Contents Basic gameplayThe goal of carpentry is to fill holes quickly and efficiently with five-block pieces, much in the same way pentominoes is played. Pirates start with a hole, and three pieces to choose from. As a pirate's improves, the game adds up to 4 different, randomly shaped holes to work on simultaneously. Working too slowly will reduce the current score. If any given hole is neglected for too long, a piece will begin to shake rapidly, chunks will fall out and a reduction will be made to the final score. A finished hole is scored based on how many extra pieces were used. There are also bonuses for grain alignment and using only one type of piece in a hole.
A pirate's standing may increase or decrease after a puzzle session depending on the score.ControlsWith a mouse:. Left-click: Select/release a piece.
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Contents.The game There are five rational (in strict order of seniority A, B, C, D and E) who found 100 gold coins. They must decide how to distribute them.The pirate world's rules of distribution say that the most senior pirate first proposes a plan of distribution. The pirates, including the proposer, then vote on whether to accept this distribution.
If the majority accepts the plan, the coins are dispersed and the game ends. In case of a tie vote, the proposer has the. If the majority rejects the plan, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again. The process repeats until a plan is accepted or if there is one pirate left.Pirates base their decisions on four factors. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins he receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.
And finally, the pirates do not trust each other, and will neither make nor honor any promises between pirates apart from a proposed distribution plan that gives a whole number of gold coins to each pirate.The result To increase the chance of his plan being accepted, one might expect that Pirate A will have to offer the other pirates most of the gold. However, this is far from the theoretical result. When each of the pirates votes, they won't just be thinking about the current proposal, but also other outcomes down the line. In addition, the order of seniority is known in advance so each of them can accurately predict how the others might vote in any scenario. This becomes apparent if we work backwards.The final possible scenario would have all the pirates except D and E thrown overboard.
Since D is senior to E, he has the; so, D would propose to keep 100 for himself and 0 for E.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 one coin in this round to win E's vote. Therefore, when only three are left the allocation is C:99, D:0, E:1.If B, C, D and E remain, B can offer 1 to D; because B has the casting vote, only D's vote is required. Thus, B proposes B:99, C:0, D:1, E:0.(In the previous round, one might consider proposing B:99, C:0, D:0, E:1, as E knows it won't be possible to get more coins, if any, if E throws B overboard. But, as each pirate is eager to throw the others overboard, E would prefer to kill B, to get the same amount of gold from C.)With this knowledge, A 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(Note: 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.)Extension The solution follows the same general pattern for other numbers of pirates and/or coins. However, the game changes in character when it is extended beyond there being twice as many pirates as there are coins. Wrote about extension to an arbitrary number of pirates in the May 1999 edition of and described the rather intricate pattern that emerges in the solution.Supposing there are just 100 gold pieces, then:. Pirate #201 as captain can stay alive only by offering all the gold one each to the lowest odd-numbered pirates, keeping none.
Pirate #202 as captain can stay alive only by taking no gold and offering one gold each to 100 pirates who would not receive a gold coin from #201. Therefore, there are 101 possible recipients of these one gold coin bribes being the 100 even-numbered pirates up to 200 and number #201. Since there are no constraints as to which 100 of these 101 he will choose, any choice is equally good and he can be thought of as choosing at random. This is how chance begins to enter the considerations for higher-numbered pirates. Pirate #203 as captain will not have enough gold available to bribe a majority, and so will die. Pirate #204 as captain has #203's vote secured without bribes: #203 will only survive if #204 also survives.
So #204 can remain safe by reaching 102 votes by bribing 100 pirates with one gold coin each. This seems most likely to work by bribing odd-numbered pirates optionally including #202, who will get nothing from #203.