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. Ian Stewart wrote about Steve Omohundro's extension to an arbitrary number of pirates in the May 1999 edition of Scientific American 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 they will choose, any choice is equally good and they 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