5 Pirates PUZZLE 100 Gold Coins 5 Pirates Game Theory based Problem
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Date: 2023-11-15
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Comments and reviews: 21
Ole
The proposed solution in the video is wrong. Acording to the rules, the condition for VOTING AGAINST a proposal is, that the pirate by doing this will gain AT LEAST one extra coin! Arguing backwards:
Pirate 4 proposes (100, 0) and is accepted.
Pirate 3 proposes (100, 0, 0, where pirate 4 vote against, but pirate 5 accept, because he will NOT gain an extra coin, if he votes against it.
The same occurs for pirate 2 and 1. Only the second-oldest pirate will vote against any proposal, and pirate 1 proposes the split (100, 0, 0, 0, 0.
In short: The oldest pirate gets all 100 coins, because all but the second-oldest pirate will get the SAME amount of coins if they reject the proposal.
If the rules were changed, such that the condition for ACCEPTING a proposal is, that the pirate by doing this will gain AT LEAST one extra coin. In this case they kill the oldest pirate UNLESS they get at least one extra coin. This is also the rule from the TED-Ed-version. With this change of rules, the logic of the video is correct, and the winning strategy for pirate 1 is (98, 0, 1, 0, 1.
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The proposed solution in the video is wrong. Acording to the rules, the condition for VOTING AGAINST a proposal is, that the pirate by doing this will gain AT LEAST one extra coin! Arguing backwards:
Pirate 4 proposes (100, 0) and is accepted.
Pirate 3 proposes (100, 0, 0, where pirate 4 vote against, but pirate 5 accept, because he will NOT gain an extra coin, if he votes against it.
The same occurs for pirate 2 and 1. Only the second-oldest pirate will vote against any proposal, and pirate 1 proposes the split (100, 0, 0, 0, 0.
In short: The oldest pirate gets all 100 coins, because all but the second-oldest pirate will get the SAME amount of coins if they reject the proposal.
If the rules were changed, such that the condition for ACCEPTING a proposal is, that the pirate by doing this will gain AT LEAST one extra coin. In this case they kill the oldest pirate UNLESS they get at least one extra coin. This is also the rule from the TED-Ed-version. With this change of rules, the logic of the video is correct, and the winning strategy for pirate 1 is (98, 0, 1, 0, 1.
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Syril
This is wrong. Pirate 3 will reject the deal because if the deal is rejected twice, he will get chance to propose new deal. Definitely he will propose for 99, 0, 1. But p4 may or may not accept the deal because if he reject the deal, he will have to propose the deal next time and p3 will reject the deal and p4 will be thrown out. There is a chance of p4 also rejecting the deal because he won't get anything. In that case p1 and p4 will be thrown out. To avoid the risk, p1 will propose a deal which will be 98, 1. 0, 0, 1. In this case, p4 will not reject because if he reject, he will have to propose next and p3 and p2 will reject it. P1 will not get anything extra even if he reject. So he will also accept. So deal will be approved.
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This is wrong. Pirate 3 will reject the deal because if the deal is rejected twice, he will get chance to propose new deal. Definitely he will propose for 99, 0, 1. But p4 may or may not accept the deal because if he reject the deal, he will have to propose the deal next time and p3 will reject the deal and p4 will be thrown out. There is a chance of p4 also rejecting the deal because he won't get anything. In that case p1 and p4 will be thrown out. To avoid the risk, p1 will propose a deal which will be 98, 1. 0, 0, 1. In this case, p4 will not reject because if he reject, he will have to propose next and p3 and p2 will reject it. P1 will not get anything extra even if he reject. So he will also accept. So deal will be approved.
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sorsocksfake
(highest for oldest)
- 2 pirates: pirate 2 votes for his own proposal and gets 50%, so he wins. 100/0
- 3 pirates: therefore, pirate 2 will always reject; pirate 3 will always approve. Pirate 1 must approve if he gets 1 gold; else he rejects. Therefore: 99/0/1
- 4 pirates: pirate 3 will always reject. Pirate 2 accepts if given anything; pirate 3 accepts if given 2 gold. Therefore 99/0/1/0
- 5 pirates: pirate 4 will always reject. Pirate 3 requires 1 gold. Pirate 2 requires 2 gold. Pirate 1 requires 1 gold. Therefore, 98/0/1/0/1.
That is to say: pirate 5 votes for it of course, to live. Pirate 4 and 2 vote against it. Pirate 3 and 1 vote for it because otherwise they get nothing from pirate 4's proposal.
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(highest for oldest)
- 2 pirates: pirate 2 votes for his own proposal and gets 50%, so he wins. 100/0
- 3 pirates: therefore, pirate 2 will always reject; pirate 3 will always approve. Pirate 1 must approve if he gets 1 gold; else he rejects. Therefore: 99/0/1
- 4 pirates: pirate 3 will always reject. Pirate 2 accepts if given anything; pirate 3 accepts if given 2 gold. Therefore 99/0/1/0
- 5 pirates: pirate 4 will always reject. Pirate 3 requires 1 gold. Pirate 2 requires 2 gold. Pirate 1 requires 1 gold. Therefore, 98/0/1/0/1.
That is to say: pirate 5 votes for it of course, to live. Pirate 4 and 2 vote against it. Pirate 3 and 1 vote for it because otherwise they get nothing from pirate 4's proposal.
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Josef
i have some doubts on the explanation: why should pirate 1 accept one coin, when he knows, that, when three pirates are left, P3 has to offer him already one, - remember he is blood thirsty! in my opinion P5 has to offer: 0 to P4, 1 to P3, 0 to P2 but 2 to P1, and 97 for himself. Then, and only then, P1 and P3 will agree.
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i have some doubts on the explanation: why should pirate 1 accept one coin, when he knows, that, when three pirates are left, P3 has to offer him already one, - remember he is blood thirsty! in my opinion P5 has to offer: 0 to P4, 1 to P3, 0 to P2 but 2 to P1, and 97 for himself. Then, and only then, P1 and P3 will agree.
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weckar
Your description of the bloodthirst does not match the solution. You say that they will kill even if they only get one more coin. In fact, per the solution, they will even kill if the outcome does not change for themselves. This is why P5 cannot give a coin to P1, because even if P5 dies, P1 would still get a coin.
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Your description of the bloodthirst does not match the solution. You say that they will kill even if they only get one more coin. In fact, per the solution, they will even kill if the outcome does not change for themselves. This is why P5 cannot give a coin to P1, because even if P5 dies, P1 would still get a coin.
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Girish
With three pirate on boat.
P3 with 99 coins P2 with 0 Coin and P1 with 1 coin.
Your conclusion is wrong.
P1 would not accept the proposal as well. bcoz if p2 and p1 reject the deal p3 will be thown out of boat.
And then p2 and p1 will share 50 50 coins each.
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With three pirate on boat.
P3 with 99 coins P2 with 0 Coin and P1 with 1 coin.
Your conclusion is wrong.
P1 would not accept the proposal as well. bcoz if p2 and p1 reject the deal p3 will be thown out of boat.
And then p2 and p1 will share 50 50 coins each.
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education
(4: 25) It's actually ANY 2 coins ditributed amoungst P3, P2 and P1 in ANY combination. It's a simple progression from oldest to youngest - the second oldest will always miss out - provided you give 1 coin to any combination of 50% or more of priates that are younger.
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(4: 25) It's actually ANY 2 coins ditributed amoungst P3, P2 and P1 in ANY combination. It's a simple progression from oldest to youngest - the second oldest will always miss out - provided you give 1 coin to any combination of 50% or more of priates that are younger.
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Josef
on second thoughts, the solution should be 98: 1: 1: 0: 0 - P4 has to accept one coin, otherwise he is removed in the next round. Only for 6 pirates an increase of the minimal -one coin- to 97: 1: 2: 0: 0: 0 is needed but the solution is not unique.
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on second thoughts, the solution should be 98: 1: 1: 0: 0 - P4 has to accept one coin, otherwise he is removed in the next round. Only for 6 pirates an increase of the minimal -one coin- to 97: 1: 2: 0: 0: 0 is needed but the solution is not unique.
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education
The true question behind this problem is that is there a pattern as to how the coins will be distributed for N pirates with the rules and conditions explained in this video? For example what if there were 6 pirates or more?
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The true question behind this problem is that is there a pattern as to how the coins will be distributed for N pirates with the rules and conditions explained in this video? For example what if there were 6 pirates or more?
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rockamania7
You only consider the next round to base the pirates' decision on it, right? In my opinion, for example, P2 will never accept until he is the oldest, if he takes more than one step into account.
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You only consider the next round to base the pirates' decision on it, right? In my opinion, for example, P2 will never accept until he is the oldest, if he takes more than one step into account.
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Guru
Nice one and gud explanation. so the idea is to distribute 1 gold coin to every alternate pirate and the eldest pilot keeps the remaining gold coins. so thats a greedy approach
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Nice one and gud explanation. so the idea is to distribute 1 gold coin to every alternate pirate and the eldest pilot keeps the remaining gold coins. so thats a greedy approach
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vishal
In solution part how it would click to us that we have left with only two pirates but it is given that there are five pirates and senior most would decide how to distribute?
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In solution part how it would click to us that we have left with only two pirates but it is given that there are five pirates and senior most would decide how to distribute?
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Josef
it should be 99: 0: 1: 0: 2 as P4 still can get 99, when offering one to P2, when he is to decide. (99: 0: 1: 0. i guess my first comment might be still valid.
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it should be 99: 0: 1: 0: 2 as P4 still can get 99, when offering one to P2, when he is to decide. (99: 0: 1: 0. i guess my first comment might be still valid.
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Harsh
Yaar kaab eshe puzzles Ka logic meh Apne life istmaal Kar paunga
problem yeh nhi h ki mujhe eshe logic pata nhi h prblm yeh h ki Kaha inkaah istmaal Karna h
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Yaar kaab eshe puzzles Ka logic meh Apne life istmaal Kar paunga
problem yeh nhi h ki mujhe eshe logic pata nhi h prblm yeh h ki Kaha inkaah istmaal Karna h
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Govardhana
Bro in 4pirates deal, if pirate 2 reject the deal how u concluded that if pirate 3 becomes head he wont get nothing, didn't understand that can u explain
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Bro in 4pirates deal, if pirate 2 reject the deal how u concluded that if pirate 3 becomes head he wont get nothing, didn't understand that can u explain
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Jovana
There is everything but logic in this one. Why wouldnt they vote off the oldest three pirates and split the money 50: 50 at the end? I call that rational.
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There is everything but logic in this one. Why wouldnt they vote off the oldest three pirates and split the money 50: 50 at the end? I call that rational.
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kuls43
Lol, people discussing whether they were able to solve it or not and here am I who couldn't even understand the puzzle ---
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Lol, people discussing whether they were able to solve it or not and here am I who couldn't even understand the puzzle ---
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Mukeshkumar
You are a great story teller.
I love way of telling and it is really interesting to listen thriller stories.
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You are a great story teller.
I love way of telling and it is really interesting to listen thriller stories.
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Amit
But what if pirate 5 proposes to give P1 & P2 1 coin each pirate 2 will also agree as he don't have any other choice
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But what if pirate 5 proposes to give P1 & P2 1 coin each pirate 2 will also agree as he don't have any other choice
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Adama
Very cool riddle! I used the same method as the explaination but it takes me few minutes to figure how to solve it.
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Very cool riddle! I used the same method as the explaination but it takes me few minutes to figure how to solve it.
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Prakhar
So It's like if there are 10 pirates. the distribution will be 96 0 1 0 1 0 1 0 1 0. wow it's like a theorem
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So It's like if there are 10 pirates. the distribution will be 96 0 1 0 1 0 1 0 1 0. wow it's like a theorem
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