5 Pirates PUZZLE (Version 2) - 100 Gold Coins 5 Pirates
video description
Even more fascinating is while x>6, the amount of pirates that can be given 2 coins is greater than the amount of remaining votes, meaning that those pirates are in a quantum superposition of getting 2 or 0 coins. Because of this uncertainty, these values can neither be counted as 0 nor 2 when considering coin distribution in a larger group. Giving this variable the value of q, the pattern takes this form: 0-1-q-0-1-q. For the individuals who would need a proposition to beat q, the only thing that could beat that value is a value of 2 or greater, as this would give the pirate a guaranteed 2 coins as opposed to a chance at 2 coins, assuming the pirates cannot strike bargains with one another. If investing a guaranteed 2 coins against q is not deemed viable (at risk of investing in a pirate who is allies with the next older brother, for example, then the pattern continues. If it does continue this way, quantum superpositions being read this way, then the following formula can be used to calculate profit based on pirate quantity: y=z-(ciel(x/3)+2(ciel(x/2)-ciel(x/3) where x=voter#>8, y=proposer profit>-1, and z=coin quantity>7
However, if that is not the case, and these are true quantum superpositions, the pattern can be broken entirely because 2 is greater than both 1 and q. At that point, when the pattern breaks, pirates who would otherwise be guaranteed 1 coin could instead will get q if investing 2 coins in someone getting 1 would be equally beneficial to the proposer. If the pattern is logically read like that, proposers will have to give younger pirates other than the next oldest pirate 2 coins each to get their respective votes once x>9. This changes the previous equation instead to y=z-(2(ciel(x/2, where x=voter#>9, y=proposer profit>-1, and z=coin quantity>7.
Date: 2023-11-15
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Comments and reviews: 29
Remy
First version of this puzzle was pretty straightforward. This one was more confusing, at least as far as the solution is concerned.
With 2 pirates, the pirate who can vote will kill the distributing pirate unless all 100 coins are relinquished but will spare the distributing pirate if all 100 coins are relinquished. I'm with you so far.
With 3 pirates, the distributing pirate can get away with giving a single coin to the pirate who is next in line to distribute. Still with you.
Where you lost me is with 4 pirates. Using the terms in the video, P4 doesn't need to give P3 anything and can get away with giving a single coin to P1, but wouldn't giving a single coin to P2 be just as effective at securing the vote while also being slightly greedier? I mean, P2 is getting one coin whether P4 lives or dies, and as we saw with 2 pirates, the voting pirate will spare the distributing pirate if all 100 coins are relinquished.
Going from where my logic has taken me, that brings us to the 5 pirate scenario. P5 doesn't need to give P4 anything because P4 won't accept anything less than 98 coins (97 in the video, but that was based on giving P2 two coins instead of one coin. P5 could get his distribution accepted by simply giving a coin to two of the following 3 pirates: P3, P2, P1. Which two of them receives a coin and which one gets nothing makes no difference, since the one who receives nothing will only be the second vote against, resulting in a 50/50 split because the pirates receiving one coin will accept, as per the logic presented by P1 not killing P2 in the scenario where P2 would give P1 all 100 coins. Thus, P5 would still end up with 98 coins, just like in the first version. and since P2 could be the unlucky pirate who (along with P4) receives nothing, the exact solution to the first puzzle is one of three valid solutions here.
Completely unrelated to either puzzle. If I were P5, I'd give everyone, myself included, 20 coins, and if I were any of the other 4 pirates, I would gladly accept a 20/20/20/20/20 distribution. I'm not -greedy- as it relates to this video, and I figure a crew of 4 subordinates (if I were P5) or 3 comrades and a leader (if I were any of the other pirates) will be beneficial in raiding merchant ships. or at least more beneficial than trying to engage in piracy alone (which I would have to do if I were P1 and killed off P2, which would only be possible if we killed off P3, which would only be possible if a majority voted to kill off P4, which would only be possible if a majority voted to kill off P5.
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First version of this puzzle was pretty straightforward. This one was more confusing, at least as far as the solution is concerned.
With 2 pirates, the pirate who can vote will kill the distributing pirate unless all 100 coins are relinquished but will spare the distributing pirate if all 100 coins are relinquished. I'm with you so far.
With 3 pirates, the distributing pirate can get away with giving a single coin to the pirate who is next in line to distribute. Still with you.
Where you lost me is with 4 pirates. Using the terms in the video, P4 doesn't need to give P3 anything and can get away with giving a single coin to P1, but wouldn't giving a single coin to P2 be just as effective at securing the vote while also being slightly greedier? I mean, P2 is getting one coin whether P4 lives or dies, and as we saw with 2 pirates, the voting pirate will spare the distributing pirate if all 100 coins are relinquished.
Going from where my logic has taken me, that brings us to the 5 pirate scenario. P5 doesn't need to give P4 anything because P4 won't accept anything less than 98 coins (97 in the video, but that was based on giving P2 two coins instead of one coin. P5 could get his distribution accepted by simply giving a coin to two of the following 3 pirates: P3, P2, P1. Which two of them receives a coin and which one gets nothing makes no difference, since the one who receives nothing will only be the second vote against, resulting in a 50/50 split because the pirates receiving one coin will accept, as per the logic presented by P1 not killing P2 in the scenario where P2 would give P1 all 100 coins. Thus, P5 would still end up with 98 coins, just like in the first version. and since P2 could be the unlucky pirate who (along with P4) receives nothing, the exact solution to the first puzzle is one of three valid solutions here.
Completely unrelated to either puzzle. If I were P5, I'd give everyone, myself included, 20 coins, and if I were any of the other 4 pirates, I would gladly accept a 20/20/20/20/20 distribution. I'm not -greedy- as it relates to this video, and I figure a crew of 4 subordinates (if I were P5) or 3 comrades and a leader (if I were any of the other pirates) will be beneficial in raiding merchant ships. or at least more beneficial than trying to engage in piracy alone (which I would have to do if I were P1 and killed off P2, which would only be possible if we killed off P3, which would only be possible if a majority voted to kill off P4, which would only be possible if a majority voted to kill off P5.
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jgoemat
In this case on day 4 pirate E has all the power. If he doesn't like
the split, pirate D will get thrown overboard. Actually since they are
bloodthirsty, pirate E will get 100% no matter what if he votes against, so
pirate D is screwed. Pirate E will always vote against the proposal, throw
pirate D overboard, and get 100%.
On day 3 pirate C knows this. He does not have to offer ANYTHING to pirate D.
If pirate D rejects the offer, he will get 0 coins AND be thrown overboard.
So pirate C offers to keep all 100 coins (X, X, 100, 0, 0) and pirate D will
accept.
On day 2 pirate B knows this. He has to get two of the other pirates on his
side. If they reject his proposal, pirate C will get 100 coins and D and E
will get nothing, but will survive. So he has to offer pirates D and E a single
coin each and shut pirate C out (X, 98, 0, 1, 1.
On day 1 pirate A knows this. He needs to get two of the other 4 pirates on
his side. The cheapest way is to offer C a single coin and offer either D or
E two coins (97, 0, 1, 2, 0) or (97, 0, 1, 0, 2.
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In this case on day 4 pirate E has all the power. If he doesn't like
the split, pirate D will get thrown overboard. Actually since they are
bloodthirsty, pirate E will get 100% no matter what if he votes against, so
pirate D is screwed. Pirate E will always vote against the proposal, throw
pirate D overboard, and get 100%.
On day 3 pirate C knows this. He does not have to offer ANYTHING to pirate D.
If pirate D rejects the offer, he will get 0 coins AND be thrown overboard.
So pirate C offers to keep all 100 coins (X, X, 100, 0, 0) and pirate D will
accept.
On day 2 pirate B knows this. He has to get two of the other pirates on his
side. If they reject his proposal, pirate C will get 100 coins and D and E
will get nothing, but will survive. So he has to offer pirates D and E a single
coin each and shut pirate C out (X, 98, 0, 1, 1.
On day 1 pirate A knows this. He needs to get two of the other 4 pirates on
his side. The cheapest way is to offer C a single coin and offer either D or
E two coins (97, 0, 1, 2, 0) or (97, 0, 1, 0, 2.
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Tech
sir, being logical, the most efficient answer would be 98, 0, 1, 0, 1-
if we analyze, -
when there are 2 persons, 0, 100. 2nd person will accept. -
when there are 3 persons, 99, 1, 0. 2nd person will accept because he gets better. and youngest person will never accept because, somehow if he gets the situation to 2 persons, he will get 100. so we can assume he will never accept even in the upcoming scenarios, so we will keep giving 0 to the youngest person. i assume this satisfies the word -BLOOD THIRSTY-. -
when there are 4 persons, 0, 99, 1, 0. because the eldest cant make the situation better than this for 2nd and 3rd persons, so 2nd and 3rd persons will accept. -
when there are 5 persons, 98, 1, 0, 1, 0. he will get 2 votes, reason is the situation is made better for 2nd person so he accepts, 4th persons situation cant be any better no matter how many times he disagrees. -
so 98, 1, 0, 1, 0 is the most optimal and efficient decision that can be made by the eldest person when there are 5 persons. -
LET ME KNOW IF MY WAY OF THINKING IS CORRECT-
Thank you.
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sir, being logical, the most efficient answer would be 98, 0, 1, 0, 1-
if we analyze, -
when there are 2 persons, 0, 100. 2nd person will accept. -
when there are 3 persons, 99, 1, 0. 2nd person will accept because he gets better. and youngest person will never accept because, somehow if he gets the situation to 2 persons, he will get 100. so we can assume he will never accept even in the upcoming scenarios, so we will keep giving 0 to the youngest person. i assume this satisfies the word -BLOOD THIRSTY-. -
when there are 4 persons, 0, 99, 1, 0. because the eldest cant make the situation better than this for 2nd and 3rd persons, so 2nd and 3rd persons will accept. -
when there are 5 persons, 98, 1, 0, 1, 0. he will get 2 votes, reason is the situation is made better for 2nd person so he accepts, 4th persons situation cant be any better no matter how many times he disagrees. -
so 98, 1, 0, 1, 0 is the most optimal and efficient decision that can be made by the eldest person when there are 5 persons. -
LET ME KNOW IF MY WAY OF THINKING IS CORRECT-
Thank you.
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James
Just a few things you got wrong according to your own rules:
They are bloodthirsty: they will kill the oldest pirates if they don-t get a BETTER deal than what they could get next round
Most importantly they don-t wish to die
Take 2 pirates for example:
0 coins and 100 coins- by your own rules the younger pirate will vote no because 100 is not better than 100 next round
3 pirates:
100/0/0 the middle pirate will vote yes no matter what because if the older pirate dies here then the middle pirate is dead too
4 pirates
In this case the next round will be accepted so pirate 3 will vote no no matter what because next round he gets it all guaranteed- so 98/0/1/1 assures his survival
And now 5 pirates:
97/0/1/0/2 or 97/0/1/2/0
Remember to follow the rules and words matter. better means more not equal to or greater -
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Just a few things you got wrong according to your own rules:
They are bloodthirsty: they will kill the oldest pirates if they don-t get a BETTER deal than what they could get next round
Most importantly they don-t wish to die
Take 2 pirates for example:
0 coins and 100 coins- by your own rules the younger pirate will vote no because 100 is not better than 100 next round
3 pirates:
100/0/0 the middle pirate will vote yes no matter what because if the older pirate dies here then the middle pirate is dead too
4 pirates
In this case the next round will be accepted so pirate 3 will vote no no matter what because next round he gets it all guaranteed- so 98/0/1/1 assures his survival
And now 5 pirates:
97/0/1/0/2 or 97/0/1/2/0
Remember to follow the rules and words matter. better means more not equal to or greater -
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jgoemat
I agree with others. The 'bloodthirsty' argument seems to imply that pirate 1 will ALWAYS reject the deal when there are only 2 pirates remaining. He will get the 100 coins either way, but he won't get a -better- deal by accepting the split and letting pirate 2 live. This has the implication that pirate 5 will still offer himself 97 coins and offer 1 coin to pirate 3. But he could offer 2 coins to either pirate 1 or pirate 2 (it doesn't matter which. It also has implications with 3 pirates remaining, pirate 3 will keep all the coins since pirate 2 knows that he will die if he rejects the offer. This is only if the -bloodthirsty- trait overrides the -wish to live- trait or he would reject the offer because the deal wouldn't get him a single extra coin even though he would lose his life. But I would consider 'living' as part of the detal.
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I agree with others. The 'bloodthirsty' argument seems to imply that pirate 1 will ALWAYS reject the deal when there are only 2 pirates remaining. He will get the 100 coins either way, but he won't get a -better- deal by accepting the split and letting pirate 2 live. This has the implication that pirate 5 will still offer himself 97 coins and offer 1 coin to pirate 3. But he could offer 2 coins to either pirate 1 or pirate 2 (it doesn't matter which. It also has implications with 3 pirates remaining, pirate 3 will keep all the coins since pirate 2 knows that he will die if he rejects the offer. This is only if the -bloodthirsty- trait overrides the -wish to live- trait or he would reject the offer because the deal wouldn't get him a single extra coin even though he would lose his life. But I would consider 'living' as part of the detal.
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LOGICALLY
-Sanket Dashpute and -Liam Swick. many thanks for your valid doubt on this puzzle. So I made a little yet logical change in the conditions and re-uploaded it.
The doubt was: In case of 3 pirates, if P3 keeps 100 coins and gives 0 coin to P2, then being a rational person P2 should accept it, because in next turn as well P2 is getting nothing.
Correction: The condition of -Bloodthirsty- is really a logical and tricky one. It took me a while to formulate this condition so that it fits the puzzle properly. Now, if you validate the above doubt with the condition of 'Bloodthirsty' you will see why P2 shouldn't be given 0 coin, rather he should be give 1 coin to get a vote in favor.
Please comment below your thoughts.
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-Sanket Dashpute and -Liam Swick. many thanks for your valid doubt on this puzzle. So I made a little yet logical change in the conditions and re-uploaded it.
The doubt was: In case of 3 pirates, if P3 keeps 100 coins and gives 0 coin to P2, then being a rational person P2 should accept it, because in next turn as well P2 is getting nothing.
Correction: The condition of -Bloodthirsty- is really a logical and tricky one. It took me a while to formulate this condition so that it fits the puzzle properly. Now, if you validate the above doubt with the condition of 'Bloodthirsty' you will see why P2 shouldn't be given 0 coin, rather he should be give 1 coin to get a vote in favor.
Please comment below your thoughts.
reply
LOGICALLY
-Sanket Dashpute and -Liam Swick. many thanks for your valid doubt on this puzzle. So I made a little yet logical change in the conditions and re-uploaded it.
The doubt was: In case of 3 pirates, if P3 keeps 100 coins and gives 0 coin to P2, then being a rational person P2 should accept it, because in next turn as well P2 is getting nothing.
Correction: The condition of -Bloodthirsty- is really a logical and tricky one. It took me a while to formulate this condition so that it fits the puzzle properly. Now, if you validate the above doubt with the condition of 'Bloodthirsty' you will see why P2 shouldn't be given 0 coin, rather he should be give 1 coin to get a vote in favor.
Please comment below your thoughts.
reply
-Sanket Dashpute and -Liam Swick. many thanks for your valid doubt on this puzzle. So I made a little yet logical change in the conditions and re-uploaded it.
The doubt was: In case of 3 pirates, if P3 keeps 100 coins and gives 0 coin to P2, then being a rational person P2 should accept it, because in next turn as well P2 is getting nothing.
Correction: The condition of -Bloodthirsty- is really a logical and tricky one. It took me a while to formulate this condition so that it fits the puzzle properly. Now, if you validate the above doubt with the condition of 'Bloodthirsty' you will see why P2 shouldn't be given 0 coin, rather he should be give 1 coin to get a vote in favor.
Please comment below your thoughts.
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varshit
Oldest will take 99 coins and second youngest will take one coin. This is the distribution.
Logic
Anyone who gives P2 at least single coin will win the deal. The youngest one will not reject because he is rational ( so he will never let to die any person without his profit) and because you say there are intelligent and rational, youngest one always knows that he will not get any coin because all other loves their life.
Still feels this puzzle is incomplete, you have to ascertain that is youngest one interprets as reducing no. of people will increase his chances or not. Although you clearly specified that they are rational and they only like to kill when there is real chance of getting one more coin.
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Oldest will take 99 coins and second youngest will take one coin. This is the distribution.
Logic
Anyone who gives P2 at least single coin will win the deal. The youngest one will not reject because he is rational ( so he will never let to die any person without his profit) and because you say there are intelligent and rational, youngest one always knows that he will not get any coin because all other loves their life.
Still feels this puzzle is incomplete, you have to ascertain that is youngest one interprets as reducing no. of people will increase his chances or not. Although you clearly specified that they are rational and they only like to kill when there is real chance of getting one more coin.
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Douglas
The view that the pirates are bloodthirsty and rational are contradictory. 'bloodthirsty' to me implies that they would vote to kill someone even if they don't end up with any extra - as long as they don't get less. Yet you argued that 'rational' implies the opposite of this. The solution you give is only correct if you have -bloodthirsty- and not -rational-. With 3 pirates you conclude the division will be 99: 0: 1 agreed. With 4 pirates that are 'rational' but not blood thirsty the division would be: 99: 0: 0: 1 as then the 2 pirates won't get any more if they vote against and the solution for 5 would then be 99: 0: 0: 0: 1, again 2 pirates vote in favour as they won't get any more if they vote against.
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The view that the pirates are bloodthirsty and rational are contradictory. 'bloodthirsty' to me implies that they would vote to kill someone even if they don't end up with any extra - as long as they don't get less. Yet you argued that 'rational' implies the opposite of this. The solution you give is only correct if you have -bloodthirsty- and not -rational-. With 3 pirates you conclude the division will be 99: 0: 1 agreed. With 4 pirates that are 'rational' but not blood thirsty the division would be: 99: 0: 0: 1 as then the 2 pirates won't get any more if they vote against and the solution for 5 would then be 99: 0: 0: 0: 1, again 2 pirates vote in favour as they won't get any more if they vote against.
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Flat
Allow me to correct you on one -teensy- thing.
P1 has to accept P2's deal, or else he dies - as per the rules, the eldest pirate can't vote, and every other pirate on board can. Thus, he'd be getting 0% of the vote and be forced to throw himself overboard. You didn't make this case an exception to your rules. P2 would logically distribute all the coins to himself knowing this. P3 can't sway P2, so he will give 1 coin to P1 and leave the other 99 for himself. P4 can't sway P3, so he will give 1 coin to P2 and 2 coins to P1, keeping the other 97 for himself. P5 can't sway P4, so he will give 1 coin to P3 and 2 coins to P2, leaving the other 97 for himself.
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Allow me to correct you on one -teensy- thing.
P1 has to accept P2's deal, or else he dies - as per the rules, the eldest pirate can't vote, and every other pirate on board can. Thus, he'd be getting 0% of the vote and be forced to throw himself overboard. You didn't make this case an exception to your rules. P2 would logically distribute all the coins to himself knowing this. P3 can't sway P2, so he will give 1 coin to P1 and leave the other 99 for himself. P4 can't sway P3, so he will give 1 coin to P2 and 2 coins to P1, keeping the other 97 for himself. P5 can't sway P4, so he will give 1 coin to P3 and 2 coins to P2, leaving the other 97 for himself.
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Paul
I disagree with the logic on this one, P1 knows that if he always votes against the worst case is the proposal is accepted and he gets whatever is proposed and the best case is it ends up with two pirates and he gets 100%. P1 will therefore always vote no. P2 knows that if it ends in two pirates he always loses, therefore he will vote in favour of anything which gives him coins, similarly with P3 who knows that he will always get zero if it goes down to 3 and so will also vote in favour of any coins. Therefore the answer is to give 1 to P3 and 1 to P2. Giving any to P1 will result in him voting against anyway as he has nothing to lose.
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I disagree with the logic on this one, P1 knows that if he always votes against the worst case is the proposal is accepted and he gets whatever is proposed and the best case is it ends up with two pirates and he gets 100%. P1 will therefore always vote no. P2 knows that if it ends in two pirates he always loses, therefore he will vote in favour of anything which gives him coins, similarly with P3 who knows that he will always get zero if it goes down to 3 and so will also vote in favour of any coins. Therefore the answer is to give 1 to P3 and 1 to P2. Giving any to P1 will result in him voting against anyway as he has nothing to lose.
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PowerStar004
The bloodthirsty arguement says -a pirate will vote against the proposal (to kill the oldest pirate) if he doesn't get a better deal than what he can get in the next turn.
100 is not more than 100, so p1 will always vote to kill p2's proposal because they get 100 coins no matter what.
Then, with 3 pirates, p3 can keep all 100 coins, because p2 will vote yes just to live.
Then with 4 pirates, p4 can give p2 and p1 one coin to buy their votes, since it's 1 more than what p3 would give them.
Finally with 5 pirates, p5 gives p3 1 coin and either p2 or p1 (doesn't matter which) 2 coins.
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The bloodthirsty arguement says -a pirate will vote against the proposal (to kill the oldest pirate) if he doesn't get a better deal than what he can get in the next turn.
100 is not more than 100, so p1 will always vote to kill p2's proposal because they get 100 coins no matter what.
Then, with 3 pirates, p3 can keep all 100 coins, because p2 will vote yes just to live.
Then with 4 pirates, p4 can give p2 and p1 one coin to buy their votes, since it's 1 more than what p3 would give them.
Finally with 5 pirates, p5 gives p3 1 coin and either p2 or p1 (doesn't matter which) 2 coins.
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Brijesh
If they ALL are truly rational, then a perfect choice would be suggested by P5 and everyone will agree with this perfect choice. They won't even need a vote. They will reject it only if they consider that oldest on board has made mistake and best possible number is not offered to them. If they know the complete distribution, then they know other's vote as well. So, puzzle can be tweaked by saying that the person concerned only know how many coins are offered to him. Does this make sense to you guys?
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If they ALL are truly rational, then a perfect choice would be suggested by P5 and everyone will agree with this perfect choice. They won't even need a vote. They will reject it only if they consider that oldest on board has made mistake and best possible number is not offered to them. If they know the complete distribution, then they know other's vote as well. So, puzzle can be tweaked by saying that the person concerned only know how many coins are offered to him. Does this make sense to you guys?
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Venkat
I had seen this puzzle after several years of this uploaded. P4 can never get much out of this as P3 and P1 will always reject any proposal from him. So it is best for P4 to have his life saved. So it is in P4s interest to accept P5 solution. Even from P2 perspective he will get max 1 when P3 becomes proposer. So thought P5: 97 P4: 1 P3: 0 P2: 2 and P1: 0.
Since P3 stands to gain rejecting P4 and P5 he is likely to reject any proposal from them. Why should P3 accept the proposal given by P5?
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I had seen this puzzle after several years of this uploaded. P4 can never get much out of this as P3 and P1 will always reject any proposal from him. So it is best for P4 to have his life saved. So it is in P4s interest to accept P5 solution. Even from P2 perspective he will get max 1 when P3 becomes proposer. So thought P5: 97 P4: 1 P3: 0 P2: 2 and P1: 0.
Since P3 stands to gain rejecting P4 and P5 he is likely to reject any proposal from them. Why should P3 accept the proposal given by P5?
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Adi
A slight change in the question that a pirate will surely kill the other one if he can't get any benefit with his life. That means if there only two pirates and older pirate proposes zero coins to himself and hundred to the other, then also the other pirate (younger one) will disagree because he knows that if older one dies then also he alone will take everything so why should older pirate will be alive. This was exactly BLOOD THIRSTY thing that you mentioned.
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A slight change in the question that a pirate will surely kill the other one if he can't get any benefit with his life. That means if there only two pirates and older pirate proposes zero coins to himself and hundred to the other, then also the other pirate (younger one) will disagree because he knows that if older one dies then also he alone will take everything so why should older pirate will be alive. This was exactly BLOOD THIRSTY thing that you mentioned.
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Zidan
It past a lot of time without seeing your puzzles bro. However, this is very easy one, I agree that the oldest pirate will defiantly keep 97 coins, but there are two different pirates could accept the deal of having one and two coins respectively that giving by the oldest pirate. For example P3 and P2.
Ala' Zayed
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It past a lot of time without seeing your puzzles bro. However, this is very easy one, I agree that the oldest pirate will defiantly keep 97 coins, but there are two different pirates could accept the deal of having one and two coins respectively that giving by the oldest pirate. For example P3 and P2.
Ala' Zayed
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David
This video is still wrong. When there are 2 pirates left P1 will still kill P2 even if P2 offers him all 100, because he is 'bloodthirsty' i. e. P1 is not getting a better deal by allowing P2 to live, so he will kill him and keep all 100 and be the only survivor. This has implications on the final answer
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This video is still wrong. When there are 2 pirates left P1 will still kill P2 even if P2 offers him all 100, because he is 'bloodthirsty' i. e. P1 is not getting a better deal by allowing P2 to live, so he will kill him and keep all 100 and be the only survivor. This has implications on the final answer
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AJIT
If the order of priorities for each pilot is:
Get as much coins,
Stay alive
Be blood thirsty by getting other killed
then P1 will always reject P2's proposal, get 100 coins and get P2 killed.
So in this thought of logic final proposal answer of P5 looks like:
97 0 1 2 0
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If the order of priorities for each pilot is:
Get as much coins,
Stay alive
Be blood thirsty by getting other killed
then P1 will always reject P2's proposal, get 100 coins and get P2 killed.
So in this thought of logic final proposal answer of P5 looks like:
97 0 1 2 0
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Fakir
No need to give second guy 2 coins in 4 pirates case. He will not get any more coins by rejecting proposal so he is rational enough to accept it without killing. Similarly no need to give one more coin to youngest in 5 pirate case as he is not getting any better
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No need to give second guy 2 coins in 4 pirates case. He will not get any more coins by rejecting proposal so he is rational enough to accept it without killing. Similarly no need to give one more coin to youngest in 5 pirate case as he is not getting any better
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Jorun
The 90 gold coins is split betweeen gruppes of ten thers 1 captain and 4 Crew members all Crew Will be split between 10 to 19 or 90 to 99 the captain Will get a 10 below the Crew so if Crew Got 20 group captain Will get 10 group Crew traits in coment above
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The 90 gold coins is split betweeen gruppes of ten thers 1 captain and 4 Crew members all Crew Will be split between 10 to 19 or 90 to 99 the captain Will get a 10 below the Crew so if Crew Got 20 group captain Will get 10 group Crew traits in coment above
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Jorun
Now British are civilsed so ther not bloodthisty nor gredy but they are powerthisty Will kill captain if they Can get and hold position AS pirate i maen captain and they want to get AS much gold AS they Can get in the systym if they Canot become captain
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Now British are civilsed so ther not bloodthisty nor gredy but they are powerthisty Will kill captain if they Can get and hold position AS pirate i maen captain and they want to get AS much gold AS they Can get in the systym if they Canot become captain
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Jorun
The piraters split op ther 100 gold coins but they are cougt by a British ship and they divide the trausre arcording to the pirate code but with tweeks to make it fair and twenty coins go to the King of britan see the tweeks in the coment blow
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The piraters split op ther 100 gold coins but they are cougt by a British ship and they divide the trausre arcording to the pirate code but with tweeks to make it fair and twenty coins go to the King of britan see the tweeks in the coment blow
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Pengcheng
When there are only two people, whatever the proposal will be voted against to kill P2. To save his life, P2
will vote for whatever P3 proposes. So the rational proposal of P3 would be he has all the gold coins, right?
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When there are only two people, whatever the proposal will be voted against to kill P2. To save his life, P2
will vote for whatever P3 proposes. So the rational proposal of P3 would be he has all the gold coins, right?
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bryce
I don-t understand, if the pirates are bloodthirsty, why would p1 vote yes for p2? He would get the same amount of coin regardless, so he would vote no just for fun
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I don-t understand, if the pirates are bloodthirsty, why would p1 vote yes for p2? He would get the same amount of coin regardless, so he would vote no just for fun
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Shubham
Since all are rational then in case of 4 pirates we should assign
98, 0, 1, 1. Since p2 is rational and will vote in favor even given 1 coin instead of 2.
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Since all are rational then in case of 4 pirates we should assign
98, 0, 1, 1. Since p2 is rational and will vote in favor even given 1 coin instead of 2.
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Vaibhav
Version #3
The 5 pirates can think upto the last scenario. Then the answer will be
P5-98 coins
P4-1coin
P3-0 coin
P2- 1 coin
P1- 0 coin
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Version #3
The 5 pirates can think upto the last scenario. Then the answer will be
P5-98 coins
P4-1coin
P3-0 coin
P2- 1 coin
P1- 0 coin
reply
Diptarka
Wrong video, been a fan for a long time. Kindly make the necessary changes. How can p2 bribe p4 for the vote when p4 manages to get 100 coins himself.
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Wrong video, been a fan for a long time. Kindly make the necessary changes. How can p2 bribe p4 for the vote when p4 manages to get 100 coins himself.
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education
Why would p1 accept anything other than 100 coins as ultimately if he negative votes he will end up with 100 coins. It doesn-t make sense
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Why would p1 accept anything other than 100 coins as ultimately if he negative votes he will end up with 100 coins. It doesn-t make sense
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Sachin
I am not agree for the solution. why the youngest one will agreed for the 1 coin. as he can get the whole. please solve the issue
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I am not agree for the solution. why the youngest one will agreed for the 1 coin. as he can get the whole. please solve the issue
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