5 Pirates PUZZLE (Version 2) | 100 Gold Coins 5 Pirates
Vložit
- čas přidán 13. 12. 2018
- 5 pirates of different ages, on a treasure hunt, come across an island, where they find a treasure box with 100 gold coins.
They must share the coins with the following approach:
The oldest pirate proposes how to share the coins. Then remaining 4 pirates vote for this proposal. They may agree or disagree with the proposal.
In version 1 of this puzzle, All 5 pirates including the oldest pirate were allowed to vote.... but in version 2, the oldest pirate who is proposing the distribution, is not allowed to vote for his own proposal.
And this leads to a whole different scenario.
Now, If 50% or more of the pirates agree on this proposal, then the coins will be shared
as proposed by the Oldest pirate.
But if the votes in favor are less than 50% , then the proposal will be rejected...
and the pirate proposing the distribution will be thrown in to the sea!
and the process is repeated with the pirates that remain.
Oldest pirate proposes how to share the coins.
Remaining pirates vote for or against it.
50% or more votes, Proposal will be accepted.
Otherwise, Proposal will be rejected.
and the pirate proposing the distribution will be thrown in to the sea.
and the process is repeated so on…
All 5 pirates are Intelligent...,they can think from the perpsctive of other pirates, and to solve this puzzle, you have to think from the perspective of each pirate.
They are Greedy... even for a single coin.
They are Bloodthirsty... so a pirate will vote against the proposal if he doesn't get a better deal than what he can get in next turn.
But they are Rational... so behind their decision, there must be a logic.And importantly... they all don't wish to die.
PUZZLE:
What would be the best distribution by the Oldest Pirate
The video has the best explanation for the Puzzle solution.
Your ""Likes"" and ""Shares"" really encourage me to make more and more videos.
Also try the following puzzles:
Can you solve the Tricky Puzzle || The Clever Commuter || Logical Puzzle
• Can you solve the Tric...
Difficult Puzzle || 25 Horses Race || Asked in Google and Microsoft Interviews
• Difficult Puzzle || 25...
Seemingly IMPOSSIBLE Fox Puzzle || Fox in a Hole || Asked in Google Interview
• Seemingly IMPOSSIBLE F...
Parking Lot Puzzle || Can you solve this Math problem
• Parking Lot Puzzle || ...
Cheryl's Birthday Puzzle || Viral Puzzle on Facebook and Whatsapp
• Cheryl's Birthday Puzz...
Dark Room and Coins Puzzle || Can You Solve this Logic Puzzle
• Dark Room and Coins Pu...
You can follow my facebook page of Logically Yours( Mohammed Ammar ):
/ mohammmedammar
Please comment below your answers and suggestions. Also LIKE the video and SUBSCRIBE to my channel if you are new.
@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.
Thanks to u sir
In reality P1 will persuade P2 to vote against all pirates and kill them. In the end when they both left alone they will divide the coins half.
How about that ? ; )
Piraters cant coperate but being blodthidty P1 Will kill P2 even when given 100 coins and P2 dont with to die so he Will accept getin 0 coins and staying alive rather than die how about that logigaly yours hmmm?
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.
No sir...if so then how about at case of p5, p3 is given 2 coins?
If someone tries to offer me 2/100 gold coins they're going for a swim.
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.
Pirate 1 will always accept, as the alternative is him following suit and throwing himself overboard, as per the rules.
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.
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
What is the implication?
@@hovikgasparyan9729Not that much actually. The only difference is that P5 gives 1 coin to P3 (same as given answer), but that he can give 2 coins to either P2 or P1, and it doesn't matter which one. P5 still ends up with the same number of coins.
@@thatreilyrandomguy Agree with this explanation.
Also, interestingly, when there are only 3 pirates, this video @4:16 explains that P3 would give P2 one coin, but, P3 can actually keep all 100 coins, and P2 would still vote YES because then P2 can at least survive. If P2 votes NO, P2 knows that he will die next round, given that P1 is bloodthirsty and would kill P2 regardless of the P2's proposal.
@@thatreilyrandomguy not exactly. In the 3 pirates scenario, P3 can keep all the coins for himself and P2 will still vote to agree, since he doesn't want to die in the 2 pirates scenario. Having 0 gold and being alive is better than having 0 gold and being dead. Following this logic, P5 can keep all the coins for himself.
@@fifolobuttons5162 Even following that logic, P5 cannot keep all coins as he needs 2 votes of other pirates - one from P3, and other from either P2 or P1.
P5 has to give 1 coin to P3 and 2 coins to P2 or P1 ( P2 or P1 will not agree on one coin because they both get 1-1 coin in 4 pirate scenario anyway, but since bloodthirsty, would prefer P5 to die for same coins).
This version is even more interesting than the first one ! Thanks to share it !
Your riddles are great.Thank you.🇮🇳🙏
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.
P1 has to accept the deal, or he throws himself overboard as per the rules in the 1 pirate round.
Out of sheer curiosity, I expanded this puzzle with the same reversed approach you used to solve this by continuing to add more elder pirates. Something very interesting happened. Note in the final puzzle how one voter has 1 coin and the other has 2 coins. The amount of pirates you need to give 1 coin is equal to ciel(x/3), and the amount of pirates you need to give 2 coins is equal to ciel(x/2)-ciel(x/3), where x is equal to the number of pirates voting on your decision. This becomes a lot clearer and formulaic as you approach x=6. At this value and beyond, the next pirate below you gets nothing. The one after gets 1 coin, and the one after that gets 2 coins. After that it continues in a 0-1-2-0-1-2 pattern until you get to P1 and P2. For some reason, the values of those pirates switches around.
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.
I quickly read your explanation and found it to be very interesting. I will prepare a video on N number of pirates which would certainly need a generic formula. I'll deeply evaluate your idea. Much appreciated :)
Really I am happy sir..
Bcz I have solved it..
This happened only bcz of the wonderful explanation in the first version tysm☺
Fantastic sir.....
So if you know this and you are the second oldest and the second youngest, don’t go for the voyage.. sit at your home 😂
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).
Pirate E has to accept, because he can't approve his own proposal, as per the rules.
Great riddle! I was getting the right answers until the last distribution (i was deceived and i thought 95-0-0-3-2)
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.
Yea!I had the same thought
P2(Bloodthirsty) will prefer to see P4 dead if he is going to get same number of coins in next round.
Njoyed ...
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.
Super and mind blowing videos bro. How you got all these ideas?....
So the oldest pirate lost 1 extra coin for note being able to vote.
This shows why voting right is important.
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 😅
Pause at 5:46. Now P4 sais to P3: "If you vote against P5, I will give you 2 coins instead of zero"
P3 declines the offer.
Lol.. there is no such thing as future commitment honor among thieves.
Just Amazing
Amazing puzzle
Friend we need some of the mind-blowing riddles
Nic approach sir please upload more complex puzzles.
Thanks Anvil... a difficult puzzle will be posted in next week.
All of ur videos are great. Which software u use for making such great vid?
Hi Dear... thanks for the appreciation :) I use Powepoint to make slides and record it in camtasia.
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).
How about the 3rd version of this puzzle where there's the same rules as this video..5 pirates and 100 gold but however the oldest pirate cannot vote and if its a tie its death for the oldest pirate..
I wont spoil the answer here because its easy
P1 has to accept P2's deal, as not doing so results in a dividing by 0 error.
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.
Who knew a difference of one vote could influence so much?
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?
They could still follow through the logic of the other pirates to determine what the entire deal is.
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.
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
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?
3:02 this is wrong according to bloodthirsty because he isnt getting a better deal he shouldve voted against. also being rational isnt described at all.
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
Can't follow this and I don't think all options are considered
P2 & P4 are fool. They didn't get anything last time also. Why did they participate this time? 😜😜😜
Nice vid, knowing the answer of the first version, it was quite straightforward, I think.
Bring version 3
Why the hell they don't distribut 20 to each and be happy 😂😂😂
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 .
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
why 2 1 to other why not 1 1 gold coins
@@umeshpathak5223 If a given pirate isn't worse off killing the one proposing, they will kill. That's what bloodthirsty means.
@@flatearthtruth1298 your comment doesn't make any sense
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
Ya but if the other person agrees, p1 will get nothing in the next turn. The key is to look for 50% votes...
As Nitin just pointed, consider, P1 rejects the lower amount of coins in case of 5 pirates and in the next case of 4 pirates. Now in the case of 3 pirates, he is getting 0 coins, even if he disagrees he is not achieving anything. So if we do reverse engineering we can understand why P1 must agree on 1 and 2 coins in the cases of 4 and 5 pirates respectively.
In the final answer why would p1 accept 2 coins. If he votes against then p5 is killed giving him a better chance at more coins.
@@JD-uh9od He'd get fewer coins if he rejected P5's deal.
Is 97:0:0:2:1 possible??
Bloodthirsty trait makes that impossible.
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
Interesting idea, but P4 and P2 will doublecross P5 on that deal.
👏🤩
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
Bloodthirsty=If murder doesn't hurt their reward, they choose to murder. This didn't happen with the P1/P2 situation because if P1 rejected, he'd have to follow the rules and throw himself overboard. The guy who created this puzzle solution probably forgot about the P1 exception that we'd definitely expect to be there.
@@flatearthtruth1298 yeah i too thought same but it can be bit. Subjective. I solved it without giving extra bonus focusing on rational criteria but yeah bloodthirsty works better
4:51 why not give only 1 coin to p2 & p1?
Coz they are bloodthirsty.A pirate will kill the oldest pirate if he doesn't get a better deal than what he can get in next turn
the creator is damn awesome 😆
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
They're all bloodthirsty. If they can murder and get a better or identical reward, they will choose murder.
Very easy mack difficult pussel for class 10 student ro improve mind
Moral of the story: Don't be greedy, Share equally. Live a happy life.
👍👍
Yeah it is but it wont really matter if the other pirates are GREEDY..
In such case 98 is better than 20..
All this assumes none of the other pirates will collaberate.
All of them are greedy, yet perfect, logicians. They will take whatever action necessary to gain as much gold as possible, even if that means being dishonest.
Part 3 - what if you need more than 50% to approve the proposal?
In a situation with just P1, it's hopeless - he can't approve his own proposal, so he has to throw himself overboard.
With P2 as the eldest, he'd keep all of the coins, as P1 doesn't want the decision to pass to himself due to that causing him to die.
With P3 as the eldest, it's hopeless - he can't sway both of them.
With P4 as the eldest, he'd give 1 coin to P1 and keep the other 99 - P3 doesn't want the decision to pass to himself due to that causing him to die, and P1 would get no coins if he rejected this deal.
With P5 as the eldest, he'd give 2 coins to P1, 1 coin to P2 and P3, and keep the other 96 - he can't sway P4, and the other 3 would get less reward if they declined.
Requiring unanimous would be a very challenging thing to analyze.
@@flatearthtruth1298 There is no voting needed if only P1 is left. Voting rule only applies with 2 or more pirates want to distribute coins.
For >50% case, major change will be in 3 pirate scenario (he needs votes of both others but P1 will still vote against even if given 100 coins) where P3 will die either way.
P1 is guaranteed to keep 100 coins and kill all other pirates when 3 or less total pirates:
-> 1 pirate - P1 keeps 100 coins
-> 2 pirates - P2 will die even if he gives 100 coins to P1 as he needs >50% vote while P1 would prefer to see him die anyway.
-> 3 pirates - P3 needs vote from both P2 and P1, but P1 will never vote even if given 100 coins, so P3 will die and P2 is happy to be just alive with 0 coins.
-> 4 pirates - P4 needs only 2 votes, so takes 99 coins, gives 0 to P3(who will vote happily as he wont die in this case), 1 coin to P2(who is also happy), 0 coins to P1(don’t need his vote).
-> 5 pirates - P5 needs 3 votes(from P3, P2 and P1), so keeps 96 coins, gives 0 to P4(don’t need his vote), 1 coin to P3(he is happy to get 1 coin instead of 0 in previous case), 2 coins to P2(1 more than previous case) and 1 coin to P1( he is happy to get compared to 0 in previous case)
@@cool-aquarian There's no point of voting if there are only 2 pirates left. The captain could just change the rules, and have only 1 person rebelling against said rule change.
For example, P2 could change the rules to the rules in version 1 and keep all the cash.
Basically, "it makes no sense" is not an argument.
@@cool-aquarian Also, it never said it only applied with 2 or more people.
@@flatearthtruth1298 Saying that P1 has to throw himself overboard when only 1 person is like saying criminal has to put himself in Jail when no Judge/Police is there. It bears no practicality and completely utter non-sense.
Now you may act stubborn and defend your statement to the end of world, but inside you know you screwed up when u said it first time.
Divide into fifths
P5 wouldn't get as much gold as possible, and P4 can't be swayed.
real ans is 100,0 0 0 0
because in 3 pirate case he will give 100 0 0 and p2 will still vote for him
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
Everybody would reject if the rule was that the distribution had to be equal.
Thoda aur logically soch le bhai, saare milke fir ek duje ko maar hi naa daalenge
Oh and if 50% vote then the captain again
Proposes new distrebusion and the Crew vote again
@@jorunholm9060 That wouldn't change the puzzle. The eldest would make the exact same proposal.
Just seing the video and i meant the coment above
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
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.
Sorry the last part should read:
With 3 pirates you conclude the division will be 99 : 1 : 0 agreed. With 4 pirates that are 'rational' but not blood thirsty the division would be: 99 : 0 : 1 : 0 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 : 1 : 0, again 2 pirates vote in favour as they won't get any more if they vote against.
why not 98 0 1 0 1 and why 97 0 1 0 2 please explain
Compare the two versions of this puzzle.. i posted both.. in version 1 the eldest pirate is also voting, so there it will be 98 0 1 0 2... but in this version 2, the pirate proposing the distribution is not allowed to vote for his own distribution... and that makes huge difference in the solutions. Pls checkout version 1.
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.
Hi Raj.. in your explanation, the case of FOUR persons is invalid... In the case of THREE the poor guys 2nd and youngest(3rd) are getting 1 and 0 coins (respectively). So, in the case of FOUR, the eldest pirate(being greedy) would think that if I give one extra coin to each of the poor guys, they would happily accept it (to avoid the situation of THREE). So, the distribution should be 97,0,2,1...
But in your explanation, in the case of FOUR, you are not letting the eldest pirate as much as he can.
Now, continue the same process for FIVE.
The eldest(5th) sees that the poor guys (in upcoming situation of FOUR) would be 3rd (with 0) and 5th(the youngest with 1)... so he gives them extra coin each... making the distribution as 97,0,1,0,2.
Please let me know if you agree.... or not.... If you don't agree, then kindly explain it on a big piece of paper (with different situations ) and please send it to my email ID : logicreloaded@gmail.com.
I will then try to revert back to you with in depth explanation.
Now i got it and understood my mistake, thank you very much. By the way u did such good videos that, i saw all of your videos in just 2 days. 😅
Many thanks Raj.... looking forward to post more and more.. :)
@@LOGICALLYYOURS The riddle solution is incorrect. If there was just P1, he could not vote, so he would be forced to throw himself overboard. This means he has to accept P2's deal. This makes the solution VERY different.
As all pirates are greedy no one will agree with 1 or 2 coins !!!!?
If they reject, they get less reward. Greedy doesn't mean being irrational and getting less reward, it means maximizing your reward.
not 20 each eh?
Fairness, while preferred, isn't a good way to maximize your profit as captain.
Agreeing in favour of a 1 coin 2 coin deal....sure those 2 pirates don't qualify as "GREEDY"
Getting as much gold as possible is what greedy means.