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番外 · 题谱 · 1917 · P4

Amusement 28 · Pilgrim’s Draw 28

算术 / 概率 · P1/P4 · 起手题

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。

Amusements in Mathematics 1917 28 arithmeticprobability

Our observation of little things is frequently defective, and our memories very liable to lapse. A certain judge recently remarked in a case that he had no recollection whatever of putting the wedding-ring on his wife's finger. Can you correctly answer these questions without having the coins in sight? On which side of a penny is the date given? Some people are so unobservant that, although they are handling the coin nearly every day of their lives, they are at a loss to answer this simple question. If I lay a penny flat on the table, how many other pennies can I place around it, every one also lying flat on the table, so that they all touch the first one? The geometrician will, of course, give the answer at once, and not need to make any experiment. He will also know that, since all circles are similar, the same answer will necessarily apply to any coin. The next question is a most interesting one to ask a company, each person writing down his answer on a slip of paper, so that no one shall be helped by the answers of others. What is the greatest number of three-penny-pieces that may be laid flat on the surface of a half-crown, so that no piece lies on another or overlaps the surface of the half-crown? It is amazing what a variety of different answers one gets to this question. Very few people will be found to give the correct number. Of course the answer must be given without looking at the coins.

我们对小事情的观察常常是有缺陷的,我们的记忆也很容易消失。某法官最近在审理一桩案件时表示,他完全不记得曾将结婚戒指戴在妻子的手指上。在看不见硬币的情况下,你能正确回答这些问题吗?一分钱的哪一面有日期?有些人是如此不细心,尽管他们几乎每天都在接触硬币,但他们却无法回答这个简单的问题。如果我将一便士平放在桌子上,我可以在它周围放置多少个其他便士,每个硬币也平放在桌子上,以便它们都接触到第一个?几何学家当然会立刻给出答案,而不需要做任何实验。他还会知道,由于所有圆圈都是相似的,因此相同的答案必然适用于任何硬币。下一个问题是问公司最有趣的一个问题,每个人都把自己的答案写在一张纸条上,这样别人的答案就不会帮助到任何人。可以将最多多少个三便士平放在半冠的表面上,使得没有任何一个硬币叠放在另一枚硬币上或与半冠的表面重叠?令人惊讶的是,人们对这个问题有多种不同的答案。很少有人会给出正确的数字。当然,必须在不看硬币的情况下给出答案。

提示 1

先写样本空间,不要直接猜答案。

提示 2

如果条件很多,先算补集或把事件拆成互斥情形。

提示 3

最后核对概率和是否为 1,并检查有没有把条件当成独立。

完整解答

先把“红”“带记号”“带记号红筹码”写成事件。只知道有一枚带记号筹码还不够;还要知道总筹码数,以及带记号筹码是否为红色。若总数为 N 且唯一带记号筹码为红色,概率才是 1/N。

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。