pimpineasy Wrote: ------------------------------------------------------- > cfagoal2 Wrote: > -------------------------------------------------- > ----- > > Still no correct answer to pirate game > > > http://www.braingle.com/brainteasers/teaser.php?op > =2;id=3551;comm=0 I stand by my answer, The problem does not say that any other prates who get 0 get killed.
A - 0 B - 0 C - 97 D - 2 E - 1
Heres the real answer (no one got it) The Result It might be expected intuitively that Pirate A will have to allocate little if any to himself for fear of being voted off so that there are fewer pirates to share between. However, this is as far from the theoretical result as is possible. This is apparent if we work backwards: if all except D and E have been thrown overboard, D proposes 100 for himself and 0 for E. He has the casting vote, and so this is the allocation. If there are three left (C, D and E) C knows that D will offer E 0 in the next round; therefore, C has to offer E 1 coin in this round to make E vote with him, and get his allocation through. Therefore, when only three are left the allocation is C:99, D:0, E:1. If B, C, D and E remain, B knows this when he makes his decision. To avoid being thrown overboard, he can simply offer 1 to D. Because he has the casting vote, the support only by D is sufficient. Thus he proposes B:99, C:0, D:1, E:0. One might consider proposing B:99, C:0, D:0, E:1, as E knows he won’t get more, if any, if he throws B overboard. But, as each pirate is eager to throw each other overboard, E would prefer to kill B, to get the same amount of gold from C. Assuming A knows all these things, he can count on C and E’s support for the following allocation, which is the final solution: A: 98 coins B: 0 coins C: 1 coin D: 0 coins E: 1 coin Also, A:98, B:0, C:0, D:1, E:1 or other variants are not good enough, as D would rather throw A overboard to get the same amount of gold from B. Source: http://en.wikipedia.org/wiki/Pirate_game
pimpineasy Wrote: ------------------------------------------------------- > cfagoal2 Wrote: > -------------------------------------------------- > ----- > > Still no correct answer to pirate game > > > http://www.braingle.com/brainteasers/teaser.php?op > =2;id=3551;comm=0 This answer makes absolutely no sense because its a different puzzle. The one here has the rule that tie votes means proposition goes
I think working backwards is the best way to approach it. With just D and E, the end game is that D will win out since he’ll always vote yes, and E gets 0. So E will be happy with anything more than 0. With C, D, and E, the end game is C will bribe E with 1 gold to get his vote. With B, C, D, and E, the end game is B will bribe E with 1 gold to get his vote. With all pirates… I think we’re going to need to bribe someone else, so I picked D. Thoughts? Edit: Oh damn the answer got posted
A ignores the rules, kills B, C, D, & E and takes all the gold. He then goes over the bond trader’s house and kills him. Eats the condoms because he thinks they are some kind of fancy candy and then nails the 3 girls bareback.
I respectfully disagree with the solution, since it doesn’t say anything about pirates making deals with other pirates under the table. Take your scenario: A: 98 coins B: 0 coins C: 1 coin D: 0 coins E: 1 coin B can say to D and E, “listen, if you both vote against this, I’ll take 97, give D 1 coin and E 2 coins”. Of course, they will both agree, out of GREED (second rule of the game is desire to maximize the gold they receive, and of course, D wants at least 1 coin), and the THIRD RULE, with all other things being equal, they vote to push over the proposing pirate (E knows they will never get less than 1 coin, so this deal makes sense for them). But, when it’s D’s turn to propose an actual allocation, he says the following: A: Dead B: 98 coins C: 0 coins D: 0 coins E: 2 coin E, wll vote for this, because if he votes against it, this is the best that could happen: A: Dead B: Dead C: 99 coins D: 0 coins E: 1 coin And that, my friends, is the right answer. Solution (no matter what the “official answer” is). A: Dead B: 98 coins C: 0 coins D: 0 coins E: 2 coin I dare any of you to argue against this one. I believe it is sound.
Sanka’s Mom Wrote: ------------------------------------------------------- > But, when it’s D’s turn to propose an actual > allocation, he says the following: > > A: Dead > B: 98 coins > C: 0 coins > D: 0 coins > E: 2 coin > > E, wll vote for this, because if he votes against > it, this is the best that could happen: > > A: Dead > B: Dead > C: 99 coins > D: 0 coins > E: 1 coin > > And that, my friends, is the right answer. > > > Solution (no matter what the “official answer” > is). > A: Dead > B: 98 coins > C: 0 coins > D: 0 coins > E: 2 coin > > I dare any of you to argue against this one. I > believe it is sound. There’s no reason why B would allocate 98-0-0-2 if he could get away with 99 coins just by bribing one pirate. This is the case any time you try to bribe any pirate with more than 1 piece of gold, shady underhand deal or not. With the pirates all knowing this, the case collapses back to the original Nash Equilibrium.
I’ll change my answer a little then, but I still think the “right” answer is wrong. And yes, i know i have said that this is my “final answer” quite a few times now. A: Dead B: 99 coins C: 0 coins D: 1 coin E: 0 coins I agree, no need to give E 2 if D will be happy with 1. In fact, no need to involve E in this at all (if you try to give E 1 coin, E will say no, because they know they will get 1 coin if B is dead) If he tries to bribe C with 1 coin, C would say no because C can proceed to push him overboard and be in the position to bribe E. But, if he tries to give D one coin, D will say yes, because he knows if C is around, C will take 99 and give E 1.
Sanka’s Mom Wrote: ------------------------------------------------------- > I’ll change my answer a little then, but I still > think the “right” answer is wrong. > > And yes, i know i have said that this is my “final > answer” quite a few times now. > > A: Dead > B: 99 coins > C: 0 coins > D: 1 coin > E: 0 coins > > I agree, no need to give E 2 if D will be happy > with 1. In fact, no need to involve E in this at > all (if you try to give E 1 coin, E will say no, > because they know they will get 1 coin if B is > dead) > > If he tries to bribe C with 1 coin, C would say no > because C can proceed to push him overboard and be > in the position to bribe E. > > But, if he tries to give D one coin, D will say > yes, because he knows if C is around, C will take > 99 and give E 1. Your logic is correct, except only if voting on B’s proposal with 4 pirates. However they are voting on A’s proposal with 5 pirates. Knowing that the equilibrium with 4 pirates is 99-0-1-0, all A has to do is bribe C and E with one coin to get 98-0-1-0-1, which is the proposed solution.
The deal making thing doesn’t make sense because it’s not in the context of the puzzle. EVEN assuming the deal making A: Dead B: 99 coins C: 0 coins D: 1 coin E: 0 coins Would NOT work, because A can offer C and E each 1 coin(the correct original solution) and they will vote for favor of it instead of coming to this solution.
wutsaCFA Wrote: ------------------------------------------------------- > Sanka’s Mom Wrote: > -------------------------------------------------- > ----- > > I’ll change my answer a little then, but I > still > > think the “right” answer is wrong. > > > > And yes, i know i have said that this is my > “final > > answer” quite a few times now. > > > > A: Dead > > B: 99 coins > > C: 0 coins > > D: 1 coin > > E: 0 coins > > > > I agree, no need to give E 2 if D will be happy > > with 1. In fact, no need to involve E in this > at > > all (if you try to give E 1 coin, E will say > no, > > because they know they will get 1 coin if B is > > dead) > > > > If he tries to bribe C with 1 coin, C would say > no > > because C can proceed to push him overboard and > be > > in the position to bribe E. > > > > But, if he tries to give D one coin, D will say > > yes, because he knows if C is around, C will > take > > 99 and give E 1. > > Your logic is correct, except only if voting on > B’s proposal with 4 pirates. However they are > voting on A’s proposal with 5 pirates. Knowing > that the equilibrium with 4 pirates is 99-0-1-0, > all A has to do is bribe C and E with one coin to > get 98-0-1-0-1, which is the proposed solution. Beat me to it
Ok, I will wave the white flag and surrender. There’s my brain workout for the day.
cfagoal2 Wrote: ------------------------------------------------------- > The deal making thing doesn’t make sense because > it’s not in the context of the puzzle. His assumption is perfectly valid, but in the end the assumption doesn’t change anything because the pirates are stated to be greedy. In fact, there’s a missing assumption that the pirates are perfectly rational, but I imagine saying that would tip people off as to the nature of the solution. If somebody stated an assumption and fully worked out a solution based on that assumption, that’s probably more useful as a brain exercise / interview question than regurgitating a known problem.
the pirate game has been posted on this forum and a lot of people had it right, it’s not THAT tricky as for Mr Pink still arguing about the condom question look: wear condom A, then over it condom B after girl 1: outside of condom B is dirty, inside of condom B is clean inside of condom A is dirty, outside of condom A is clean you take off B, proceed with girl 2 by using outside of A which is clean because it was covered by B the entire time after girl 2, A is dirty on both sides, so you take B and flip it inside out because it’s inside is still clean, and you wear the dirty side over your condom A so you aren’t affected either, proceed with girl 3
kblade Wrote: ------------------------------------------------------- > the pirate game has been posted on this forum and > a lot of people had it right, it’s not THAT > tricky > > > as for Mr Pink still arguing about the condom > question > > look: > wear condom A, then over it condom B > > after girl 1: > outside of condom B is dirty, inside of condom B > is clean > inside of condom A is dirty, outside of condom A > is clean > > you take off B, proceed with girl 2 by using > outside of A which is clean because it was covered > by B the entire time > > after girl 2, A is dirty on both sides, so you > take B and flip it inside out because it’s inside > is still clean, and you wear the dirty side over > your condom A so you aren’t affected either, > proceed with girl 3 I still say go with the shocker. Although typically referring to the use of certain fingers, it applies perfectly well here as well.
kblade Wrote: ------------------------------------------------------- > the pirate game has been posted on this forum and > a lot of people had it right, it’s not THAT > tricky > > > as for Mr Pink still arguing about the condom > question > > look: > wear condom A, then over it condom B > > after girl 1: > outside of condom B is dirty, inside of condom B > is clean > inside of condom A is dirty, outside of condom A > is clean > > you take off B, proceed with girl 2 by using > outside of A which is clean because it was covered > by B the entire time > > after girl 2, A is dirty on both sides, so you > take B and flip it inside out because it’s inside > is still clean, and you wear the dirty side over > your condom A so you aren’t affected either, > proceed with girl 3 Condom A better have one heck of a volume tip to accomodate 3 servings of essence.
Exactly higgmond. kblade, even in your analysis you say condom A is dirty inside after girl 1. Plus I’ve never tried to put a condom on that I’ve previously worn and taken off, but I’m guessing it’s f’n difficult.
> Plus I’ve never tried to put a condom on that I’ve > previously worn and taken off, but I’m guessing > it’s f’n difficult. did you even go to college?
Mr. Pink Wrote: ------------------------------------------------------- > Exactly higgmond. > > kblade, even in your analysis you say condom A is > dirty inside after girl 1. > > Plus I’ve never tried to put a condom on that I’ve > previously worn and taken off, but I’m guessing > it’s f’n difficult. it’s dirty inside but that doesn’t matter because you never take it off in the first place