题面摘自 Henry E. Dudeney 公版文本；英文为古腾堡原文整理，中文为本站自译，提示、解答骨架和闲谈保留本站原创结构。
Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards, Francis, and Gudgeon, were recently engaged in play. The name of the particular game is of no consequence. They had agreed that whenever a player won a game he should double the money of each of the other players—that is, he was to give the players just as much money as they had already in their pockets. They played seven games, and, strange to say, each won a game in turn, in the order in which their names are given. But a more curious coincidence is this—that when they had finished play each of the seven men had exactly the same amount—two shillings and eightpence—in his pocket. The puzzle is to find out how much money each man had with him before he sat down to play.
七名男子，他们的名字分别是亚当斯、贝克、卡特、多布森、爱德华兹、弗朗西斯和古吉恩，最近正在玩耍。特定游戏的名称并不重要。他们同意，每当一名球员赢得一场比赛，他就应该将其他球员的钱加倍——也就是说，他要给球员们的钱与他们口袋里已有的钱一样多。他们打了七场比赛，奇怪的是，按照名字的顺序，每人依次赢得一场比赛。但更奇怪的巧合是，当比赛结束时，七个人的口袋里都有完全相同的金额——两先令和八便士。谜题是找出每个人坐下来玩之前带了多少钱。
提示 1
先把每一步允许做什么写成状态表。
提示 2
找一个不会随操作改变的量，或把对象分成互斥类别。
提示 3
检查构造是否覆盖全部对象，而不是只给出一个漂亮例子。
完整解答
固定主人面向门口，就消除了旋转重复。先数其余人任意排列，再减去两位指定朝圣者相邻的情形。相邻时把这两人看成一个双人块，块内部有 2 种顺序；最后用总数减去坏排列。
Dudeney 的趣题常把难点藏在“看起来可以试”的地方。别急着猜答案；先把图、表或状态画出来，再问哪些限制一直没有变。这也是它和 Carroll 逻辑题互补的地方：一个拆句子，一个拆结构。