5 Pirates PUZZLE || 100 Gold Coins 5 Pirates || Game Theory based Problem
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- Опубликовано: 26 авг 2024
- 5 pirates of different ages, on a treasure hunt, come across a small island, where they find a treasure box with 100 gold coins.
On their ship, they decide to split the coins using this scheme:
The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it.
If 50% or more of the pirates vote for it, then the coins will be shared that way.
But if the votes are less than 50%, the proposal will be rejected...
and the pirate proposing the scheme will be thrown overboard !
again the process is repeated with the pirates that remain.
All the 5 pirates are Intelligent, Greedy, Bloodthirsty, Rational and importantly... they all don't wish to die.
PUZZLE : What would be the best distribution by the Oldest Pirate ?
The video explains the Puzzle's solution and the Logic behind.
Please comment below your answers and suggestions. Also LIKE the video and SUBSCRIBE to my channel if you are new.
Never seen a person who is bloodthirsty and rational 😂
Good explanation
Mankind is full of them unfortunately.
This is simply the best way to explain an extremely complicated puzzle. You are a genius
I all ready know the answer to the riddle
Lol, people discussing whether they were able to solve it or not and here am I who couldn't even understand the puzzle 😂😂😂
@@kuls43 I also belong to your category🤣
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
bahut badhiya, you catch the point!
Good one! The narration really helps in grasping the concept...!
diak812 ... thank u !! I hope you'd enjoy the other puzzles too ☺
Lol, people discussing whether they were able to solve it or not and here am I who couldn't even understand the puzzle 😂😂😂
Why there is not 20-20-20-20-20?!!
That background music made it even more better😂😂
(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.
Actually the solution can be any one of the following:
98:0:1:0:1
98:0:0:1:1
98:0:1:1:0
Because the oldest pirate only has to please 2 other pirates to avoid getting killed. The 4th pirate cannot be pleased anyway, so dont give him anything. Now the 5th pirate has to please any 2 of P1, P2, P3. P3 and P1 would be content to get 1 coin as they wont get it if P5 gets killed. However P2 knows that he will get only 1 coin even if P5 gets killed. So P5 can select any of the the 2 pirates among P1, P2 and P3 to give 2 coins and win the vote, keep himself alive and get max benefit
You are right
really very simple, painless and clear explanation .
Different way of approaching to the solution......which is really fantastic
Superb explanation bro
Very cool riddle ! I used the same method as the explaination but it takes me few minutes to figure how to solve it.
Four people start at the corners of a square room, 10 meters on each side.
Each person moves at the same speed directly towards the next person counterclockwise as seen from above.
As a result, they spiral in until they all eventually collide at the center of the square. How far has each person traveled?
Can you make a video on this?
Dear Tanay... I have noted your question... will give my best to make a video on this.
@@LOGICALLYYOURS you could,also,divide the coins as 25 and 25 and 25 among the okdest abd two others and then spmit the remaining 25 among the teoblosers right..,they still,might reject because they are getting less than the others.,thst is also a valid answer correct?
@@leif1075 No because the older pirate is greedy and knows he can do better and get 98 coins. He will do the split that guarantees him the most.
Each person will have traveled 7.85 meters. Or one quarter of a 10 meter diameter circles circumference .
This Question Is from RD SHARMA Physics Book Class 11
Classy man... perfect explaination
Thanks bro : )
superb explanation
You are a great story teller....
I love way of telling and it is really interesting to listen thriller stories...
Amazing puzzle! I'm impressed!
Wow, just wow!
Thanks a lot :) .. Also check out it's version 2 which is also interesting.
Pirate5 is Jack sparrow for sure
This same riddle was uploaded by ted ed a few months after you uploaded with the same explanation 😕 they got millions of views just because more people know about them ☹️
Ya, I saw that one....
Hardest test of life is to be patient... I hope i'll pass it :)
@@LOGICALLYYOURS Best Wishes for you Mate♥️
Woow...nice explaination
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?
Hi SD... your question is perfect.. i will note it down and certainly answer it once we have Q&A.
For 6 pirates,oldest pirate will get highest no of coins and 2 more players will get 1 coin each. For even no of pirates,oldest pirate will try to achieve 50% proposal acceptance and hence he will get more coins in that case when compared to odd no of pirates coz in that case the old pirate is forced to achieve >50% proposal acceptance. For eg. When there r seven pirates,oldest pirate will get most no of coins and he has to give 1 coin each to three other pirates. If there were N coins then for 6 pirates,oldest pirate will get N-2 coins and for 7 pirates,he will get N-3 coins. Same logic can be applied to any no of pirates
Amazing analysis!
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.
Damn... amazing bro.
Nice explaination
Your video is amazing ...grt job u are doing..keep going
got goosebumps just by hearings pirates of the Caribbean theme song
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 ?
Bhai pirates of carribean ka bgm bhi use kiya gajab
Nice... very interesting.
Mona Tam
thanks a lot
I think that the oldest pirate (A) should offer C and D 1 coin each and keep the other 98 coins for himself. Let's note the pirates in order of descending age as A, B, C, D, and E. To denote the offers, we mark pirates that have been thrown overboard with X and other pirates with the number of coins they were offered. For instance if the oldest pirate was thrown overboard and B offered to keep 50 coins, give 30 to C, 15 to D, and 5 to E it would be (X,50,30,15,5). So starting backwards, if there are two pirates left there is no reason for pirate D to offer pirate E anything because 1 of 2 votes is 50% and they will divide the gold according to D's plan. So D offers (X,X,X,100,0) and E gets nothing. C knows this so if there were only three pirates left he would offer E a single coin and take the rest, giving E more money than he would otherwise have (X,X,99,0,1). B knows this so if there were only four pirates left he would offer D a single coin and keep the rest (X,99,0,1,0) because if D votes against this he will get nothing. So if there are 5 pirates left, A should offer (98,0,1,0,1), giving C and E more money than they would otherwise have and keeping the rest.
Fantastic puzzle sir
Thanks Sukanth :) whenever you have any doubt or need any clarification on any puzzle, anytime send me email.
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.
Hey Josef... Your answer would be perfect only when the pirate who is distributing the coins doesn't participate in voting.
Since you said P1 is accepting 1 coin in case of 3 pirates, so with 5 he should ask for at least 2.
But the problem is that, you have to consider the case with 4 pirates. With 4 pirates, P1 will get nothing. So, with 5 pirates, P1 will be thankful to P5 for giving him 1 coin.
Request you to revisit the Puzzle once again.
I am glad to see that you came up with different possibilities :)
Amazing Man!
Awesome question and answer...
👏👏👌👌🇮🇳🇮🇳
Amazing explanation!
Best explanation..
Thank you Suman 🙂
OHHHHHH How actually do you get these?
AWESOME
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.
Thanks for giving a easy explaination
Is it only me? If all the pirates are smart why would they agree to such a scheme in the first place? Especially pirates 2 and 4 who both get nothing.
Great Puzzle👍👍
Very cool
Just amazing ❤❤
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.
Well,
1)There r only 100 coins
2) There r 5 pirates
3) Min of 50% votes are needed for proposal acceptance
I understand the logic, but I don't think this would fly with real pirates: ;)
(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.
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.
The 4th guy may betray the 5th one and make it 100:0
Also I understand thar when last 2 are remaining they may break the rules and start dueling lol.
this is a wonderful question
Thanks Hari:)
Nice buddy keep going
Very complicated puzzle to understand
Wrong. P5 would have to offer P1 TWO coins, because otherwise he would be indifferent, because he'd be getting one from P3 anyway. And P5 really, REALLY doesn't want P1 to randomly vote NAY.
best explanation
Vry vry nycc
Thanks Kittu :)
Didnt you say they get 1 coin per kill on the next turn? How did you mean that?
Which country are you from?
India
Proud to be an Indian😄💪
P5 is the pirate of the pirates
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
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.
In the final distribution why can't it 1 coin for p2 instead of p1
Recursion in disguise
99:0:1:0:1 is correct as only the immediate future counts: Pirate one has to accept one coin
You are missing something !! 99:0:1:0:1 should add up to result 101 coins.... but we have just 100.
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
It all goes back to the 2 pirate deal. In that scenario, P2 would get 100 and P1 gets 0. So P1 has everything to gain by avoiding a 2 pirate situation. With 3 pirates if P1 gets even 1 coin, he will be happy and give his vote. If P3 and P1 are both voting yes, then they don't care about P2's vote so he doesn't need to be given any coins (resulting in 99 - 0 - 1). Therefore P2 would want to avoid the 3 pirate situation and will vote yes in the 4 pirate deal even with only 1 coin. Remember that they are all completely logical so P2 knows that a 2 pirate scenario is just a pipe dream.
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....
p1 and p2 wont share 50-50 afterwards,they will repeat same process as stated,p2 will propose as 100 for him and 0 coin for p1,he will vote for himself and p1 will be against it which is 50-50% votes so deal will be accepted and p1 will get 0,therefore thinking ahead p1 will accept the deal of 1 coin to avoid getting 0 in future.
Well,after p3 is thrown away,p2 gets to decide coin distribution and it is already mentioned that he is not as stupid as accepting 50 coins when he could easily get 100
Who are thinking to create this type of questions
You are AWESOME
At 3:33 why the 1 coin cant be given to p1
Mind-blowing your brain is awesome if you can solve such problems yourself
Why you always shows ted-ed riddles
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
Hi Amit... i truly appreciate your point of view, you really spent time on thinking on this puzzle. Well, the pirate P5 can give one coin to P3(as per the solution) or instead he can give it to P2(as per you). But you also need to consider that they are blood thirsty. so P2 may prefer to decline the deal so as to kill P5, and P2 will later get 1 coin from P4.... so to make it safer, P5 will give one coin to P3 (and not to P2).
LOGICALLY YOURS yeah I guess than your logic is correct but it's just little tough to digest
Amit... I realized that your doubt is mind blowing. Well this is a standard game theory puzzle, but if we pay close attention to the condition of blood thirstiness, your counter question is absolutely powerful. I must attend your query very precisely. It was really a great contribution, thank you :)
Mindfuck !! Amazing
Hey we need in description format
Ramesh, please let me know bro if you are looking for any additional explanation. I'll keep in mind when I post the next version of this puzzle.
@@LOGICALLYYOURS but we not have much time for watch the whole video and we are feel difficulty. So u can give the answering in theory format at the description .
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
You have 240 barrels of wine, one of which has been poisoned. After drinking the poisoned wine, one dies exactly 24 hours. You have 5 slaves whom you are willing to sacrifice in order to determine which barrel contains the poisoned wine. How do you achieve this in 48 hours?
Can you make one more video on this as well?
I don't think its possible in 48 hours!
Divide 240 barrels in group of 5 as G1, G2,G3,G4 and G5 for each slave to try. Each slave have to taste 48 wines. Each slave can sample of its own group every 15 mins. He would be done drinking 48 samples in 12 hours. whatever time slot one of the slave dies, you can figure out the barrel.
@@jc5318 if slave is dying in exactly 24 hours then action of poison is a slow one. Which implies that dilution of the poisonous wine has to affect that individual more slowly as compared to effect which we would have observed if servant would have drank only the poisonous wine. There would be 1 poisonous and 47 normal wine barrels in any set and thinking realistically, there will be highest dilution of contents of poisonous wine if any one of the servants consume the poisonous wine in his very first attempt. So I think in that case we would face several problems with your solution cause I thought of the same solution before. After taking into account these considerations,I commented that it won't be possible in 48 hours in general scenario.There will be cases when your solution will work perfectly.
That's my point of view
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).
Oh God....I can't think this much of selfishness..
It's an example of game theory. What you did is called backward induction.
Q
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.
I think it would also work if the oldest pirate give 1 coin each to the two youngest pirates
Based on the rules given, I don't see an issue with this. I suppose the only potential problem outside the rules is that it introduces a little bit of chance. If P2 is going to end up with 1 coin regardless of whether P5 lives or dies, there might be a 50/50 chance of voting against. At least if the coins are given to P3 and P1, we can be certain that P5 will live. All that aside, you're not wrong.
Pirates are bloodthirsty, so P2 has a choice:
- take 1 coin from P5 and let him live
- take 1 coin from P4 and let P5 die.
The same coin gain, however he will choose second option because he is bloodthirsty :)
bhai music link please
Can't we say that the 2nd youngest lives and gets all the coins?
Make it in Hindi language...
Translate in. Tamil
98:0:1:1:0
CAN YOU STOP COPYING ALL OF TED ED’s VIDEO
Stolen from TED-ED
GOOD WAY TO SPLIT THE COINS???
CALL JACK SPARROW
I really don't understand why they accept the deal even with 1 coin!!
Coz they will either get 1 coin or nothing.
Worst
Copied from ted-ed
How easy it is to say "copied" !! Compare the dates published and comeback here again to share your precious thoughts.