, 0, 1, 0)
10: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1)
11: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1, 0)
...
196: (1, 0, 1, 2, 0, 1, 0, 1, ... 0, 1)
197: (1, 0, 1, 2, 0, 1, 0, 1, ... 1, 0)
198: (0, 2, 1, 0, 0, 1, 0, 1, ... 0, 1)
199: (0, 1, 0, 2, 1, 1, 0, 1, ... 1, 0)
200: (0, 1, 2, 1, 0, 0, 0, 1, ... 0, 1) (Pirate #200 can only afford 99 votes, and is killed.)
201: (0, 0, 2, 0, 0, 1, 1, 1, 0, ... 1, 0) (Pirate #201 can only afford 99 votes, but survives because #200 accepts the proposal in order to stay alive.)
202: (0, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 1) (Pirate #202 can only afford 100 votes, and is killed.)
203: (0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 0, 1, 0) (Pirate #203 gets #202's vote for free, and survives 101-101.)
204: (0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0) (Pirate #204 gets no free votes and is killed.)
205: (0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0) (Pirate #205 gets #204's vote for free, but still loses 101-102.)
206: (0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0, 0, 0) (Pirate #206 gets #204 and #205, but loses 102-103.)
207: (0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0, 0, 0) (Pirate #207 get #204-206, and survives 103-103.)
Pirates #208-214 die, but #215 survives 107-107.
Pirates #216-230 die, but #231 survives 115-115.
Pirates #232-262 die, but #263 survives 131-131.
There is a pattern.
Each pirate is separated from the previous one by a power of two. In general, pirate $199+2^n$ survives.
There are also patterns in the distribution of the coins, such that it is possible to find, for any pirate, what they would get with x number of pirates.
In general, for x number of pirates and G of coins, the pirates will run out of coins at $x=2G-2$, and from 2G onwards, only pirates equal to $2G-1+2^n$ will survive, where n is a natural number 1, 2, 3, ...
answered Aug 12 at 4:33
Josh P
63
Before a definitive answer can be given, two questions need to be answered:
If the pirates are indifferent to a proposal, will they accept it or will they kill the head pirate?
If the head pirate could equally well pay multiple pirates, which will be paid? Might the gold, perhaps, be distributed randomly?
Different answers to these questions will yield different results.
As an example of how to analyse this game, I am going to consider what I take to be the best interpretation: the pirates might kill, and the head pirate, when given a choice, will pay the highest-ranked pirates first. I will call the lowest-ranked pirate #1, the second-lowest #2, and so on.
2: Doesn't matter, pirate #1 gets 100.
3: (100, 0, 0) (pirate #2 may die if this proposal doesn't go through, and will therefore accept it.)
4: (98, 0, 1, 1)
5: (97, 0, 1, 2, 0) (#1 or #2 could have been offered 2 coins, but #2 is ranked higher.)
6: (96, 0, 1, 2, 0, 1)
7: (96, 0, 1, 2, 0, 1, 0)
8: (95, 0, 1, 2, 0, 1, 0, 1)
9: (95, 0, 1, 2, 0, 1, 0, 1, 0)
10: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1)
11: (94, 0, 1, 2, 0, 1, 0, 1, 0, 1, 0)
...
196: (1, 0, 1, 2, 0, 1, 0, 1, ... 0, 1)
197: (1, 0, 1, 2, 0, 1, 0, 1, ... 1, 0)
198: (0, 2, 1, 0, 0, 1, 0, 1, ... 0, 1)
199: (0, 1, 0, 2, 1, 1, 0, 1, ... 1, 0)
200: (0, 1, 2, 1, 0, 0, 0, 1, ... 0, 1) (Pirate #200 can only afford 99 votes, and is killed.)
201: (0, 0, 2, 0, 0, 1, 1, 1, 0, ... 1, 0) (Pirate #201 can only afford 99 votes, but survives because #200 accepts the proposal in order to stay alive.)
202: (0, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 1) (Pirate #202 can only afford 100 votes, and is killed.)
203: (0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 0, 1, 0) (Pirate #203 gets #202's vote for free, and survives 101-101.)
204: (0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0) (Pirate #204 gets no free votes and is killed.)
205: (0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0) (Pirate #205 gets #204's vote for free, but still loses 101-102.)
206: (0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, ... 1, 0, 0, 0, 0) (Pirate #206 gets #204 and #205, but loses 102-103.)
207: (0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, ... 1, 0, 0, 0, 0, 0) (Pirate #207 get #204-206, and survives 103-103.)
Pirates #208-214 die, but #215 survives 107-107.
Pirates #216-230 die, but #231 survives 115-115.
Pirates #232-262 die, but #263 survives 131-131.
There is a pattern.
Each pirate is separated from the previous one by a power of two. In general, pirate $199+2^n$ survives.
There are also patterns in the distribution of the coins, such that it is possible to find, for any pirate, what they would get with x number of pirates.
In general, for x number of pirates and G of coins, the pirates will run out of coins at $x=2G-2$, and from 2G onwards, only pirates equal to $2G-1+2^n$ will survive, where n is a natural number 1, 2, 3, ...
answered Aug 12 at 4:33
Josh P
63
answered Aug 12 at 4:33
Josh P
63
answered Aug 12 at 4:33
Josh P
63
answered Aug 12 at 4:33
Josh P
63
63
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What happens in case of a tie?
– Hagen von Eitzen
Jul 18 at 4:08
The plan of A still accepted when he has 2/4 vote "yes" from B C D E (A can't vote for his plan)
– Hâm Zon
Jul 18 at 4:10
Thank you. I've edited it
– Hâm Zon
Jul 18 at 4:23
1
@joriki, "casting vote" is a phrase meaning "tie-breaking vote".
– Gerry Myerson
Jul 18 at 9:30
1
@GerryMyerson: Thanks, I should have checked -- I'd never heard that before.
– joriki
Jul 18 at 9:45