I think I have an alternative and better solution to the Pirates problem:
Orginal problem is:
Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).
The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.
If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.
What is the maximum number of coins the captain can keep without risking his life?
SOLUTION: Hey this is just like our PL/GL's Rating distribution in TCS, we are all pirates and no matter how hard or good work we do, our PL/GL will cheat us....
Plz. read full solution (My friend says correct ans is 98 but I do not find my sol to be wrong)
The captain can keep all 100 coins by playing politics.
first he distributes 50 coins to 2 pirates except himself---he will have majority of 3 including himself, now kill the two pirates who recieved the money (walk the plank)
now do redistribution (since two pirates who had money are dead)------give 100 coins to 2nd pirate ank kill him (using majority of 3rd one)
now do redistribution---give 0 coins to the only left pirate, and keep 100 with himself....(He still has 50 % majority)
i think the answer is 98, and not 100 as trueromesh said.. because there is no redistribution if the captain's proposal prevailed, redistribution will be done if the captain is killed.
the 2 pirates will definitely agree (will be given 1 coin each)on the captain's proposal since they don't want to end up empty handed..
I think that this one is really dumb. Why? Because if you were the one that got one coin, would that make you feel OK? I'd want my own equal share. I'd want 20 freakin coins, or I'm going to get the others to gang up on him, kill him, and then we'd all get 25 coins. 1 coin? MAYBE 15 would be ok, but it's just too hard to say. You just can't say how they'd react.
for those of you who don't understand the solution, I'm going to attempt to make you understand, since I also didn't get it at first. this is how the captain predicts the responses of the other pirates:
the puzzles states that all the pirates are extremely intelligent, therefore they should think about what would happen if they disagreed to the proposal. We know that if the majority of the pirates (3 out of 5) disagree to the proposal, the captain will have to walk the plank and there will be 4 pirates remaining with the most senior captain as the new captain.
To better explain, let's assign each pirate a ranking, 5 being the most senior (the current captain making the proposal) and 1 being the "least senior" (this guy would never be captain). In this situation, the only way a pirate is killed is if he is captain and makes a proposal that the majority of the pirates disagree to. If 5 dies, 4 becomes captain, if 4 dies, 3 becomes captain, and so on.
You also have to remember that all of these pirates are greedy, if 4,3, or 2 become captain you will have to assume that they will divide it in a way where they would get the most money.
The solution explains it by working backwards, this seems like the best way but I'll try to clear up on some parts they left out.
Consider the situation if there were only 2 pirates. That is, 3 are killed and 2 are remaining. 2, being the most senior, would be captain and would make a proposal that all 100 coins go to him. He has 50% majority so it is distributed as he says. 1's vote does not matter in this situation as either way 2 automatically wins majority. Obviously 1 would not be happy with this but he cannot do anything about it.
Now consider the situation if there were only 3 pirates. 2 are killed and 3 are remaining. 3, being the most senior, would be captain and would propose that 99 coins go to him and 1 coin goes to pirate 1. He would not propose to give 1 coin to pirate 2 because pirate 2 could disagree along with pirate 1 and pirate 3, having lost the majority, would be killed and there would be 2 pirates remaining (previous paragraph explains what happens in this situation). So pirate 3 would give 1 coin to pirate 1, who would obviously say "yes" to avoid 3 being killed and him getting nothing when 2 becomes captain. 2 misses out on getting any gold.
Now consider the situation if there were only 4 pirates. 4, being the most senior, would be captain and would propose that 99 coins go to him and 1 coin goes to pirate 2. He knows that 2 will not say "no" because if he did it will become like the situation mentioned in the previous paragraph and 2 would miss out on getting any gold. The captain and pirate 2 agreeing would be a 50% majority and the distribution would go as said. 1 and 3 would miss out on getting any gold
Now we can look at the situation that we have here. 5, the current captain, would propose that 98 coins go to him and 1 coin goes to pirate 1 and 1 coin goes to pirate 3. These pirates are forced to agree because if they didn't, the captain would be killed and the situation would continue as mentioned in the previous paragraph where they would be the ones missing out on gold. The captain would win a 60% majority and the distribution would continue as said.
The only reason why any of the pirates agree to the proposals is to avoid the captain being killed and the next senior captain making a proposal where they would receive no gold.
I hope my long explanation clears it up for people who don't understand the solution.
Consider the following table (hope it renders well...), where the columns are the number of pirates left, and each row represents a pirate, by rank, 1 being the highest (captain), and 5 being lowest. The values show how much each pirate will get under each scenario:
Since the rules stipulate that the pirates are each infinitely intelligent (logical), each knows how this will play out. The key here is to work backwards, as shown: if 2 pirates remain, the higher ranking pirate will get all 100. Therefore, if there are 3 pirates, the highest ranking pirate need only offer the lowest ranking pirate 1 gold, because this is more than the 0 gold the pirate will get if s/he votes no... etc.
THIS EXPLANATION IS STUPID AND IGNORANT, ITS MORE FICTIONAL THEN FACT, IF U HAVE A 100 GOLD PIECE LOOT AND U JUST GIVE 1% TO TWO PEOPLE, WHAT THE F*** ARE THEY GONNA BUY WITH THAT WHEN THEY GET BACK TO SHORE, THERE WILL 100% BE MUTINY AS ALL THE PIRATES WOULD BE PISSED AND KILL THAT SELFISH CAPTAIN, THEREFORE THIS LATERAL THINKING QUESTION IS STUPID IT IS NOT SMART!!! ASK YOURSELF THIS QUESTION A PIRATE WHAT WOULD HE RATHER DO CHOOSE DEATH, OR ONE COIN, HE KNOWS THE OTHER PIRATES BE PISSED TOO SO LOGICAL TO KILL THE CAPTAIN AND GET 25COINS
Suppose pirates 1 and 3 conspire together to go 50/0/50. Pirate 3 plans to go back on his promise and offer 99/0/1. Pirate 1 agrees because he gets one coin either way, and there's still an off chance that pirate 3 is being sincere (or stupid). The captain foresees this and has to offer 2 coins to pirate 1, bringing the total down to 97.
These 3 points are quite disruptive. If the pirates are all of the above and the captain who is especial of all these 3 aspects then... Its clear the captain is in his seat because of this. In real life dumb sailors will just will just follow the mob theory. They'll just kick his b***s but since its stated that they're all of the above, then it'll a different story. There are simple answers to this question but we're just trying to guess the maximum profit the captain's gonna make.
Okay, we have 5 pirates. Im going to use A-E. Seniorest-Juniorest.
100 coins.
A proposes 98-1-1-0-0 distribution. He would have majority. However, Pirates B and C should be smart enough to disagree with just 1 coin, and D and E with 0. So, the 4 pirates total rebel.
We have B-E now.
B proposes 99-1-0-0
Pirates C-E disagree and rebel.
C-E now.
C proposes 99-1-0
D and E rebel.
D-E now.
D proposes 100-0
E cannot rebel.
Provided all pirates are aware of this, the captain will want to ensure no one rebels.
So, expected profit from each would be this:
A: 34-33-33-0-0
B: 50-50-0-0
C: 50-50-0
D: 100-0
E: 0
To keep his life, A actually needs to give all the coins to B and C, split 50-50.
Because B and C can join with D and E and rebel, and then it takes just B and C to gain majority.
This is more like the Unexpected Hanging Paradox than the completely determined puzzle-with-answer Ant presents.
The Unexpected Hanging Paradox: a judge says the prisoner will be hanged this week on a day that will be a surprise to him. The prisoner argues, well, if we get all the way to Sunday without my being hanged, then I can't be surprised when it's Sunday, so you can't do it then. But if you absolutely can't do it Sunday, then by the time we get to Saturday, I know _that's_ the last possible day, so I won't be surprised when you do it Saturday, so you can't do it then. But if you can't do it Saturday, then on Friday, etc., etc. so you can't even do it Monday.
It seems like an unbreakable chain of logic, but at the same time, the end result is untenable.
Because if the prisoner gives the argument Monday to the Judge: "So therefore I expect you to do it today, therefore it can't be unexpected today either, and you can't ever do it," the Judge can reply, "Interesting," and if he goes ahead and holds the hanging Wednesday or Thursday, it will actually be unexpected; the prisoner can claim he was expecting it, but after claiming that on Monday and Tuesday too, his "expectation" just isn't very meaningful or believable. If the Judge refuses to accept the logic that the hanging can't be done at all, the Prisoner doesn't really know when it will happen, and it will be unexpected after all.
What destroys the logical certainty of the Unexpected Hanging is the human uncertainty of "expectation."
And that destroys the logic of this puzzle, too. Pirate 5, the most senior, cannot expect no risk with the proposed solution.
Let's try it. Pirate 5 lays out the argument exactly as Ant does, and says, therefore, I'm giving one coin to #3 and one to #1 --
But suddenly #1 says, "I don't think so. I think you're going to give 50 coins to me and 50 coins to #3 and be happy to walk away with your life."
#5: Nonsense. I won't go along with that, and you'll lose --
#1: I'll lose one lousy gold piece, while you will lose your life.
#5: But you are selfish, and must prefer one gold piece to no gold piece.
#1: I'm selfish and much prefer 50 gold pieces to one gold piece, just as I know that you are selfish and must prefer life to death. If #3 feels the same way I do --
#3: And why wouldn't I? Yeah, I'm selfish: I'm in for 50 too.
#1: And #2 will also be happy with my plan, because he now realizes that if you die, #5, then #4 will be in the same hot seat you're in now, #2 can demand all the money to ransom his life.
#4: Hey!
#1, #2, #3 (in unison): But it's true.
#5: It's still just a game of chicken. Whether I live or I die, your selfishness should prefer 1 coin to no coins.
#1: First of all, you forget my treacherousness, which would much prefer a solution that screws you, my captain. But quite aside from that, I might do better than nothing even if you tough it out. If you tough it out and we kill you, #4 is in the Captain's hot seat. After seeing for a fact that one gold piece won't hold loyalty, he can't offer #2 just one piece. And when #2 demands all the pieces, maybe I'll jump in and offer to let him live for only 90 pieces. #3 might do the same. Who knows how the ensuing bidding-war/negotiation will turn out? There's no guarantee that #3 or I will actually come up zero. I may well get more than the one piece you're offering now. So you have no strong expectation that I (or #3) will chicken out if you play chicken with me now. Look into my intelligent, treacherous, and selfish eyes and believe me when I tell you that if you stiff me with one gold piece, you're going down.
In the face of this, it is absurd to say that #5 can keep 98 coins without risking his life. Indeed, it's not clear that he can keep _any_ coins without shouldering some risk, _especially_ if he's the logical machine Ant assumes all characters are. But we could argue that if #5 bids #1 and #3 up from one coin, he will reach a point where the certainty of getting 20 (or whatever) coins is preferable to both the one coin originally offered and the unknown second-round possibilities if #5 dies. So #5 can probably keep some coins. But how many is not susceptible to an absolutely determined answer.
#5's logic is in some sense impeccable, but that doesn't make it _irresistible_. #1 and #3 simply don't have to play along. This is a paradox, not a puzzle with an inarguable correct answer.
Notice, I'm not throwing away the premises and just saying, "Aaaah, they're pirates, man, they aren't going to stand for this one-coin garbage." I am strictly sticking to the premises that they are intelligent, treacherous, and selfish (looking for the best deal) and that the Captain wants _no_ risk. If it comes down to the Captain risking his life (and 98 gold pieces, by the way), versus #1 risking one gold piece, it is not reasonable to assume that the risks are felt with anywhere near equal strength. Given that corollary to the original premises, #5's logic-chopping would not be sufficient to carry the day.
I grant you, in the scenario I just suggested, #1 isn't likely to make money on round two despite what I have him say above (though it may usefully confuse the others). But if, against the odds, #5 chooses to die, #1 may be the first to say to #4, "I'll let you live for one gold piece" -- which nobody else can undercut. He'll have lost no money _and_ sated his treacherousness against the original captain. And if he's not the quickest to make that offer, he's still out only one coin and can still enjoy the treachery. The main point remains: #5 can see #1 has little to lose, while he has everything to lose; he is likely to cave, and that likelihood strengthens #1's resolve to risk the lousy one piece.
But the more I think about, the more I see that the Ant solution could work after all. The problem with the counteproposal I have written for #1 above is that #2 would jump in and offer to undercut #1 or #3 and back the Captain for less than 50 coins. So the bidding war would break out in this round, with downward pressure on what the Captain would have to pay. (#4 would not participate, because he can almost certainly get better than half the coins on the second round.)
Still, some of the force of the argument above still holds: by the time they low-bid each other down to about one gold coin, #'s 1,2, and 3 are more likely to say, hey, the selfish victory is not winning this lowball auction, it's getting the most money, let's stick together and demand 33 apiece. They won't all get it; if the Captain then decides to give only two of them 33 coins each, those two will surely go along with it, they have no loyalty to the group of three. And if he preemptively says, no, I'm giving #3 and #2 only twenty apiece -- they might well go along with it as better than the uncertainty of round two. Certainly, whichever pirate pops up and explains to #5 that his logic is no good, that's the one who will be left out of the final split. So there's not much incentive to even pop up. But I'm not convinced the downward pressure goes all the way down to one gold piece. If all three of them look #5 in the eye and say they won't take less than, say, 20, he has the choice of giving 2 of them 20 and keeping sixty _with no risk_ or giving two of them less than that and just praying that one of them isn't so pissed off as to reject it and take his chances in round two. Of course, there will be downward pressure in round two too...
It still seems to me that once someone points out that "one/no coins" is a much smaller risk than "life/death," Captain #5 is going to wind up paying out more than one coin each. Maybe not a helluva lot more. Not a determined and fixed number, though. Not a puzzle with a fixed solution. Again, granted, if #1 argues as I say he does, he certainly isn't in a strong position to make money. But the fact that he doesn't care much increases #5's risk and the amount #5 is likely to pay out to feel risk-free.
Firstly, the puzzle can be saved and the official solution as elaborated by Ant can be made bulletproof by making one more premise explicit: no bargaining. If pirate law dictates that no one but the Captain can speak before the divvy, and if the Captain's divvy is irrevocable, take it or leave it, then yes, #1 and #3 simply can't do any better than to take one coin apiece. I concur.
However, if bargaining or negotiation is possible before a _final_ divvy offer, the whole situation changes, and more radically than I realized in my first post. Basically, no Captain above #2 can keep any coins at all.
Let's walk it back.
Two pirates: yes, #2 gets everything, as per official solution.
Three pirates: #3 offers #1 one gold piece, and #1 says, "I don't think so. I think you're going to give me all the coins and be happy to walk away with your life."
#3: Nonsense. I won't go along with that, and you'll lose --
#1: I'll lose one lousy gold piece, while you will lose your life.
#3: But you are selfish, and must prefer one gold piece to no gold piece.
#1: I'm selfish and much prefer 100 gold pieces to one gold piece, just as I know that you are selfish and must prefer life to death.
#3: It's still just a game of chicken. Whether I live or I die, your selfishness should prefer 1 coin to no coins.
#1: It's a game of chicken where you have much more to lose than I do. I insist on 100 coins; if you don't back down and I don't back down, I lose one lousy coin, you lose 100 coins _and your life._ Whereas you can be absolutely sure of keeping your life (though losing the coins) if you back down. Want to avoid risk of your life? Give me the 100 coins.
#3: Well, you say that, but here, I'm putting 50 coins on the table for you. That's a sure thing, which you will take...
#1: No, as a selfish guy I'm holding out for the full 100. I'm standing up now and walking toward you with #2 (who is happy to back me, even though he gets nothing this time, because once you're dead he gets 100 percent of all _future_ divisions). It's your money or your life, and it would be neither selfish nor intelligent of you to give up your life when the only other choice I leave you is to keep it. Give or die.
#3: I give, I give.
So with three pirates, #1 gets everything. This affects the next case. Four pirates:
#4 can't offer anything less than 100 coins to #1, of course, since #1 will get everything if negotiations break down and #4 is killed. So if he doesn't give 100 to 1, he has to win over one of the other two. And now #3 says, "I get nothing if I become Captain, so I want something if I help keep you alive. I want the full 100 coins."
#2: Don't pay him 100, I'll take 90 instead and leave you with ten.
#3: Okay, I'll underbid #2 and take 80.
#2: Don't you see where this is headed, #3? We'll bid each other down to one coin. As intelligent selfish guys, we can do better than that by cooperating. Let's each demand that #4 pay both of us 50 coins.
#3: Yeah, but suppose I make that demand, but then you cut me out by saying you'll accept 45? You're treacherous.
#2: But I'm also selfish. Why should I take 45 when I know I can get 50 by sticking with you?
#3: But suppose he offers you 60 (and nothing to me)?
#2: Then you offer to side with him for 50, and he selfishly would prefer to pay out 50 instead of 60, so I quickly drop to 45. Faced with another competitive bidding race to the bottom, we realize again that our best bet is to stick together.
#3: It's true. I can't expect to get better than 50 with a competitor bidding, and we can guarantee 50 if we stick together. Let's advance on him. #1 will go with us, since he gets everything if the mutiny is completed.
#1: They're right, #4. To avoid risk of your life, you have to either give it all to the two of them, evenly divided, or give it all to me.
#4: No, if I give it all to you, you'll mutiny anyway, because that way you'll be in the three-pirate game where you not only get it all this time, but every _subsequent_ time something is divided up. I have to deal with them. #2 and #3, I offer you each 40, all I ask is 20. If you play chicken with me, you no longer risking one lousy coin, you're each risking 40.
#2 and #3 (advancing on him): But we're confident your selfish interest in your own life is much more imperative. We can get the full 50 each by saying give or die. So, give or die.
#4: I give, I give.
Okay, so the four-pirate game ends with #2 and #3 getting all the money. That affects the five-pirate game.
Five pirates:
#5 can't afford to offer #2 or #3 less than fifty, because they know they can get that much for sure on the second round if #5 is killed. So he has to win over the other two. #4 says, "I get nothing if I become Captain, so I want something if I help keep you alive. I want the full 100 coins."
#2: Don't pay him 100, pay him one coin (you need two of us), and I'll take 90 and leave you with 9.
#3: Okay, I'll underbid #2 and take 75.
#1: I'll underbid #3 and take 50.
#2 and 3#: I'll bid under 50.
#5: Well, I can't pay #2 or #3 less than 50 because they'll mutiny anyway to get the 50 they get when there are four pirates. So it's up to #1 and #4, and remember I need both of you.
#4:Don't you see where this is headed, #1? We'll bid each other down to two coins and one coin. As intelligent selfish guys, we can do better than that by cooperating. Let's each demand that #5 pay both of us 50 coins.
#1: Yeah, but suppose I make that demand, but then you cut me out by saying you'll accept 45? You're treacherous.
#4: But I'm also selfish. Why should I take 45 when I know I can get 50 by sticking with you?
#1: But suppose he offers you 60 (and nothing to me)?
#4: Then you offer to side with him for 50, and he selfishly would prefer to pay out 50 instead of 60, so I must quickly drop to 45. Faced with another competitive bidding race to the bottom, we realize again that our best bet is to stick together.
#1: It's true. I can't expect to get better than 50 with a competitor bidding, and we can guarantee 50 if we stick together. Let's advance on him. #2 and #3 will go with us, since they get everything if the mutiny is completed.
#2 and #3: They're right, #5. You have to either give it all to the two of them, evenly divided, or give it all to us.
#5: No, if I give it all to you, you'll mutiny anyway, because that way you'll be in the four-pirate game where you not only get it all this time, but every _subsequent_ time something is divided up. I have to deal with them. #1 and #4, I offer you each 40, all I ask is 20. If you play chicken with me, you aren't just risking one coin, you're each risking 40.
#1 and #4 (advancing on him with #2 and #3): But we're confident your selfish interest in your own life is much more imperative. We can get the full 50 each by saying give or die. So, give or die.
#5: I give, I give.
It's a similar iterative pattern, but a different outcome. If the other pirates can negotiate and bargain, Captain #5 is suddenly at risk, and cannot follow his original plan, but has to give everything up. If negotiating and bargaining are _not_ allowed, Captain #5 can indeed keep 98 coins.
I was wrong about the solution degrading toward paradox. But there are two different solutions, depending on whether bargaining is allowed.
You logic breaks down at 3 pirates. (P1 for Pirate 1, etc)
3 Pirates
P3 would take all gold pieces.
P3 tells P2, if you don't agree with the decision and mutiny with P1, after I die, P1 will simply kill you and take all gold pieces. So you better agree with me for your life.
4 Pirates
P4 takes 99 and give P2 1 gold.
P4 tells P2, after I die either P3 or P2 takes all money, so 1 gold is better than nothing.
5 Pirates
P5 takes 98 gold and give P1 & P3 1 gold.
P5 tells P1 & P2, if I die neither one of you get anything from P4, so 1 gold is better than nothing
ok, let's assume P1 offer's anything P2, P3 and P4 can all say lets take 33.3 coins and not give anything to P1 and P5 so they decide to kill P1.
then the next step goes to P2 any offer p2 takes will be turned down because P3 and P4 are going to take 50-50. so lets say also P2 dies.
ok, now its a bit different because this time P4 and P5 cant take 50-50 and kill P3 because then P5 will kill P4 and take 100 coins.(because there is no mojarity for P4 against P5) so in this case P3 can offer P4 1 coin because he knows that if P4 turns down the coin he will end up with 0 and P5 will have 100 coins.
so P3 gives P4 one coin and takes 99.
now lets work are way up!
p2 can make a deal with P4 and offer him 2 coins instead of 1(like in the privious deal) and this time P2 will offer P5 one coin which is more then he will get in the privios deal which is zero.
now P1 can make a deal where he offers P4 3 coins instead of 2 which he make in the previos deal. and he offers P5 2 coins which is more then 1 that he gets in the previous deal.
Your making the assumption that all pirates have a different rank. No where in the question does it say this. Just kill the capitalist bastard cheating captain, then there's no more captain, share the loot properly and 4 pirates will be happy ( for now)
In order of captaincy: Capn: Bob , joe, Harsha, Koushik, Kartik
Ans) Let us start with the base case of 2 men (since one man is not possible) and work our way upward. If 2 men are present
Captain Koushik and Kartik , then the captain can keep 100 coins and give the other guy 0 because both have 50% vote. Working our
way upwards to 3 men, Kartik knows that if he votes against Harsha, then in the next step he will be left with 0 coins, so harsha
has the automatic vote even if he keeps 99 and gives Kartik 1 coin. Working our way up to 4 people. Koushik knows that if he
votes against Captain joe then he will get 0 coins, so joe gives Koushik 1 coin and keeps 99. Back to 5 people. Harsha and
Kartik know that if they vote against bob then that they will get 0 so, bob gives both Harsha and Kartik one apiece and keeps 98.
One of the pirates given 1 gold coin must realise that the 2 pirates with no Gold coins will rebel so if he does likewise then the captains walking the plank.
Therefore the pirate offered 1 Gold coin would be making more money if he rejects.
Anyway there is never 'no risk'.
He could offer them 25 each and there would still be some risk of the pirates being, well, pirates and voting against it anyway. Human decisions are never certain.
The real answer to this riddle should be 60 coins. All the pirates are greedy bastards but they are smart enough to realize that 100 split 5 ways is 20 coins and that is the MINIMUM they should receive. If captain hook over here thinks he is the boss and says he is going to take 98 coins and give two other pirates one coin each, the pirates (being their greedy selves) will just say "screw that" because they know if they all vote againt the captain, their will be 4 left, thus leaving them with 25 coins each.
Based on this logic, the captain will allow two pirates to receive their equal share (20 coins each) and take everyone elses coins - leaving him with 60. That is the only way these greedy pirates will allow the captain to do this. Why settle for one coin and let some a-hole take 98 when they can easily just vote him off the boat. (technically you can use this logic and say the captain is going to be killed either way because if these pirates are so greedy they will keep voting the senior captain off, leaving two pirates. The older pirate gets all 100)
Since I am not here to play the devils advocate, and since human decisions are too much of a variable to figure this out, the only logical answer is the Captain can take 60 coins and not risk being overthrown.
My very, very, very carefully thought out solution
Bill's solution is partly right, but at 4 pirates his logic is off.
Let the Captain be P1, with P2, P3, P4, and P5 next in line.
Going from Bill's final solution, with 3 pirates P5 gets all the coins by ransoming P3's life.
At 4 pirates, when P3 and P4 agree with each other to get 50 coins each and leave P2 with nothing, P2 is going to say, "Wait a moment, P3 (or P4)*! I don't need both of you, so how 'bout I give you 51 coins for supporting me, keeping 49 myself and giving your friend P4 (or P3) nothing.
*NOTE: P2 CAN CHOOSE P3 OR P4 TO MAKE A DEAL WITH. FROM NOW ON WE WILL ASSUME THAT HE CHOOSES P3 (WITH P4 IN PARENTHESIS). P4 (P3 IN PARENTHESIS) WILL BE USED TO INDICATE THE PIRATE THAT P2 DOESN'T CHOOSE.
P3 (or P4) suspiciously says, "But if I do that it's not a definite chance that P4 (or P3) won't offer you 50 coins for his support, cutting me out of the deal instead. You're too greedy a pirate to not want 50 instead of 49 coins."
P2: "I'm not stupid. If I foolishly accept P4's (or P3)'s offer of 50 coins, you will jump in with 49, and on your downward bidding war you two will decide it's better to get 50 coins each as a sure thing. Even if I refuse your offer of 49 coins, why would P4 (or P3) be loyal enough to me to care whether I get the other 50 coins or you do. He may make me give you the other 50 coins out of spite. And besides, if you refuse my current offer of 51 coins, I'll go to him with the same offer, and as a treacherous, greedy pirate he will show more sense than you and accept."
P4 (or P3): "He's right, you know. If I see an opportunity for 51 coins instead of 50, I'll let go of our deal and you will get nothing, P3 (or P4)."
P3 (or P4): "In that case I accept your offer of 51 coins, P2."
And this changes the very nature of the 5 pirate situation. In the 4 pirate situation P3 and P4 don't know if they will get 51 coins or 0. This is counter intuitive that they would rather be unsure of getting 51 coins or 0, but in reality they rather would get 50 coins as a sure thing. The problem is, they are treacherous pirates and even if they make a deal with each other to FOR SURE get 50 coins each, when one of them is offered 51 he will accept every time. So, in the 5 pirate situation they don't want to bid under 25.5 coins, since they will AVERAGE getting 25.5 coins in the 4 pirate situation. Many pirates are gamblers, but it is illogical to gamble less than the average amount they would have gotten in the 4 pirate situation. But P2 and P5 will get no coins in the 4 pirate situation, so they are fine with bidding 25 or lower. If either of them bids 26 or higher, then they have the danger of being undercut by P3 or P4. So they will each ask for 25 coins, with no bidding against each other because they BOTH need to support P1. So P1 will get exactly 50 coins.
Ignore my last few lines. In the scenario with 5 pirates, P1 should not give money to P2 BECAUSE P2 WOULD GET 49 COINS IF P1 DIED. I meant to say that P1 should give P5 25 coins and either P3 or P4 26 coins, more than the average amount they would get if P1 died. So P1 would really only keep 49 coins.
i think Bill Adams got it almost right... read his last example. this is my scenario continuing from Bills 5-pirate game example:
.
..
...
#1 and #4 (advancing on him with #2 and #3): But we're confident your selfish interest in your own life is much more imperative. We can get the full 50 each by saying give or die. So, give or die.
#5 My final word is i give u (#1, #4) 1 coin each. now i`ve made my proposal and i cant take it back. so what do u want more; 1 coin (since you get nothing if u go 4-pirate game) or watch me die?
[#4 thinks: ]
[#1 thinks: ]
so #1 and #4 say: whatever give us the f...n coin.
#4 (if i mutiny and take it to round 4 I either get nothing or I try #5`s strategy. But if #3 thinks the same way then im dead and #3 gets 99 coins with this strategy in round 3. So im good with 1 coin and my life.)
#1 (if I mutiny all i can achieve is take it to round 3 where i still win just 1 coin. So i`ll take it now leavin` my crew with 5 members. why would i want to weaken my crew? im not gonna be captain anyway...)
Orginal problem is:
Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).
The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.
If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.
What is the maximum number of coins the captain can keep without risking his life?
SOLUTION: Hey this is just like our PL/GL's Rating distribution in TCS, we are all pirates and no matter how hard or good work we do, our PL/GL will cheat us....
Plz. read full solution (My friend says correct ans is 98 but I do not find my sol to be wrong)
The captain can keep all 100 coins by playing politics.
first he distributes 50 coins to 2 pirates except himself---he will have majority of 3 including himself, now kill the two pirates who recieved the money (walk the plank)
now do redistribution (since two pirates who had money are dead)------give 100 coins to 2nd pirate ank kill him (using majority of 3rd one)
now do redistribution---give 0 coins to the only left pirate, and keep 100 with himself....(He still has 50 % majority)
how dd you know if the other pirates would agree or disagree?
the 2 pirates will definitely agree (will be given 1 coin each)on the captain's proposal since they don't want to end up empty handed..
the puzzles states that all the pirates are extremely intelligent, therefore they should think about what would happen if they disagreed to the proposal. We know that if the majority of the pirates (3 out of 5) disagree to the proposal, the captain will have to walk the plank and there will be 4 pirates remaining with the most senior captain as the new captain.
To better explain, let's assign each pirate a ranking, 5 being the most senior (the current captain making the proposal) and 1 being the "least senior" (this guy would never be captain). In this situation, the only way a pirate is killed is if he is captain and makes a proposal that the majority of the pirates disagree to. If 5 dies, 4 becomes captain, if 4 dies, 3 becomes captain, and so on.
You also have to remember that all of these pirates are greedy, if 4,3, or 2 become captain you will have to assume that they will divide it in a way where they would get the most money.
The solution explains it by working backwards, this seems like the best way but I'll try to clear up on some parts they left out.
Consider the situation if there were only 2 pirates. That is, 3 are killed and 2 are remaining. 2, being the most senior, would be captain and would make a proposal that all 100 coins go to him. He has 50% majority so it is distributed as he says. 1's vote does not matter in this situation as either way 2 automatically wins majority. Obviously 1 would not be happy with this but he cannot do anything about it.
Now consider the situation if there were only 3 pirates. 2 are killed and 3 are remaining. 3, being the most senior, would be captain and would propose that 99 coins go to him and 1 coin goes to pirate 1. He would not propose to give 1 coin to pirate 2 because pirate 2 could disagree along with pirate 1 and pirate 3, having lost the majority, would be killed and there would be 2 pirates remaining (previous paragraph explains what happens in this situation). So pirate 3 would give 1 coin to pirate 1, who would obviously say "yes" to avoid 3 being killed and him getting nothing when 2 becomes captain. 2 misses out on getting any gold.
Now consider the situation if there were only 4 pirates. 4, being the most senior, would be captain and would propose that 99 coins go to him and 1 coin goes to pirate 2. He knows that 2 will not say "no" because if he did it will become like the situation mentioned in the previous paragraph and 2 would miss out on getting any gold. The captain and pirate 2 agreeing would be a 50% majority and the distribution would go as said. 1 and 3 would miss out on getting any gold
Now we can look at the situation that we have here. 5, the current captain, would propose that 98 coins go to him and 1 coin goes to pirate 1 and 1 coin goes to pirate 3. These pirates are forced to agree because if they didn't, the captain would be killed and the situation would continue as mentioned in the previous paragraph where they would be the ones missing out on gold. The captain would win a 60% majority and the distribution would continue as said.
The only reason why any of the pirates agree to the proposals is to avoid the captain being killed and the next senior captain making a proposal where they would receive no gold.
I hope my long explanation clears it up for people who don't understand the solution.
2 3 4 5
1 98
2 99 0
3 99 0 1
4 100 0 1 0
5 0 1 0 1
Since the rules stipulate that the pirates are each infinitely intelligent (logical), each knows how this will play out. The key here is to work backwards, as shown: if 2 pirates remain, the higher ranking pirate will get all 100. Therefore, if there are 3 pirates, the highest ranking pirate need only offer the lowest ranking pirate 1 gold, because this is more than the 0 gold the pirate will get if s/he votes no... etc.
100 coins.
A proposes 98-1-1-0-0 distribution. He would have majority. However, Pirates B and C should be smart enough to disagree with just 1 coin, and D and E with 0. So, the 4 pirates total rebel.
We have B-E now.
B proposes 99-1-0-0
Pirates C-E disagree and rebel.
C-E now.
C proposes 99-1-0
D and E rebel.
D-E now.
D proposes 100-0
E cannot rebel.
Provided all pirates are aware of this, the captain will want to ensure no one rebels.
So, expected profit from each would be this:
A: 34-33-33-0-0
B: 50-50-0-0
C: 50-50-0
D: 100-0
E: 0
To keep his life, A actually needs to give all the coins to B and C, split 50-50.
Because B and C can join with D and E and rebel, and then it takes just B and C to gain majority.
The solution is wrong.
The Unexpected Hanging Paradox: a judge says the prisoner will be hanged this week on a day that will be a surprise to him. The prisoner argues, well, if we get all the way to Sunday without my being hanged, then I can't be surprised when it's Sunday, so you can't do it then. But if you absolutely can't do it Sunday, then by the time we get to Saturday, I know _that's_ the last possible day, so I won't be surprised when you do it Saturday, so you can't do it then. But if you can't do it Saturday, then on Friday, etc., etc. so you can't even do it Monday.
It seems like an unbreakable chain of logic, but at the same time, the end result is untenable.
Because if the prisoner gives the argument Monday to the Judge: "So therefore I expect you to do it today, therefore it can't be unexpected today either, and you can't ever do it," the Judge can reply, "Interesting," and if he goes ahead and holds the hanging Wednesday or Thursday, it will actually be unexpected; the prisoner can claim he was expecting it, but after claiming that on Monday and Tuesday too, his "expectation" just isn't very meaningful or believable. If the Judge refuses to accept the logic that the hanging can't be done at all, the Prisoner doesn't really know when it will happen, and it will be unexpected after all.
What destroys the logical certainty of the Unexpected Hanging is the human uncertainty of "expectation."
And that destroys the logic of this puzzle, too. Pirate 5, the most senior, cannot expect no risk with the proposed solution.
Let's try it. Pirate 5 lays out the argument exactly as Ant does, and says, therefore, I'm giving one coin to #3 and one to #1 --
But suddenly #1 says, "I don't think so. I think you're going to give 50 coins to me and 50 coins to #3 and be happy to walk away with your life."
#5: Nonsense. I won't go along with that, and you'll lose --
#1: I'll lose one lousy gold piece, while you will lose your life.
#5: But you are selfish, and must prefer one gold piece to no gold piece.
#1: I'm selfish and much prefer 50 gold pieces to one gold piece, just as I know that you are selfish and must prefer life to death. If #3 feels the same way I do --
#3: And why wouldn't I? Yeah, I'm selfish: I'm in for 50 too.
#1: And #2 will also be happy with my plan, because he now realizes that if you die, #5, then #4 will be in the same hot seat you're in now, #2 can demand all the money to ransom his life.
#4: Hey!
#1, #2, #3 (in unison): But it's true.
#5: It's still just a game of chicken. Whether I live or I die, your selfishness should prefer 1 coin to no coins.
#1: First of all, you forget my treacherousness, which would much prefer a solution that screws you, my captain. But quite aside from that, I might do better than nothing even if you tough it out. If you tough it out and we kill you, #4 is in the Captain's hot seat. After seeing for a fact that one gold piece won't hold loyalty, he can't offer #2 just one piece. And when #2 demands all the pieces, maybe I'll jump in and offer to let him live for only 90 pieces. #3 might do the same. Who knows how the ensuing bidding-war/negotiation will turn out? There's no guarantee that #3 or I will actually come up zero. I may well get more than the one piece you're offering now. So you have no strong expectation that I (or #3) will chicken out if you play chicken with me now. Look into my intelligent, treacherous, and selfish eyes and believe me when I tell you that if you stiff me with one gold piece, you're going down.
In the face of this, it is absurd to say that #5 can keep 98 coins without risking his life. Indeed, it's not clear that he can keep _any_ coins without shouldering some risk, _especially_ if he's the logical machine Ant assumes all characters are. But we could argue that if #5 bids #1 and #3 up from one coin, he will reach a point where the certainty of getting 20 (or whatever) coins is preferable to both the one coin originally offered and the unknown second-round possibilities if #5 dies. So #5 can probably keep some coins. But how many is not susceptible to an absolutely determined answer.
#5's logic is in some sense impeccable, but that doesn't make it _irresistible_. #1 and #3 simply don't have to play along. This is a paradox, not a puzzle with an inarguable correct answer.
Notice, I'm not throwing away the premises and just saying, "Aaaah, they're pirates, man, they aren't going to stand for this one-coin garbage." I am strictly sticking to the premises that they are intelligent, treacherous, and selfish (looking for the best deal) and that the Captain wants _no_ risk. If it comes down to the Captain risking his life (and 98 gold pieces, by the way), versus #1 risking one gold piece, it is not reasonable to assume that the risks are felt with anywhere near equal strength. Given that corollary to the original premises, #5's logic-chopping would not be sufficient to carry the day.
It's embarrassing...
But the more I think about, the more I see that the Ant solution could work after all. The problem with the counteproposal I have written for #1 above is that #2 would jump in and offer to undercut #1 or #3 and back the Captain for less than 50 coins. So the bidding war would break out in this round, with downward pressure on what the Captain would have to pay. (#4 would not participate, because he can almost certainly get better than half the coins on the second round.)
Still, some of the force of the argument above still holds: by the time they low-bid each other down to about one gold coin, #'s 1,2, and 3 are more likely to say, hey, the selfish victory is not winning this lowball auction, it's getting the most money, let's stick together and demand 33 apiece. They won't all get it; if the Captain then decides to give only two of them 33 coins each, those two will surely go along with it, they have no loyalty to the group of three. And if he preemptively says, no, I'm giving #3 and #2 only twenty apiece -- they might well go along with it as better than the uncertainty of round two. Certainly, whichever pirate pops up and explains to #5 that his logic is no good, that's the one who will be left out of the final split. So there's not much incentive to even pop up. But I'm not convinced the downward pressure goes all the way down to one gold piece. If all three of them look #5 in the eye and say they won't take less than, say, 20, he has the choice of giving 2 of them 20 and keeping sixty _with no risk_ or giving two of them less than that and just praying that one of them isn't so pissed off as to reject it and take his chances in round two. Of course, there will be downward pressure in round two too...
It still seems to me that once someone points out that "one/no coins" is a much smaller risk than "life/death," Captain #5 is going to wind up paying out more than one coin each. Maybe not a helluva lot more. Not a determined and fixed number, though. Not a puzzle with a fixed solution. Again, granted, if #1 argues as I say he does, he certainly isn't in a strong position to make money. But the fact that he doesn't care much increases #5's risk and the amount #5 is likely to pay out to feel risk-free.
I guess I still believe that . . .
Firstly, the puzzle can be saved and the official solution as elaborated by Ant can be made bulletproof by making one more premise explicit: no bargaining. If pirate law dictates that no one but the Captain can speak before the divvy, and if the Captain's divvy is irrevocable, take it or leave it, then yes, #1 and #3 simply can't do any better than to take one coin apiece. I concur.
However, if bargaining or negotiation is possible before a _final_ divvy offer, the whole situation changes, and more radically than I realized in my first post. Basically, no Captain above #2 can keep any coins at all.
Let's walk it back.
Two pirates: yes, #2 gets everything, as per official solution.
Three pirates: #3 offers #1 one gold piece, and #1 says, "I don't think so. I think you're going to give me all the coins and be happy to walk away with your life."
#3: Nonsense. I won't go along with that, and you'll lose --
#1: I'll lose one lousy gold piece, while you will lose your life.
#3: But you are selfish, and must prefer one gold piece to no gold piece.
#1: I'm selfish and much prefer 100 gold pieces to one gold piece, just as I know that you are selfish and must prefer life to death.
#3: It's still just a game of chicken. Whether I live or I die, your selfishness should prefer 1 coin to no coins.
#1: It's a game of chicken where you have much more to lose than I do. I insist on 100 coins; if you don't back down and I don't back down, I lose one lousy coin, you lose 100 coins _and your life._ Whereas you can be absolutely sure of keeping your life (though losing the coins) if you back down. Want to avoid risk of your life? Give me the 100 coins.
#3: Well, you say that, but here, I'm putting 50 coins on the table for you. That's a sure thing, which you will take...
#1: No, as a selfish guy I'm holding out for the full 100. I'm standing up now and walking toward you with #2 (who is happy to back me, even though he gets nothing this time, because once you're dead he gets 100 percent of all _future_ divisions). It's your money or your life, and it would be neither selfish nor intelligent of you to give up your life when the only other choice I leave you is to keep it. Give or die.
#3: I give, I give.
So with three pirates, #1 gets everything. This affects the next case. Four pirates:
#4 can't offer anything less than 100 coins to #1, of course, since #1 will get everything if negotiations break down and #4 is killed. So if he doesn't give 100 to 1, he has to win over one of the other two. And now #3 says, "I get nothing if I become Captain, so I want something if I help keep you alive. I want the full 100 coins."
#2: Don't pay him 100, I'll take 90 instead and leave you with ten.
#3: Okay, I'll underbid #2 and take 80.
#2: Don't you see where this is headed, #3? We'll bid each other down to one coin. As intelligent selfish guys, we can do better than that by cooperating. Let's each demand that #4 pay both of us 50 coins.
#3: Yeah, but suppose I make that demand, but then you cut me out by saying you'll accept 45? You're treacherous.
#2: But I'm also selfish. Why should I take 45 when I know I can get 50 by sticking with you?
#3: But suppose he offers you 60 (and nothing to me)?
#2: Then you offer to side with him for 50, and he selfishly would prefer to pay out 50 instead of 60, so I quickly drop to 45. Faced with another competitive bidding race to the bottom, we realize again that our best bet is to stick together.
#3: It's true. I can't expect to get better than 50 with a competitor bidding, and we can guarantee 50 if we stick together. Let's advance on him. #1 will go with us, since he gets everything if the mutiny is completed.
#1: They're right, #4. To avoid risk of your life, you have to either give it all to the two of them, evenly divided, or give it all to me.
#4: No, if I give it all to you, you'll mutiny anyway, because that way you'll be in the three-pirate game where you not only get it all this time, but every _subsequent_ time something is divided up. I have to deal with them. #2 and #3, I offer you each 40, all I ask is 20. If you play chicken with me, you no longer risking one lousy coin, you're each risking 40.
#2 and #3 (advancing on him): But we're confident your selfish interest in your own life is much more imperative. We can get the full 50 each by saying give or die. So, give or die.
#4: I give, I give.
Okay, so the four-pirate game ends with #2 and #3 getting all the money. That affects the five-pirate game.
Five pirates:
#5 can't afford to offer #2 or #3 less than fifty, because they know they can get that much for sure on the second round if #5 is killed. So he has to win over the other two. #4 says, "I get nothing if I become Captain, so I want something if I help keep you alive. I want the full 100 coins."
#2: Don't pay him 100, pay him one coin (you need two of us), and I'll take 90 and leave you with 9.
#3: Okay, I'll underbid #2 and take 75.
#1: I'll underbid #3 and take 50.
#2 and 3#: I'll bid under 50.
#5: Well, I can't pay #2 or #3 less than 50 because they'll mutiny anyway to get the 50 they get when there are four pirates. So it's up to #1 and #4, and remember I need both of you.
#4:Don't you see where this is headed, #1? We'll bid each other down to two coins and one coin. As intelligent selfish guys, we can do better than that by cooperating. Let's each demand that #5 pay both of us 50 coins.
#1: Yeah, but suppose I make that demand, but then you cut me out by saying you'll accept 45? You're treacherous.
#4: But I'm also selfish. Why should I take 45 when I know I can get 50 by sticking with you?
#1: But suppose he offers you 60 (and nothing to me)?
#4: Then you offer to side with him for 50, and he selfishly would prefer to pay out 50 instead of 60, so I must quickly drop to 45. Faced with another competitive bidding race to the bottom, we realize again that our best bet is to stick together.
#1: It's true. I can't expect to get better than 50 with a competitor bidding, and we can guarantee 50 if we stick together. Let's advance on him. #2 and #3 will go with us, since they get everything if the mutiny is completed.
#2 and #3: They're right, #5. You have to either give it all to the two of them, evenly divided, or give it all to us.
#5: No, if I give it all to you, you'll mutiny anyway, because that way you'll be in the four-pirate game where you not only get it all this time, but every _subsequent_ time something is divided up. I have to deal with them. #1 and #4, I offer you each 40, all I ask is 20. If you play chicken with me, you aren't just risking one coin, you're each risking 40.
#1 and #4 (advancing on him with #2 and #3): But we're confident your selfish interest in your own life is much more imperative. We can get the full 50 each by saying give or die. So, give or die.
#5: I give, I give.
It's a similar iterative pattern, but a different outcome. If the other pirates can negotiate and bargain, Captain #5 is suddenly at risk, and cannot follow his original plan, but has to give everything up. If negotiating and bargaining are _not_ allowed, Captain #5 can indeed keep 98 coins.
I was wrong about the solution degrading toward paradox. But there are two different solutions, depending on whether bargaining is allowed.
Whew!
3 Pirates
P3 would take all gold pieces.
P3 tells P2, if you don't agree with the decision and mutiny with P1, after I die, P1 will simply kill you and take all gold pieces. So you better agree with me for your life.
4 Pirates
P4 takes 99 and give P2 1 gold.
P4 tells P2, after I die either P3 or P2 takes all money, so 1 gold is better than nothing.
5 Pirates
P5 takes 98 gold and give P1 & P3 1 gold.
P5 tells P1 & P2, if I die neither one of you get anything from P4, so 1 gold is better than nothing
P1 will be the oldest P5 is the yougest!
ok, let's assume P1 offer's anything P2, P3 and P4 can all say lets take 33.3 coins and not give anything to P1 and P5 so they decide to kill P1.
then the next step goes to P2 any offer p2 takes will be turned down because P3 and P4 are going to take 50-50. so lets say also P2 dies.
ok, now its a bit different because this time P4 and P5 cant take 50-50 and kill P3 because then P5 will kill P4 and take 100 coins.(because there is no mojarity for P4 against P5) so in this case P3 can offer P4 1 coin because he knows that if P4 turns down the coin he will end up with 0 and P5 will have 100 coins.
so P3 gives P4 one coin and takes 99.
now lets work are way up!
p2 can make a deal with P4 and offer him 2 coins instead of 1(like in the privious deal) and this time P2 will offer P5 one coin which is more then he will get in the privios deal which is zero.
now P1 can make a deal where he offers P4 3 coins instead of 2 which he make in the previos deal. and he offers P5 2 coins which is more then 1 that he gets in the previous deal.
so all together
P1(the oldest) gets 95 coins!
P4 gets 3 coins!
P5 (the youngest) gets 2 coins!
thank you!
Ans) Let us start with the base case of 2 men (since one man is not possible) and work our way upward. If 2 men are present
Captain Koushik and Kartik , then the captain can keep 100 coins and give the other guy 0 because both have 50% vote. Working our
way upwards to 3 men, Kartik knows that if he votes against Harsha, then in the next step he will be left with 0 coins, so harsha
has the automatic vote even if he keeps 99 and gives Kartik 1 coin. Working our way up to 4 people. Koushik knows that if he
votes against Captain joe then he will get 0 coins, so joe gives Koushik 1 coin and keeps 99. Back to 5 people. Harsha and
Kartik know that if they vote against bob then that they will get 0 so, bob gives both Harsha and Kartik one apiece and keeps 98.
Therefore the pirate offered 1 Gold coin would be making more money if he rejects.
Anyway there is never 'no risk'.
He could offer them 25 each and there would still be some risk of the pirates being, well, pirates and voting against it anyway. Human decisions are never certain.
Based on this logic, the captain will allow two pirates to receive their equal share (20 coins each) and take everyone elses coins - leaving him with 60. That is the only way these greedy pirates will allow the captain to do this. Why settle for one coin and let some a-hole take 98 when they can easily just vote him off the boat. (technically you can use this logic and say the captain is going to be killed either way because if these pirates are so greedy they will keep voting the senior captain off, leaving two pirates. The older pirate gets all 100)
Since I am not here to play the devils advocate, and since human decisions are too much of a variable to figure this out, the only logical answer is the Captain can take 60 coins and not risk being overthrown.
Let the Captain be P1, with P2, P3, P4, and P5 next in line.
Going from Bill's final solution, with 3 pirates P5 gets all the coins by ransoming P3's life.
At 4 pirates, when P3 and P4 agree with each other to get 50 coins each and leave P2 with nothing, P2 is going to say, "Wait a moment, P3 (or P4)*! I don't need both of you, so how 'bout I give you 51 coins for supporting me, keeping 49 myself and giving your friend P4 (or P3) nothing.
*NOTE: P2 CAN CHOOSE P3 OR P4 TO MAKE A DEAL WITH. FROM NOW ON WE WILL ASSUME THAT HE CHOOSES P3 (WITH P4 IN PARENTHESIS). P4 (P3 IN PARENTHESIS) WILL BE USED TO INDICATE THE PIRATE THAT P2 DOESN'T CHOOSE.
P3 (or P4) suspiciously says, "But if I do that it's not a definite chance that P4 (or P3) won't offer you 50 coins for his support, cutting me out of the deal instead. You're too greedy a pirate to not want 50 instead of 49 coins."
P2: "I'm not stupid. If I foolishly accept P4's (or P3)'s offer of 50 coins, you will jump in with 49, and on your downward bidding war you two will decide it's better to get 50 coins each as a sure thing. Even if I refuse your offer of 49 coins, why would P4 (or P3) be loyal enough to me to care whether I get the other 50 coins or you do. He may make me give you the other 50 coins out of spite. And besides, if you refuse my current offer of 51 coins, I'll go to him with the same offer, and as a treacherous, greedy pirate he will show more sense than you and accept."
P4 (or P3): "He's right, you know. If I see an opportunity for 51 coins instead of 50, I'll let go of our deal and you will get nothing, P3 (or P4)."
P3 (or P4): "In that case I accept your offer of 51 coins, P2."
And this changes the very nature of the 5 pirate situation. In the 4 pirate situation P3 and P4 don't know if they will get 51 coins or 0. This is counter intuitive that they would rather be unsure of getting 51 coins or 0, but in reality they rather would get 50 coins as a sure thing. The problem is, they are treacherous pirates and even if they make a deal with each other to FOR SURE get 50 coins each, when one of them is offered 51 he will accept every time. So, in the 5 pirate situation they don't want to bid under 25.5 coins, since they will AVERAGE getting 25.5 coins in the 4 pirate situation. Many pirates are gamblers, but it is illogical to gamble less than the average amount they would have gotten in the 4 pirate situation. But P2 and P5 will get no coins in the 4 pirate situation, so they are fine with bidding 25 or lower. If either of them bids 26 or higher, then they have the danger of being undercut by P3 or P4. So they will each ask for 25 coins, with no bidding against each other because they BOTH need to support P1. So P1 will get exactly 50 coins.
.
..
...
#1 and #4 (advancing on him with #2 and #3): But we're confident your selfish interest in your own life is much more imperative. We can get the full 50 each by saying give or die. So, give or die.
#5 My final word is i give u (#1, #4) 1 coin each. now i`ve made my proposal and i cant take it back. so what do u want more; 1 coin (since you get nothing if u go 4-pirate game) or watch me die?
[#4 thinks: ]
[#1 thinks: ]
so #1 and #4 say: whatever give us the f...n coin.
in the end its still a paradox :)
#1 (if I mutiny all i can achieve is take it to round 3 where i still win just 1 coin. So i`ll take it now leavin` my crew with 5 members. why would i want to weaken my crew? im not gonna be captain anyway...)