Weekly puzzle 4 – solution
Pirates Solution
Pirate 5 should propose to give two coins to pirate 1 (or 2), one coin to pirate 3, and the remaining 997 coins to himself!
The idea behind the solution to this puzzle is that a pirate will accept a proposal only if he knows that in the case where he does not accept, he gets less of the treasure.
If pirate 1 is the last pirate left, he would get all the gold coins.
If pirates 1 and 2 are the only ones left, pirate 2 would die for sure since pirate one is bloodthirsty and will reject all proposals of pirate 2. He then keeps all of the gold coins.
When pirate 3 is still alive, he needs the agreement of one of the other two to get a 2-1 majority to carry the proposal. Pirate 2 will agree with every proposal since, as we have seen, he would die if he didn’t (and he certainly wants to live!). So pirate 3 should propose to keep everything for himself.
When we have four pirates, pirate 4 needs two of the other three to agree to win a 3-1 majority. So he proposes to give one coin to pirate 1, one to pirate 2 and the rest to himself. Pirates 1 and 2 will accept since they are greedy, and if they didn’t agree, they would get less.
But as we know, there are five pirates. If pirate 5 gives both pirate 1 (or 2) and pirate 3 one coin more than in the previous case, they are willing to accept the proposal. Then a majority (three out of five) of the pirates will support the proposal, and pirate 5 can keep the rest of the treasure to himself.
Thanks for all your comments this week. Next puzzle on Wednesday
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You’ll always fare well with puzzles of pirates… or my favorite on-line game wouldn’t be YoYo Puzzle Pirates :-)
I look for a 6th pirate proposal because the puzzle appears to offer a very strange, unpredictable way of generating series. Who would guess that to 1000, 0, 1000, 998 and 997 would succeed 994?
You really have thought about this! 6 pirates takes it to a whole new level that I’ve not really thought about, but your logic above seems sound, hmm, maybe I’ll save this for another week!
Hooray!… I was right in my solution (:-p) But it was close… almost gave up a wrong answer. What puzzles me is how we could attack the problem if we had 6 pirates. One “sure” solution appears to be the 6th pirate to take 994 to himself and give 1 to the 4th, 2 to the 3th and 3 to one of the first twos. Can he do better? If he knew which one of the first two was going to be bribed by the 5th, he could have gave only 1 gold to the other, but he don’t knows. The one to receive 1 gold don’t knows either, so, will he think he is receiving one more ore one less than with 5th proposal? Until know the reasoning was solid logic but now, it breaks, demanding more kinds of quality to the pirates than cleverness, bloodthirsty and greediness, if the 6th pirate is going to have more than 994 golds.