I think you might be referring to this.
You have five pirates, ranked from 5 to 1 in descending order. The highest ranked pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. And the process begins again. How should he allocate the gold in order to maximize his share but live to enjoy it? (Hint: One pirate ends up with 98 percent of the gold.)
He would keep all the coins to himself.
The reason for this is that pirates 1 and 2 (half the remaining pirates) would agree because they would know that they could never get more then 0 coins anyway...
This is because if one or both disagreed then that would leave pirate 4 in charge, who would then propose keeping all the money as well (for the same reason). Which in the result of disagreement (for the sake of this explanation) would leave pirate 3 in control with 2 having to agree to them having all the coins in fear for their life (for pirate 1 would just disagree to pirate 2 were they in charge to keep all the coins for themselves).
I enjoyed this puzzle =)
---The highest ranked pirate gets 98 gold coins
---Two pirates get 1 gold coin each
---The other 2 pirates get nothing.
Those who get nothing will complain but they are less than half of the 5 pirates, so no problem
Those who got only one coin each may not complain since they at least get one coin each.
Not foolproof to me.
Why won't the two pirates who get one coin each complain also?
This is a more plausible solution.
---the highest ranked pirate and two more pirates get 96/3 = 32 gold coins each. Fair for them.
---the other two pirates get 2 gold coins each. And they can complain all they want.
I suggest that, if all the pirates were sufficiently intelligent and had thought the situation through thoroughly, the lowest two ranks would simply agree to the first pirate taking all the coins to save time as they would know that they were never to receive more then nothing anyway, as explained in my above post, (the only reason they would have to disagree would be in the hope that the over pirates hadnít their complete understanding of the predicament, but Iím choosing to ignore this possibility for the sake of a logical answer).
What do you think of this though (I'm very confident in my answer but it would be cool if someone could prove it wrong)?
Actually, I just realized, the 3rd ranking pirate would also agree based on this logic (funnily enough, technically, the 4th may as well too ). I know it sounds ridiculous at first, but it's logical!
If you understand my meaning, please reply so that I know Iím not deluded. Thank you!
If the number of pirates remaining is reduced to two, pirate 2 will be killed and pirate 1 will be left with all the gold. Therefore pirate 2 must accept a solution that leaves at least three pirates in the running.
If it comes down to three pirates, pirate 3 can keep all the gold, since pirate 2 will vote with him to avoid dying in the next round. Therefore pirate 3 wants the scenario to reduce to three pirates, and pirates 1 and 2 will prefer a four-pirate solution if pirate 4 offers them any gold at all.
If it comes down to four pirates, then even if pirate 4 offers all the gold to pirate 3, there's no guarantee that pirate 3 will accept the deal, since he will win either way. Therefore pirate 4 must concentrate on pirates 1 and 2. He should offer 1 gold each to pirates 1 and 2, and keep 98 for himself. They will both vote for him then because it leaves them with something, while if they kill him pirate 3 will take all the gold. (Pirate 3 can promise more to pirates 1 and 2 at this stage in an attempt to get them to hold out, but he's a pirate, so his comrades should realize that if it comes down to three people he will take all the gold.)
Pirate 5 must pre-empt this strategy from pirate 4 by offering a better deal to at least two of his comrades. This he can accomplish most cheaply by offering 1 gold to pirate 3 and 2 gold to either pirate 1 or pirate 2. Pirate 5 can thus end up with 97% of the loot... and probably a dagger in his back that evening. (Personally, in his boots I'd split the loot evenly: pirates 1, 2 and 3 accept the deal gladly, and I'd be less likely to have to deal with daggers.)
This problem was recently posed (but not answered) at Job interview brainteasers - Aug. 30, 2007, which hints that one pirate should get 98% of the loot. But using the logic above it seems to me that a 97/0/1/0/2 or 97/0/1/2/0 split is the logical endpoint.
Note also that this puzzle is subtly different to other formulations like Five pirates problem@Everything2.com or Math Is Fun Forum / Five Pirates, in which a split vote including the vote of highest-ranking surviving pirate means the proposal succeeds.
Working backward again, but with a bit of a different twist.
Two pirates. Pirate #2 must give all to #1 to save his life.
Three pirates. #3 must have one vote to keep the dough. Knows #1 will vote against because he gets all in a 2-pirate split. Hence, #3 keeps 99 and gives 1 to #2. #2 votes for the decision, knowing that's the best he can hope for. #1 seethes.
Four pirates. #4 must get two votes. Knows #3 will vote against, as he gets 99 in a 3-pirate split. Hence, #4 keeps 98, gives 1 to #1 and #2 each. #2 votes for, as it's no difference to him if he gets his 1 coin from #3 or #4. #1 votes for, as he knows he would get zilch in a 3-pirate split.
Five pirates. #5 needs two votes, knows #4 will vote against, as he'd get 98 in a 4-way split. He can give 1 each to any two-pirate subset of #1, #2, #3: #3 knows this is better than he can hope for from #4, and the other two don't mind whether they get their 1 coin from #4 or #5.
Pirates be nasty folk (Arrgh!), and for that reason #5 can't count on votes from #1 and #2 unless he actually gives them a better deal than they can get from #4. They might sadistically decide to enjoy #5's demise before collecting their loot. So he can give just 1 gold to#3, but he has to give 2 to #1 or #2.
Also, #3 doesn't have to give any gold to #2 in the 3-pirate scenario, because if #2 doesn't vote for #3's proposal, #2 will die in the next round.
In some formulations of this puzzle, the ambiguity is resolved by assigning pirates' motives, in descending order of importance:
- seeing others killed
Why must pirate 4 give 1 and 2 any of the gold, they already have all the incentive they need to vote for him because they will end up with nothing anyway. Why would they disagree to getting nothing only to be offered nothing again, which they would have to take?
Okay I understand that you’re valuing the pirates’ bloodlust, in which case your answer is correct OldFogie (provided that 1 gold coin would suffice to repress the desire to kill another pirate). But say you were to neglect that impulse and just considered wealth, what would your answer be?
If the pirates don't care one way or the other about seeing their comrades killed, so that they vote randomly yes or no when presented with a choice between two otherwise equal rewards, then my solution is still best because it guarantees pirate 5 will survive even if the random votes go against him.
To allow him to give away only 2 gold and still be sure that his solution will be accepted, we must change the pirates' motivations to
Dividing the loot as equally as possible in 3 (or 5) portions doesn't meet the puzzle's constraint that the pirate in control should maximize his personal return. I take your point, though: this particular pirate society isn't going to survive very long unless they change their rules.
- shedding as little blood as possible